Search Results for “” – Page 183

Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium

Leave a Comment / Class 10 / By Prasanna

Students can Download Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium Pdf, Samacheer Kalvi 10th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium

Instructions

The question paper comprises of four parts.
You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
All questions of Part I, II, III and IV are to be attempted separately.
Question numbers 1 to 14 in Part I are Multiple Choice Quèstions of one-mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and.writing the option code and the corresponding answer.
Question numbers 15 to 28 in Part II àre two-marks questions. These are to be answered in about one or two sentences.
Question numbers 29 to 42 in Part III are five-marks questions. These are to be answered in about three to five short sentences.
Question numbers 43 to 44 in Part IV are eight-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 100

PART – 1

I. Choose the correct answer. Answer all the questions. [14 × 1 = 14]

Question 1.
If there are 10 relations from a set A = {1, 2, 3,4, 5} to a set B, then the number of elements in B is ………….. .
(1) 3
(2) 2
(3) 4
(4) 8
Answer:
(2) 2

Question 2.
If g = {(1,1),(2, 3),(3,5),(4,7)} is a function given by g(x) = αx + β then the values of α and β are ………….. .
(1) (-1,2)
(2) (2,-1)
(3) (-1,-2)
(4) (1,2)
Answer:
(2) (2,-1)

Question 3.
If 7^4k = ………….. (mod 100)
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(1) 1

Question 4.
The next term ot the sequence $\frac{3}{16}, \frac{1}{8}, \frac{1}{12}, \frac{1}{18}$ is ………….. .
(1) $\frac{1}{24}$
(2) $\frac{1}{27}$
(3) $\frac{2}{3}$
(4) $\frac{1}{81}$
Answer:
(2) $\frac{1}{27}$

Question 5.
If (x – 6) is the HCF of x² – 2x – 24 and x² – kx – 6 then the value of K is ………….. .
(1) 3
(2) 5
(3) 6
(4) 8
Answer:
(2) 5

Question 6.
Find the matrix X if 2X + $\left[ \begin{matrix} 1 & 3 \\ 5 & 7 \end{matrix} \right] =\left[ \begin{matrix} 5 & 7 \\ 9 & 5 \end{matrix} \right]$ ………….. .
(1) $\left[ \begin{matrix} -2 & -2 \\ 2 & -1 \end{matrix} \right] $
(2) $\left[ \begin{matrix} 2 & 2 \\ 2 & -1 \end{matrix} \right] $
(3) $\left[ \begin{matrix} 1 & 2 \\ 2 & 2 \end{matrix} \right] $
(4) $\left[ \begin{matrix} 2 & 1 \\ 2 & 2 \end{matrix} \right] $
Answer:
(2) $\left[ \begin{matrix} 2 & 2 \\ 2 & -1 \end{matrix} \right] $

Question 7.
The two tangents from an external points P to a circle with centre at O are PA and PB. If ∠APB = 70° then the value of ∠AOB is ………….. .
(1) 100°
(2) 110°
(3) 120°
(4) 130°
Answer:
(2) 110°

Question 8.
The equation of a line passing through the origin and perpendicular to the line 7x – 3y + 4 = 0 is ………….. .
(1)7x – 3y + 4 = 0
(2) 3x – 7y + 4 = 0
(3) 3x + 7y = 0
(4) 7x – 3y = 0
Answer:
(3) 3x + 7y = 0

Question 9.
If x = a tan θ and y = b sec θ then ………….. .
(1) $\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1$
(2) $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
(3) $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=0$
(4) $\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=0$
Answer:
(1) $\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1$

Question 10.
The ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height is ………….. .
(1) 1:2:3
(2) 2:1:3
(3) 1:3:2
(4) 3:1:2
Answer:
(4) 3:1:2

Question 11.
The probability of getting a job for a person is $\frac { x }{ 3 }$. If the probability of not getting the job is $\frac { 2 }{ 3 }$ then the value of x is ………….. .
(1) 2
(2) 1
(3) 3
(4) 1.5
Answer:
(2) 1

Question 12.
Variance of the first 11 natural numbers is ………….. .
(1) √5
(2) √10
(3) 5√2
(4) 10
Answer:
(4) 10

Question 13.
If α and β are the roots of the equation ax² + bx + c = 0 then (α + β)² is ………….. .
(1) $-\frac{b^{2}}{a^{2}}$
(2) $\frac{c^{2}}{a^{2}}$
(3) $\frac{b^{2}}{a^{2}}$
(4) $\frac{b c}{a}$
Answer:
(3) $\frac{b^{2}}{a^{2}}$

Question 14.
If K(x) = 3x – 9 and L(x) = 7x – 10 then Lok is ………….. .
(1) 21x + 73
(2) -21x + 73
(3) 21x – 73
(4) 22x – 73
Answer:
(3) 21x – 73

PART – II

II. Answer any ten questions. Question No. 28 is compulsory. [10 × 2 = 20]

Question 15.
Let A = {1,2,3,4,…, 45} and R be the relation defined as “is square of ” on A. Write R as a subset of A × A. Also, find the domain and range of R.
Answer:
A = {1,2,3,4. . . .45} .
The relation is defined as “is square of’
R = {(1, 1) (2, 4) (3, 9) (4, 16) (5, 25) (6, 36)} .
Domain of R = {1, 2, 3, 4, 5, 6}
Range of R = {1,4,9,16,25,36}

Question 16.
If f(x) = 3x – 2, g(x) = 2x + k and if fog = gof, then find the value of k.
Answer:
f(x) = 3x – 2, g(x) = 2x + k
fog(x) = f(g(x)) = f(2x + k) = 3(2x + k) – 2 = 6x + 3k – 2
Thus, fog(x) = 6x + 3k – 2.
gof(x) = g(3x – 2) = 2(3x – 2) + k
Thus, gof(x) = 6x – 4 + k.
Given that fog = gof
Therefore, 6x + 3k – 2 = 6x – 4 + k
6x – 6x + 3k – k = – 4 + 2 ⇒ k = – 1

Question 17.
Find the rational form of the number $0 . \overline{123}$.
Answer:
Letx = $0 . \overline{123}$
= 0.123123123….
= 0.123 + 0.000123 + 000000123 + ….
This is an infinite G.P
Here a = 0.123, r = $\frac{0.000123}{0.123}$ = 0.001
S_n = $\frac{a}{1-r}=\frac{0.123}{1-0.001}=\frac{0.123}{0.999}=. \frac{41}{333}$

Question 18.
How many consecutive odd integers beginning with 5 will sum to 480?
Answer:
First term (a) = 5
Common difference (d) = 2 (consecutive odd integer)
S_n = 480

$\frac { n }{ 2 }$ [2a + (n – 1)d] = 480
$\frac { n }{ 2 }$ [10 + (n – 1)2] = 480
$\frac { n }{ 2 }$ [10 + 2n – 2] = 480
$\frac { n }{ 2 }$ (8 + 2 n) = 480
n( 4 + n) = 480
4n + n² – 480 = 0
n² + 4n – 480 = 0
(n + 24) (n – 20) = 0
n + 24 = 0 or n – 20 = 0
n = -24 or n = 20 [number of terms cannot be negative]
∴ Number of consecutive odd integers is 20

Question 19.
Simplify $\frac{5 t^{3}}{4 t-8} \times \frac{6 t-12}{10 t}$
Answer:

Question 20.
Solve the following quadratic equations by factorization method √2x² + 7x + 5 √2 = 0
Answer:
√2x² + 7x + 5√2 = 0
√2x² + 2x + 5x + 5√2 = 0
√2x(x + √2) + 5(x + √2) = 0
(x + √2) + 5(x + √2) = 0

(x + √2) or 5(x + √2) = 0 (equate the product of factors to zero)
x = -√2 or √2x = -5 x ⇒ x = $\frac{-5}{\sqrt{2}}$
The roots are -√2, $\frac{-5}{\sqrt{2}}$

Question 21.
Find the value of a, b, c, d from the equation $\begin{pmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{pmatrix}=\begin{pmatrix} 1 & 5 \\ 0 & 2 \end{pmatrix}$
Answer:
The given matrices are equal. Thus all corresponding elements are equal.
Therefore, a – b = 1 …(1)
2a + c = 5 …(2)
2a – b = 0 ….(3)
3c + d = 2 …(4)
(3) gives 2a – b = 0 …(4)
2 a = b …(5)
Put 2a = b in equation (1), a – 2a = 1 gives a = -1
Put a = -1 in equation (5), 2(-1) = b gives b = -2 .
Put a = -1 in equation (2), 2(-1) + c = 5 gives c = 7
Put c = 7 in equation (4), 3(7) + d = 2 gives = -19
Therefore, a = -1, b = -2, c = 7, d = -19

Question 22.
In ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC If AD = 8x -7 , DB = 5x – 3 , AE = 4x – 3 and EC = 3x – 1, find the value of x.
Answer:

Given AD = 8x – 7; BD = 5x – 3; AE = 4x – 3; EC = 3x – 1
In ∆ABC we have DE || BC
By Basic proportionality theorem
$\frac{A D}{D B}=\frac{A E}{E C}$
$\frac{8 x-7}{5 x-3}=\frac{4 x-3}{3 x-1}$
(8x – 7) (3x -1) = (4x -3) (5x -3)
24x² – 8x – 21x + 7 = 20x² – 12x – 15x + 9
24x² – 20x² – 29x + 27x + 7 – 9 = 0
4x² – 2x – 2 = 0 .
2x² – x – 1 = 0 (Divided by 2)
2x² – 2x + x – 1 = 0

2x(x – 1) + 1 (x – 1) = 0
(x – 1) (2x + 1) = 0
x – 1 = 0 or 2x + 1 = 0
x = 1 or 2x = -1 ⇒ x = $\frac { 1 }{ 2 }$ (Negative value will be omitted)
The value of x = 1

Question 23.
The hill in the form of a right triangle has its foot at (19, 3) . The inclination of the hill to the ground is 45°. Find the equation of the hill joining the foot and top.
Answer:

Slope of AB (m) = tan 45° = 1
Equation of the hill joining the foot and the top is 45°
y – y₁ = m(x – x₁
y – 3 = 1(x – 19)
y – 3 = x – 19
– x + y – 3 + 19 = 0
– x + y + 16 = 0
x – y – 16=0
The required equation is x – y – 16 = 0

Question 24.
Prove that $\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}$ = sec θ + tan θ
Answer:

Question 25.
If the total surface area of a cone of radius 7cm is 704 cm², then find its slant height.
Answer:
Given that, radius r = 7 cm
Now, total surface area of the cone = πr(l + r)sq. units
T.S.A = 704 cm²
704 = $\frac { 22 }{ 7 }$ × 7(l + 7)
32 = l + 7 implies l = 25 cm
Therefore, slant height of the cone is 25 cm.

Question 26.
The first term of an A.P is 6 and the common difference is 5. Find the A.P and its general term.
Answer:
Given a = 6, d = 5
General term t_n = a + (n – 1) d
= 6 + (n – 1)5
= 6 + 5n – 5
= 5n + 1
The general form of the A.P is a, a + d, a + 2d ………
The A.P. is 6, 11, 16, 21 …. 5n + 1

Question 27.
If θ is an acute angle and tan θ + cot θ = 2 find the value of tan⁷θ + cot⁷θ
Answer:
Given tan θ + cot θ = 2
tan θ + $\frac{1}{\tan \theta}$ = 2
$\frac{\tan ^{2} \theta+1}{\tan \theta}$ = 2
tan²θ + 1 = 2 tan θ
tan²θ – 2 tan θ + 1 = 0
(tan θ – 1)² = 0
∴ tanθ – 1 = 0
tanθ = 1
tanθ = tan45 ⇒ θ = 45°
tan⁷θ + cot⁷θ = tan⁷45° + cot ⁷45°
= (1)⁷ + (1)^{7

= 2}

Question 28.
Cards marked with the numbers 2 to 101 are placed in a box and mixed thoroughly one card is drawn from this box. Find the probability that the number on the card is a number which is a perfect square.
Answer:
Sample space = {2, 3, 4,… 101}
n(s) = 100
Let A be the event of getting perfect square numbers
A= {4, 9, 16, 25, 36, 49, 64, 81, 100}
n(A) = 9
P(A) = $\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{9}{100}$
Probability of getting a card marked with a number which is a perfect square is $\frac{9}{100}$

PART – III

III. Answer any ten questions. Question No. 42 is compulsory. [10 × 5 = 50]

Question 29.
Consider the functions f(x) = x², g(x) = 2x and h(x) = x + 4 Show that (fog)oh = fo(goh)
Answer:
f(x) = x² ; g(x) = 2x and h(x) = x + 4
(fog) x = f[g(x)]
= f( 2X)
= (2x)²
= 4x²
(fog) oh (x) = fog [h(x)]
= fog(x + 4)
= 4(x + 4)²
= 4[x² + 8x + 16]
= 4x² + 32x + 64 ….(1)
goh (x) = g[h(x)]
= g(x + 4)
= 2(x + 4)
= 2x + 8
fo(goh)x = fo[goh(x)}
= f[2x + 8}
= (2x + 8)²
= 4x² + 32x + 64 ……. (2)
From (1) and (2) we get (fog) oh = fo(goh)

Question 30.

(i) f(4)
(ii) f(-2)
(iii) f(4) + 2f(1)
(iv) $\frac{f(1)-3 f(4)}{f(-3)}$
Answer:
The function f is defined by three values in intervals I, II, III as shown by the side

For a given value of x = a, find out the interval at which the point a is located, there after find f(a) using the particular value defined in that interval.
(i) First, we see that, x = 4 lie in the third interval.
Therefore, f(x) = 3x – 2 ; f(4) = 3(4) – 2 = 10

(ii) x = -2 lies in the second interval.
Therefore, f(x) = x² – 2 ; f(-2) = (-2)² – 2 = 2

(iii) From (i), f(4) = 10.
To find f(1), first we see that x = 1 lies in the second interval.
Therefore,f(x) = x² – 2 => f(1) = 1² – 2 = -1
So, f(4) + 2f(1) = 10 + 2 (-1) = 8

(iv) We know that f(1) = -1 and f(4) = 10.
For finding f(-3), we see that x = -3 , lies in the first interval.
Therefore, f(x) = 2x + 7; thus, f(-3) = 2(-3) + 7 = 1
Hence, $\frac{f(1)-3 f(4)}{f(-3)}=\frac{-1-3(10)}{1}=-31$

Question 31.
If (m + 1)^th term of an A.P. is twice the (n + 1)^th> term, then prove that (3m + 1)^th term is twice the (m + n + 1)^th term.
Answer:
t_n = a + (n – 1)d
Given t_m+1 = 2 t_n+1
a + (m + 1 – 1)d = 2[a + (n + 1 – 1)d]
a + md = 2(a + nd) ⇒ a + md = 2a + 2nd
md – 2nd = a
d(m – 2n) = a ……. (1)
To prove that t_{3m + 1} = 2(t_{3m + n + 1})
L.H.S. = t_{3m + 1}
= a + (3m + 1 – 1 )d
= a + 3md
= d(m – 2n) + 3md (from 1)
= md – 2nd + 3md
= 4md – 2nd
= 2d (2m – n)

R.H.S. = 2 (t_{m + n + 1})
= 2[a + (m + n + 1 – 1) d]
= 2 [a + (m + n)d]
= 2 [d (m – 2n) + md + nd)] (from 1)
= 2 [dm – 2nd + md + nd]
= 2 [2md – nd] = 2d (2m – n)
R.H.S = L.H.S
∴ t_{(3m + 1)}= 2t_{(m + n + 1)}

Question 32.
Find the sum to n terms of the series 5 + 55 + 555 + ….
Answer:
The series is neither Arithmetic nor Geometric series. So it can be split into two series and then find the sum.
5 + 55 + 555 + …. + n terms = 5[1 + 11 + 111 + …. + n terms]
= $\frac { 5 }{ 9 }$[9 + 99 + 999 + …. + « terms]
= $\frac { 5 }{ 9 }$[(10 – 1) + (100 – 1) + (1000 – 1) + …. + n terms)]
= $\frac { 5 }{ 9 }$[(10 + 100 + 1000 + …. +n terms) -n]
= $\frac{5}{9}\left[\frac{10\left(10^{n}-1\right)}{(10-1)}-n\right]=\frac{50\left(10^{n}-1\right)}{81}-\frac{5 n}{9}$

Question 33.
Vani, her father and her grand father have an average age of 53. One-half of her grand father’s age plus one-third of her father’s age plus one fourth of Vani’s age is 65. If 4 years ago Vani’s grandfather was four times as old as Vani then how old are they all now?
Answer:
Let the age of Vani be”x” years
Vani father age = “y” years
Vani grand father = “z” years
By the given first condition.
$\frac{x+y+z}{3}=53$
x + y + z = 159….(1)
By the given 2nd condition.
$\frac{1}{2} z+\frac{1}{3} y+\frac{1}{4} x=65$
Multiply by 12
6z + 4y +3x = 780
3x + 4y + 6z = 780 ….(2)
By the given 3^rd condition
z – 4 = 4 (x – 4) ⇒ z – 4 = 4x – 16
-4x + z = -16 + 4
4x – z = 12 ….(3)

Substitute the value of x = 24 in (3)
4 (24) – z = 12
96 – z = 12
-Z = 12 – 96
z = 84
Substitute the value of
x = 24 and z = 84 in (1)
24 + y + 84 = 159
y + 108 = 159
y = 159 – 108
= 51
Vani age = 24 years
Vani’s father age = 51 years
Vani – grand father age = 84 years

Question 34.
If A = $\left( \begin{matrix} 1 & 2 & 1 \\ 2 & -1 & 1 \end{matrix} \right)$ and B = $\left( \begin{matrix} 2 & -1 \\ -1 & 4 \\ 0 & 2 \end{matrix} \right)$ show that (AB)^T = B^TA^T
Answer:

Question 35.
In figure, O is the centre of the circle with radius 5 cm. T is a point such that OT = 13 cm and OT intersects the circle E, if AB is the tangent to the circle at E, find the length of AB.
Answer:

In the right ∆ OTP,
PT² = OT² – OP²
= 13² – 5²
= 169 – 25 = 144
PT = √144 = 12 cm
Since lengths of tangent drawn from a point to circle are equal.
∴ AP = AE = x .
AT = PT – AP
= (12 – x) cm
Since AB is the tangent to the circle E.
∴ OE ⊥ AB .
∠OEA= 90°
∠AET = 90°
In ∆AET, AT² = AE² + ET²
In the right triangle AET,
AT² = AE² + ET²
(12 – x)² = x² + (13 – 5)²
144 – 24x + x² = x² + 64
24x = 80 ⇒ x = $\frac{80}{24}=\frac{20}{6}=\frac{10}{3}$
BE = $\frac{10}{3}$ cm
AB = AE + BE
= $\frac{10}{3}+\frac{10}{3}=\frac{20}{3}$
∴ Length of AB = $\frac{20}{3}$ cm

Question 36.
Find the area of the quadrilateral whose vertices are at (-9, 0), (-8, 6), (-1, -2) and (-6,-3)
Answer:

Let the vertices A(-9, 0), B(-8, 6), C(-1, -2) and D(-6, -3)
Plot the vertices in a graph and take them in counter – clock wise order.
Area of the Quadrilateral DCB
= $\frac { 1 }{ 2 }$ [(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄ )]

= $\frac { 1 }{ 2 }$ [27 + 12- 6 + 0 -(0 + 3 + 16 – 54)]
= $\frac { 1 }{ 2 }$ [33 -(-35)]
= $\frac { 1 }{ 2 }$ [33 + 35] = $\frac { 1 }{ 2 }$ × 68 = 34 sq. units.
Area of the Quadrilateral = 34 sq. units

Question 37.
A pole 5 m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point ‘A’ on the ground is 60° and the angle of depression to the point ‘A’ from the top of the tower is 45°. Find the height of the tower. (√3 = 1.732)
Answer:
Let BC be the height of the tower and CD be the height of the pole.
Let‘A’be the point of observation.
Let BC = x and AB = y.
From the diagram,

∠BAD = 60° and ∠XCA = 45° = ∠BAC ,
In right triangle ABC, tan 45° = $\frac{B C}{A B}$
gives 1 = $\frac { x }{ y }$ so, x = y …… (1)
In right triangle ABD, tan60° = $\frac{B D}{A B}=\frac{B C+C D}{A B}$
gives √3 =  so, √3y = x + 5
we get √3x = x + 5 [From (1)]

Hence, height of the tower is 6.83 m.

Question 38.
A shuttle cock used for playing badminton has the shape of a frustum of a cone is mounted on a hemisphere. The diameters of the frustum are 5 cm and 2 cm. The height of the entire shuttle cock is 7 cm. Find its external surface area.
Answer:

Radius of the lower end of the frustum (r) = 1 cm
Radius of the upper end of the frustum (R) = 2.5 cm
Height of the frustum (h) = 6 cm
Let “l” be the slant height of the frustum
l = $\sqrt{h^{2}+(\mathrm{R}-r)^{2}}$
= $\sqrt{6^{2}+(2.5-1)^{2}}$
= $\sqrt{36+2.25}$ = $\sqrt{38.25}$
= 6.18 cm
External surface area of shuttle cock = C.S.Aof the frustum + C.S.Aof a hemisphere
= πl(R + r) + 2 πr²
= π [6.18 (2.5 + 1) + 2 × 1²] cm²
= $\frac { 22 }{ 7 }$[6.18 × 3.5 + 2]
= $\frac { 22 }{ 7 }$ × (21.63 + 2)
= $\frac { 22 }{ 7 }$ × 23.63 cm² = 74.26 cm²
External surface area = 74.26 cm²

Question 39.
The mean and variance of seven observations are 8 and 16 respectively. If five of these are 2, 4,10,12 and 14, then find the remaining two observations.
Answer:
Let the missing two observation be ‘a’ and ‘b’
Arithmetic mean = 8

$\frac{2+4+10+12+14+a+b}{7}$ = 8 ⇒ $\frac{42+a+b}{7}$ = 8
a + b + 42 = 56
a + b = 56 – 42
a + b = 14 ……… (1)
Variance = 16
Variance = $\frac{\Sigma x_{i}^{2}}{n}-\left(\frac{\Sigma x_{i}}{n}\right)^{2}$

560 – 460 = a² + b²
a² + b²= 100 ⇒ (a + b)² – 2ab = 100 [a² + b² = (a+b)² – 2 ab]
14² – 2 ab = 100 ⇒ 196 – 2 ab = 100 [a + b = 14(from (1)]
196 – 100 = 2ab
96 = 2ab ⇒ ab = $\frac{96}{2}$ = 48
∴ b = $\frac{48}{a}$ …… (2)
Substitute thr value of b = $\frac{48}{a}$ in (1)
a + $\frac{48}{a}$ = 14 ⇒ a² + 48 = 14a
a² – 14a + 48 = 0 ⇒ (a – 6) (a – 8) = 0
a = 6 or 8
when a = 6
b = $\frac{48}{a}=\frac{48}{6}=8$ = 8

when a = 8
b = $\frac{48}{a}=\frac{48}{6}=8$ = 6

∴ Missing observation is 8 and 6 (or) 6 and 8

Question 40.
From a solid circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and same base is removed. Find (i) the volume of the remaining solid also, find (ii) the whole surface area.
Answer:
Circular cylinder
Radius of the base (r) = 6 cm
Height of the cylinder (h) = 10 cm

Circular Cone
Radius of the base (R) = 6 cm
Height of the cone (H) = 10 cm

(i) Volume of the remaining solid = Volume of the cylinder – Volume of the cone

= πr²h – $\frac { 1 }{ 3 }$πR²H
= π(r²h – $\frac { 1 }{ 3 }$R²H)
= $\frac { 22 }{ 7 }$ [6 × 6 × 10 – $\frac { 1 }{ 3 }$ × 6 × 6 × 10] cm²
= $\frac { 22 }{ 7 }$ [360 – 120] cm³
= $\frac { 22 }{ 7 }$ × 240 cm³
= $\frac { 5280 }{ 7 }$ cm³ = 754.29 cm³

(ii) Slant height of a cone = $\sqrt{h^{2}+r^{2}}$
= $\sqrt{10^{2}+6^{2}}$
= $\sqrt{100+36}$
= $\sqrt{136}$ = 11.66 cm
Whole surface area of the solid = curved surface area of the cylinder + curved surface area of the cone + base area
= 2 πrh + πRI + πr²
= π[2 × 6 × 10 + 6 × 11.66 + 6 × 6]
= $\frac { 22 }{ 7 }$[120 + 69.96 + 36] cm²
= $\frac { 22 }{ 7 }$ × 225.96 cm²
= $\frac { 4971.12 }{ 7 }$ cm²
= 710.16 cm²
(i) Volume of the remaining solid = 754.29 cm³
(ii) whole surface area = 710.16 cm²

Question 41.
If a and p are the roots of the equation 3x² – 6x + 1=0 form the equation whose roots are 2α + β and 2β + α
Answer:
α and β are the roots of 3x² – 6x + 1 = 0
α + β = $\frac { 6 }{ 3 }$ = 2
αβ = $\frac { 1 }{ 3 }$
Given the roots are 2α + β; 2β + α
Sum of the roots = 2α + β + 2β + α
= 2(α + β) + (α + β)
= 2(2) + 2
= 6
Product of roots = (2α + β) (2β + α)
= 4αβ + 2α² + 2β² + αβ
= 5αβ + 2(α² + β²)
5αβ + 2[(α + β)² – 2αβ]
= 5 × $\frac { 1 }{ 3 }$ + 2 (4 – 2 × $\frac { 1 }{ 3 }$)
= $\frac { 5 }{ 3 }$ + $2\left(\frac{12-2}{3}\right)$
= $\frac { 5 }{ 3 }$ + 2 × $\frac { 10 }{ 3 }$
= $\frac{5}{3}+\frac{20}{3}$
= $\frac { 25 }{ 3 }$
The quadratic polynomial is x² – (sum of the roots)x – product of the roots = 0
x² – 6x + $\frac { 25 }{ 3 }$ = 0
3x² – 18x + 25 = 0

Question 42.
Tw o dice are rolled simultaneously. Find the probability that that sum of the numbers on the faces is neither divisible by 3 nor by 4.
Answer:
Sample space = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3.1) , (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5.1) , (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
n(S) = 36
Let A be the event of getting the sum is divisible by 3 ‘
A = { (1,2) (2,1) (1,5) (5,1) (2,4) (4,2) (3,3) (3,6) (6,3) (4,5) (5,4) (6,6)}
n (A) = 12
P(A) = $\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{12}{36}$
Let B be the event of getting a sum is divisible by 4.
B = {(1,3) (2,2) (2,6) (3,1) (3,5) (4,4) (5,3) (6,2) (6,6)} n (B) = 9
n(B) = 9
P(B) = $\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{9}{36}$
A ∩ B = {(6,6)}
n(A ∩ B) = 1
P(A ∩ B) = $\frac{n(\mathrm{A} \cap \mathrm{B})}{n(\mathrm{S})}=\frac{1}{36}$
P(A ∪ B) = P(A) + P(B) – P (A ∩ B)
$=\frac{12}{36}+\frac{9}{36}-\frac{1}{36}$
$=\frac{12+9-1}{36}=\frac{20}{36}$
Neither divisible by 3 nor by 4
P(A’ ∩ B’) = P(A ∪ B)’
= 1 – p(A ∪ B) = 1 – $\frac{20}{36}=\frac{36-20}{36}$
= $\frac{16}{36}=\frac{4}{9}$

PART-IV

IV. Answer all the questions. [2 × 8 = 16]

Question 43.
(a) Draw the two tangents from a point which is 10 cm away from the centre of a circle of radius 5 cm. Also, measure the lengths of the tangents.
Answer:

Steps of construction:

With. O as centre, draw a circle of radius 5 cm.
Draw a line OP = 10 cm.
Draw a perpendicular bisector of OP, which cuts OP at M.
With M as centre and MO as radius draw a circle which cuts previous circle at A and B.
Join AP and BP. AP and BP are the required tangents.

Verification: In the right ∆ OAP
PA² = OP² – OA²
= 10² – 5² = $\sqrt{100-25}$ = √75 = 8.7 cm.
Length of the tangent is 8.7 cm

[OR]

(b) Construct a ∆PQR which the base PQ = 4.5 cm, R = 35° and the median from R to RG is 6 cm.
Answer:

Steps of construction

Draw a line segment PQ = 4.5 cm
At P, draw PE such that ∠QPE = 60°
At P, draw PF such that ∠EPF = 90°
Draw the perpendicular bisect to PQ, which intersects PF at O and PQ at G.
With O as centre and OP as radius draw a circle.
From G mark arcs of radius 5.8 cm on the circle. Mark them at R and S
Join PR and RQ. PQR is the required triangle.

Question 44.
(a) Draw the graph of y = x² and hence solve x² – 4x – 5 = 0.
Answer:
Given equations are y = x² and x² – 4x – 5 = 0
(i) Assume the values of x from – 4 to 5.

(ii) Plot the points (-4, 16), (- 3, 9), (-2,4), (-1, 1), (0, 0), (1, 1), (2,4), (3, 9), (4, 16), (5, 25).
(iii) Join the points by a smooth curve.
(iv) Solve the given equations

(v) The points of intersection of the line and the parabola are (-1, 1) and (5, 25).
The x-coordinates of the points are -1 and 5.
Thus solution set is {- 1, 5}.

[OR]

(b) Draw the graph of y – x² -5x – 6 and hence solve x² – 5x – 14 = 0
Answer:
Let y = x² – 5x – 6
(i) Draw the graph of y = x² – 5x – 6 by preparing the table of values as below.

(ii) Plot the points (-3, 18), (-2, 8), (-1, 0), (0, -6), (1, -10), (2, -12), (3, -12), (4, -10), (5, -6), (6, 0) and (7, 8).
(iii) Join the points by a free hand to get smooth curve.
(iv) To solve x² – 5x – 14 = 0, subtract x² – 5x – 14 = 0 from y = x² – 5x – 6.

The equation y = 8 represent a straight line draw a straight line through y = 8 intersect the curve at two places. From the two points draw perpendicular line to the X – axis it will intersect at -2 and 7.
The solution is -2 and 7

Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Samacheer Kalvi 11th Accountancy Bank Reconciliation Statement Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the Correct Answer

Question 1.
A bank reconciliation statement is prepared by ………………
(a) Bank
(b) Business
(c) Debtor to the business
(d) Creditor to the business
Answer:
(b) Business

Question 2.
A bank reconciliation statement is prepared with the help of ………………
(a) Bank statement
(b) Cash book
(c) Bank statement and bank column of the cash book
(d) Petty cash book
Answer:
(c) Bank statement and bank column of the cash book

Question 3.
Debit balance in the bank column of the cash book means ………………
(a) Credit balance as per bank statement
(b) Debit balance as per bank statement
(c) Overdraft as per cash book
(d) None of the above
Answer:
(a) Credit balance as per bank statement

Question 4.
A bank statement is a copy of ………………
(a) Cash column of the cash book
(b) Bank column of the cash book
(c) A customer’s account in the bank’s book
(d) Cheques issued by the business
Answer:
(c) A customer’s account in the bank’s book

Question 5.
A bank reconciliation statement is prepared to know the causes for the difference between:
(a) The balance as per the cash column of the cash book and bank column of the cash book
(b) The balance as per the cash column of the cash book and bank statement
(c) The balance as per the bank column of the cash book and the bank statement
(d) The balance as per petty cash book and the cash book
Answer:
(c) The balance as per the bank column of the cash book and the bank statement

Question 6.
When money is withdrawn from bank, the bank ………………
(a) Credits customer’s account
(b) Debits customer’s account
(c) Debits and credits customer’s account
(d) None of these
Answer:
(b) Debits customer’s account

Question 7.
Which of the following is not the salient feature of bank reconciliation statement?
(a) Any undue delay in the clearance of cheques will be shown up by the reconciliation
(b) Reconciliation statement will discourage the accountant of the bank from embezzlement
(c) It helps in finding the actual position of the bank balance
(d) Reconciliation statement is prepared only at the end of the accounting period
Answer:
(d) Reconciliation statement is prepared only at the end of the accounting period

Question 8.
Balance as per cash book is ₹ 2,000. Bank charge of ₹ 50 debited by the bank is not yet shown in the cash book. What is the bank statement balance now?
(a) ₹ 1,950 credit balance
(a) ₹ 1,950 credit balance
(b) ₹ 1,950 debit balance
(c) ₹ 2,050 debit balance
(d) ₹ 2,050 credit balance
Answer:
(a) ₹ 1,950 credit balance

Question 9.
Balance as per bank statement is ₹ 1,000. Cheque deposited, but not yet credited by the bank is ₹ 2, 000. What is the balance as per bank column of the cash book?
(a) ₹ 3,000 overdraft
(b) ₹ 3,000 favourable
(c) ₹ 1,000 overdraft
(d) ₹ 1,000 favourable
Answer:
(b) ₹ 3,000 favourable

Question 10.
Which one of the following is not a timing difference?
(a) Cheque deposited but not yet credited
(b) Cheque issued but not yet presented for payment
(c) Amount directly paid into the bank
(d) Wrong debit in the cash book
Answer:
(d) Wrong debit in the cash book

II. Very Short Answer Questions

Question 1.
What is meant by bank overdraft?
Answer:
It is not possible to have unfavourable cash balance in the cash book. But, it is possible to have unfavourable balance in the bank account. When the business
is not having sufficient money in its bank account, it can borrow money from the bank. As a result of this, amount is overdrawn from bank.

Question 2.
What is bank reconciliation statement?
Answer:
If every entry in the cash book matches with the bank statement, then bank balance will be the same in both the records. But, practically it may not be possible. When the balances do not agree with each other, the need for preparing a statement to explain the causes arises. This statement is called bank reconciliation statement (BRS).

Question 3.
State any two causes of disagreement between the balance as per bank column of cash book and bank statement.
Answer:
(a) Cheques issued but not yet presented for payment.
(b) Cheques deposited into bank but not yet credited.

Question 4.
Give any two expenses which may be paid by the banker as per standing instruction.
Answer:
Insurance premium, loan instalment, etc., paid as per standing instructions.

Question 5.
Substitute the following statements with one word/phrase
(a) A copy of customer’s account issued by the bank.
(b) Debit balance as per bank statement.
(c) Statement showing the causes of disagreement between the balance as per cash book and balance as per bank statement.
Answer:
(a) Pass book
(b) Pass book favourable
(c) (1) Timing difference, (2) Errors in recording

Question 6.
Do you agree on the following statements? Write “yes” if you agree, and write “no” if you disagree.
(a) Bank reconciliation statement is prepared by the banker.
(b) Adjusting the cash book before preparing the bank reconciliation statement is compulsory.
(c) Credit balance as per bank statement is an overdraft.
(d) Bank charges debited by the bank increases the balance as per bank statement.
(e) Bank reconciliation statement is prepared to identify the causes of differences between balance as per bank column of the cash book and balance as per cash column of the cash book.
Answer:
(a) No
(b) No
(c) No
(d) No
(e) Yes

III. Short Answer Questions

Question 1.
Give any three reasons for preparing bank reconciliation statement.
Answer:
The main reasons for preparing bank reconciliation statement are:

To identify the reasons for the difference between the bank balance as per the cash book and bank balance as per bank statement.
To identify the delay in the clearance of cheques.
To ascertain the correct balance of bank column of cash book.

Question 2.
What is meant by the term “cheque not yet presented?”
Answer:
When the cheques are issued by the business, it is immediately entered on the credit side of the cash book by the business. But, this may not be entered in the bank statement on the same day. It will be entered in the bank statement only after it is presented with the bank.

Question 3.
Explain why does money deposited into bank appear on the debit side of the cash book, but on the credit side of the bank statement?
Answer:
When the cheques are deposited into bank, the amount is debited in the cash book on the same day. But, these may not be shown in the bank pass book on the same day because these will be entered in the bank statement only after the collection of the cheques.

Question 4.
What will be the effect of interest charged by the bank, if the balance is an overdraft?
Answer:
The bank has to cover the cost of running the customer’s account. So debit is given to the account of the business towards bank charges. Also, if the business had taken any loan or overdrawn, interest has to be paid by the business. These entries for bank charges and interest are made in the bank statement. But, the entry is made in the cash book only when the bank statement is received by the business. Till then, the Cash book shows more balance than bank statement.

Question 5.
State the timing differences in BRS with examples.
Answer:
The timing differences in BRS are:
(a) cheques issued but not yet presented for payment
(b) cheques deposited into bank but not yet credited
(c) bank charges and interest on loan and overdraft
(d) interest and dividends collected by the bank
(e) dishonour of cheques and bills
(f) amount paid by parties directly into the bank
(g) payment made directly by the bank to others
(h) bills collected by the bank on behalf of its customer

IV. Exercises

Question 1.
From the following particulars prepare a bank reconciliation statement of Jayakumar as on 31^st December, 2016. (3 Marks)
(a) Balance as per cash book ₹ 7,130
(b) Cheque deposited but not cleared ₹ 1,000
(c) A customer has deposited ₹ 800 into the bank directly
Answer:
Bank reconciliation statement of Jayakumar as on 31^st December, 2016

Question 2.
From the following particulars of Kamakshi traders, prepare a bank reconciliation statement as on 31^st March, 2018. (3 Marks)
(a) Debit balance as per cash book ₹ 10,500
(b) Cheque deposited into bank amounting to ₹ 5,500 credited by bank, but entered twice in the cash book
(c) Cheques issued and presented for payment amounting to ₹ 7,000 omitted in the cash book
(d) Cheque book charges debited by the bank ₹ 200 not recorded in the cash book.
(e) Cash of ₹ 1,000 deposited by a customer of the business in cash deposit machine not recorded in the cash book.
Answer:
Bank reconciliation statement of Kamakshi as on 31^st March, 2018

Question 3.
From the following information, prepare bank reconciliation statement to find out the bank statement balance as on 31^st December, 2017. (5 Marks)

Answer:
Bank reconciliation statement as on 31^st December, 2017

Question 4.
On 31^st March, 2017, Anand’s cash book showed a balance of ₹ 1,12,500. Prepare bank reconciliation statement. (5 Marks)
(a) He had issued cheques amounting to ₹ 23,000 on 28.3.2017, of which cheques amounting to ₹ 9,000 have so far been presented for payment.
(b) A cheque for ₹ 6,300 deposited into bank on 27.3.2017, but the bank credited the same only on 5th April 2017.
(c) He had also received a cheque for ₹ 12,000 which, although entered by him in the cash book, was not deposited in the bank.
(d) Wrong credit given by the bank on 30th March 2017 for ₹ 2,000.
(e) On 30th March 2017, a bill already discounted with the bank for ₹ 3,000 was dishonoured, but no entry was made in the cash book.
(f) Interest on debentures of ₹ 700 was received by the bank directly.
(g) Cash sales of ₹ 4,000 wrongly entered in the bank column of the cash book.
Answer:
Bank reconciliation statement of Anand as on 31^st March, 2017

Question 5.
From the following particulars of Siva and Company, prepare a bank reconciliation statement as on 31^st December, 2017. (2 Marks)
(a) Credit balance as per cash book ₹ 12,000.
(b) A cheque of ₹ 1,200 issued and presented for payment to the bank, wrongly credited in the cash book as ₹ 2,100.
(c) Debit side of bank statement was undercast by ₹ 100.
Answer:
Bank reconciliation statement of Siva and Company as on 31^st December, 2017

Question 6.
From the following particulars of Raheem traders, prepare a bank reconciliation statement as on 31^st March, 2018. (3 Marks)
(a) Overdraft as per cash book ₹ 2,500
(b) Debit side of cash book was undercast by ₹ 700
(c) Amount received by bank through RTGS amounting to ₹ 2,00,000, omitted in the cash book.
(d) Two cheques issued for ₹ 1,800 and ₹ 2,000 on 29^th March 2018. Only the second cheque is presented for payment.
(e) Insurance premium on car for ₹ 1,000 paid by the bank as per standing instruction not recorded in the cash book.
Answer:
Bank reconciliation statement of Raheem as on 31^st March, 2018

Question 7.
From the following information, prepare bank reconciliation statement as on 31^st December, 2017 to find out the balance as per bank statement. (5 Marks)

Answer:
Bank reconciliation statement as on 31^st December, 2017

Question 8.
Prepare bank reconciliation statement from the following data. (5 Marks)

Answer:

Question 9.
From the following particulars of Veera traders, prepare a bank reconciliation statement as on 31^st December, 2017. (2 Marks)
(a) Credit balance as per bank statement ₹ 6,000
(b) Amount received by bank through NEFT for ₹ 3,500, entered twice in the cash book.
(c) Cheque dishonoured amounting to ₹ 2,500, not entered in cash book.
Answer:
Bank reconciliation statement of Veera Traders as on 31^st December, 2017

Question 10.
Prepare bank reconciliation statement from the following data and find out the balance as per cash book as on 31^st March, 2018. (3 Marks)

Answer:
Bank reconciliation statement as on 31^st March, 2018

Question 11.
Ascertain the cash book balance from the following particulars as on 31^st December, 2017: (5 Marks)

Credit balance as per bank statement ₹ 2,500
Bank charges of ₹ 60 have not been entered in the cash book
Cheque deposited on 28^th December 2017 for ₹ 1,000 was not yet credited by the bank
Cheque issued on 24^th December 2017 for ₹ 700, not yet presented for payment
A dividend of ₹ 400 collected by the bank directly but not entered in the cash book
A cheque of ₹ 600 had been dishonoured, but no entry was made in the cash book
Interest on term loan ₹ 1,200 debited by bank but not accounted in cash book
No entry had been made in the cash book for a trade subscription of ₹ 500 paid vide banker’s order on 23^rd December 2014

Answer:
Bank reconciliation statement as on 31^st December, 2017

Question 12.
From the following particulars of Raja traders, prepare a bank reconciliation statement as on 31^st January, 2018. (5 Marks)
(a) Balance as per bank statement ₹ 5,000
(b) Cheques amounting to ₹ 800 had been recorded in the cash book as having been deposited into the bank on 25^th January 2018, but were entered in the bank statement on 2^nd February 2018.
(c) Amount received by bank through NEFT amounting to ₹ 3,000, omitted in the cash book.
(d) Two cheques issued for ₹ 3,000 and ₹ 2,000 on 29^th March 2018. Only the first cheque is presented for payment.
(e) Insurance premium on motor vehicles for ₹ 1,000 paid by the bank as per standing instruction not recorded in the cash book.
(f) Credit side of cash book was undercast by ₹ 700
(g) Subsidy received directly by the bank from the state government amounting to ₹ 10,000, not entered in cash book.
Answer:
Bank reconciliation statement of Raja Traders as on 31^st January, 2018

Question 13.
From the following particulars of Simon traders, prepare a bank reconciliation statement as on 31^st March, 2018. (2 Marks)
(a) Debit balance as per bank statement ₹ 2,500
(b) Cheques deposited amounting to ₹ 10,000, not yet credited by bank.
(c) Payment through net banking for ₹ 2,000, omitted in the cash book
Answer:
Bank reconciliation statement of Simon Traders as on 31^st March, 2018

Question 14.
From the following particulars, ascertain the cash book balance as on 31^st December, 2016.

Overdraft balance as per bank statement ₹ 1,26,640 (3 Marks)
Interest on overdraft entered in the bank statement, but not yet recorded in cash book ₹ 3,200
Bank charges entered in bank statement, but not found in cash book ₹ 600
Cheques issued, but not yet presented for payment ₹ 23,360
Cheques deposited into the bank but not yet credited ₹ 43,400
Interest on investment collected by the bank ₹ 24,000

Answer:
Bank reconciliation statement as on 31^st December, 2016

Question 15.
From the following particulars of John traders, prepare a bank reconciliation statement as on 31^st March, 2018. (5 Marks)
(a) Bank overdraft as per bank statement ₹ 4,000
(b) Cheques amounting to ₹ 2,000 had been recorded in the cash book as having been deposited into the bank on 26^th March 2018, but were entered in the bank statement on 4^th April 2018.
(c) Amount received by bank through cash deposit machine amounting to ₹ 5,000, omitted in the cash book.
(d) Amount of ₹ 3,000 wrongly debited to John traders account by the bank, for which no details are available.
(e) Bills for collection credited by the bank till 29th March 2017 amounting to ₹ 4,000, but no advice received by John traders.
(f) Electricity charges made through net banking for ₹ 900 was wrongly entered in cash column of the cash book instead of bank column.
(g) Cash sales wrongly recorded in the bank column of the cash book for ₹ 4,000.
Answer:
Bank reconciliation statement of John Traders as on 31^st March, 2018

Question 16.
Prepare bank reconciliation statement from the following data. (3 Marks)

Answer:
Bank reconciliation statement

Question 17.
Prepare bank reconciliation statement as on 31^st March, 2017 from the following extracts of cash book and bank statement.

Bank Statement

Answer:

Question 18.
A trader received his bank statement on 31^st December, 2017 which showed an overdraft balance of ₹ 12,000. On the same day, his cash book showed a debit balance of ₹ 2,000.
Analyse the following transactions. Choose the possible causes and prepare a bank reconciliation statement to show the causes of differences.
(a) Cheque deposited for ₹ 2,000 on 21^st December, 2017. Bank credited the same on 26^th December, 2017.
(b) Cheque issued for payment on 26^th December, 2017 amounting to ₹ 2,500, not yet presented until 31^st, December, 2017.
(c) Bank charges amounting to ₹ 200 not yet entered in the cash book.
(d) Online payment for ₹ 1,500 entered twice in the cash book.
(e) Cheque deposited amounting to ₹ 1,000, but omitted in the cash book. The same cheque was dishonoured by bank, but not yet entered in cash book.
(f) Cheque deposited, not yet credited by bank amounting to ₹ 17,800.
Answer:
Bank reconciliation statement as on 31^st December, 2017

Note: Transactions (a) and (c) have been entered in both cash book and bank statement also. So we need not false in to accounts.

Textbook Case Study Solved

Magesh, an enthusiastic young entrepreneur, started a business on 1^stDecember, 2017. He opened a current account with a nationalised bank for his business transaction. In the same bank, he maintains his personal savings bank account too. He did not find time to maintain his cash book. So he appointed a person called Dinesh to take care of bank transactions. But that person was inexperienced.

Question 1.
On 1^st December, 2017, the opening balance as per cash book and bank record was the same. On 2^nd December, Magesh issued a cheque for ₹ 2,000 to a supplier, but the same was entered in the credit side of the cash book as ₹ 200.
Answer:
Credit balance as per bank statement is ₹ 19,700.

Question 2.
On 3^rd, December, Magesh issued his savings bank account cheque for his personal expenses amounting to ₹ 2,500, but Dinesh assumed this as current account cheque and the same was entered in the cash book as drawings.
Answer:
Over casting of debit side of bank column of the cash book is ₹ 1800.

Question 3.
Dinesh was asked to deposit cash of ₹ 1,000 in cash deposit machine in order to make a payment to one of the business’ supplier. He credited the same in the bank column of the cash book.
Answer:
Wrong debit in cash book is ₹ 2,500.

Question 4.
On 15^th December, one of his customers made online payment to Magesh’s current account, amounting to ₹ 1,000. There was no entry in the cash book for this.
Answer:
Wrong credit in cash book is ₹ 1,000.
Instead of personal bank account he can open business bank account (i.e.) current account.

Question 5.
Dinesh received his salary in cash for ₹ 5,000. He credited this amount in the bank column of cash book.
Answer:
Online payment no recorded in cash book is ₹ 1,000.

Question 6.
Bank made payment on 23^rd December, amounting to ₹ 2,500, as per standing instruction. But, there is no entry in the cash book for the same.
Answer:
Wrong credit in the cash book of bank column is ₹ 5,000.

Question 7.
On 31^st, December 2017, Magesh received a bank statement from his bank, which showed a credit balance of ₹ 19,700. He instructed Dinesh to check the statement with the cash book. On comparing both, Dinesh found that the cash book showed a balance of ₹ 14,500. He was puzzled. He needs your help to reconcile the balances.
Answer:
Insurance premium paid but not entered in cash book is ₹ 2,500.
Bank reconciliation cash book as on 31^st December 2017

Samacheer Kalvi 12th Computer Applications Solutions Chapter 5 PHP Function and Array

Leave a Comment / Class 12 / By Prasanna

Students can Download Computer Applications Chapter 5 PHP Function and Array Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Computer Applications Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Computer Applications Solutions Chapter 5 PHP Function and Array

Samacheer Kalvi 12th Computer Applications PHP Function and Array Text Book Back Questions and Answers

I. Choose The Correct Answer

Question 1.
Which one of the following is the right way of defining a function in PHP?
(a) function {function body }
(b) data type functionName(parameters) {function body}
(c) functionName(parameters) {function body }
(d) function functionName(parameters) { function body }
Answer:
(d) function functionName(parameters) { function body }

Question 2.
A function in PHP which starts with ………. (double underscore) is known as …………………………
(a) Magic Function
(b) Inbuilt Function
(c) Default Function
(d) User Defined Function
Answer:
(a) Magic Function

Question 3.
PHP’s numerically indexed array begin with position ………………………..
(a) 1
(b) 2
(c) 0
(d) -1
Answer:
(c) 0

Question 4.
Which of the following are correct ways of creating an array?
(i) state[0] = “Tamilnadu”;
(ii) $state[ ] = arfay(“Tamilnadu”);
(iii) $state[0] = “Tamilnadu”;
(iv) $state = array(“Tamilnadu”);
(a) (iii) and (iv)
(b) (ii) and (iii)
(c) Only (i)
(d) (ii), (iii) and (iv)
Answer:
(a) (iii) and (iv)

Question 5.
What will be the output of the following PHP code? <?php
$a=array(“A”,“Cat”,“Dog”,“A”,“Dog”);
$b=array(“A”,“A”,“Cat”,“A”,“Tiger”);
$c=array_combine($a,$b);
print_r(array_count_values($c));
?>
(а) Array ([A] => 5 [Cat] => 2 [Dog] => 2 [Tiger] => 1)
(b) Array ([A] => 2 [Cat] => 2 [Dog] => 1 [Tiger] => 1)
(c) Array ([A] => 6 [Cat] => 1 [Dog] => 2 [Tiger] => 1)
(d) Array ([A] => 2 [Cat] => 1 [Dog] => 4 [Tiger] => 1)
Answer:
(а) Array ([A] => 5 [Cat] => 2 [Dog] => 2 [Tiger] => 1)

Question 6.
For finding nonempty elements in array we use
(a) is_array ( ) function)
(b) sizeof( ) function
(c) array_count ( ) function
(d) count ( ) function
Answer:
(d) count ( ) function

Question 7.
Indices of arrays can be either strings or numbers and they are denoted as
(a) $my_array {4}
(b) $my array [4]
(c) $my_array |4|
(d) None of these
Answer:
(b) $my array [4]

Question 8.
PHP arrays are also called as
(a) Vector arrays
(b) Perl arrays
(c) Hashes
Answer:
(c) Hashes

Question 9.
As compared to associative arrays, vector arrays are much
(a) Faster
(b) Slower
(c) Stable
(d) None of them
Answer:
(b) Slower

Question 10.
What functions count elements in an array?
(a) count
(b) both a and b
(c) Array Count
(d) Count_array
Answer:
(b) both a and b

PART – II
I. Short Answer

Question 1.
Define Function in PHP?
Answer:
In most of the programming language, a block of segment in a program that performs a specific operation tasks (Insert, Execute, Delete, Calculate, etc.). This segment is also known as Function. A Function is a type of sub routine or procedure in a program.

Question 2.
Define User defined Function?
Answer:
User Defined Function:
User Defined Function (UDF) in PHP gives a privilege to user to write own specific operation inside of existing program module.

Function Declaration:
A user-defined Function declaration begins with the keyword “function”.

Question 3.
What is parameterized Function?
Answer:
Parameterized Function:

PHP Parameterized functions are the functions with parameters or arguments.
Required information can be shared between function declaration and function calling part inside the program.
The parameter is also called as arguments, it is like variables.
The arguments are mentioned after the function name and inside of the parenthesis.
There is no limit for sending arguments, just separate them with a comma notation.

Question 4.
List out System defined Functions?
Answer:

round
cos
tan
is_number
rand etc
sqrt
sin
pi
number_format

Question 5.
Define Array in PHP?
Answer:
Array is a concept that stores more than one value of same data type (homogeneous) in single array variable. They are 3 types of array in PHP.

Indexed Arrays
Associative Array and
Multi-Dimensional Array

‘

Question 7.
Usage of Array in PHP?
Answer:
One of the most useful aspects of using arrays in PHP is when combined with the foreach statement. This allows you to quickly loop though an array with very little code.

Question 8.
List out the types of array in PHP?
Answer:

Question 9.
Define associative array?
Answer:
Associative arrays are a key-value pair data structure. Instead of having storing data in a linear array, with associative arrays you can store your data in a collection and assign it a unique key which you may use for referencing your data.

Question 10.
Write array Syntax in PHP?
Answer:
Array defines with the keyword array( )

PART – III
III. Explain in Brief Answer

Question 1.
Write the features of System defined Functions?
Answer:
System Defined Functions: A function is already created by system it is a reusable piece or block of code that performs a specific action. Functions can either return values when called or can simply perform an operation without returning any value.
Features of System defined functions:

System defined functions are otherwise called as predefined or built-in functions.
PHP has over 700 built in functions that performs different tasks.
Built in functions are functions that exists in PHP installation package.

Question 2.
Write the purpose of parameterized Function?
Answer:

Required information can be shared between function declaration and function calling part inside the program.
The parameter is also called as arguments, it is like variables. The arguments are mentioned after the function name and inside of the parenthesis.
There is no limit for sending arguments, just separate them with a comma notation.

Question 3.
Differentiate user define and system defined Functions?
Answer:
System defined:

These functions are already present in the system (comes along with installation package).
Cannot be edited.
Name is provided by the developer Eg. isnum( ), isalpha( ).

System defined:

These functions are already present in the system (comes along with installation package).
Cannot be edited.
Name is provided by the developer Eg. isnum( ), isalpha( ).

Question 4.
Write Short notes on Array?
Array in PHP
Answer:
Array is a concept that stores more than one value of same data type (homogeneous) in single array variable. They are 3 types of array concepts in PHP.

Indexed Arrays,
Associative Array and
Multi-Dimensional Array.

Question 5.
Differentiate Associate array and Multidimensional array?
Answer:
Associate array:
Associative arrays are a key-value pair data structure. Instead of having storing data in a* linear array, with associative arrays you can store your data in a collection and assign it a unique key which you may use for referencing your data.

Multidimensional array:
A multidimensional array is an array containing one or more arrays. PHP understands multidimensional arrays that are two, three, four, five, or more levels deep. However, arrays more than three levels deep are hard to manage for most people.

PART – IV
IV. Explain in detail

Question 1.
Explain Function concepts in PHP?
Answer:
Functions in PHP
In most of the programming language, a block of segment in a program that performs a specific operation tasks (Insert, Execute, Delete, Calculate, etc.). This segment is also known as Function. A Function is a type of sub routine or procedure in a program.

Function will be executed by a call to the Function and the Function returns any data type values or NULL value to called Function in the part of respective program.
The Function can be divided into three types as follows:

User defined Function,
Pre-defined or System or built-in Function, and
Parameterized Function.

1. User Defined Function:
User Defined Function (UDF) in PHP gives a privilege to user to write own specific operation inside of existing program module. Two important steps the Programmer has to create for users define Functions are: Function declaration and Function calling.

2. System Define Functions:
A function is already created by system it is a reusable piece or block of code that performs a specific action. Functions can either return values when called or can simply perform an operation without returning any value.

3. Parameterized Function:
PHP Parameterized functions are the functions with parameters or arguments.

Question 2.
Discuss in detail about User defined Functions?
Answer:
User Defined Function:

User can create their own functions.
User Defined Function (UDF) in PHP gives a privilege to user to write own specific operation inside of existing program module.
Two important steps the Programmer has to create for users define Functions are:

Function Declaration:
A user-defined Function declaration begins with the keyword “function”. User can write any custom logic inside the function block.
Syntax:
function functionName( )
{
Custom Logic code to be executed;
}
Function Calling:
A function declaration part will be executed by a call to the function. Programmer has to create Function Calling part inside the respective program.
Syntax:
functionName( );
Example:
<?php
function insertMsg( )
{
echo “Student Details Inserted Successfully!”;
}
insertMsg( ); // call the function
?>
Parameterized Function:

Required information can be shared between function declaration and function calling part inside the program.
The parameter is also called as arguments, it is like variables.
The arguments are mentioned after the function name and inside of the parenthesis. There is no limit for sending arguments, just separate them with a comma notation.

Question 3.
Explain the Multidimensional Array?
Answer:
Multidimensional Arrays:

A multidimensional array is an array containing one or more arrays.
PHP understands multidimensional arrays that are two, three, four, five, or more levels deep.
However, arrays more than three levels deep are hard to manage for most people.

Example:
<?php
// A two-dimensional array Sstudent-array
(
array(“Iniyan”, 100,96),
array(“Kavin”,60,59),
array(“Nilani”,1313,139)
);
echo $$student[0][0].“: Tamil Mark: “.$student [0][1]English mark: “.$student [0] [2].” ”;

echo $$student[1][0].“: Tamil Mark: “.$student [1][1].”. English mark: “.$student [1] [2].” ”;

echo $$student[2][0].“: Tamil Mark: “.$student [2][1]English mark: “.$student [2] [2].” ”;
?>

Question 4.
Explain Array concepts and their types?
Answer:
Arrays in PHP:
Array is a concept that stores more than one value of same data type (homogeneous) in single array variable. They are 3 types of array concepts in PHP.

Indexed Arrays,
Associative Array and
Multi-Dimensional Array.

SYNTAX:
Array Syntax:
Array defines with the keyword array( )

1. Indexed Arrays:
Arrays with numeric index for the available values in array variable which contains key value pair as user / developer can take the values using keys.

2. Associative Arrays:

Associative arrays are a key-value pair data structure.
Instead of having storing data in a linear array, with associative arrays you can store your data in a collection and assign it a unique key which you may use for referencing your data.

Question 3.
Multidimensional Arrays?
Answer:

A multidimensional array is an array containing one or more arrays. .
PHP understands multidimensional arrays that are two, three, four, five, or more levels deep.
However, arrays more than three levels deep are hard to manage for most people.

Question 5.
Explain Indexed array and Associate array in PHP?
Answer:
Indexed Arrays:
Arrays with numeric index for the available values in array variable which contains key value pair as user / developer can take the values using keys.
Example:
<?php
$teacher_name=array(“Iniyan”, “Kavin”, “Nilani”);
echo “The students name are “ . $teacher_name[0]. “, “ . $$teacher_name[l]. “ and” . $teacher_name[2].
?>

Associative Arrays:

Associative arrays are a key-value pair data structure.
Instead of having storing data in a, linear array, with associative arrays you can store your data.

Example:
<?php
$Marks=array(“Studentl”=>“35”,“Student2”==>“17”,“Student3”=>“43”);
echo “Studentl mark is” . $Marks[‘Studentl’]. “ is eligible for qualification”;
echo “Student2 mark is” . $Marks[‘Student2’]. “ is not eligible for qualification”;

Samacheer Kalvi 12th Computer Applications Solutions PHP Function and Array Additional Questions and Answers

I. Choose the Best Answer

Question 1.
………………………. are functions that exist in PHP installation package.
Answer:
Built-in functions

Question 2.
PHP has over ………………………. built in functions.
(a) 200
(b) 500
(c) 700
(d) 900
Answer:
(c) 700

Question 3.
A function is a type of …………………………… in a program.
(a) sub routine
(b) procedure
(c) both a & b
(d) array
Answer:
(c) both a & b

Question 4.
How many classification of functions are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 5.
Find the wrong statement from the following?
(a) pre-defined functions are called as built-in functions
(b) pre-defined functions are called as system functions
(c) parameterized functions are called system functions
Answer:
(c) parameterized functions are called system functions

Question 6.
UDF stands for …………………………..
Answer:
User defined functions

Question 7.
Pick the odd one out related to functions.
(a) userdefined
(b) pre-defined
(c) system
(d) builtin
Answer:
(a) userdefined

Question 8.
Find the statement which is correct.
I: A user-defined function declaration begins with the keyword “user”
II: user can write any custom logic inside the function block
(a) I-True, II-False
(b) I-Flase, II-True
(c) I, II-True
(d) I, II-False
Answer:
(b) I-Flase, II-True

Question 9.
The parameter is also called as
(a) Arguments
(b) Value
(c) Calling function
(d) return
Answer:
(a) Arguments

Question 10.
THe arguments are separated by
(a) .
(b) :
(c) ,
(d) ;
Answer:
(c) ,

Question 11.
The arguments have to be enclosed within
(a) [ ]
(b) <>
(c) { }
(d) ( )
Answer:
(d) ( )

Question 12.
The function blocks are given by
(a) [ ]
(b) <>
(c) { }
(d) ( )
Answer:
(c) { }

Question 13.
How many arguments can be send to return a value in a function?
(a) 2
(b) 3
(c) 4
(d) No limit
Answer:
(d) No limit

Question 14.
Which statement can be used to return a value in a function?
(a) pass
(b) call
(c) return
(d) give
Answer:
(c) return

Question 15.
Which is used to store more than one value in a single variable of same data type?
(a) Array
(b) Functions
(c) Class
(d) Object
Answer:
(a) Array

Question 16.
How many types of arrays are there?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 17.
Pick the odd one out.
Indexed, Associative, Multi functional, Multi dimensional
Answer:
Multi functional

Question 18.
What is the keyword for creating an array?
(a) Arr( )
(b) array( )
(c) a( )
(d) a[ ]
Answer:
(b) array( )

Question 19.
One of the most useful aspects of using array is when it is combined with the …………………………. statement.
(a) if
(b) while
(c) for
(d) foreach
Answer:
(d) foreach

Question 20.
…………………………. specifies the value in the syntax of Association Array.
(a) value
(b) val
(c) value at
(d) value of
Answer:
(a) value

Question 21.
The key in the Associative Array may be of …………………………… or …………………………..
Answer:
numeric or string

Question 22.
………………………… arrays are a key value pair data structure.
(a) Indexed
(b) Numeric
(c) Associative
(d) Multidimensional
Answer:
(c) Associative

Question 23.
In PHP ………………………….. are Information hiding.
(a) Functions
(b) Variables
(c) Keywords
(d) datatypes
Answer:
(a) Functions

II. Short Answer

Question 1.
Classify functions?
Answer:
The Function can be divided in to three types as follow

User defined Function,
Pre-defined or System or built-in Function, and
Parameterized Function.

Question 2.
Define Magic functions?
Answer:

A function in PHP which starts with underscore ( ) is called as magic function.
Magic function is present inside the class.

III. Explain in Brief Answer

Question 1.
Write a PHP program for function with one arguments?
Answer:
<?php
function School_Name($sname) {
echo $sname.“in Tamilnadu. ”;
}
SchoolName (“Government Higher Secondary School Madurai”);
SchoolName (“Government Higher Secondary School Trichy”);
School Name (“Government Higher Secondary School Chennai”);
School Name (“Government Higher Secondary School Kanchipuram”);
School Name (“Government Higher Secondary School Tirunelveli”);
?>

Question 2.
Give an example for function with two arguments?
Answer:
<?php
function School_Name($sname,$Stfength) {
echo $sname.“in Tamilnadu and Student Strength is”.$Strength;
}
School Name (“Government Higher Secondary School Madurai”,200);
School Name (“Government Higher Secondary School Trichy”,300);
School Name (“Government Higher Secondary School Chennai”,250);
School Name (“Government Higher Secondary School Kanchipuram”, 100);
School Name (“Government Higher Secondary School Tirunelveli”,200);
?>

Question 3.
Write a PHP program to explain function with return values?
Answer:
For a function to return a value, use the return statement
<?php
function sum($x, $y) {
$z = $x + $y;
return $z;
}
echo “5 + 10 = “ . sum(5, 10). “ ”;
echo “7 + 13 = “ . sum(7, 13). “ ”;
echo “2 + 4 = “ . sum(2, 4);
?>

Question 4.
Give the Syntax for Associative Arrays?
Answer:
Associative Arrays Syntax:
array(key ⇒ value,key ⇒ value,key ⇒ value,etc.);
key = Specifies the key (numeric or string)
value = Specifies the value.

Question 5.
What is the use of array combine( ) in PHP?
Answer:
The array_combine( ) is an inbuilt function in PHP.
It is used to combine two arrays and create a new array by using one array for keys and another for values.
Eg. $c = array_combine($a, $b);

IV. Explain in detail

Question 1.
Give important characteristics of PHP functions?
Answer:

PHP Functions are Reducing duplication of code.
PHP Functions are Decomposing complex problems into simpler pieces.
PHP Functions are Improving clarity of the code.
PHP Functions are Reuse of code.
PHP Functions are Information hiding.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation

Samacheer Kalvi 12th Bio Zoology Principles of Inheritance and Variation Text Book Back Questions and Answers

Question 1.
Haemophilia is more common in males because it is a ____________
(a) Recessive character carried by Y-chromosome
(b) Dominant character carried by Y-chromosome
(c) Dominant trait carried by X-chromosome
(d) Recessive trait carried by X-chromosome
Answer:
(d) Recessive trait carried by X-chromosome

Question 2.
ABO blood group in man is controlled by ____________
(a) Multiple alleles
(b) Lethal genes
(c) Sex linked genes
(d) Y-linked genes
Answer:
(a) Multiple alleles

Question 3.
Three children of a family have blood groups A, AB and B. What could be the geno types of their parents?
(a) I^AI^B and ii
(b) I^A1^O and I^BI^O
(c) I^B I^B and I^AI^A
(d) I^AI^A and ii
Answer:
(b) I^AI^O and I^BI^O

Question 4.
Which of the following is not correct?
(a) Three or more alleles of a trait in the population are called multiple alleles.
(b) A normal gene undergoes mutations to form many alleles
(c) Multiple alleles map at different loci of a chromosome
(d) A diploid organism has only two alleles out of many in the population
Answer:
(c) Multiple alleles map at different loci of a chromosome

Question 5.
Which of the following phenotypes in the progeny are ____________
(a) A and B only
(b) A,B only and AB
(c) AB only
(d) A,B,AB and O
Answer:
(d) A,B,AB and O

Question 6.
Which of the following phenotypes is not possible in the progeny of the parental genotypic combination I^AI^O × l^Al^B ?
(a) AB
(b) O
(c) A
(d) B
Answer:
(b) O

Question 7.
Which of the following is true about Rh factor in the offspring of a parental combination DdXDd (both Rh positive)?
(a) All will be Rh positive
(b) Half will be Rh positive
(c) About 3/4 will be Rh negative
(d) About one fourth will be Rh negative
Answer:
(d) About one fourth will be Rh negative

Question 8.
What can be the blood group of offspring when both parents have AB blood group?
(a) AB only
(b) A, B and AB
(c) A, B, AB and O
(d) A and B only
Answer:
(b) A, B and AB

Question 9.
If the childs blood group is ‘O’ and fathers blood group is ‘A’ and mother’s blood group is ‘B’ the genotype of the parents will be _________
(a) I^AI^A and I^AI^O
(b) I^AI^O and I^BI^O
(c) I^AI^O and I^OI^O
(d) I^OI^O and I^BI^B
Answer:
(b) I^AI^O and I^BI^O

Question 10.
XO type of sex determination and XY type of sex determination are examples of _________
(a) Male heterogamety
(b) Female heterogamety
(c) Male homogamety
(d) Both (b) and (c)
Answer:
(a) Male heterogamety

Question 11.
In an accident there is great loss of blood and there is no time to analyse the blood group which blood can be safely transferred?
(a) ‘O’ and Rh negative
(b) ‘O’ and Rh positive
(c) ‘B’ and Rh negative
(d) ‘AB’ and Rh positive
Answer:
(a) ‘O’ and Rh negative

Question 12.
Father of a child is colour blind and mother is carrier for colour blindness, the probability of the child being colour blind is _________
(a) 25%
(b) 50%
(c) 100%
(d) 75%
Answer:
(b) 50%

Question 13.
A marriage between a colour blind man and a normal woman produces _________
(a) All carrier daughters and normal sons
(b) 50% carrier daughters, 50% normal daughters
(c) 50% colour blind sons, 50% normal sons
(d) All carrier off springs
Answer:
(a) All carrier daughters and normal sons

Question 14.
Mangolism is a genetic disorder which is caused by the presence of an extra chromosome number.
(a) 20
(b) 21
(C) 4
(d) 23
Answer:
(b) 21

Question 15.
Klinefelters’ syndrome is characterized by a karyotype of _________
(a) XYY
(b) XO
(c) XXX
(d) XXY
Answer:
(d) XXY

Question 16.
Females with Turners’syndrome have _________
(a) Small uterus
(b) Rudimentary ovaries
(c) Underdeveloped breasts
(d) All of these
Answer:
(d) All of these

Question 17.
Pataus’ syndrome is also referred to as _________
(a) 13-Trisomy
(b) 18-Trisormy
(c) 21-Trisormy
(d) None of these
Answer:
(a) 13-Trisomy

Question 18.
Who is the founder of Modem Eugenics movement?
(a) Mendel
(b) Darwin
(c) Fransis Galton
(d) Karl pearson
Answer:
(c) Fransis Galton

Question 19.
Improvement of human race by encouraging the healthy persons to marry early and produce large number of children is called _________
(a) Positive eugenics
(b) Negative eugenics
(c) Positive euthenics
(d) Positive euphenics
Answer:
(a) Positive eugenics

Question 20.
The _________ deals with the control of several inherited human diseases especially inborn errors of metabolism.
(a) Euphenics
(b) Eugenics
(c) Euthenics
(d) All of these
Answer:
(a) Euphenics

Question 21.
“Universal Donor” and “Universal Recipients” blood group are _________ and _________ respectively.
(a) AB, O
(b) O, AB
(c) A, B
(d) B, A
Answer:
(b) O, AB

Question 22.
ZW-ZZ system of sex determination occurs in _________
(a) Fishes
(b) Reptiles
(c) Birds
(d) All of these
Answer:
(d) All of these

Question 23.
Co-dominant blood group is _________
(a) A
(b) AB
(c) B
(d) O
Answer:
(b) AB

Question 24.
Which of the following is incorrect regarding ZW-ZZ type of sex determination?
(a) It occurs in birds and some reptiles
(b) Females are homogametic and males are heterogametic
(c) Male produce two types of gametes
(d) It occurs in gypsy moth
Answer:
(b) Females are homogametic and males are heterogametic

Question 25.
What is haplodiploidy?
Answer:
In haplodiploidy, the sex of the offspring is determined by the number of sets of chromosomes it receives. Fertilized eggs develop into females (Queen or Worker) and unfertilized eggs develop into males (drones) by parthenogenesis. It means that the males have half the number of chromosomes (haploid) and the females have double the number (diploid).

Question 26.
Distinguish between heterogametic and homogametic sex determination systems
Answer:
Heterogametic Sex:

Organisms producing two different types of gametes.
Example: Human male.
Sperm with X chromosome Sperm with a Y chromosome

Heterogametic Sex:

Organisms producing two different types of gametes.
Example: Human female.
Every egg produced contain X chromosomes.

Question 27.
What is Lyonisation?
Answer:
Lyonisation is a process of inactivation of one of the X chromosomes in some females.

Question 28.
What is criss-cross inheritance?
Answer:

Inheritance of genes from a male parent to female child and then to male grandchild or female parent to male child and then to female grandchild.
E.g., X-linked gene inheritance.

Question 29.
Why are sex-linked recessive characters more common in male human beings?
Answer:
Sex-linked inherited traits are more common in males than females because males are hemizygous and therefore express the trait when they inherit one mutant allele.

Question 30.
What are holandric genes?
Answer:
The genes present in the differential region of Y chromosome are called Y- linked or holandric genes. The Y linked genes have no corresponding allele in X chromosome.

Question 31.
Mention the symptoms of Phenylketonuria.
Answer:
Severe mental retardation, light pigmentation of skin and hair. Phenylpyruvic acid is excreted in the urine.

Question 32.
Mention the symptoms of Down’s syndrome.
Answer:
Severe mental retardation, defective development of the central nervous system, increased separation between the eyes, flattened nose, ears are malformed, mouth is constantly open and the tongue protrudes.

Question 33.
Differentiate Intersexes from Supersexes.
Answer:
Intersexes :
Intersexes refers to the individuals having the characteristics of both female and male sexes and their sexual anatomy does not seem to fit the typical definition of male or female.
Example: Super males in humans human beings have 44+XYY chromosomes.

Supersexes:
Supersexes are formed as a result of an abnormal combination of sex chromosomes.
Example: Super males in humans human beings have 44+XYY chromosomes.

Question 34.
Explain the genetic basis of ABO blood grouping in man.
Multiple allele inheritance of ABO blood groups
Answer:
Blood differs chemically from person to person. When two different incompatible blood types are mixed, agglutination (clumping together) of erythrocytes (RBC) occurs. The basis of these chemical differences is due to the presence of antigens (surface antigens) on the membrane of RBC and epithelial cells. Karl Landsteiner discovered two kinds of antigens called antigen ‘A’ and antigen ‘B’ on the surface of RBC’s of human blood. Based on the presence or absence of these antigens three kinds of blood groups, type ‘A’, type ‘B’, and type ‘O’ (universal donor) were recognized. The fourth and the rarest blood group ‘AB’ (universal recipient) was discovered in 1902 by two of Landsteiner’s students Von De Castelle and Sturli.

Bernstein in 1925 discovered that the inheritance of different blood groups in human beings is determined by a number of multiple allelic series. The three autosomal alleles located on chromosome 9 are concerned with the determination of blood group in any person. The gene controlling blood type has been labeled as ‘L’ (after the name of the discoverer, Landsteiner) or I (from isoagglutination). The I gene exists in three allelic forms, IA, IB and 1°. IA specifies A antigen. IB allele determines B antigen and 1° allele specifies no antigen. Individuals who possess these antigens in their fluids such as the saliva are called secretors.

Each allele (IA and IB) produces a transferase enzyme. IA allele produces N-acetyl galactose transferase and can add N-acetyl galactosamine (NAG) and IB allele encodes for the enzyme galactose transferase that adds galactose to the precursor (i.e. H substances). In the case of I^O/I^O allele no terminal transferase enzyme is produced and therefore called “null” allele and hence cannot add NAG or galactose to the precursor.

From the phenotypic combinations it is evident that the alleles I^A and I^B are dominant to 1°, but co-dominant to each other (I^A = I^B). Their dominance hierarchy can be given as (I^A=I^B> 1^O). A child receives one of three alleles from each parent, giving rise to six possible genotypes and four possible blood types (phenotypes). The genotypes are I^AI^A, I^AI^O, I^BI^B, I^BI^O, I^AI^B and I^OI^O

Question 35.
How is sex determined in human
Answer:

Genes determining sex in human beings are located on two sex chromosomes, called allosomes.
In mammals, sex determination is associated with chromosomal differences between the two sexes, typically XX females and XY males. 23 pairs of human chromosomes include 22 pairs of autosomes (44A) and one pair of sex chromosomes (XX or XY). Females are homogametic producing only one type of gametes (egg), each containing one X chromosome while the males are heterogametic producing two types of sperms with X and Y chromosomes. An independently evolved XX: XY system of sex chromosomes also exist in Drosophila.

Question 36.
Explain male heterogamety.
Answer:
Male heterogamety (XY males) is a type of sex determination in which males produce two different types of gametes. For example, human males produce two kinds of sperms that is sperm with X-chromosome and sperms with Y-chromosome.

Question 37.
Brief about female heterogamety.
Answer:
Female heterogamety (ZO females) refers to the condition, where the female produces two types of egg cells. Some with Z chromosome and some without the Z chromosome.

Question 38.
Give an account of genetic control of Rh factor?
Genetic control of Rh factor
Answer:

Fisher and Race hypothesis: Rh factor involves three different pairs of alleles located on three different closely linked loci on the chromosome pair. This system is more commonly in use today, and uses the ‘Cde’ nomenclature. In the given figure, three pairs of Rh alleles (Cc, Dd and Ee) occur at 3 different loci on homologous chromosome pair-1.

The possible genotypes will be one C or c, one D or d, one E or e from each chromosome. For e.g. CDE/cde; CdE/cDe; cde/cde; CDe/CdE etc. All genotypes carrying a dominant ‘D’ allele will produce Rh+positive phenotype and double recessive genotype ‘dd’ will give rise to Rh negative phenotype.

Wiener Hypothesis:
Wiener proposed the existence of eight alleles (R¹, R², R⁰, R^Z, r, r¹, r¹¹, r^y) at a single Rh locus. All genotypes carrying a dominant ‘R allele’ (R¹, R² ,R⁰ ,R^z) will produce ‘Rh-positive’ ^phenotype and double recessive genotypes (rr, rr¹, rr¹¹, rr^y) will give rise to Rh-negative phenotype.

Question 39.
Explain the mode of sex determination in honeybees.
Haplodiploidv in Honeybees:
Answer:
Haplodiploidv in Honeybees:
In hymenopteran insects such as honeybees, ants and wasps, a mechanism of sex determination called haplodiploidy mechanism of sex determination is common. In this system, the sex of the offspring is determined by the number of sets of chromosomes it receives. Fertilized eggs develop into females (Queen or Worker) and unfertilized eggs develop into males (drones) by parthenogenesis. It means that the males have half the number of chromosomes (haploid) and the females have double the number (diploid), hence the name haplodiplody for this system of sex determination.

This mode of sex determination facilitates the evolution of sociality in which only one diploid female becomes a queen and lays the eggs for the colony. All other females which are diploid having developed from fertilized eggs help to raise the queen’s eggs and so contribute to the queen’s reproductive success and indirectly to their own, a phenomenon known as Kin Selection. The queen constructs their social environment by releasing a hormone that suppresses fertility of the workers.

Question 40.
Discuss the genic balance mechanism of sex determination with reference to Drosophila?
Answer:
XX-XY type (Lygaeus Type) sex determination is seen in Drosophila. The females are homogametic with XX chromosome, while the males are heterogametic with X and Y

chromosome. Homogametic females produce only one kind of egg, each with one X chromosome, while the heterogametic males produce two kinds of sperms some with X chromosome and some with Y chromosome.

The sex of the embryo depends on the fertilizing sperm. An egg fertilized by an ‘X’ bearing sperm produces a female, if fertilized by a ‘Y’ bearing sperm, a male is produced.

Question 41.
What are the applications of Karyotyping?
Answer:

Karyotyping helps in gender identification.
It is used to detect chromosomal aberrations like deletion, duplication, translocation, non-disjunction of chromosomes.
It helps to identify the abnormalities of chromosomes like aneuploidy.
It is also used in predicting the evolutionary relationships between species.
Genetic diseases in human beings can be detected by this technique.

Question 42.
Explain the inheritance of sex-linked characters in human beings.
Answer:
Haemophilia is commonly known as bleeder’s disease, which is more common in men than women. This hereditary disease was first reported by John Cotto in 1803. Haemophilia is caused by a recessive X-linked gene. A person with a recessive gene for haemophilia lacks a normal clotting substance (thromboplastin) in blood, hence minor injuries cause continuous ’bleeding, leading to death. The females are carriers of the disease and would transmit the disease to 50% of their sons even if the male parent is normal. Haemophilia follows the characteristic criss-cross pattern of inheritaitce.

Question 43.
What is extrachromosomal inheritance? Explain with an example.
Answer:
The cytoplasmic extranuclear genes have a characteristic pattern of inheritance which does not resemble genes of nuclear chromosomes and are known as Extrachromosomal/ Cytoplasmic inheritance.

Question 44.
Comment on the methods of Eugenics.
Answer:
Eugenics refers to the study of the possibility of improving the qualities of human population. Methods of Eugenics:

Sex-education in school and public forums.
Promoting the uses of contraception.
Compulsory sterilization for mentally retarded and criminals.
Egg donation.
Artificial insemination by donors.
Prenatal diagnosis of genetic disorders and performing MTP
Gene therapy
Cloning
Egg/sperm donation of healthy individuals.

Samacheer Kalvi 12th Bio Zoology Principles of Inheritance and Variation Additional Questions and Answers

1 – Mark Questions

Question 1.
If a colorblind female marries a normal male, their sons will be _________
(a) All normal visioned
(b) All color blinded
(c) One-half normal visioned another half colorblind
(d) Three fourth colorblind one fourth normal
Answer:
(c) One-half normal visioned another half colorblind

Question 2.
Excess hair growth on pinna is a feature noticed only in males because _________
(a) Males produce more testosterone
(b) a gene responsible for the character is located in Y-chromosome
(c) Estrogen suppresses the character in females
(d) females act only as a carrier for this character
Answer:
(b) gene responsible for the character is located in Y-chromosome

Question 3.
ABO blood group is a classical example for _________
(a) Multiple allelism
(b) Pleotropism
(c) Incomplete dominance
(d) Polygenic mechanism
Answer:
(a) Multiple allelism

Question 4.
Unit of heredity is _________
(a) allele
(b) allelomorph
(d) gene
Answer:
(d) gene

Question 5.
Identify the proper dominance hierarchy.
(a) I^B = 1° > I^B
(b) I^A = I^B> 0
(c) I^O = I^B >
(d) I^B = I^A > O
Answer:
(b) I^A = I^B> 0

Question 6.
Haemophilia is more common in human males than human females. The reason is due to ______
(а) X-linked dominant gene
(b) X-linked recessive gene
(c) Y-linked recessive gene
(d) Allosomal abnormality
Answer:
(b) X-linked recessive gene

Question 7.
Identify the correct statement.
(a) Homozygous sex chromosome (XX) produce males in Drosophila
(b) Homozygous sex chromosome (ZZ) determine female sex in birds
(c) Heterozygous sex chromosome (XO) determine male sex in grasshopper
(d) Heterozygous sex chromosome (ZW) determine male sex in gypsy moth
Answer:
(c) Heterozygous sex chromosome (XO) determine male sex in grasshopper

Question 8.
Which blood group does not possess antibodies?
(a) I^AI^B
(b) I^OI^O
(c) I^AO
(d) I^BI^B
Answer:
(a) I^AI^B

Question 9.
Assertion (A): On diagnosis, Ramu is reported to have underdeveloped testis and gynecomastia.
Reason (R): His karyotype reveals XXY condition.
(а) A is right but R is wrong
(b) R explains A
(c) Both A and R are wrong
(d) Both and R are right but R is not the correct explanation of A
Answer:
(b) R explains A

Question 10.
Pick out the odd man.
(a) Klinefelter’s syndrome
(b) Turner’s syndrome
(c) Huntington’s chorea
(d) 13-Trisomy
Answer:
(c) Huntington’s chorea

Question 11.
Pick the odd one out regarding Mendelian disorder.
(a) Thalassemia
(b) phenylketonuria
(c) Albinism
(d) Huntington’s chorea
Answer:
(d) Huntington’s chorea

Question 12.

Answer:
(a) A – iv, B – ii, D – i, D – iii

Question 13.
Identify the proper ratio of normal visioned individuals against colorblind individuals, if colorblind carrier female marries a normal male.
(a) 1 : 1
(b) 3 : 1
(c) 1 : 3
(d) All four are normal visioned
Answer:
(c) 1 : 3

Question 14.
Pick out the correct statement.
(i) Karyotyping helps in gender identification
(ii) Holandric genes are located on X-chromosome
(iii) Trisomy-21 is an allosomal abnormality
(iv) Cooley’s anaemia is an autosomal recessive disorder
(a) i, iii, iv are correct
(c) i and iv are correct
Answer:
(c) i and iv are correct

Question 15.
DOPA stands for _________
(a) 3,4 – dihydroxy phenylacetate
(b) 3,4 – dihydroxy phenyle alanine
(c) 3,4 – dihydroxy phenyl asparate
(d) 3,4 – dihydroxy phenyle aldehyde
Answer:
(b) 3,4 – dihydroxy phenyle alanine

Question 16.
The type of antibody generated against Rh antigen is _________
(a) IgE
(b) IgG
(c) IgA
Answer:
(b) IgG

Question 17.
Which of the following symbol is used in pedigree analysis to represent unspecified sex?

Answer:

Question 18.
A colorblind man marries a woman with normal sight who has no history of color blindness in her family. What is the probability of their grandson being colorblind?
(a) 1/4
(b) 3/4
(c) 2/4
(d) 4/4
Answer:
(a) 1/4

Question 19.
Multiple alleles are located _________
(a) at different loci on homologous chromosome
(b) at same locus on homologous chromosome
(c) at different loci on non-homologous chromosome
(d) at different chromosomes
Answer:
(b) at same locus on homologous chromosome

Question 20.
Identify the incorrect statement regarding haplodiploidy.
(g) Haplodiploidy is noticed in honeybees and drosophila
(b) Unfertilized eggs develop into drones
(c) Fertilized eggs develop into queen and worker bees
(d) Males have half the total chromosomal number
Answer:
(a) Haplodiploidy is noticed in honeybees and drosophila

Question 21.
IA and IB genes of ABO blood group are _________
(a) Co-dominant
(b) Pleotropic
(c) Dominant and recessive
(d) Epistatic
Answer:
(a) Co-dominant

Question 22.
Which one of the following crosses show 3 : 1 ratio of normal visioned versus carrier blind?
(a) X^C X^C × X⁺Y
(b) X⁺ X^C × X^C Y
(c) X⁺X^C × X⁺Y
(d) X⁺X⁺× X^CY
Answer:
(c) X⁺X^C × X⁺Y

2 – Mark Questions

Question 1.
Define multiple allelism.
Answer:
When three or more alleles of a gene that control a particular trait occupy the same locus on the homologous chromosome of an organism, they are called multiple alleles and their inheritance is called multiple allelism.

Question 2.
Name the discoverers of antigen A, B and AB.
Answer:
Antigens A and Antigen B was discovered by Karl Landsteiner. Antigen AB was discovered by Von De Castelle and Sturli.

Question 3.
What happens if type A blood is injected to a person having B blood group? Explain the reason.
Answer:
When two different incompatible blood types are mixed, agglutination (clumping together) of erythrocytes (RBC) occurs. The basis of these chemical differences is due to the presence of antigens (surface antigens) on the membrane of RBC and epithelial cells.

Question 4.
State the allelic forms of I gene and mention its chromosomal location.
Answer:
The I gene exists in three forms: IA, IB and I°. The alleles are located on chromosome 9.

Question 5.
Write the possible genotypes for a person having B-blood group.
Answer:
The possible genotypes of a B-blood group person are IBIB or IBI°.

Question 6.
State Wiener Hypothesis on Rh-factor.
Answer:
Wiener proposed the existence of eight alleles (R¹, R², R⁰, R^z, r, r¹, r¹¹, r^y) at a single Rh locus. All genotypes carrying a dominant ‘R allele’ (R¹, R² ,R⁰ ,R^z) will produce ‘Rh-positive’ phenotype and double recessive genotypes (rr, rr¹, rr¹¹, rr^y) will give rise to Rh-negative phenotype.

Question 7.
Distinguish between homogametic and heterogametic condition with example.
Answer:
Homogametic organism:

Organism producing only one type of gametes.
e.g. Human female (Only X)

Heterogametic organism:

Organism producing two different types of gametes.
e.g. Human Male (X and Y)

Question 8.
Name any four organism expressing ZW-ZZ type of sex determination.
Answer:
Gypsy moth, fishes, reptiles and birds.

Question 9.
Expand

Answer:

SRY – Sex Determining region
TDF – Testes Determining Factor

Question 10.
Define Barr body.
Answer:
In 1949, Barr and Bertram first observed a condensed body in the nerve cells of female cat which was absent in the male. This condensed body was called sex chromatin by them and was later referred as Barr body.

Question 11.
Based on Lyon’s hypothesis, mention the number of Barr bodies in XXY males, XO females.
Answer:

XXY males – One Barr body.
XO females – No Barr body.

Question 12.
State Lyon’s hypothesis.
Answer:
Lyon’s hypothesis states that in mammals the necessary dosage compensation is accomplished by the inactivation of one of the X chromosome in females so that both males and females have only one functional X chromosome per cell.

Mary Lyon suggested that Barr bodies represented an inactive chromosome, which in females becomes tightly coiled into a heterochromatin, a condensed and visible form of chromatin (Lyon’s hypothesis). The number of Barr bodies observed in cell was one less than the number of X-Chromosome. XO females have no Barr body, whereas XXY males have one Barr body.

Question 13.
Mention few X-linked inherited diseases.
Answer:
Red-green colour blindness or daltonism, haemophilia and Duchenne’s muscular dystrophy.

Question 14.
Define Karyotyping.
Answer:
Karyotyping is a technique through which a complete set of chromosomes is separated from a cell and the chromosomes are arranged in pairs. An idiogram refers to a diagrammatic representation of chromosomes.

Question 15.
Explain the inheritance pattern of Y-linked genes with example.
Answer:
Genes in the non-homologous region of the Y-chromosome are inherited directly from male to male. In humans, the Y-linked or holandric genes for hypertrichosis (excessive development of hairs on pinna of the ear) are transmitted directly from father to son, because males inherit the Y chromosome from the father. Female inherits only X chromosome from the father and are not affected.

Question 16.
Observe the symbol used in pedigree analysis and give the proper terms they represent.

Answer:

Question 17.
Write a brief note on pedigree analysis.
Answer:
Pedigree is a “family tree”, drawn with standard genetic symbols, showing the inheritance pathway for specific phenotypic characters. Pedigree analysis is the study of traits as they have appeared in a given family line for several past generations.

Question 18.
What do you mean by ‘Mendelian disorder’.
Answer:
Alteration or mutation in a single gene causes Mendelian disorders. These disorders are transmitted to the off springs on the same line as the Mendelian pattern of inheritance.
E.g., Thalassemia.

Question 19.
Name any four Mendelian disorders.
Answer:

Thalassemia
Albinism
sickle cell anaemia
Huntington’s chorea

Question 20.
What is the phenotype of

I^AI^O
I^OI^O

Answer:

I^AI^O – A blood group person
I^OI^O – O blood group person

Question 21.
On which chromosomes does HBA1 gene and HBB genes are located?
Answer:

HBA1 gene is located on chromosome 16.
HBB gene is located on chromosome 11.

Question 22.
Complete the equation.

Answer:
(a) A = Phenylalanine hydroxylase
(b) B = Tyrosinase

Question 23.
Write a note on Huntington’s chorea.
Answer:
Huntington’s chorea is inherited as an autosomal dominant lethal gene in man. It is characterized by involuntary jerking of the body and progressive degeneration of the nervous system, accompanied by gradual mental and physical deterioration. The patients with this disease usually die between the age of 35 and 40.

Question 24.
Comment on Trisomy-21.
Answer:
Trisomic condition of chromosome – 21 results in Down’s syndrome. It is characterized by severe mental retardation, defective development of the central nervous system, increased separation between the eyes, flattened nose, ears are malformed, mouth is constantly open and the tongue protrudes.

Question 25.
Mention the genetic makeup of Turner’s syndrome person and Klinefelter’s syndrome , person.
Answer:
Klinefelter’s syndrome – 44AA+XXY Turner’s syndrome – 44AA+XO

Question 26.
List out any four clinical symptoms of Klinefelter’s syndrome.
Answer:
Gynaecomastia, high pitched voice, under developed genetalia and tall with long limbs.

3 – Mark Questions

Question 27.
Write the types of sex-determination mechanisms does the following crosses as shown. Give an example for each.

Female XX with Male XO
Female ZW with Male ZZ

Answer:

Male heterogamety. e.g., Human beings.
Female heterogamety. e.g., Birds.

Question 28.
What are the enzymes encoded by the alleles I^A, I^B and I°?
Answer:
I^A allele produces N-acetyl galactose transferase and can add N-acetyl galactosamine (NAG) and I^B allele encodes for the enzyme galactose transferase that adds galactose to the precursor (i.e. H substances). In the case of I^O/I^O allele no terminal transferase enzyme is produced and therefore called “null” allele and hence cannot add NAG or galactose to the precursor.

Question 29.
Draw a tabular column representing various types of blood group in human beings, their phenotypes, genotypes, antigens and respective antibodies.
Answer:
Genetic basis of the human ABO blood groups:

Question 30.
Give an account on Rhesus factor.
Answer:
Rhesus or Rh – Factor: The Rh factor or Rh antigen is found on the surface of erythrocytes. It was discovered in 1940 by Karl Landsteiner and Alexander Wiener in the blood of rhesus monkey, Macaca rhesus and later in human beings. The term ‘Rh factor’ refers to “immunogenic D antigen of the Rh blood group system. An individual having D antigen are Rh D positive (Rh+) and those without D antigen are Rh D negative (Rh”)”. Rhesus factor in the blood is inherited as a dominant trait. Naturally occurring Anti D antibodies are absent in the plasma of any normal individual. However if an Rh” (Rh negative) person is exposed to Rh+ (Rh positive) blood cells (erythrocytes) for the first time, anti D antibodies are formed in the blood of that individual. On the other hand, when an Rh positive person receives Rh negative blood no effect is seen.

Question 31.
How Erythroblastosis foetalis can be prevented?
Answer:
If the mother is Rh negative and foetus is Rh positive, anti D antibodies should be administered to the mother at 28th and 34th week of gestation as a prophylactic measure. If the Rh negative mother delivers Rh positive child then anti D antibodies should be administered to the mother soon after delivery. This develops passive immunity and prevents the formation of anti D antibodies in the mothers blood by destroying the Rh foetal RBC before the mother’s immune system is sensitized. This has to be done whenever the woman attains pregnancy.

Question 32.
Explain XX-XO type of sex determination.
Answer:

XX-XO method of sex determination is seen in bugs, some insects such as cockroaches and grasshoppers. Pi The female with two X chromosomes are homogametic Gametes (XX) while the males with only one X chromosome are heterogametic (XO). The presence of an unpaired X chromosomes determines the male sex. The males PI Generation with unpaired ‘X’ chromosome produce two types of sperms, one half with X chromosome and other half without X chromosome. The sex of the offspring depends upon the sperm that fertilizes the egg.

Question 33.
Name the type of sex-determination mechanism of the following organisms.

Gypsy moth
Human beings
Butterflies

Answer:

Gypsy moth -ZW – ZZ type (ZW-females, ZZ – males)
Human beings – XX – XY type (XX-females, XY – males)
Butterflies – ZO – ZZ type (ZO-females, ZZ – males)

Question 34.
Complete the following cross.
AAZW × AAZZ (female) (male)
Answer:

Question 35.
Role of Y- chromosome is crucial for maleness – Justify.
Answer:
Current analysis of Y chromosomes has revealed numerous genes and regions with potential genetic function; some genes with or without homologous counterparts are seen on the X. Present at both ends of the Y chromosome are the pseudoautosomal regions (PARs) that are similar with regions on the X chromosome which synapse and recombine during meiosis. The remaining 95% of the Y chromosome is referred as the Non-combining Region of the Y (NRY).

The NRY is divided equally into functional genes (euchromatic) and non-functional genes (heterochromatic). Within the euchromatin regions, is a gene called Sex determining region Y (SRY). In humans, absence of Y chromosome inevitably leads to female development and this SRY gene is absent in X chromosome. The gene product of SRY is the testes determining factor (TDF) present in the adult male testis.

Question 36.
Color blindness is a perfect example for criss-cross of inheritance – Justify the statement.
Answer:
A marriage between a colour blind man and a normal visioned woman will produce normal visioned male and female individuals in F₁ generation but the females are carriers. The marriage between a F₁ normal visioned carrier woman and a normal visioned male will produce one normal visioned female, one carrier female, one normal visioned male and one colour blind male. Colour blind trait is inherited from the male parent to his grandson through carrier daughter, which is an example of criss-cross pattern of inheritance.
image 12

Question 37.
How the Karyotype of lymphocytes was prepared by Tjio and Levan?
Answer:
Preparation of Karyotype Tjio and Levan (1960) described a simple method of culturing lymphocytes from the human blood. Mitosis is induced followed by addition of colchicine to arrest cell division at metaphase stage and the suitable spread of metaphase chromosomes is photographed. The individual chromosomes are cut from the photograph and are arranged in an orderly fashion in homologous pairs. This arrangement is called a karyotype. Chromosome banding permits structural definitions and differentiation of chromosomes.

Question 38.
What is a genetic disorder? Mention its types?
Answer:
A genetic disorder is a disease or syndrome that is caused by an abnormality in an individual -DNA. Abnormalities can range from a small mutation in a single gene to the addition or subtraction of an entire chromosome or even a set of chromosomes. Genetic disorders are of two types namely, Mendelian disorders and chromosomal disorders.

Question 39.
Explain the genetic basis of Phenylketonuria.
Answer:
Phenylketonuria is an inborn error of Phenylalanine metabolism caused due to a pair of autosomal recessive genes. It is caused due to mutation in the gene PAH (phenylalanine hydroxylase gene) located on chromosome 12 for the hepatic enzyme “phenylalanine hydroxylase”.

This enzyme is essential for the conversion of phenylalanine to tyrosine. Affected individual lacks this enzyme, so phenylalanine accumulates and gets converted to phenylpyruvic acid and other derivatives. It is characterized by severe mental retardation, light pigmentation of skin and hair. Phenylpyruvic acid is excreted in the urine.

Question 40.
Give an account on Patau’s syndrome.
Answer:
Trisomic condition of chromosome 13 results in Patau’s syndrome. Meiotic non-disjunction is thought to be the cause for this chromosomal abnormality. It is characterized by multiple and severe body malformations as well as profound mental deficiency. Small head with small eyes, cleft palate, malformation of the brain and internal organs are some of the symptoms of this syndrome.

Question 41.
Define aneuploidy.
Answer:
Failure of chromatids to segregate during cell division resulting in the gain or loss of one or more chromosomes is called aneuploidy. It is caused by non-disjunction of chromosomes.

Question 42.
What do you mean by “syndrome”? Give two examples.
Answer:
Group of signs and symptoms that occur together and characterize a particular abnormality is called a syndrome, e.g., Down’s syndrome and Turner’s syndrome.

5 – Mark Questions

Question 42.
Explain in detail about Erythroblastosis foetalis.
Answer:
Rh incompatability has great significance in child birth. If a woman is Rh negative and the man is Rh positive, the foetus may be Rh positive having inherited the factor from its father. The Rh negative mother becomes sensitized by carrying Rh positive foetus within her body. Due to damage of blood vessels, during child birth, the mother’s immune system recognizes the Rh antigens and gets sensitized. The sensitized mother produces Rh antibodies. The antibodies are IgG type which are small and can cross placenta and enter the foetal circulation. By the time the mother gets sensitized and produce anti ‘D’ antibodies, the child is delivered.

Usually no effects are associated with exposure of the mother to Rh positive antigen during the first child birth, subsequent Rh positive children carried by the same mother, may be exposed to antibodies produced by the mother against Rh antigen, which are carried across the placenta into the foetal blood circulation. This causes haemolysis of foetal RBCs resulting in haemolytic jaundice and anaemia. This condition is known as Erythroblastosis foetalis or Haemolytic disease of the new bom (HDN).

Question 43.
Decribe female heterogamy and its types.
Answer:
Heterogametic Females:
In this method of sex determination, the homogametic male possesses two ‘X’ chromosomes as in certain insects and certain vertebrates like fishes, reptiles and birds producing a single type of gamete; while females produce dissimilar gametes. The female sex consists of a single ‘X’ chromosome or one ‘X’ and one ‘Y’ chromosome. Thus the females are heterogametic and produce two types of eggs. Heterogametic females are of two types, ZO-ZZ type and ZW-ZZ type.

ZO-ZZ Type:
This method of sex determination is seen in certain moths, butterflies and domestic chickens. In this type, the female possesses
single ‘Z’ chromosome in its body cells and is heterogametic (ZO) producing two kinds of eggs some with ‘Z’ chromosome and some without ‘Z’ chromosome, while the male possesses two ‘Z’ chromosomes and is homogametic (ZZ).

ZW-ZZ type:
This method of sex determination occurs in certain insects (gypsy moth) and in vertebrates such as fishes, reptiles and birds. In this method the female has one ‘Z’ and one ‘ W’ chromosome (ZW) producing two types of eggs, some carrying the Z chromosomes and some carry the W chromosome. The male sex has two ‘Z’ chromosomes and is homogametic (ZZ) producing a single type of sperm.

Question 44.
Write elaborately about the following Mendelian disorders.
Answer:
(a) Thalassemia
(b) Albinism
Answer:
(a) Thalassemia: Thalassemia is an autosomal recessive disorder. It is caused by gene mutation resulting in excessive destruction of RBC’s due to the formation of abnormal haemoglobin molecules. Normally haemoglobin is composed of four polypeptide chains, two alpha and two beta globin chains. Thalassemia patients have defects in either the alpha or beta globin chain causing the production of abnormal haemoglobin molecules resulting in anaemia.

Thalassemia is classified into alpha and beta based on which chain of haemoglobin molecule _ is affected. It is controlled by two closely linked genes HBA1 and HBA2 on chromosome 16. Mutation or deletion of one or more of the four alpha gene alleles causes Alpha Thalassemia. In Beta Thalassemia, production of beta globin chain is affected. It is controlled by a single gene (HBB) on chromosome 11. It is the most common type of Thalassemia and is also known as Cooley’s anaemia. In this disorder, the alpha chain production is increased and damages the membranes of RBC.

(b) Albinism: Albinism is an inborn error of metabolism, caused due to an autosomal recessive gene. Melanin pigment is responsible for skin colour. Absence of melanin results in a condition called albinism. A person with the recessive allele lacks the tyrosinase enzyme system, which is required for the conversion of dihydroxyphenyl alanine (DOPA) into melanin pigment inside the melanocytes. In an albino, melanocytes are present in normal numbers in their skin, hair, iris, etc., but lack melanin pigment.

Question 45.
Discuss any two Allosomal anomalies in human.
Allosomal abnormalities in human beings
Answer:
Mitotic or meiotic non-disjunction of sex chromosomes causes allosomal abnormalities. Several sex chromosomal abnormalities have been detected.
E.g. Klinefelter’s syndrome and Turner’s syndrome.

1. Klinefelter’s Syndrome (XXY Males): This genetic disorder is due to the presence of an additional copy of the X chromosome resulting in a karyotype of 47,XXY. Persons with this syndrome have 47 chromosomes (44AA+XXY). They are usually sterile males, tall, obese, with long limbs, high pitched voice, under developed genetalia and have feeble breast (gynaecomastia) development.

2. Turner’s Syndrome (XO Females): This genetic disorder is due to the loss of a X chromosome resulting in a karyotype of 45,X. Persons with this syndrome have 45 chromosomes (44 autosomes and one X chromosome) (44AA+XO) and are sterile females. Low stature, webbed neck, under developed breast, rudimentary gonads lack of menstrual cycle during puberty, are the main symptoms of this syndrome.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
On analysis, a person’s karyotype reveals an extra one chromosome of twenty first pair. What does this condition represents? which type of symptoms can be noticed in the person?
Answer:

Trisomy-21 or Down’s syndrome.
Symptoms – Mental retardation, malformed ears, protruded tongue, mouth is constantly open etc.

Question 2.
A female whose blood group is AB- got conceived and later it is diagnoised that her – foetus possess B+. What measures would be taken to prevent the foetus from Haemolytic
disease of Newborn (HDN)
Answer:
If the mother is Rh negative and foetus is Rh positive, anti D antibodies should be administered to the mother at 28th and 34th week of gestation as a prophylactic measure. If the Rh negative mother delivers Rh positive child then anti D antibodies should be administered to the mother soon after delivery. This develops passive immunity and prevents the formation of anti D antibodies in the mothers blood by destroying the Rh foetal RBC before the mother’s immune system is sensitized. This has to be done whenever the woman attains pregnancy.

Question 3.
The following table shows the genotypes for ABO blood grouping and this phenotypes. Complete the table by filling the gaps.
Answer:

Question 4.
Give one example for each of the following group of drugs,

Stimulants
Analgesic
Hallucinogens

Answer:

Stimulants – Eg : Nicotine
Analgesic – Eg : Opium
Hallucinogens – Phencyclidine

Samacheer Kalvi 11th Bio Botany Solutions Chapter 3 Vegetative Morphology

Leave a Comment / Class 11 / By Prasanna

Students can Download Bio Botany Chapter 3 Vegetative Morphology Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 3 Vegetative Morphology

Samacheer Kalvi 11th Bio Botany Vegetative Morphology Text Book Back Questions and Answers

Choose the correct answer
Question 1.
Which of the following k polycarpic plant?
(a) Mangifera
(b) Bambusa
(c) Musa
(d) Agave
Answer:
(a) Mangifera

Question 2.
Roots are …………… .
(a) descending. negatively geotrophic, positively phototrophic
(b) descending, positively geotrophic, negatively phototrophic
(c) ascending, positively geotrophic, negatively phototrophic
(d) ascending, negatively geotrophic, positively phototrophic
Answer:
(b) descending, positively geotrophic, negatively pholotrophic

Question 3.
Bryophyllum and Dioscorea are example for …………… .
(a) foliar bud, apical bud
(b) foliar bud, cauline bud
(c) cauline bud, apical bud
(d) cauline bud, foliar bud
Answer:
(b) foliar bud, cauline bud

Question 4.
Which of the following is correct statement?
(a) In Pisum sativum leaflets modified into tendrils
(b) In Atalantia terminal bud is modified into thorns
(c) In Nepenthes midrib is modified into lid
(d) In Smilax inflorescence axis is modified into tendrils
Answer:
(b) In Pisum sativum leaflets modified into tendrils

Question 5.
Select the mismatch pair.

Answer:
(d) Allamanda – (iv) Ternate phyllotaxy

Question 6.
Draw and label the parts of regions of root.
Answer:
The parts of regions of root:

Question 7.
Write the similarities and differences between
1. Avicennia and Trapa
2. Radical buds and foliar buds
3. Phylloclade and cladode
Answer:
1. Avicennia and Trapa
2. Radical buds and foliar buds
3. Phylloclade and cladode

Question 8.
How root climbers differ from stem climbers?
Answer:
Root climbers:

Plants climbing with the help of adventitious roots (arise from nodes)
Example: Piper betel

Stem climbers (twiners):

These climbers lack specialised structure for climbing and the stem itself coils around the support.
Example: Ipomoea

Question 9.
Compare sympodial branching with monopodial branching.
Answer:
Sympodial branching:
The terminal bud cease to grow after a period of growth and the further growth is taken care by successive or several lateral meristem or buds. This type of growth is also known as sympodial branching. Example: Cycas.

Monopodial branching:
The terminal bud grows uninterrupted and produce several lateral branches. This type of growth is also known as monopodial branching. Example: Polyalthia.

Question 10.
Compare pinnate unicostate venation and palmate multicostate venation.
Answer:
Pinnate Unicostate Venation and Palmate Multicostate Venation:

Pinnate unicostate venation: In pinnate unicostate there is only one prominent midrib.
Palmate multicostate venation: In palmate multicostate there are many midribs running parallel to each other.

Text Book Activities Solved

Question 1.
Collection of medicines prepared from root, stem, leaf of organic plants.
Answer:
Medicines Prepared From Root, Stem, Leaf Of Organic Plants:

Question 2.
Prepare a report of traditional medicines.
Answer:
Traditional medicines: Ayurvcda, Siddha, Unani, Acupuncture, Homeopathy, Naturopathy, Chinese or oriental medicine.

Question 3.
Growing micro greens In class room – project work. (Green seed sprouts)
Answer:
Definition: Micro green are miniature plants of greens, herbs or vegetables. They arc rich nutrient source and also added as flavouring agent in food. Growing micro greens in classroom:

Step 1. Take a shallow tray.
Step 2. Place a inch of organic potting soil in the bottom of tray.
Step 3. Scatter the seeds (celery, lettuce, etc.) over the soil surface.
Step 4. Sprinkle some water. Cover the seeds with a thin layer of soil.
Step 5. Place the whole setup near the lighted window.
Step 6. Within a day or two, seeds with germinate.
Step 7. Maintain the moisture of soil. Greens can be harvested within 2 – 3 weeks.

Samacheer Kalvi 11th Bio Botany Vegetative Morphology Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer
Question 1.
The study about external features of an organism is known as …………… .
(a) morphology
(b) anatomy
(c) physiology
(d) taxonomy
Answer:
(a) morphology

Question 2.
…………… deals with the study of shape. site and structure of plant parts.
(a) External morphology
(b) Internal morphology
(c) External anatomy
(d) Internal anatomy
Answer:
(a) External morphology

Question 3.
The branch of science that deals with the classification of organisms is called as …………… .
(a) taxonomy
(b) morphology
(c) physiology
(d) anatomy
Answer:
(a) taxonomy

Question 4.
…………… deals with the study about the root and shoot system.
(a) Vegetative morphology
(b) Reproductive morphology
(c) External morphology
(d) Internal morphology
Answer:
(a) Vegetative morphology

Question 5.
…………… deals with the study about flowers, fruits and seeds of a plant.
(a) Reproductive morphology
(b) Vegetative morphology
(c) External morphology
(d) Internal morphology
Answer:
(a) Reproductive morphology

Question 6.
The general form of a plant is referred to as …………… .
(a) habitat
(b) structure
(c) habit
(d) shape and size
Answer:
(c) habit

Question 7.
…………… are soft stemmed plants with less wood or no wood.
(a) Shrubs
(b) Trees
(c) Herbs
(d) Climbers
Answer:
(c) Herbs

Question 8.
Perennial herbs having a bulb, corm. rhizome or tuber as the underground stem are termed as …………… .
(a) thallophytes
(b) rhodophytes
(c) geophytes
(d) cyanophytes
Answer:
(c) geophytes

Question 9.
Climbers are also called as …………… .
(a) herbs
(b) trees
(c) vines
(d) shrubs
Answer:
(c) vines

Question 10.
…………… is a perennial, woody plant with several main stems arising from the ground level.
(a) Herb
(b) Runner
(c) Climber
(d) Shrub
Answer:
(d) Shrub

Question 11.
…………… is an example for vines.
(a) Hibiscus
(b) Phyllanthus
(c) Palmyra
(d) Ventilago
Answer:
(d) Ventilago

Question 12.
…………… grows on rocks.
(a) Mesophytes
(b) Lithophytes
(c) Xerophytes
(d) Psammophytcs
Answer:
(b) Lithophytes

Question 13.
The plants that grows on dry habitats are called …………… .
(a) meophytes
(b) xerophytes
(c) lithophytes
(d) psammophytes
Answer:
(b) xerophytes

Question 14.
Mesophytes grows on …………… .
(a) sand
(b) soils with sufficient water
(c) rocks
(d) dry habitats
Answer:
(b) soils with sufficient water

Question 15.
…………… grows on sand.
(a) Mesophytes
(b) Psammophytes
(c) Lithophytes
(d) Xerophytes
Answer:
(b) Psammophytes

Question 16.
Azadirachta indica is an example for …………… .
(a) psammophytes
(b) mesophytes
(c) lithophy es
(d) xerophytes
Answer:
(b) mesophytes

Question 17.
Ipomoea pes – caprae is an example for …………… .
(a) mesophyles
(b) psammophytes
(c) lithophytes
(d) serophytes
Answer:
(b) psammophytes

Question 18.
…………… is an example for xerophytes.
(a) Lichens
(b) Euphorbia
(c) Ficus
(d) Ipomoea
Answer:
(b) Euphorbia

Question 19.
Plants growing emergent in marshy saline habitat are called as …………… plants.
(a) free floating
(b) submerged
(c) emergent
(d) mangroves
Answer:
(d) mangroves

Question 20.
…………… type of aquatic plants with roots or stems are anchored to the substrate under water with aerial shoots growing above water.
(a) Submerged
(b) Free floating
(c) Emergent
(d) Mangroves
Answer:
(c) Emergent

Question 21.
Hydrilla and Vallisneria are the examples of …………… type of aquatic plants.
(a) emergent
(b) free floating
(c) submerged
(d) mangroves
Answer:
(c) submerged

Question 22.
Plants that grows completely under water are called as ……………. type of aquatic plants.
(a) emergent
(b) free floating
(c) submerged
(d) mangroves
Answer:
(c) submerged

Question 23.
Plants growing on the water surface are called as …………… type of aquatic plants.
(a) emergent
(b) submerged
(c) free floating
(d) mangroves
Answer:
(c) free floating

Question 24.
Therophyte is a plant that completes its life cycle in …………… growing season.
(a) one
(b) two
(c) three
(d) several
Answer:
(a) one

Question 25.
A plant that grows vegetatively during the first season and undergoes flowering and fruiting during the second season is called as …………… .
(a) biennial
(b) therophyte
(c) perennial
(d) geophyte
Answer:
(a) biennial

Question 26.
Spinach, carrot and lettuce are the examples of …………… .
(a) biennial
(b) annual
(c) geophytes
(d) ephemerals
Answer:
(a) biennial

Question 27.
A plant that grows for many years that flowers and set fruits for several season during the life span is called as …………… .
(a) geophytes
(b) annual
(c) biennial
(d) ephemerals
Answer:
(a) geophytes

Question 28.
When perennial plants bear fruits every year they are called as …………… .
(a) polycarpic
(b) ephemerals
(c) annual
(d) therophyte
Answer:
(a) polycarpic

Question 29.
Talipot palm, Bamboo and Agave are examples of …………… .
(a) polycarpic geophytes
(b) therophytes
(c) monocarpic geophytes
(d) biennial
Answer:
(c) monocarpic geophytes

Question 30.
Watermelon is a / an …………… plant.
(a) biennial
(b) perennial
(c) geophytic
(d) ephemeral
Answer:
(d) ephemeral

Question 31.
…………… is an example for polycarpic perennial.
(a) Peas
(b) Fennel
(c) Agave
(d) Sapota
Answer:
(d) Sapota

Question 32.
Flowering plants are also called as …………… .
(a) sporophytes
(b) thallophytes
(c) magnoliophytes
(d) phaeophytes
Answer:
(c) magnoliophytes

Question 33.
The part of a plant that arises from radicle is …………… .
(a) stem
(b) root
(c) bud
(d) flower
Answer:
(b) root

Question 34.
Root cap is made up of …………… cells.
(a) parenchyma
(b) collenchyma
(c) sclerenchyma
(d) chlorenchyma
Answer:
(a) parenchyma

Question 35.
In …………… plant multiple root cap is seen.
(a) Pandanus
(b) Pistia
(c) Bougainvillea
(d) Pea
Answer:
(a) Pandanus

Question 36.
Root pockets are seen in …………… .
(a) Pandanus
(b) Pistia
(c) Bougainvillea
(d) Pea
Answer:
(b) Pistia

Question 37.
In …………… zone of the root, the cells get differentiated.
(a) Root hair zone
(b) Elongation zone
(c) Meristematic zone
(d) Maturation zone
Answer:
(d) Maturation zone

Question 38.
Roots developing from any part of the plant other than the radicle is called …………… root.
(a) fibrous
(b) adventitious
(c) tap
Answer:
(b) adventitious

Question 39.
Match the following topics with correct examples.

Answer:
(d) a – ii, b – iv, c – i, d – iii.

Question 40.
The cells of this region undergo active cell division …………… .
(a) root hair zone
(b) maturation zone
(c) elongation zone
(d) meristematic zone
Answer:
(d) meristematic zone

Question 41.
Tap root system develops from this part of embryo …………… .
(a) plumule
(b) coleoptile
(c) node
(d) radicle
Answer:
(d) radicle

Question 42.
Negatively geotropic roots are seen in plant like …………… .
(a) Beta vulgaris
(b) Hibiscus
(c) Rhizophora
(d) Euphorbia
Answer:
(c) Rhizophora

Question 43.
Respiratory roots are found in …………… .
(a) Sweet potato
(b) Bruguiera
(c) Mango
(d) Dahlia
Answer:
(b) Bruguiera

Question 44.
Beaded root are also called as …………… root.
(a) annulated
(b) moniliform
(c) tuberous
(d) fasciculated
Answer:
(b) moniliform

Question 45.
Match the following.

Answer:
(c) a – iv, b – iii, c – ii, d – i.

Question 46.
Match the following.

Answer:
(d) a – v, b – i, c – ii, d – iii, e – iv.

Question 47.
The roots that grows vertically downwards from the lateral branches into the soil is called as …………… roots
(a) brace
(b) climbing
(c) buttress
(d) prop
Answer:
(d) prop

Question 48.
…………… roots are thick roots growing obliquely from the basal nodes of their main stem.
(a) pillar
(b) stilt
(c) buttress
(d) prop
Answer:
(b) stilt

Question 49.
Clasping roots are also called as …………… .
(a) Pillar
(b) Stilt
(c) Clinging
(d) Buttress
Answer:
(c) Clinging

Question 50.
Match the following.

Answer:
(c) a – iii, b – iv, c – i, d – ii.

Question 51.
Which part of embryo develops into stem?
(a) Radicle
(b) Micropyle
(c) Ostia
(d) Plumule
Answer:
(d) Plumule

Question 52.
The point from which leaf arises is called as …………… .
(a) internode
(b) intranode
(c) node
(d) code
Answer:
(c) node

Question 53.
Which is not a character of stem?
(a) Exogenous branches
(b) Descending portion
(c) Nodes
(d) buds
Answer:
(b) Descending portion

Question 54.
Which is not a function of stem?
(a) Support
(b) transport of food
(c) Transport of water
(d) Absorption of water
Answer:
(d) Absorption of water

Question 55.
Which is the primary function of stem?
(a) Storage
(b) Perennation
(c) Photosynthesis
(d) Water transport
Answer:
(d) Water transport

Question 56.
Collateral bud is a …………… bud.
(a) terminal
(b) lateral
(c) extra axillary
(d) accessory
Answer:
(d) accessory

Question 57.
Cauline buds arise from …………… .
(a) root
(b) stem
(c) leaf
(d) nodes
Answer:
(b) stem

Question 58.
Which of the following plant produces bulbils?
(a) Bryophyllum
(b) Begonia
(c) Allium proliferum
(d) Solanum americanum
Answer:
(c) Allium proliferum

Question 59.
…………… plants produce foliar buds.
(a) Allium
(b) Solanum
(c) Citrus
(d) Begonia
Answer:
(d) Begonia

Question 60.
Radical buds develop from …………… .
(a) root
(b) stem
(c) leaf
(d) plumule
Answer:
(a) root

Question 61.
In Polyalthia longifolia, the stem is …………… .
(a) decurrent
(b) caudex
(c) excurrent
(d) culm
Answer:
(c) excurrent

Question 62.
Which of the following is not a creeper?
(a) Cynodon
(b) Oxalis
(c) Indigofera
(d) Centella
Answer:
(c) Indigofera

Question 63.
Clockwise coiling climbers are called …………… .
(a) hexose
(b) dextrose
(c) ministrose
(d) sinistrose
Answer:
(b) dextrose

Question 64.
In Artabotrys, …………… is modified into hook.
(a) leaflets
(b) inflorescence axis
(c) petiole
(d) axillary buds
Answer:
(b) inflorescence axis

Question 65.
…………… are woody stem climbers.
(a) Lianas
(b) Tendrils
(c) Phylloclade
(d) Phyllode
Answer:
(a) Lianas

Question 66.
……………. is a characteristic adaptation of xerophytes.
(a) Hook
(b) Thorn
(c) Cladode
(d) Phylloclade
Answer:
(d) Phylloclade

Question 67.
Flattened cladode is present in …………… .
(a) Asparagus
(b) Atalantia
(c) Carissa
(d) Ruscus
Answer:
(d) Rusus

Question 68.
Musa is an example for …………… .
(a) climber
(b) runner
(c) stolon
(d) sucker
Answer:
(d) sucker

Question 69.
Eichhornia exhibit …………… type of stem modification.
(a) stolon
(b) offset
(c) runner
(d) sucker
Answer:
(b) offset

Question 70.
Underground stems are generally called as …………… .
(a) root caps
(b) root stocks
(c) root pockets
(d) root modification
Answer:
(b) root stocks

Question 71.
Which is not a character of root stock?
(a) Nodes
(b) Internodes
(c) Terminal bud
(d) Root cap
Answer:
(d) Root cap

Question 72.
Which is an example for compound tunicated bulb?
(a) Allium cepa
(b) Allium sativum
(c) Tulipa sps.
(d) Bulbophyllum
Answer:
(b) Allium sativum

Question 73.
…………… is a pseudobulb.
(a) Allium cepa
(b) Allium sativum
(c) Tulipa sps.
(d) Bulbophyllum
Answer:
(d) Bulbophyllum

Question 74.
…………… is a horizontally growing underground stem.
(a) Corm
(b) Rhizome
(c) Bulb
(d) Tuber
Answer:
(b) Rhizome

Question 75.
All the leaves of a plant together called as …………… .
(a) phyllome
(b) phyllode
(c) phylloclade
(d) Phyllanthus
Answer:
(a) phyllome

Question 76.
Which Is not a primary function of leaf?
(a) Photosynthesis
(b) Transpiration
(c) Storage
(d) Gas exchange
Answer:
(c) Storage

Question 77.
Lamina of leaf is called as …………… .
(a) hypopodium
(b) mesopodium
(c) endopodiurn
(d) epipodium
Answer:
(d) epipodium

Question 78.
Swollen, broad leaf base Is called as …………… .
(a) phyttome
(b) pulvinus
(c) stipule
(d) bract
Answer:
(b) pulvinus

Question 79.
Pulvinus is a characteristic feature of …………… .
(a) Malvaceae
(b) Fabaceae
(c) Musaceae
(d) Solanaceae
Answer:
(b) Fabaceae

Question 80.
Stalk of leaf is called as …………… .
(a) pedicel
(b) pctiole
(c) phyllum
(d) perianth
Answer:
(b) petiote

Question 81.
Poaceac is seen in …………… family.
(a) Malvaceae
(b) Fabaceae
(c) Musaccac
(d) Sotanaccac
Answer:
(c) Musacese

Question 82.
Which of the following plant possess sessile leases?
(a) Hibiscus
(b) mangifera
(c) Psidiun
(d) Gloriosa
Answer:
(d) Gloriosa

Question 83.
Arrangement of veins on lamina is called …………… .
(a) venalion
(b) aesivation
(c) placentation
(d) phyllotaxy
Answer:
(a) venation

Question 84.
Parallel venatlon is the characteristic feature of …………… .
(a) angiospertns
(b) gymnosperms
(c) dicots
(d) monocots
Ans.
(d) monocots

Question 85.
In greek, ‘taxis’ means …………… .
(a) crowding
(b) spreading
(c) arrangement
(d) attachment
Answer:
(c) arrangement

Question 86.
Palmately parallel divergent venation is seen in …………… .
(a) Carica papaya
(b) Borassus flabellifer
(c) Zizyphus
(d) Cinnamomum
Answer:
(b) Borassus flabellifer

Question 87.
Spiral arrangement of leaves show vertical rows called …………… .
(a) decussate
(b) bifarious
(c) orthostichies
(d) distichous
Answer:
(c) orthostichies

Question 88.
Nerium exhibits …………… phyllotaxy.
(a) ternate
(b) whorled
(c) decussate
(d) alternate
Answer:
(a) ternate

Question 89.
In Allamanda, …………… phyllotaxy is seen.
(a) ternate
(b) verticillate
(c) alternate
(d) opposite
Answer:
(b) verticillate

Question 90.
The secondary rachii are called as …………… .
(a) stipel
(b) ligule
(c) petiole
(d) pinnae
Answer:
(d) pinnae

Question 91.
Tripinnate compound leaves are seen in …………… .
(a) Citrus
(b) Oxalis
(c) Oroxylum
(d) Allamanda
Answer:
(c) Oroxylum

Question 92.
In Gloriosa, …………… is modified into tendril.
(a) apical leaflet
(b) entire leaf
(c) leaf tip
(d) Petiole
Answer:
(c) leaf tip

Question 93.
In Euphorbia, …………… are modified into spines.
(a) stipels
(b) ligules
(c) stipules
(d) Petiole
Answer:
(c) stipules

Question 94.
Storage leaves are observed in …………… family.
(a) Solanaceae
(b) Cucurbitaceae
(c) Crassulaceae
(d) Musaceae
Answer:
(c) Crassulaceae

Question 95.
Phyllodes are the modification of …………… .
(a) pedicel
(b) pinnae
(c) petiole
(d) stipule
Answer:
(c) petiole

Question 96.
…………… is an example for pitcher.
(a) Sarracenia
(b) Acacia
(c) Parkinsonia
(d) Sedum
Answer:
(a) Sarracenia

Question 97.
Stamens are equivalent to …………… .
(a) megasporophyll
(b) microsporophyll
(c) microsporangium
(d) megasporangium
Answer:
(a) megasporophyll

Question 98.
Rolling or folding of individual leaves is called …………… .
(a) venation
(b) phyllotaxy
(c) ptyxis
(d) branching
Answer:
(c) ptyxis

Question 99.
In Mimusops, the leaves are …………… .
(a) fagacious
(b) evergreen
(c) deciduous
(d) marcescent
Answer:
(b) evergreen

Question 100.
Heterophylly is mainly seen in …………… .
(a) xerophytes
(b) mesophytes
(c) lithophytes
(d) hydrophytes
Answer:
(d) hydrophytes

Question 101.
Isobilateral leaf is seen in …………… .
(a) onion
(b) pine
(c) tridax
(d) grass
Answer:
(d) grass

II. Very Short Answer Type Questions (2 Marks)

Question 1.
Define morphology.
Answer:
The study of various external features of the organism is known as morphology.

Question 2.
What is plant morphology?
Answer:
Plant morphology also known as external morphology deals with the study of shape, size and structure of plants and their parts like (roots, stems, leaves, flowers, fruits and seeds).

Question 3.
Differentiate between vegetative morphology & reproductive morphology.
Answer:
between vegetative morphology & reproductive morphology:

Vegetative morphology: Vegetative morphology deals with the shoot system and root system.
Reproductive morphology: Reproductive morphology deals with flowers, inflorescence, fruits & seals.

Question 4.
Define shrub. Give an example.
Answer:
A shrub is a perennial, woody plant with several main stems arising from the ground level. e.g.. Hibiscus

Question 5.
Classify plants based on habitat.
Answer:
Depending upon where plants grow habitats may be classified into major categories:

Terrestrial and
Aquatic.

Question 6.
Distinguish between terrestrial and aquatic plants.
Answer:
Between terrestrial and aquatic plants:

Terrestrial plants: Plants growing on land are called terrestrial plants.
Aquatic plants: Plants living in water environment are called aquatic plants.

Question 7.
Classify the terrestrial plants based on their adaptation.
Answer:
Mcsophytes, serophytes, psammophytes and lithophytes.

Question 8.
What do you mean by caudex?
Answer:
If the trunk of a plant remains unbranched it is said to be caudex. e.g.. Palmyra and coconut.

Question 9.
What is a liana? Mention its importance.
Answer:
Liana is a vine that is perennial and woody. Lianas are major components in the tree canopy layer of some tropical forests. e.g. Ventilago.

Question 10.
What are therophytes?
Answer:
Therophyte is a plant that completes its life cycle in one growing season. e.g. Peas.

Question 11.
Name the two primary functions of roots.
Answer:
The two primary functions of roots:

Absorption of water and nutrients from soil.
Anchorage of plants in soil.

Question 12.
Where does the roots develop from?
Answer:
Root develops from the radicle which is the first structure that comes out when a seed is placed in the soil.

Question 13.
Give an example for
(a) Free floating plants
(b) Submerged plants
(c) Mangroves and
(d) Emergent plants?
Answer:
(a) Free floating plant – e.g Pistia.
(b) Submerged plant – e.g: Hydrilla
(c) Mangroves – e.g: Avicennia
(d) Emergent – e.g: Typha

Question 14.
What is root cap? Mention its role.
Answer:
Root tip is covered by a dome shaped parenchymatous cells called root Cap. It protects the meristematic cells in the apex.

Question 15.
Explain the nature of root based on tropism.
Answer:
Based on tropism. roots are positively geotropic and negatively phototropic.

Question 16.
What are adventitious roots?
Answer:
Roots developing from any part of the plant other than radicle is called adventitious roots.

Question 17.
Why do roots modify their structure? Name the types of root modification.
Answer:
Roots modify their structure to perform secondary functions. The two types of root modification are tap root modification and fibrous root modification.

Question 18.
Name the places from which adventitious roots arise.
Answer:
Adventitious roots may rise from base of stem, nodes or internodes.

Question 19.
What are prop roots? Give an example.
Answer:
Prop (Pillar) roots grow vertically downward from the lateral branches into the soil, e.g., Ficus benghalensis (banyan tree) and Indian rubber.

Question 20.
How velamen helps the Vanda plant?
Answer:
Velamen is a spongy tissue developed by epiphytic roots of Vanda. It helps in absorption of moisture from atmosphere.

Question 21.
Which type of adventitious roots are seen in Bryophyllum plants?
Answer:
Bryophyllum produces foliar roots which arise from the veins or lamina of leaf for the formation of new plants.

Question 22.
How Cuscuta survives?
Answer:
Cuscuta is a parasitic plant which produces haustorial roots from stem that penetrates into the tissue of host plant and suck nutrients. Thus Cuscuta survives.

Question 23.
“Roots perform photosynthesis” – Justify with example.
Answer:
Roots of some climbing or epiphytic plants develop chlorophyll and turn green which help in photosynthesis. These roots are called as photosynthetic roots. e.g.,Tinospora.

Question 24.
Which part of embryo gives rise to root and shoot?
Answer:
Radicle gives rise to root. Plumule gives rise to shoot.

Question 25.
Define node & internode.
Answer:
Node & Internode:

Node: The point from which the leaf arises is called node.
Internode: The region between two adjacent nodes is called intemode.

Question 26.
Explain stem based on tropism.
Answer:
Stem is positively phototropic & negatively geotropic.

Question 27.
Classify buds based on their origin & function.
Answer:
Buds based on their origin & functio:

Based on origin, buds are classified into (a) Terminal or Apical bud and (b) Lateral or Axillary or Axil bud.
Based on function, buds are classified into (a) Vegetative bud and (b) Floral or Reproductive bud.

Question 28.
What are adventitious buds? Give an example.
Answer:
Buds arising at any part other than stem are known as adventitious bud. e.g., Begonia.

Question 29.
Mention various types of stem seen in angiosperms.
Answer:
Majority of angiosperm possess upright, vertically growing erect stem. They are:

Excurrent
Decurrent
Caudex and
Culm.

Question 30.
Why do certain plants climb?
Answer:
Climbing helps to display the leaves towards sunlight and to position the flower for effective pollination.

Question 31.
What are creepers? Give example.
Answer:
Creepers are plants growing horizontally closer to the ground and produces roots at each node. e.g., Oxalis.

Question 32.
Distinguish between dextrose & sinistrose climbers.
Answer:
Between dextrose & sinistrose climbers:

Dextrose: Clockwise coiling climbers are called dextrose, e.g. Dioscorea alata.
Sinistrose: Anti – clockwise coiling climbers are called sinistrose. e.g. Dioscorea bulbifera.

Question 33.
What are root stocks? What are its function?
Answer:
Perennial and some biennial herbs have underground stems, which are generally known as root stocks. Root stock functions as a storage and protective organ.

Question 34.
What is a bulb? Mention its types.
Answer:
Bulb is a condensed, conical or convex stem surrounded by fleshy scale leaves. There are two types:

Tunicated (coated) bulb and
Scaly bulb.

Question 35.
Define branching. Mention its types.
Answer:
The mode of arrangement of branches on a stem is known as branching. There are two main types of branching:

Lateral branching and
Dichotomous branching.

Question 36.
Define phyllome.
Answer:
All the leaves of a plant together are referred as phyllome.

Question 37.
List out any four primary functions of leaves.
Answer:
Photosynthesis, transpiration, gaseous exchange and protection of buds.

Question 38.
What is a leaf base?
Answer:
The part of the leaf attached to the node of the stem is called leaf base. It protects growing buds at its axil.

Question 39.
What is pulvinus?
Answer:
In legumes leaf base become broad, thick and swollen which is known as pulvinus. e.g., Clitoria.

Question 40.
Which leaf part acts a bridge between leaf & stem? Define.
Answer:
Petiole is the bridge between leaf and stem. Petiole or leaf stalk is a cylindrical or subcylindrical or flattened structure of a leaf which joins the lamina with the stem.

Question 41.
Mention the types of leaves based on petiole.
Answer:
A leaf with petiole is said to be petiolate. e.g. Hibiscus. Leaves that do not possess petiole is said to be sessile, e.g. Calotropis.

Question 42.
What are stipules? State its functions.
Answer:
Stipules are the two lateral appendages develop at the leaf base of dicot plants. Stipules protects the leaf in bud condition.

Question 43.
Define venation. Mention its types.
Answer:
The arrangement of veins and veinlets on the leaf blade or lamina is called venation. Venation is classified into two types:

Reticulate venation and
Parallel venation.

Question 44.
Define ligule.
Answer:
In some grasses (Monocots) an additional out growth is present between leaf base and lamina. It is called ligule.

Question 45.
What are stipulate & exstipulate leaves?
Answer:
Stipulate & Exstipulate Leaves:

Leaves with stipules are called stipulate leaves.
Leaves without stipules are called exstipulate or estipulate leaves.

Question 46.
Define phyllotaxy. Mention its types.
Answer:
The mode of arrangement of leaves on the stem is known as phyllotaxy. There are 4 types of phyllotaxy:

Alternate
Opposite
Temate and
Whorled.

Question 47.
How phyllotaxy helps the plants?
Answer:
Phyllotaxy is to avoid over crowding of leaves and expose the leaves maximum to the sunlight for photosynthesis.

Question 48.
You are given a mango leaf. Which type does it belongs to? Define it.
Answer:
Mango leaf is a simple leaf. A simple leaf is the one, where the petiole bears a single lamina.

Question 49.
Define heterophylly. Which type of plants show this adaptation?
Answer:
Occurence of two different kinds of leaves in the same plants is called heterophylly. Heterophylly is an adaptation of aquatic plants.

Question 50.
When a leaf is said to be centric?
Answer:
When the leaf is more or less cylindrical and directed upwards or downwards, as in pine and onion, etc., the leaf is said to be centric.

Question 51.
Which type of leaf is common among monocots? Define it.
Answer:
When the leaf is directed vertically upwards, as in many monocotyledons, it is said to be ‘ isobilateral leaf. Example: Grass.

Question 52.
Define ptyxis.
Answer:
Rolling or folding of individual leaves is called as ptyxis.

Question 53.
Classify plants based on habit.
Answer:
Based on habit plants are classified into herbs, shrubs, climbers (vines) and trees.

Question 54.
Musa is a monocarpic perennial. Give possible reason.
Answer:
Musa is a monocarpic perennial, since it produces flowers and fruits only once and die after a vegetative growth of several years.

Question 55.
What are the parts that constitute the typical leaf?
Answer:
There are three main parts in a typical leaf:

Leaf base (Hypopodium)
Petiole (Mesopodium) and
Lamina (Epipodium).

Question 56.
What is a pseudobulb?
Answer:
Pseudobulb is a short erect aerial storage or propagating stem of certain epiphytic and terrestrial sympodial orchids, e.g. Bulbophyllum.

Question 57.
Which factor determines the branching patterns?
Answer:
Branching pattern is determined by the relative activity of apical meristem.

Question 58.
Mention the secondary functions of leaf with an example for each.
Answer:
Functions:

Storage
Protection
Support
Reproduction

Examples:

Aloe
Opuntia
Nepenthes
Bryophyllum

Question 59.
List out the families that possess sheathing leafbase.
Answer:
Arecaceae, Musaceae, Zingiberaceae and Poaceae.

Question 60.
What are stipels?
Answer:
Sometimes, small stipule like outgrowths are found at the base of leaflets of a compound leaf. They are called stipels.

Question 61.
Compare the stem nature of Corm and Rhizome
Answer:
Corm:

Stem is succulent underground
Presence of erect growing tips.

Rhizome:

Stem is horizontal underground
Presence of lateral growing tips.

III. Short Answer Type Questions (3 Marks)

Question 1.
Morphological study is important in Taxonomy. Why?
Answer:
Morphological features are important in determining productivity of crops. Morphological characters indicate the specific habitats of living as well as the fossil plants and help to correlate the distribution in space and time of fossil plants. Morphological features are also significant for phylogeny.

Question 2.
Differentiate between polycarpic and monocarpic perennial.
Answer:
Polycarpic perennial:

When they bear fruits every year, they are called polycarpic.
e.g. mango, sapota, etc.

Monocarpic perennial:

Some plants produce flowers and fruits only once and die after a vegetative growth of several years. These plants are called monocarpic.
e.g. Bamboo, Agave, Musa, Talipot palm.

Question 3.
List down the key difference between roots and shoots.
Answer:
Roots:

Positively geotropic
Negatively phototropic
Non – green in colour
Nodes, intemodes and buds are absent

Shoots:

Negatively geotropic
Positively phototropic
Green in colour
Nodes, intemodes and buds are present

Question 4.
Name the three distinct zones of root.
Answer:
The three distinct zones of root:

Meristematic Zone
Zone of Elongation
Zone of Maturation

Question 5.
Draw a simplified diagram showing the various regions of root.
Answer:
The various regions of root:

Question 6.
Give a brief account on tap root system.
Answer:
Primary root is the direct prolongation of the radicle. When the primary root persists and continues to grow as in dicotyledons, it forms the main root of the plant and is called the tap root. Tap root produces lateral roots that further branches into finer roots. Lateral roots along with its branches together called as secondary roots.

Question 7.
How does the fibrous root system develops?
Answer:
Most of the monocots the primary root of the seedling is short lived and lateral roots arise from various regions of the plant body. These are bunch of thread – like roots equal in size which are collectively called fibrous root system generally found in grasses. Example: Oryza sativa.

Question 8.
Briefly explain the development process of leaf primordium.
Answer:
The plumule of the embryo of a germinating seed grows into stem. The epicotyl elongates after embryo growth into the axis (the stem) that bears leaves from its tip, which contain the actively dividing cells of the shoot called apical meristem. Further cell divisions and growth result in the formation of mass of tissue called a leaf primordium.

Question 9.
Name the three types of Adventitious buds.
Answer:
The three types of Adventitious buds:

Radical buds
Foliar buds
Cauline buds

Question 10.
Write a brief note on Bulbils.
Answer:
Bulbils (or specialized buds): Bulbils are modified and enlarged bud, meant for propagation. When bulbils detach from parent plant and fall on the ground, they germinate into new plants and serve as a means of vegetative propagation, e.g., Agave.

Question 11.
Distinguish between root climbers and stem climbers.
Answer:
Root climbers:

Plants climbing with the help of adventitious roots (arise from nodes).
e.g. Piper betel, Piper nigrum, Hedera helix, Pothos, Hoya.

Stem climbers:

These climbers lack specialised structure for climbing and the stem itself coils around the support.
e.g. Ipomoea, Convolvulus, Dolichos, Clitoria, Quisqualis.

Question 12.
Explain the three different types of trailers with an example.
Answer:
Types of Trailers:

Prostrate (Procumbent): A stem that grows flat on the ground, e.g. Evolvulus alsinoides, Indigofera prostrata.
Decumbent: A stem that grows flat but becomes erect during reproductive stage. e.g., Portulaca, Tridax, Lindenbergia.
Diffuse: A trailing stem with spreading branches, e.g. Boerhaavia diffusa, Merremia tridentata.

Question 13.
Cladode is a stem modification. Comment on it.
Answer:
Cladode is a flattened or cylindrical stem similar to Phylloclade but with one or two inteENodes only. Their stem nature is evident by the fact that they bear buds, scales and flowers. e.g. Asparagus (cylindrical cladode), Ruscus (flattened Cladode).

Question 14.
Comment on Corm with an example.
Answer:
Corm is a succulent underground stem with an erect growing tip. The corm is surrounded by scale leaves and exhibit nodes and intemodes. e.g., Amorphophallus, Gladiolus, Colacasia, Crocus, Colchicum.

Question 15.
Differentiate between monopodial and sympodial branching.
Answer:
Monopodial branching:

The terminal bud grows uninterrupted and produce several lateral branches. This type of growth is also known as monopodial branching.
e.g. Polyalthia, Swietenia, Antiaris.

Sympodial branching:

The terminal bud caese to grow after a period of growth and the further growth is taken care by successive or several lateral meristem or buds. This type of growth is also known as sympodial branching.
e.g. Cycas.

Question 16.
Draw and label the parts of a leaf.
Answer:
The parts of a leaf:

Question 17.
Describe the pattern of leaf arrangement in mosaic leaf.
Answer:
In mosaic leaf, leaves tend to fit in with one another and adjust themselves in such a way that they may secure the maximum amount of sunlight with minimum amount of overlapping. The lower leaves have longer petioles and successive upper leaves possess decreasing length petioles, e.g., Acalypha, Begonia.

Question 18.
How the leaf hooks helps the Bignonia plant?
Answer:
In cat’s nail (Bignonia unguiscati) an elegant climber, the terminal leaflets become modified into three, very sharp, stiff and curved hooks, very much like the nails of a cat. These hooks cling to the bark of a tree and act as organs of support for climbing.

Question 19.
Which types of plants develop tendril? How does it helps the plant?
Answer:
In some plants Stem is very weak and hence they have some special organs for attachment to the support. So some leaves are partially or wholly modified into tendril. Tendril is a slender wiry coiled structure which helps in climbing the support.

Question 20.
Rosa species plants are not eaten by herbivores. Why?
Answer:
Rosa species plants develop Prickles. Prickles are small, sharp structure which are the outgrowths from epidermal cells of stem or leaf. It helps the plant in scrambling over other plants. It is also protective against herbivory.

Question 21.
Certain plants like Aloe and Agave can survive in extreme dry conditions. How ?
Answer:
Aloe and Agave are Xerophytes. They develop fleshy and swollen leaves. These succulent leaves store water, mucilage and food. They also resist desiccation.

Question 22.
Write a brief note on Phyllode.
Answer:
Phyllodes are flat, green – coloured leaf – like modifications of petioles or rachis. The leaflets or lamina of the leaf are highly reduced or caducous. The phyllodes perform photosynthesis and other functions of leaf.
Example: Acacia auriculiformis (Australian acacia), Parkinsonia.

Question 23.
Briefly describe the leaf modification in Nepenthes.
Answer:
The leaf becomes modified into a pitcher in Nepenthes. In Nepenthes the basal part of the leaf is laminar and the midrib continues as a coiled tendrillar structure. The apical part of the leaf as modified into a pitcher the mouth of the pitcher is closed by a lid which is the modification of leaf apex.

Question 24.
How the leaves of Utricularia helps in its nourishment?
Answer:
In bladderwort (Utricularia), a rootless free – floating, aquatic plant the leaf is very much segmented. Some of these segments are modified to form bladder – like structures, with a trap – door entrance that traps aquatic animalcules.

Question 25.
What do you understand by the term developmental heterophylly.
Answer:
In plants like Ficus heterophylla leaves vary from entire to variously lobed structures during different developmental stages. Young leaves are usually lobed or dissected and the mature leaves are entire. Such type is known as developmental heterophylly.

Question 26.
List out few secondary functions of stem.
Answer:
Foods storage, perennation, water storage, photosynthesis, buoyancy, protection and support.

Question 27.
Make a tabular column showing types of terrestrial plants and their environmental adaptation with examples.
Answer:
Types of terrestrial plants and their environmental adaptation with examples:

IV. Long Answer Type Questions (5 Marks)

Question 1.
Make a list of aquatic plants and their environmental adaptation with an example.
Answer:
Aquatic plants and their environmental adaptation with an example:

Question 2.
Compare the location, cellular types and the functions of different zones of root.
Answer:
The location, cellular types and the functions of different zones of root:

Question 3.
Draw a flow chart depicting the various types of root modification.
Answer:
The various types of root modification:

Question 4.
Describe the tap root modification for storage purpose with diagram.
Answer:
Tap root modification – Storage roots

Conical Root – These are cone like, broad at the base and gradually tapering towards the apex. e.g. Daucus carota.
Fusiform root – These roots are swollen in the middle and tapering towards both ends. e.g. Raphanus sativus
Napiform root – It is very broad and suddenly tapers like a tail at the apex. e.g. Beta vulgaris

Question 5.
Define bud. Explain the types of buds based on location.
Answer:
Buds are the growing points surrounded by protective scale leaves:

Terminal bud or Apical bud: These buds are present at the apex of the main stem and at the tips of the branches.
Lateral bud or Axillary bud: These buds occur in the axil of the leaves and develop into a branch or flower.
Extra axillary bud: These buds are formed at nodes but outside the axil of the leaf as in Solanum americanum.
Accessory bud: An extra bud on either side (collateral bud) or above (superposed bud or serial bud) the axillary bud. e.g: Citrus and Duranta.

Question 6.
Draw a flow chart illustrating stem modifications.
Answer:
A flow chart illustrating stem modifications:

Question 7.
List out the characteristics of leaf.
Answer:
The characteristics of leaf:

Leaf is a lateral appendage of the stem.
It is borne at the node of the stem.
It is exogenous in origin.
It has limited growth.
It does not possess apical bud.
It has three main parts namely, leaf base, petiole and lamina.
Lamina of the leaf is traversed by vascular strands, called veins.

Question 8.
Define phyllotaxy. Explain its types
Answer:
The mode of arrangement of leaves on the stem is known as phyllotaxy (Greek. Phyllon = leaf; taxis = arrangement). Phyllotaxy is to avoid over crowding of leaves and expose the leaves maximum to the sunlight for photosynthesis. The four main types of phyllotaxy are:

Alternate
Opposite
Temate and
Whorled.

1. Alternate phyllotaxy: In this type there is only one leaf per node and the leaves on the successive nodes are arranged alternate to each other. Spiral arrangement of leaves show vertical rows are called orthostichies. They are two types:

Alternate spiral: In which the leaves are arranged alternatively in a spiral manner. e.g. Hibiscus and Ficus.
Alternate distichous or Bifarious: In which the leaves are organized alternatively in two rows on either side of the stem, e.g. Monoon longifolium (Polyalthia longifolia).

2. Opposite phyllotaxy: In this type each node possess two leaves opposite to each other. They are organized in two different types:

Opposite superposed: The pair of leaves arranged in succession are in the same direction, that is two opposite leaves at a node lie exactly above those at the lower node. e.g. Psidium (Guava), Eugenia jambolana (Jamun) and Quisqualis (Rangoon creeper).
Opposite decussate: In this type of phyllotaxy one pair of leaves is placed at right angles to the next upper or lower pair of leaves, e.g., Calotropis, Zinnia and Ocimum

3. Ternate phyllotaxy: In this type there are three leaves attached at each node. e.g. Nerium.

4. Whorled (verticillate) type of phyllotaxy: In this type more than three leaves are present in a whorl at each node forming a circle or whorl, e.g. Allamanda and Alstonia scholaris.

Question 9.
Define ptyxis & explain its types.
Answer:
Rolling or folding of individual leaves is called ptyxis. There are seven types of ptyxis as follows:

Reclinate – When the upper half of the leaf blade is bent upon the lower half as in loquat (Eriobotrya japonica).
Conduplicate – When the leaf is folded lengthwise along the mid – rib, as in guava, sweet potato and camel’s foot tree (Bauhinia).
Plicate or Plaited – When the leaf is repeatedly folded longitudinally along ribs in a zig – zag manner, as in Borassus flabellifer.
Circinate – When the leaf is rolled from the apex towards the base like the tail of a dog, as in ferns.
Convolute – When the leaf is rolled from one margin to the other, as in banana, aroids and Indian pennywort. Musa and members of Araceae.
Involute – When the two margins are rolled on the upper surface of the leaf towards the mid – rib or the centre of the leaf, as in water lily, lotus, Sandwich Island Climber (Antigonon) and Plumbago.
Crumpled – When the leaf is irregularly folded as in cabbage.

Question 10.
How the duration of leaf is determined? Classify leaves according to duration.
Answer:
Leaves may stay and function for few days to many years, largely determined by the adaptations to climatic conditions.

Cauducuous (Fagacious): Falling off soon after formation, e.g., Opuntia and Cissus quadrangularis.
Deciduous: Falling at the end of growing season so that the plant (tree or shrub) is leafless in winter / summer season, e.g., Maple, Plumeria, Launea and Erythrina.
Evergreen: Leaves persist throughout the year, falling regularly so that tree is never leafless. e.g., Mimusops and Calophyllum.
Marcescent: Leaves not falling but withering on the plant as in several members of Fagaceae.

V. Higher Order Thinking Skills (HOTs)

Question 1.
Roots are non – green coloured. Is there is any green coloured root? Explain.
Answer:
Yes, roots of certain epiphytic & climbing plants develop chlorophyll and turn green to perform the function of photosynthesis. Such a root is called photosynthetic root or assimilatory root. E.g., Tinospora.

Question 2.
Which part of ginger and onion are edible?
Answer:
The edible part of onion is stem covered by fleshy leaves (bulb). The edible part of ginger is underground stem (rhizome).

Question 3.
Name the body parts of the following plants which is modified for food storage.
Answer:
Plant:

Carrot
Colocasia
Aloe

Modified part for food storage:

Root
Stem
Leaves

Question 4.
Give two examples for angiospermic plants producing adventitious roots.
Answer:
Two examples for angiospermic plants producing adventitious roots:

Buttress root of Bombax.
Prop (Pillar) root of Ficus benghalensis.

Question 5.
Rhizome of ginger is like roots of other plants grown underground. Despite this fact ginger is a stem not a root – Justify.
Answer:
Rhizome of ginger is a underground stem not a root because, it possess nodes, internodes, scale leaves & buds, which are the characteristics of stem.

Question 6.
Vanda is an epiphyte. Epiphytes are the plants growing on branches of trees. They do not have direct contact with soil. How they obtain water for its photosynthetic activity?
Answer:
Epiphytic plants like Vanda develop special type of roots containing sponge – like tissue called velamen. These spongy tissue helps in absorbing the atmospheric moisture and utilize it for their photosynthetic activity.

Question 7.
How does a pneumatopore work?
Answer:
Pneumatopores are the special above-ground roots growing above the surface of water seen in plants growing in water logged soils. These pneumatopores has small pores that facilitate the intake of oxygen by roots.

Question 8.
Carnivorous plants like Nepenthes have nutritional adaptations. Which part of Nepenthes plant is modified to solve this problem?
Answer:
The apical part of the leaf is modified into pitcher and the leaf tip is modified into lid of pitcher.

Question 9.
Why do we use the term ‘monocarpic perennial’ for Musa?
Answer:
Musa is a monocarpic perennial because it grows for several years but produces flowers and fruits once in its life time.

Question 10.
Mention any two morphological characters to differentiate monocots from dicots.
Answer:
Two morphological characters to differentiate monocots from dicots:

Monocots have fibrous roots and dicots have tap root system.
Monocot leaves show parallel venation, whereas dicot leaves show reticulate venation.

Question 11.
Why potato tuber is considered as a stem? Although it is an underground plant part.
Answer:
Though potato tuber is found underground, it is a stem since it possess axillary buds & scale leaves.

Question 12.
Fibrous roots are adventitious in origin – Explain.
Answer:
Adventitious roots are those arising from plant parts other than radicle. Fibrous roots are seen in monocots. In monocots, the primary root arising from radicle is short – lived & soon replaced by the lateral roots arising in bunches from the base of stem. Thus fibrous roots are adventitious in origin.

Samacheer Kalvi 11th Bio Botany Solutions Chapter 2 Plant Kingdom

Leave a Comment / Class 11 / By Prasanna

Students can Download Bio Botany Chapter 2 Plant Kingdom Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 2 Plant Kingdom

Samacheer Kalvi 11th Bio Botany Plant Kingdom Text Book Back Questions and Answers

Choose the correct answer
Question 1.
Which of the plant group has gametophyte as a dominant phase?
(a) Pteridophytes
(b) Bryophytes
(c) Gymnosperm
(d) Angiosperm
Answer:
(b) Bryophytes

Question 2.
Which of following represent gametophytic generation in pteridophytes?
(a) Prothallus
(b) Thallus
(c) Cone
(d) Rhizophore
Answer:
(a) Prothallus

Question 3.
The haploid number of chromosome for an Angiosperm is 14, the number of chromosome in its endosperm would be …………… .
(a) 7
(b) 14
(c) 42
(d) 28
Answer:
(c) 42

Question 4.
Endosperm in Gymnosperm is formed …………… .
(a) at the time of fertilization
(b) before fertilization
(c) after fertilization
(d) along with the development of embryo
Answer:
(b) before fertilization

Question 5.
Differentiate halpontic and diplontic life cycle.
Answer:
Halpontic cycle:

Gametophyte is dominant
Sporophyte is represented by zygote

Diplontic cycle:

Sporophyte is dominant
Gametophyte is represented by few cell gametophyte

Question 6.
What is Plectostele? Give example.
Answer:
Plectostele: Xylem plates alternates with phloem plates. Example: Lycopodiurn clavatum.

Question 7.
What do you infer from the term Pycnoxylic?
Answer:
Pycnoxylic wood is compact with narrow medullary ray. Example: Pinus

Question 8.
Mention two characters shared by Gymnosperms and Angiosperms.
Answer:
Gymnosperms:

Vessels are absent (except Gnetales)
Phloem lacks companion cells

Angiosperms:

Vessels are present
Companion cells are present

Question 9.
Do you think shape of chloroplast is unique for algae? Justify your answer.
Answer:
Variation among the shape of the chloroplast is found in members of algae. It is Cup shaped (chlamydomonas). Discoid ((Chara), Girdle shaped (Ulothrix), reticulate (Oedogonium), spiral (Spirogyra), stellate (Zygnema) and plate like (Mougeoutia).

Question 10.
Do you agree with the statement ‘Bryophytes need water for fertilization’? Justify your answer.
Answer:
Yes, in Bryophytes. water plays a vital role in fertilisation, since water film is needed for the transfer of spermatium (male sex cell) to the egg cell.

Entrance Examination Questions Solved

Question 1.
Which of the following are found in extreme saline conditions? (NEET – 2017)
(a) Archaebacteria
(b) Eubacteria
(c) Cyanobacteria
(d) Mycobacteria
Answer:
(a) Archaebacteria

Question 2.
Select the mismatch …………… . (NEET – 2017)

Answer:
(b) Rhodospirillum – Mycorrhiza

Question 3.
Which among the following are the smallest living cells, known without a definite cell wall, pathogenic to plants as well as animals and can survive without oxygen? (NEET – 2017)
(a) Bacillus
(b) Pseudomonas
(c) Mycoplasma
(d) Nostoc
Answer:
(c) Mycoplasma

Question 4.
Read the following statements (A to E) and select the option with all correct statements: (AIPMT – 2015).
A. Mosses and Lichens are the first organisms to colonise a bare rock.
B. Selaginella is a homosporous pteridophyte,
C. Coralloid roots in Cycas have VAM.
D. Main plant body in bryophytes is gametophytic, whereas in pteridophytes it is sporophytic.
E. In gymnosperms, male and female gametophytes arc present within sporangia located on sporophyte.

(a) B, C and E
(b) A, C and D
(c) B, C and D
(d) A, D and E
Answer:
(d) A, D and E

Question 5.
An example of colonial algae is …………… . (NEET – 2017)
(a) Chlorella
(b) Volvox
(c) Ulothrix
(d) Spirogyra
Answer:
(b) Volvox

Question 6.
Five kingdom system of classification suggested by RH. Whittaker is not based on …………… . (AIPMT- 2014)
(a) Presence or absence of a well defined nucleus
(b) Mode of reproduction
(c) Mode of nutrition
(d) Complexity of body organisation
Answer:
(a) Presence or absence of a well defined nucleus

Question 7.
Mycorrhizae are the example of …………… . (NEET – 2017)
(a) Fungitasis
(b) Amensalism
(c) Antibiosis
(d) Mutualism
Answer:
(d) Mutualism

Question 8.
Which of the following shows coiled RNA strand and capsomeres? (AIPMT – 2014)
(a) Polio virus
(b) Tobacco mosaic virus
(c) Measles virus
(d) Retrovirus
Answer:
(b) Tobacco mosaic virus

Question 9.
Viroids differ from viruses in having: (NEET – 2017)
(a) DNA molecules with protein coat
(b) DNA molecules without protein coat
(c) RNA molecules with protein coat
(d) RNA molecules without protein coat
Answer:
(d) RNA molecules without protein coat

Question 10.
Select the mismatch: (NEET – 2017)
(a) Pinus – Dioecious
(b) Cycas – Dioecious
(c) Salvinia – Heterosporous
(d) Equisetum – Homosporous
Answer:
(a) Pinus – Dioecious

Question 11.
Life cycle of Ectocarpus and Fucus respectively are …………… . (NEET – 2017)
(a) Haplontic, Diplontic
(b) Diplontic, Haplodiplontic
(c) Haplodiplontic, Diplontic
(d) Haplodiplontic, Halplontic
Answer:
(c) Haplodiplontic, Diplontic

Question 12.
Zygote meiosis Is characterisitic of …………… . (NEET – 2017)
(a) Marchantia
(b) Fucus
(c) Funaria
(d) Chlamydomonas
Answer:
(d) Chlamydomonas

Question 13.
Which of the following is correctly matched for the product produced by them? (NEET – 2017)
(a) Acetobacter aceti : Antibiotics
(b) Merthanobacterium : Lactic acid
(c) Penicillium natatum : Acetic acid
(d) Saccharomyces cerevisiae : Ethanol
Answer:
(d) Saccharomyces cerevisiae : Ethanol

Question 14.
Which of the following components provides sticky character to the bacterial cell? (NEET – 2017)
(a) Cell wall
(b) Nuclear membrane
(e) Plasma membrane
(d) Glycocalyx
Answer:
(d) Glycocalyx

Question 15.
Which of the following statements is wrong for viroids? (NEET – 2016)
(a) They lack a protein coat
(b) They are smaller than viruses
(c) They causes infections
(d) Their RNA is a high molecular weight
Answer:
(d) Their RNA Is a high molecular weight

Question 16.
In bryophytes and pteridophytes, transport of male gametes require …………… . (NEET – 20L6)
(a) Wind
(b) Insects
(c) Birds
(d) Water
Answer:
(d) Water

Question 17.
How many organisms in the list below are autotrophs? (AIPMT Mains 2012)
Answer:
Lactobacillus, Nostoc, Chora, Nitrosomonas, Nitrobacter, Streptomyces, Saccharomyces, Trypanosoma, Porphyra, Wolffia.
(a) Four
(b) Five
(c) Six
(d) Three
Answer:
(c) Six

Question 18.
Which of the following would appear as the pioneer organisms on bare rocks? (NEET – 2016)
(a) Lichens
(b) Liverworts
(c) Mosses
(d) Green algae
Answer:
(a) Lichens

Question 19.
Monoecious plant of Chara shows occurrence of …………… .(NEET – 2013)
(a) Stamen and carpel on the same plant
(b) Upper antheridium and lower oogonium on the same plant
(c) Upper oogonium and lower antheridium on the same plant
(d) Antheridiophore and archegoniophore on the same plant
Answer:
(c) Upper oogonium and lower antheridium on the same plant

Question 20.
Read the following five statement (A – E) and answer as asked next to them …………… . (AIPMT Prelims – 2012)
(a) In Equisetum, the female gametophyte is retained on the parent sporophyte
(b) In Ginkgo, male gametophyte is not independent
(c) The sporophyte in Riccia is more developed than that in Polytrichum
(d) Sexual reproduction in Volvox is isogamous
(e) The spores of slime moulds lack cell walls
How many of the above statement are correct? (AIPMT Prelims – 2012)
(a) Two
(b) Three
(c) Four
(d) One
Answer:
(d) One

Question 21.
One of the major components of cell wall of most fungi is …………… . (NEET – 2016)
(a) Chitin
(b) Peptidoglycan
(c) Cellulose
(d) Hemicellulose
Answer:
(a) Chitin

Question 22.
Which one of the following statements is wrong? (NEET – 2016)
(a) Cyanobacteria are also called blue – green algae
(b) Golden algae are also called desmids
(c) Eubacteria are also called false bacteria
(d) Phycomycetes are also called algal fungi
Answer:
(c) Eubacteria are also called false bacteria

Question 23.
Flagellated male gametes are present in all the three of which one of the following sets? (AIPMT Prelims – 2007)
(a) Riccia, Dryopteris and Cycas
(b) Anthoceros, Funaria and Spirogyra
(c) Zygnema, Saprolegnia and Hydrilla
(d) Fucus, Marsilea and Calotropis
Answer:
(a) Riccia, Dryopteris and Cycas

Question 24.
Ectophloic siphonostele is found in …………… . (AIPMT Prelims – 2005)
(a) Adiantum and Cucurbitaceae
(b) Osmunda and Equisetum
(c) Marsilea and Botrychium
(d) Dicksonia and maiden hair fern
Answer:
(b) Osmunda and Equisetum

Question 25.
Which part of the tobacco plant is infected by Meloidogyne incognita? (NEET – 2016)
(a) Flower
(b) Leaf
(c) Stem
(d) Root
Answer:
(d) Root

Question 26.
Select the correct statement: (NEET – 2016)
(a) Gymnosperms are both homosporous and heterosporous
(b) Salvinia, Ginkgo and Pinus all are gymnosperms
(c) Sequoia is one of the tallest trees
(d) The leaves of gymnosperms are not well adapted to extremes of climate
Answer:
(c) Sequoia is one of the tallest trees

Question 27.
Seed formation without fertilization in flowering plants involves the process of …………… . (NEET – 2016)
(a) Sporulation
(b) Budding
(c) Somatic hybridization
(d) Apomixis
Answer:
(d) Apomixis

Question 28.
Chrysophytes, Euglenoids, Dinoflagellates and Slime moulds are included in the kingdom …………… . (NEET – 2016)
(a) Animalia
(b) Monera
(c) Protista
(d) Fungi
Answer:
(b) Monera

Question 29.
The primitive prokaryotes responsible for the production of biogas from the dung of ruminant animals, include the …………… . (NEET – 2016)
(a) Halophiles
(b) Thermoacidophiles
(c) Methanogens
(d) Eubacteria
Answer:
(c) Methanogens

Samacheer Kalvi 11th Bio Botany Plant Kingdom Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer
Question 1.
Gametophytic phase is …………… .
(a) triploid
(b) tetraploid
(c) haploid
(d) diploid
Answer:
(c) haploid

Question 2.
Haplodiplontic life cycle is seen in …………… .
(a) algae
(b) gymnosperm
(c) bryophytes
(d) angiosperm
Answer:
(c) bryophytes

Question 3.
Which algae leads an endozoic life in Hydra?
(a) Chlorella
(b) Gracilaria
(c) Ulothrix
(d) Chlamydomonas
Answer:
(a) Chlorella

Question 4.
Study of algae is called …………… .
(a) biology
(b) mycology
(c) bryology
(d) phycology
Answer:
(d) phycology

Question 5.
Siliceous walls are present in …………… .
(a) Chara
(b) Chlamydomonas
(c) Dunaliella
(d) Diatoms
Answer:
(d) Diatoms

Question 6.
In Chara, thallus is encrusted with …………… .
(a) calcium carbonate
(b) hydrogen sulphate
(c) silica
(d) ammonium carbonate
Answer:
(a) calcium carbonate

Question 7.
Pyrenoids are present in …………… .
(a) mitochondrion
(b) chloroplast
(c) ribosomes
(d) lysosomes
Answer:
(b) chloroplast

Question 8.
Type of vegetative reproduction seen in ulothrix is …………… .
(a) bulbils
(b) fission
(c) fragmentation
(d) tubers
Answer:
(c) fragmentation

Question 9.
…………… are thin walled non – motile spores.
(a) Zoospores
(b) Akinetes
(c) Aplanospores
(d) Genunae
Answer:
(c) Aplanospores

Question 10.
Fusion of either morphologically or physiologically dissimilar gametes is called as …………… .
(a) isogamy
(b) anisogamy
(c) syngamy
(d) oogamy
Answer:
(b) anisogamy

Question 11.
According to Fritsch, the algae are classified into …………… classes.
(a) 10
(b) 12
(c) 11
(d) 9
Answer:
(c) 11

Question 12.
Photosynthetic part of the phaeophyceae thallus Is called as
(a) holdfast
(b) stipes
(c) lamina
(d) fronds
Answer:
(d) fronds

Question 13.
A characteristic pigment of phaeophyceae is …………… .
(a) xanthophyle
(b) carotenoid
(c) fucoxanthin
(d) chlorophyll
Answer:
(c) fucoxanthin

Question 14.
…………… is used as single cell protein.
(a) Chlorella
(b) Kelps
(c) Chlamydomonas
(d) Spirogyra
Answer:
(a) Chlorella

Question 15.
Gelidium belongs to …………… members.
(a) Rhodophyccae
(b) Phaeophyceae
(c) Cyanophyccae
(d) Dinophyceae
Answer:
(a) Rhodophyceae

Question 16.
Carrageenan is obtained from …………… .
(a) Chlorella
(b) Chara
(c) Chondrus
(d) Chlamydomonas
Answer:
(c) Chondrus

Question 17.
…………… are the amphibians of the plant kingdom.
(a) Pteridophytes
(b) Algae
(c) Gymnosperms
(d) Bryophytes
Answer:
(d) Bryophytes

Question 18.
Marchantia vegetatively propagates by …………… .
(a) tubers
(b) gemmae
(c) buds
(d) brood bodies
Answer:
(b) gemmae

Question 19.
Peat is obtained from …………… .
(a) Anthoceros
(b) Dendroceros
(c) Sphagnum
(d) Funaria
Answer:
(c) Sphagnum

Question 20.
…………… is a bryophyte used to cure pulmonary tuberculosis.
(a) Marchantia polymorpha
(b) Polytrichum
(c) Sphagnum
(d) Bryum
Answer:
(a) Marchantia polymorpha

Question 21.
Type of stele seen in Marsilea is …………… .
(a) Protostele
(b) Siphonostele
(c) Adiantum
Answer:
(b) Siphonostele

Question 22.
Which of the following pteridophyte is used as a biofertiliser?
(a) Marsilea
(b) Pteridium
(c) Pteris
(d) Azolla
Answer:
(d) Azolla

Question 23.
Which of the following is naked seed producing plant?
(a) Angiosperm
(b) Gymnosperm
(c) Pteridophytes
(d) Bryophytes
Answer:
(b) Gymnosperm

Question 24.
Amber is obtained from …………… .
(a) Angiosperm
(b) Gymnosperm
(c) Pteridophytes
(d) Bryophytes
Answer:
(b) Gymnosperm

Question 25.
Coralloid roots of cycas have symbiotic association with …………… .
(a) Blue green algae
(b) Mycorrhiza
(c) Euglena
(d) Rhizobium
Answer:
(a) Blue green algae

Question 26.
Pinus roots are in symbiotic relationship with …………… .
(a) Blue green algae
(b) Mycorrhiza
(c) Euglena
(d) Rhizobium
Answer:
(b) Mycorrhiza

Question 27.
Which is not a class of Gymnosperm?
(a) Lycopodia
(b) Cycadopsida
(c) Coniferopsida
(d) gnetopsida
Answer:
(a) Lycopodia

Question 28.
The endosperm of gymnosperm is …………… .
(a) haploid
(b) triploid
(c) diploid
(d) Poliploidy
Answer:
(a) haploid

Question 29.
Shiwalik fossil park is located at …………… .
(a) Madhya Pradesh
(b) Himachal Pradesh
(c) Rajmahal hills
(d) Jharkhand
Answer:
(b) Himachal Pradesh

Question 30.
When does the angiosperm appeared on Earth?
(a) Devonian
(b) Cambrian
(c) Early cretaceous
Answer:
(c) Early cretaceous

Question 31.
Which is also called as vascular cryptogam?
(a) Gymnosperms
(b) Pteridophytes
(c) Bryophytes
(d) Algae
Answer:
(b) Pteridophytes

Question 32.
Which is not a cryptogam?
(a) Algae
(b) Bryophytes
(c) Pteridophyta
(d) Angiospermae
Answer:
(d) Angiospermae

Question 33.
…………… is a haiophytic alga.
(a) Chlamydomonas nivalis
(b) Dunaliella salina
(c) Coleochaete
(d) Volvox
Answer:
(b) Dunaliella salina

Question 34.
Who is called as the Father of Indian Phycology?
(a) M.O. Parthasarathy
(b) Y. Bharadwaja
(c) V.S. Sundaralingam
(d) V. Desikachary
Answer:
(a) M.O. Parthasarathy

Question 35.
Wedge shaped modified branches developed by Sphacelaria are called as …………… .
(a) Buds
(b) Akinetes
(c) Tubers
(d) Bulbils
Answer:
(d) Bulbils

Question 36.
Pteridophytes were abundant in the …………… period.
(a) Cambrian
(b) Precambrian
(c) Devonian
(d) Cretaceous
Answer:
(c) Devonian

Question 37.
Heterospory is originated in …………… .
(a) Gymnosperms
(b) Pteridophytes
(c) Bryophytes
(d) Algae
Answer:
(b) Pteridophytes

Question 38.
Sago is obtained from …………… .
(a) Cycas revoluta
(b) Pinus roxburghii
(c) Finus insularis
(d) Cedrus deodara
Answer:
(a) Cycas revoluta

II. Very Short Answer Type Questions (2 Marks)

Question 1.
Define alternation of generation.
Answer:
Alternation of the haploid gametophytic phase (n) with diploid sporophytic phase (2n) during the life cycle is called alternation of generation.

Question 2.
Name any two marine algae.
Answer:
Two marine alga:

Gracilaria and
Sargassum

Question 3.
Name any two fresh water algae.
Answer:
Two fresh water algae:

Oedogonium and
Ulothrix

Question 4.
Mention any two endozoic algae.
Answer:
Two endozoic algae:

Chlorella and
Cladophora crispata.

Question 5.
Define Phycology.
Answer:
The study of algae is called as phycology or algology.

Question 6.
Define epiphytic algae with an example.
Answer:
Algae growing on the surface of aquatic plants are called as epiphytic algae.
Example: Coleochaete.

Question 7.
Name few eminent algologist.
Answer:
F.E. Fritsch, F.E. Round, Y. Bharadwaja and T.V. Desikachary

Question 8.
Write the chemical composition of algae cell wall.
Answer:
Cellulose and hemicellulose.

Question 9.
List out the criteria involved in algal classification.
Answer:
Pigmentation, reserve food materials and flagellation pattern.

Question 10.
What are pyrenoids? Mention its role.
Answer:
Pyrenoids are proteinaceous bodies found in chromatophores of algae and assist in the synthesis and storage of starch.

Question 11.
Distinguish between Isogamy and Oogamy with example.
Answer:
Between Isogamy Oogamy with example:

Isogamy	Oogamy
Fusion of morphologically and physiologically similar gametes. e.g. Ulothrix	Fusion of both morphologically and physiologically dissimilar gametes. e.g. Sargassum

Question 12.
Which is the reserve food material of phaeophyceae members.
Answer:
Mannitol and Laminarin starch.

Question 13.
Name the male & female sex organ of Rhodophyceae members.
Answer:
Male sex organ is called spermatangium. Female sex organ is called carpogonium.

Question 14.
Which is responsible for pigmentation of Brown algae?
Answer:
A golden brown pigment called fucoxanthin is present and it gives shades of colour from olive green to brown to the algal members of Phaeophyceae.

Question 15.
Mention any two algae members used in Agar – Agar production.
Answer:
Two algae member used in Agar – Agar production:

Gracilaria and
Gigartina.

Question 16.
Bryophytes are amphibians of plant kingdom – Justify.
Answer:
Bryophytes are called as ‘amphibians of plant kingdom’ because they need water for completing their life cycle.

Question 17.
Why bryophytes are called as Non – vascular cryptogam?
Answer:
Vascular tissue like xylem and phloem are completely absent in bryophytes, hence called as ‘Non – vascular cryptogams’.

Question 18.
Which type of sexual reproduction occurs in Bryophytes. Name the male & female parts.
Answer:
Sexual reproduction is Oogamous. Male sex organ is called as Antheridia. Female sex organ is called as Archegonia.

Question 19.
What are sporophylls?
Answer:
Sporophylls are the special leaves on which spore bearing sporangia are borne. Sporophylls organize to form strobilus or cone.

Question 20.
Compare Eusporangiate and Leptosporangiate.
Answer:
Eusporangiate and Leptosporangiate:

Eusporangiate	Leptosporangiate
Development of sporangium from group of initials	Development of sporangium from single initial

Question 21.
Differentiate homospory and heterospory with example.
Answer:
Homospory and Heterospory With Example:

Homospory	Heterospory
Production of one type of spores. e.g. Lycopodium	Production of two types of spores. e.g. Selaginella

Question 22.
Which period, does the pteridophytes dominate the surface?
Answer:
Devonian period of Paleozoic era.

Question 23.
List out the ways of vegetative propagation by Pteridophytes.
Answer:
Fragmentation, resting buds, root tubers and adventitious buds.

Question 24.
Define Stele & mention its types.
Answer:
Stele refers to the central cylinder of vascular tissues consisting of xylem, phloem. pericycle and sometimes medullary rays with pith. There are two types of steles:

Protostele
Siphonostele

Question 25.
Distinguish between Protostele & Siphonostele.
Answer:

Protostele	Siphonostele
In protostele, xylem surrounds phloem.	In siphonostele, phloem surrounds xylem.

Question 26.
DefIne Eustele?
Answer:
The stele is split into distinct collateral vascular bundles around the pith. Example: Dicot stem.

Question 27.
What Is amber? Which group of plants produce amber?
Answer:
Amber is a plant secretion that is a efficient preservative that doesn’t get degraded and hence can preserve remains of extinct life forms. The amber is produced by Pinites succinifera, a Gymnosperm.

Question 28.
Which period, does the gymnosperm dominate the Earth?
Answer:
Jurassic and Cretaceous periods of Mesozoic era.

Question 29.
Distinguish between Manoxylic & Pycnoxylic.
Answer:

Manoxylic	Pycnoxylic
Porous, soft wood	Compact hard wood
More parenchyma with wide medullary rays.	Compact with narrow medullary rays.

Question 30.
Define Siphonogamous condition.
Answer:
Siphonogainy refers to the development of pollen tubes for the transfer of male nuclei to egg cell.

Question 31.
Mention any two common features for both gymnosperm & angiosperm.
Answer:
Two common features for both gymnosperm & angiosperm.:

Production of seeds
Presence of Eustele

Question 32.
What is Canada balsam. Add a note on it.
Answer:
Canada balsam is a resin obtained from Abies balsamea. It is used as mounting medium in permanent slide preparation.

Question 33.
Why do we use the term ‘form genera’ for fossil plants?
Answer:
The term ‘form genera’ is used to name the fossil plants because the whole plant is not recovered as fossils instead organs or parts of the extinct plants are obtained in fragments.

Question 34.
Name few fossil sites of india.
Answer:
few fossil sites of india:

1. Shiwalik fossil park	1. Himachal Pradesh
2. Mandla fossil park	2. Madhya Pradesh
3. Rajmahal hills	3. Jharkhand
4. Ariyalur	4. Tamil Nadu

Question 35.
Mention the names of any two fossil gymnosperm.
Answer:
Medullosa, Lepidocarpon and Lepidodendron.

Question 36.
Which group of plants dominate the Earth today? Define it.
Answer:
Angiosperms are the group of plants producing ovules enclosed by ovary.

Question 37.
What is an open vascular bundle?
Answer:
A vascular bundle is open when it has Cambium.

Question 38.
What is a closed vascular bundle?
Answer:
A vascular bundle is closed when it does not have Cambium.

Question 39.
Mention any two morphological differences between Dicot & Monocot.
Answer:
Two morphological differences between Dicot:

Leaves show reticulate venation
Flowers are tetramerous or pentamerous

Two morphological differences between Monocot:

Leaves show parallel venation
Flowers are trimerous

Question 40.
Name the two divisions of spermatophyta?
Answer:
The two divisions are:

Gymnospermae and
Angiospermae

Question 41.
What are brood bodies?
Answer:
Brood bodies are the small detachable branches which help in vegetative propagation.
e.g., Bryopteris fruticulosa.

Question 42.
What are gemmae?
Answer:
Gemmae are small propagative structures which help in asexual reproduction.
e.g., Marchantia.

III. Short Answer Type Questions (3 Marks)

Question 1.
What are cryptogam? Mention its division.
Answer:
Cryptogams are non – flowering or non – seed producing plants. It has been divided into Algae, Bryophytes and Pteridophytes.

Question 2.
In which group of plants we can observe Haplodiplontic life cycle? Draw a diagram of Haplodiplontic life cycle.
Answer:
Bryophytes and Pteridophytes:

Question 3.
Name the 3 types of life cycles seen in plants?
Answer:
The 3 types of life cycles seen in plant:

Haplontic life cycle
Diplontic life cycle
Haplodiplontic life cycle

Question 4.
Where can we see cryophytic & halophytic algae? Give example.
Answer:
Cryophytic & Halophytic Algae:

Cryophytic algae grow on snow. e.g., Chlamydomonas nivalis.
Halophytic algae grow in salt pans, e.g., Dunaliella salina.

Question 5.
List out the various types of vegetative reproduction seen in algae.
Answer:
Fission, fragmentation, budding, bulbils, tubers.

Question 6.
List out the various asexual spores produced by algae.
Answer:
Zoospores, aplanospores, autospores, hypnospores, tetraspores and akinetes.

Question 7.
Write any three differences between chlorophyceae and phaeophyceae members?
Answer:

Question 8.
Define exoscopic embryogeny.
Answer:
In exoscopic embryogeny, the first division of the zygote is transverse & form inner and outer cell. The apex of the embryo develops from outer cell.

Question 9.
Name the three classes of Bryophytes, according to Proskauer.
Answer:
Three Classes of Bryophytes, According to Proskauer:

Hepaticopsida
Anthocerotopsida and
Bryopsida.

Question 10.
How peat is obtained? Write its economic value.
Answer:
A large amount of dead thallus of Sphagnum gets accumulated and compressed, hardened to form peat. It is used as fuel in commercial scale (Netherlands). Nitrates, brown dye and tanning materials are derived from peat. Sphagnum and peat are also used in horticulture as packing material because of their water holding capacity.

Question 11.
Mention any three Pteridophytes and their economic value.
Answer:
Economic Importance of Pteridophyte:
Pteridophyte:

Marsilea
Azolla
Pteris vittata

Uses:

Food
Biofertilizer
Removal of heavy metals from soils – Bioremediation

Question 12.
How the vascular plants dominate the Earth?
Answer:
The success and dominance of vascular plants is due to the development of,

Extensive root system.
Efficient conducting tissues.
Cuticle to prevent desiccation.
Stomata for effective gaseous exchange.

Question 13.
Name the three classes of gymnosperms.
Answer:
Three Classes of Gymnosperms:

Cycadospsida
Coniferopsida and
Gnetopsida.

Question 14.
Name any three economically important products & uses of the gymnosperm plants.
Answer:
Three economically important products & uses of the gymnosperm plants:

Question 15.
Compare the anatomical features between Dicots & Monocots.
Answer:

IV. Long Answer Type Questions (5 Marks)

Question 1.
Explain in detail about the various life cycle patterns in plants.
Answer:
Life cycle patterns in plants: Alternation of Generation: Alternation of generation is common in all plants. Alternation of the haploid gametophytic phase (n) with diploid sporophytic phase (2n) during the life cycle is called alternation of generation. Following type of life cycles are found in plants:

(a) Haplontic life cycle: Gametophytic phase is dominant. photosynthetic and independent, whereas sporophytic phase is represented by the zygote. Zygote undergoes meiosis to restore haploid condition. Example: Volvox and Spimgyra.

(b) Diplontic life cycle: Sporophytic phase (2n) is dominant, photosynthetic and independent. The gametophytic phase is represented by the single to few celled gametophyte. The gametes fuse to form zygote which develops into sporophyte. e.g., Fucus, gymnosperms and angiosperms.

(c) Haplodiplontic life cycle: This type of life cycle is found in Bryophytes and pteridophytes which is intermediate between haplontic and diplontic type. Both the phases are multicellular, but they differ in their dominant phase.

In Bryophytes dominant independent phase is gametophyte and it alternates with short – lived multicellular sporophyte totally or partially dependent on the gametophyte. In Pteridophytes sporophyte is the independent phase. It alternates with multicellular saprophytic or autotrophic, independent, short lived gametophyte (n).

Question 2.
Write a note on diversified thallus organisation seen in algae with examples.
Answer:
A wide range ofthallus organisation is found in algae. Unicellular motile (Chlamydomonas), unicellular non-motile (Chlorella), Colonial motile (Volvox), Colonial non-motile (Hydrodictyon), siphonous (Vaucheria), unbranched filamentous (Spirogyra), branched filamentous (Cladophora), discoid (Coleochaete) heterotrichous (Fritschiella), Foliaceous (Ulva) to Giant Kelps (Laminaria and Macrocystis).

Question 3.
Describe the various types of sexual reproduction observed in algae.
Answer:
Sexual reproduction in algae are of three type:

Isogamy: Fusion of morphologically and Physiologically similar gametes E.g. Ulothrix.
Anisogamy: Fusion of either morphologically or physiologically dissimilar gametes E.g. Pandorina
Oogamy: Fusion of both morphologically and physiologically dissimilar gametes. E.g. Sargassum.

The life cycle shows distinct alternation of generation.

Question 4.
Describe the salient features of Chlorophyceae members.
Answer:
The salient features of Chlorophyceae members:

Chlorophyceae commonly called as green algae.
Mostly aquatic (fresh water or marine), few terrestrial.
Shape of chloroplast differs. It may be cup shaped (Chlamydomonas) or girdle – shaped or reticulate, or stellate etc.
Chlorophyll ‘a’ and ‘b’ are photosynthetic pigments.
Pyrenoids store starch & also proteins.
Outer cell wall is made of pectin and inner is cellulose.
Vegetative reproduction is by fragmentation.
Asexual reproduction by zoospores, aplanospores and akinetes.
Sexual reproduction may be isogamous, anisogamous or oogamous. E.g. Chlamydomonas, Volvox and Spirogyra.

Question 5.
Describe the salient features of Phaeophyceae members.
Answer:
The salient features of Phaeophyceae members:

Phaeophyceae commonly called as Brown algae.
Majority are marine habitats. Pleurocladia is a fresh water form.
Thallus may be filamentous, frond – like or giant kelps.
Thallus is differentiated into photosynthetic part-frond, stalk – like structure – stipe and a holdfast for attachment.
Chlorophyll ‘a’ and ‘c’, carotenoids and Xanthophylls are photosynthetic pigments.
A golden brown fucoxanthin pigment gives olive green to brown colour.
Mannitol and Laminarin starch is the storage material.
Motile spores with unequal flagella (one whiplash and one tinsel) are present.
Oogamous is the major type of sexual reproduction. Isogamy is also seen.
Alternation of generation is seen. Example: Sargassum, Fucus, Laminaria and Dictyota.

Question 6.
Describe the salient features of Rhodophyceae.
Answer:
The salient features of Rhodophyceae:

Rhodophyceae commonly called as red algae.
Mostly marine habitats.
The thallus is multicellular, macroscopic, and may be filamentous, ribbon – like etc.
Chlorophyll ‘a’, r-phycoerythrin and r-phycocyanin are photosynthetic pigments.
Asexual reproduction is by means of monospores, neutral spores and tetraspores.
Floridean starch is the storage material
Sexual reproduction in oogamous.
Male sex organ is spermatangium producing spermatium.
Female sex organ is carpogonium.
Spermatium is carried by water and fuses with egg forming zygote.
Zygote undergoes meiosis forming carpospores.
Alternation of generation is seen. Example: Ceramium, Gelidium and Gigartina.

Question 7.
Tabulate the economic importance of algae.
Answer:
Economic importance of Algae:

Question 8.
Enumerate the general character of Bryophytes.
Answer:
The general character of Bryophytes:

Bryophytes are non – vascular cryptogams due to absence of xylem & phloem.
The plant body is a gametophyte and it is conspicuous, long – lived.
Plant body is undifferentiated into root, stem & leaves. Thalloid forms with rhizoids are seen in liverworts & hornworts. Leaf – like and stem – like structures are seen in mosses.
Vegetative reproduction is by adventitious buds, tubers, brood bodies or by gemmae.
Sexual reproduction is oogamous producing Antheridia & Archegonia in multicellular protective coverings.
Antheridia produces biflagellate antherozoids which swims in water & fuse with egg forming diploid zygote.
Water is essential for fertilization.
Zygote is the first cell of sporophyte. Zygote undergoes mitotics forming undifferentiated embryo, forming sporophyte. The embryogeny is exoscopic.
Sporophyte is dependent on gametophyte.
Sporophyte is differentiated into foot, seta & capsule.
Capsule of Sporophyte produces haploid spores by meiosis.
Bryophtyes are homosporous which are dispersed by elaters.
Spores germinate producing haploid gametophyte.
Heterologous alternation of generation.
Proskauer classified bryophytes into 3 classes, Hepaticopsida(Riccia),Anthocerotopsida (Anthoceros) and Bryopsida (Funaria).

Question 9.
List out the general characters of Pteridophytes.
Answer:
General characteristic features of Pteridophytes:

Plant body is sporophyte (2n) and it is the dominant phase. It is differentiated into root, stem and leaves.
Roots are adventitious.
Stem shows monopodial or dichotomous branching.
Leaves may be microphyllous or megaphyllous.
Stele is protostele but in some forms siphonostele is present (Marsilea)
Tracheids are the major water conducting elements but in Selaginella vessels are found.
Sporangia, spore bearing bag like structures are borne on special leaves called sporophyll. The sporophylls gets organized to form cone or strobilus. e.g., Selaginella and Equisetum.
They may be homosporous (produce one type of spores – Lycopodium) or Heterosporous (produce two types of shorts – Selaginella). Heterospory is the origin for seed habit.
Development of sporangia may be eusporangiate (development of sporangium from group of initials) or leptosporangiate (development of sporangium from single initial).
Spore mother cells undergo meiosis and produce spores (n).
Spore germinates to produce haploid, multicellular green, cordate shaped independent gametophytes called prothallus.
Fragmentation, resting buds, root tubers and adventitious buds help in vegetative reproduction.
Sexual reproduction is Oogamous. Sex organs, namely antheridium and archegonium are produced on the prothallus.
Antheridium produces spirally coiled and multiflagellate antherozoids.
Archegonium is flask shaped with broad venter and elongated narrow neck. The venter possesses egg or ovum and neck contain neck canal cells.
Water is essential for fertilization. After fertilization a diploid zygote is formed and undergoes mitotic division to form embryo.
Pteridophytes show apogamy and apospory.

Question 10.
Write a note on economic importance of Pteridophytes.
Answer:
A note on economic importance of Pteridophytes:
Pteridophyte:

Rumohra adiantiformis (leather leaf fem)
Marsilea
Azolla
Dryopteris filix – mas
Pteris vittata
Pteridium sp.
Equisetum sp.
Psilotum, Lycopodium, Selaginella, Angiopteris, Marattia.

Uses:

Cut flower arrangements
Food
Biofertilizer
Treatment for tapeworm
Removal of heavy metals from soils – Bioremediation
Leaves yield green dye
Stems for scouring
Ornamental plants

Question 11.
What is protostele? Explain its types.
Answer:
In protostele xylem surrounds phloem. The type includes Haplostele, Actinostele, Plectostele and mixed protostele.

Haplostele: Xylem surrounded by phloem is known as haplostele. E.g. Selaginella.
Actinostele: Star shaped xylem core is surrounded by phloem is known as actinostele. E.g. Lycopodium serratum.
Plectostele: Xylem plates alternates with phloem plates. E.g. Lycopodium clavatum.
Mixed prototostele: Xylem groups uniformly scattered in the phloem. E.g. Lycopodium cernuum.

Question 12.
Define Siphonostele. Explain its types.
Answer:
In siphonostele xylem is surrounded by phloem with pith at the centre. It includes Ectophloic siphonostele, Amphiphloic siphonostele, Solenostele, Eustele, Atactostele and Polycylic stele.

Ectophloic siphonostele: The phloem is restricted only on the external side of the xylem. Pith is in centre. E.g. Osmunda.
Amphiphloic siphonostele: The phloem is present on both the sides of xylem. The pith is in the centre. E.g. Marsilea.
Solenostele: The stele is perforated at a place or places corresponding the origin of the leaf trace.
- Ectophloic solenostele – Pith is in the centre and the xylem is surrounded by phloem. E.g. Osmunda.
- Amphiphloic solenostele – Pith is in the centre and the phloem is present on both sides of the xylem. E.g. Adiantum pedatum.
- Dictyostele – The stele is separated into several vascular strands and each one is called meristele. E.g. Adiantum capillus – veneris.
Eustele: The stele is split into distinct collateral vascular bundles around the pith. E.g. Dicot stem.
Atactostele: The stele is split into distinct collateral vascular bundles and are scattered in the ground tissue. E.g. Monocot stem.
Polycyclic stele: The vascular tissues are present in the form of two or more concentric cylinders. E.g. Pteridium.

Question 13.
Point out the general characters of Gymnosperms.
Answer:
General characteristic features:

Most of the gymnosperms are evergreen woody trees or shrubs. Some are lianas (Gnetum)
The plant body is sporophyte and is differentiated into root, stem and leaves.
A well developed Tap root system is present. Coralloid Roots of Cycas have symbiotic association with blue green algae. In Pinus the roots have mycorrhizae.
The stem is aerial, erect and branched or unbranched (Cycas) with leaf scars.
In conifers two types of branches namely branches of limited growth (Dwarf shoot) and Branches of unlimited growth (Long shoot) is present.
Leaves are dimorphic, foliage and scale leaves are present. Foliage leaves are green, photosynthetic and borne on branches of limited growth. They show xerophytic features.
The xylem consists of tracheids but in Gnetum and Ephedra vessels are present.
Secondary growth is present. The wood may be Manoxylic (Porous, soft, more parenchyma with wide medullary ray – Cycas) or Pycnoxylic (compact with narrow medullary ray – Pinus).
They are Heterosporous. The plant may be monoecious (Pinus) or dioecious (Cycas).
Microsporangia and Megasporangia are produced on Microsporophyll and Megasporophyll respectively.
Male and female cones are produced.
Anemophilous pollination is present.
Fertilization is siphonogamous and pollen tube helps in the transfer of male nuclei.
Sporne (1965) classified gymnosperms into 3 classes, 9 orders and 31 families. The classes include
- Cycadospsida
- Coniferopsida
- Gnetopsida.

Question 14.
List out the features common for both Gymnosperms & Angiosperms.
Answer:
Gymnosperms resemble with angiosperms in the following features:

Presence of well organised plant body which is differentiated into roots, stem and leaves
Polyembryony (presence of many embryo). The naked ovule develops into seed. The endosperm is haploid and develop before fertilization.
The life cycle shows alternation of generation. The sporophytic phase is dominant and gametophytic phase is highly reduced.
Presence of cambium in gymnosperms as in dicotyledons.
Flowers in Gnetum resemble to the angiosperm male flower. The Zygote represent the first cell of sporophyte.
Presence of integument around the ovule.
Both plant groups produce seeds
Pollen tube helps in the transfer of male nucleus in both.
Presence of Eustele.

Question 15.
Differentiate the characters of Gymnosperm & Angiosperm.
Answer:
Difference between Gymnosperms and Angiosperms:
Gymnosperms:

Vessels are absent [except Gnetales]
Phloem lacks companion cells
Ovules are naked
Wind pollination only
Double fertilization is absent
Endosperm is haploid
Fruit formation is absent
Flowers absent

Angiosperms:

Vessels are present
Companion cells are present
Ovules are enclosed within the ovary
Insects, wind, water, animals etc., act as pollinating agents
Double fertilization is present
Endosperm is triploid
Fruit formation is present
Flowers present

Question 16.
List out the economic importance of Gymnosperms.
Answer:
Economic importance of Gymnosperms:

Question 17.
List Out the salient’ features of Anglosperms.
Answer:
Salient features of Anglosperms:

Vascular tissue (Xylem and Phloem) is well developed.
Flowers are produced instead of cone.
The embryosac (Ovule) remains enclosed in the ovary.
Pollen tube helps in fertilization, so water is not essential for fertilization.
Double fertilization is present. The endosperm is triploid.
Angiosperms are broadly classified into two classes namely Dicotyledons and Monocotyledons.

Question 18.
Distinguish between Dicotyledons and Monocotyledons.
Answer:
Angiosperms:

V. Higher Order Thinking Skills (HOTs)

Question 1.
State which were the first true land plants? Mention two Characteristics features of these plants.
Answer:
Pteridophytes are the first true land plants. Pteridophytes are the first plants to acquire vascular tissue. Heterosporous condition was developed from Pteridophytes.

Question 2.
Give a comparative account of the following:
(a) Marchantia and Marsilea
(b) Cycas and rose
Answer:
(a) Marchantia is a Bryophyte whereas Marsilea is a Pteridophyte.
(b) Cycas is a gymnospermic plant and rose is a angiospermic plant.

Question 3.
Why are angiosperms so called? In which structures do the seeds develop?
Answer:
Angiosperms are so called because these plants have covered seeds. Seed of angiosperms develop within the ovary which later modify into fruit.

Question 4.
Name the gymnosperms that are exception with regard – to vascular tissue.
Answer:
All the gymnosperms possess tracheids as conducting tissues whereas gymnosperms like Gnetum & Ephedra possess vessels as their conducting tissues.

Question 5.
Both gymnosperms & angiosperms are seed bearers. Yet they are classified separately. Why?
Answer:
Gymnosperms and angiosperms are classified separately because the seeds of the angiosperms are enclosed by ovary (fruit wall) whereas the gymnospermic seeds are naked (not covered by ovary).

Question 6.
Bryophytes maintain soil texture – comment.
Answer:
Bryophytes play a major role in soil formation through succession and help in soil conservation.

Question 7.
Why heterosporous condition is advanced?
Answer:
Heterospory refers to the development of two different types of spores. Heterospory is the origin for seed habit.

Question 8.
Associate the following features with groups in which they first appeared.
(a) Vascular tissues
(b) Seeds inside fruits
(c) Heterospore production
Answer:
(a) Vascular tissues – Pteridophytes
(b) Seeds inside fruits – Angiosperms
(c) Heterospory – Pteridophytes.

Samacheer Kalvi 10th Science Model Question Paper 4 English Medium

Leave a Comment / Class 10 / By Prasanna

Students can Download Samacheer Kalvi 10th Science Model Question Paper 4 English Medium Pdf, Samacheer Kalvi 10th Science Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamil Nadu Samacheer Kalvi 10th Science Model Question Paper 4 English Medium

General Instructions:

The question paper comprises of four parts
You are to attempt all the questions in each part. An internal choice of questions is provided wherever applicable.
All questions of Part I, II, III and IV are to be attempted separately.
Question numbers 1 to 12 in Part I are Multiple Choice Questions of one mark each.
These are to be answered by writing the correct answer along with the corresponding option code.
Question numbers 13 to 22 in Part II are of two marks each. Any one question should be answered compulsorily.
Question numbers 23 to 32 in Part III are of four marks each. Any one question should be answered compulsorily.
Question numbers 33 to 35 in Part IV are of seven marks each. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 75

Part – I

(i) Answer all the questions. [12 × 1 = 12]
(ii) Choose the most suitable answer and write the code with the corresponding answer.

Question 1.
Impulse is equal to _________.
(a) rate of change of momentum
(b) rate of force and time
(c) change of momentum
(d) rate of change of mass
Answer:
(c) change of momentum

Question 2.
The value of universal gas constant ________.
(a) 3.81 mol^-1 KJ^-1
(b) 8.03 J mol^-1 K^-1
(c) 1.38 mol^-1 KJ^-1
(d) 8.31 J mol^-1 K^-1
Answer:
(d) 8.31 J mol^-1 K^-1

Question 3.
The safe limit of receiving the radiation is about _______.
(a) 1R
(b) 0.1 R
(c)100 R
(d) 10 R
Answer:
(b) 0.1 R

Question 4.
The gram molecular mass of oxygen molecule is ________.
(a) 16 g
(b) 18 g
(c) 32 g
(d) 17 g
Answer:
(c) 32 g

Question 5.
The basis of modem periodic law is _________.
(a) atomic number
(b) atomic mass
(c) Isotopic mass
(d) number of neutrons
Answer:
(a) atomic number

Question 6.
A solution is a ______ mixture.
(a) homogeneous
(b) heterogeneous
(c) homogeneous and heterogeneous
(d) non-homogeneous
Answer:
(a) homogeneous

Question 7.
Which is the sequence of correct blood flow _______.
(a) ventricle – atrium – vein – arteries
(b) atrium – ventricle – veins – arteries
(c) atrium – ventricle – arteries – vein
(d) ventricles – vein – atrium – arteries
Answer:
(c) atrium – ventricle – arteries – vein

Question 8.
Polyphagia is a condition seen in ______.
(a) Diabetes insipidus
(b) Diabetes mellitus
(c) Obesity
(d) AIDS
Answer:
(b) Diabetes mellitus

Question 9.
The xylem and phloem arranged side by side on same radius is called _______.
(a) radial
(b) amphivasal
(c) conjoint
(d) None of these
Answer:
(c) conjoint

Question 10.
Which is formed during anaerobic respiration?
(a) Carbohydrate
(b) Ethyl alcohol
(c) Acetyl CoA
(d) Pyruvate
Answer:
(b) Ethyl alcohol

Question 11.
Which is used to build scripts?
(a) Script area
(b) Block palette
(c) stage
(d) Sprite
Answer:
(a) Script area

Question 12.
Rice normally grows well in alluvial soil, but _______ is a rice variety produced by mutation breeding that grows well in saline soil.
(a) Atlas 66
(b) Triticale
(c) Protina
(d) Atomita-2-rice
Answer:
(d) Atomita-2-rice

Part – II

Answer any seven questions. (Q.No: 22 is compulsory) [7 × 2 = 14]

Question 13.
Why a spanner with a long handle is preferred to tighten screws in heavy vehicles?
Answer:
This is because turning effect to tighten the screws depends upon the perpendicular distance of the applied force from the axis of rotation in power arm. Larger the power arm less is the force required to turn the screws. So spanner is provided with a long handle.

Question 14.
State Snell’s law.
Answer:
The ratio of the sine of the angle of incidence and sine of the angle of refraction is equal to the ratio of refractive indices of the two media. This law is also known as Snell’s law.
$\frac{\sin i}{\sin r}=\frac{\mu_{2}}{\mu_{1}}$

Question 15.
What is co-efficient of cubical expansion?
Answer:
The ratio of increase in volume of the body per degree rise in temperature to its unit volume is called as coefficient of cubical expansion.

Question 16.
Define Relative atomic mass.
Answer:
Relative atomic mass of an element is the ratio between the average mass of its isotopes to
l/12th part of the mass of a carbon-12 atom. It is denoted as Ar.

Question 17.
Differentiate aqueous and non-aqueous solution:
Answer:
Aqueous solution:

Water acts as a solvent
E.g. Common salt in water

Non -Aqueous solution:

Any liquid, other than water, acts as a solvent.
E.g. Sulphur dissolved in carbon disulphide

Question 18.
How is diastema formed in rabbit?
Answer:
Diastema is a gap between the incisors and premolar. It is formed due to the absence of canine. It helps in mastication and chewing of food in herbivorous animals.

Question 19.
What is the role of parathormone?
Answer:
The parathormone regulate calcium and phosphorus metabolism in the body. They act on bone, kidney and intestine to maintain blood calcium levels.

Question 20.
Identify the parts A, B, C and D?
Answer:

A – Head
B – Acrosome
C – Nucleus
D – Mitochondria

Question 21.
Write any two importance of Ethnobotany.
Answer:

It provides traditional uses of plant.
It gives information about certain unknown and known useful plants.

Question 22.
The focal length of the concave lens is 7 m. Calculate the power of lens.
Answer:
Given f = -7 m
Power of lens = $\frac{1}{f}=\frac{1}{-7}$ =
= -0.14 dioptre or -0.14 D

Part – III

Answer any seven questions (Q.No: 32 is compulsory) [7 × 4 = 28]

Question 23.
Write the applications of universal law of gravitation.
Answer:

Dimensions of the heavenly bodies can be measured using the gravitation law. Mass of the Earth, radius of the Earth, acceleration due to gravity, etc. can be calculated with a higher accuracy. .
Helps in discovering new stars and planets.
One of the irregularities in the motion of stars is called ‘Wobble’ that lead to the disturbance in the motion of a planet nearby. In this condition the mass of the star can be calculated using the law of gravitation.
Helps to explain germination of roots is due to the property of geotropism, which is the property of a root responding to the gravity.
Helps to predict the path of the astronomical bodies.

Question 24.
(i) Differentiate convex lens and concave lens.
Answer:

Convex lens	Concave lens
The lens which is thicker at the centre than at the edges..	The lens which is thinner at the centre than at the edges
A beam of light passing through it is converged to a point.	A beam of light passing through it, is diverged or spreadout.
It is called as converging lens.	It is called as diverging lens.

(ii) Define dispersion of light.
Answer:
When a beam of white light or composite light is refracted through any transparent media such as glass or water, it is split into its component colours. This phenomenon is called as‘dispersion of light’.

Question 25.
(i) Three resistors of resistance 5 ohm, 3 ohm and 2 ohm are connected in series with 10 V battery. Calculate their effective resistance and the current flowing through the circuit.
Answer:
R₁ = 5 Ω, R₂ = 3 Ω, R₃ = 2 Ω, V = 10 V
R_s = R₁ + R₂ + R₃
= 5 + 3 + 2= 10 Ω
Hence Rs = 10 Ω
The current, I = $\frac{V}{R_{s}}=\frac{10}{10}$ = IA

(ii) An alloy of nickel and chromium is used as the heating element. Why?
Answer:
An alloy of nickel and chromium is used as the heating element. Because:

It has high resistivity
It has a high melting point
It is not easily oxidized.

Question 26.
State and explain the applications of Avogadro’s law.
Answer:
Avogadro’s law : Equal volumes of all gases under the same conditions of temperature and pressure contain an equal number of molecules.
Applications of Avogadro’s law:

It is used to determine the atomicity of gases.
It is helpful in determining the molecular formula of gaseous compounds.
It establishes the relationship between the vapour density and molecular mass of a gas.
It gives the value of molar volume of gases at STP.
Molar volume of a gas at STP = 22.4 litres.
It explains Gay lussac’s law effectively.

Question 27.
Explain the types of double displacement reactions with examples.
Answer:
There are two major classes of double displacement reactions. They are,
(i) Precipitation Reactions:
When aqueous solutions of two compounds are mixed, if they react to form an insoluble compound and a soluble compound, then it is called precipitation reaction.
pb(NO₃)_2(aq) + 2KI_(aq) → PbI_2(s)↓+ 2KNO₃_(aq)

(ii) Neutralisation Reactions:
Another type of displacement reaction in which the acid reacts with the base to form a salt and water. It is called ‘neutralisation reaction’ as both acid and base neutralize each other.
NaOH_(aq) + HCl_1(aq) → NaCl_(aq) + H₂O_(l)

Question 28.
(i) Explain the structure of an atom bomb.
Answer:
Principle:

Uncontrolled chain reaction
Release large amount of energy in short time

Structure:

Fissile material whose mass is sub – critical.
Cylindrical fissile material is fit into void.
Cylinder is injected into the void, using conventional explosive
Two pieces together to form super – critical mass, leads to explosion.
Tremendous amount of energy is released in the form of heat, light and radiation.
γ radiation, affects the living creatures
This type of atom bombs were exploded at Hiroshima and Nagasaki in Japan during Second World War.

(ii) Calculate the pH of 1.0 × 10^-4 molar solution of HNO3.
Answer:
[H⁺] = 1.0 × 10^-4 m, pH = ?
pH = – log₁₀ [H⁺]
= – log₁₀ 1 × 10^-4
[M⁺] = – log₁₀ ¹– log₁₀ 10^-4
= 0 – (-4) log₁₀ ¹⁰ ∴ log₁₀ ¹⁰= 1
= 4 × 1
pH =4

Question 29.
With a neat labelled diagram explain the structure of a neuron.
Answer:

A neuron or nerve cell is the structural and functional unit of the nervous system. It consists of Cyton, Dendrites and Axon.

(a) Cyton: Cyton has a central nucleus with the abundant – cytoplasm called neuroplasm. The cytoplasm has large granular body called Nissel’s granules and the other cell organelles like mitochondria, ribosomes, lysosomes, and endoplasmic reticulum.

(b) Dendrites: These are the numerous branched cytoplasmic processes, that project from the surface of the cell body. They conduct nerve impulses, towards the cyton.

(c) Axon: The axon is a single, elongated, slender projection. The end of axon terminates as fine branches, which terminate into knob like swellings called synaptic knob. The plasma membrane of axon is called axolemma, while the cytoplasm is called axoplasm. It carries impulses away from the cyton.

The axon may be covered by a protein sheath called myelin sheath, which is further covered by a layer of Schwann cells called Neurilemma. Myelin sheath breaks at intervals, by depressions called Nodes of Ranvier. The region between the nodes is called as internode.

Question 30.
(i) What are the various routes by which transmission of human immuno deficiency virus takes place?
Answer:
AIDS is not transmitted by touch or any physical contact. HIV is generally transmitted by

Sexual contact with infected person.
Use of contaminated or infected blood or blood products, needles or syringes.
By the use of contaminated needles or syringes.
From infected mother to her child through placenta.

(ii) Name three improved characteristics of wheat that helped India to achieve high productivity.
Answer:

High yielding
Semi-dwarf
Fertilizer responsive

Question 31.
(i) Kavitha gave birth to a female baby. Her family members say that she can give birth to only female babies because of her family history. Is the statement given by her family members true. Justify your answer.
Answer:
No, the statement is not true. Sex determination is a chance of probability as to which category of sperm fuses with egg.
If the egg [X] is fused by the X-bearing sperm, an XX individual (female) is produced.

If the egg [X] is fused by the Y-bearing sperm an XY individual (male) is produced.

The sperm, produced by the father, determines the sex of the child. The mother is not responsible in determining the sex of the child.

(ii) Who discovered Rh factor? Why was it named so?
Answer:
The Rh factor was discovered by Landsteiner and Wiener in 1940 in Rhesus monkey. Hence the name Rh factor.

Question 32.
Describe and name three stages of cellular respiration that aerobic organisms use to obtain energy?
Answer:
Stages of cellular respiration
(i) Glycolysis: It takes place in cytoplasm of the cell. It is the break down of one molecule of Glucose (6 Carbon) into two molecules of Pyruvic acid (3 Carbon).

(ii) Kreb’s cycle: This occurs in mitochondria matrix. During this cycle, oxidation of pyruvic acid result into CO₂ and water.

(iii) Electron transport chain(ETC):
ETC is located on the inner membrane of the mitochondria.

(a) NADH₂ and FADH₂ molecules formed during Glycolysis and Kreb’s cycle are oxidised to NAD⁺ and FAD⁺ to release the energy via electrons.

(b) The electrons, as they move through the system, release energy which is trapped by ADP to synthesize ATP. This is called Oxidative Phosphorylation. During this process, oxygen which is the acceptor of electrons get reduced to water.

Part – IV

(1) Answer all the questions. [3 × 7 = 21]
(2) Each question carries seven marks.
(3) Draw diagram wherever necessary.

Question 33.
(a) (i) State Newton’s third law of motion.
Answer:
‘For every action, there is an equal and opposite reaction. They always act on two different bodies’.

(ii) Deduce the equation of a force using Newton’s second law of motion.
Answer:
“The force acting on a body is directly proportional to the rate of change of linear momentum of the body and the change in momentum takes place in the direction of the force”.
Let, ‘m’ be the mass of a moving body, moving along a straight line with an initial speed ‘u’. After a time interval of ‘t’ the velocity of the body changes to ‘v’ due to the impact of an unbalanced external force F.

Initial momentum of the body P_i = mu
Final momentum of the body P_f = mv
Change in momentum Δp = P_f – P_i
= mv – mu

By Newton’s second law of motion,
Force, F ∝ rate of change of momentum
F ∝ change in momentum / time
F ∝ $\frac{m v-m u}{t}$
F = $\frac{k m(v-u)}{t}$

Here, k is the proportionality constant, k = 1 in all systems of units. Hence,
F = $\frac{m(v-u)}{t}$
Since, acceleration = change in velocity / time, a = (v – u)/t. Hence, we have
F = m × a
Force = mass × acceleration

No external force is required to maintain the motion of a body moving with uniform velocity.
When the net force acting on a body is not equal to zero, then definitely the velocity of the body will change.
Thus, change in momentum takes place in the direction of the force. The change may take place either in magnitude or in direction or in both.

[OR]

(b) A piece of wire of resistance 10 ohm is drawn out so that its length is increased to three times its original length. Calculate the new resistance.
Answer:
Specific resistance (ρ) = $\frac{\mathrm{RA}}{l}$
R = $\frac{\rho l}{\mathrm{A}}$
When the length is increased by three and then the area of cross section is reduced by three
Resistance of wire R = 10 Ω
New resistance R’ = $\frac{\rho(3 l)}{(A / 3)}$
= ρ (3l) x $\frac{3}{\mathrm{A}}$
R’ = $\frac{9 \rho l}{\mathrm{A}}$
R’ = 9 x R = 9 x 10
R’ = 90 Ω

Question 34.
(a) (i) Calculate the mass of the 0.023 x 10²⁰ molecules of H₂O.
Answer:
Molecular mass of H₂O = H × 2 + 0 × 1
= 1 × 2 + 16 × 1
= 2 + 16 = 18

= 10 × 18
Mass of an atom = 0.018 g

(ii) Calculate the moles of 12 g of magnesium.
Answer:
Number of moles = $\frac{\text { Given mass }}{\text { molecular mass }}$
= $\frac{12}{24}=\frac{1}{2}$ = 0.5
∴ 12 g magnesium makes 0.5 moles.

(iii) Calculate the number of molecules present in 50 g of Fe.
Answer:

= 5.387 × 1023
∴ 50 g Fe contain 5.387 × 10²³ atoms/molecules

[OR]

(b) Explain smelting process..
Answer:
Smelting: The charge consisting of roasted ore, coke and limestone in the ratio 8 : 4 : 1 is smelted in a blast furnace by introducing it through the cup and cone arrangement at the top.
There are three important regions in the furnace.

(a) The Lower Region (Combustion Zone): The temperature is at 1500°C. In this region, coke bums
with oxygen to form CO₂ when the charge comes in contact with a hot blast of air.

It is an exothermic reaction since heat is liberated.

(b) The middle region (Fusion Zone): The temparature prevails at 1000° C. In this region, CO₂ is reduced to CO.

Limestone decomposes to calcium oxide and CO₂

These two reactions are endothermic due to absorption of heat. Calcium oxide combines with silica to form calcium silicate slag.
CaO + SiO₂ → CaSiO₃

(c) The Upper Region (Reduction Zone): The temperature prevails at 400°C . In this region carbon monoxide reduces ferric oxide to form a fairly pure spongy iron.

The molten iron is collected at the bottom of the furnace after removing the slag.
The iron thus formed is called pig iron. It is remelted and cast into different moulds. This iron is called cast iron.

Question 35.
(a) (i) Write any two physiological effects of ethylene.
Answer:
Physiological effects of ethylene:

It promotes the ripening of fruits. E.g., Tomato, Apple, Mango, Banana, etc.
It breaks the dormancy of buds,seeds and storage organs.

(ii) What is Autogamy?
Answer:
Autogamy or self pollination is the transfer of pollen grains from the anther to the stigma of same flower or another flower borne on the same plant. Eg. Hibiscus.

(iii) How will you prevent soil erosion?
Answer:
Prevention of soil erosion:

Retain vegetation cover, so that soil is not exposed.
Cattle grazing should be controlled.
Crop rotation and soil management improve soil organic matter.
Run off water should be stored in the catchment.
Reforestation, terracing and contour ploughing.
Wind speed can be controlled by planting trees in the form of a shelter belt.

[OR]

(b) (i) How do you differentiate Homologous organs from Analogous organs.
Answer:

Homologous organs	Analogous organs
1. Homologous organs are those inherited from common ancestors, with similar developmental pattern in embryos.	1. The analogous organs look similar and perform similar functions but they have different origin and development pattern.
2. A human hand, a front leg of cat, flipper of a whale and a bat’s, wing look dissimilar and adapted for different functions.	2. The function of the wings of a bat, the wings of a bird and wings of an insect are similar, but their basic structures are different.

(ii) Why are the rings of cartilage found in trachea of rabbit?
Answer:
The tracheal walls of rabbit are supported by rings of cartilage which help in the free passage of air.

Samacheer Kalvi 10th Maths Model Question Paper 3 English Medium

Leave a Comment / Class 10 / By Prasanna

Students can Download Samacheer Kalvi 10th Maths Model Question Paper 3 English Medium Pdf, Samacheer Kalvi 10th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Model Question Paper 3 English Medium

Instructions

The question paper comprises of four parts.
You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
All questions of Part I, II, III and IV are to be attempted separately.
Question numbers 1 to 14 in Part I are Multiple Choice Quèstions of one-mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and.writing the option code and the corresponding answer.
Question numbers 15 to 28 in Part II àre two-marks questions. These are to be answered in about one or two sentences.
Question numbers 29 to 42 in Part III are five-marks questions. These are to be answered in about three to five short sentences.
Question numbers 43 to 44 in Part IV are eight-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 100

PART – I

I. Choose the correct answer. Answer all the questions. [14 × 1 = 14]

Question 1.
A = {a, b, p}, B = {2, 3}, C = {p, q, r, 5} then n[(A ∪ C) × B] is ……………. .
(1) 8
(2) 20
(3) 12
(4) 16
Answer:
(3) 12

Question 2.
If a, b, c are in G.P., then $\frac{a-b}{b-c}$ is equal to ……………. .
(1) $\frac{a}{b}$
(2) $\frac{b}{a}$
(3) $\frac{a}{c}$
(4) $\frac{c}{b}$
Answer:
(1) $\frac{a}{b}$

Question 3.
If k + 2, 4k – 6, 3k – 2 are the 3 consecutive terms of an A.P, then the value of k is ……………. .
(1) 2
(2) 3
(3) 4
(4) 5
Answer:
(2) 3

Question 4.
If (x – 6) is the HCF of x² – 2x – 24 and x² – kx – 6 then the value of k is ……………. .
(1) 3
(2) 5
(3) 6
(4) 8
Answer:
(2) 5

Question 5.
If A is a 2 × 3 matrix and B is a 3 × 4 matrix, how many columns does AB have ……………. .
(1) 3
(2) 4
(3) 2
(4) 5
Answer:
(2) 4

Question 6.
In a ∆ABC, AD is the bisector of ∠BAC. If AB = 8 cm, BD = 6 cm and DC = 3 cm. The length of the side AC is ……………. .
(1) 6 cm
(2) 4 cm
(3) 3 cm
(4) 8 cm
Answer:
(2) 4 cm

Question 7.
(2, 1) is the point of intersection of two lines.
(1) x – y – 3 = 0; 3x – y – 7 = 0
(2) x + y = 3; 3x + y = 7
(3) 3x + y = 3; x + y = 7
(4) x + 3y – 3 = 0; x – y – 1 = 0
Answer:
(2) x + y = 3; 3x + y = 7

Question 8.
If the ratio of the height of a tower and the length of its shadow is √3 : 1 , then the angle of elevation of the sun has measure ……………. .
(1) 45°
(2) 30°
(3) 90°
(4) 60°
Answer:
(4) 60°

Question 9.
A spherical ball of radius r₁ units is melted to make 8 new identical balls each of radius r₂ units. Then r₁ : r₂ is ……………. .
(1) 2:1
(2) 1:2
(3) 4:1
(4) 1:4
Answer:
(1) 2:1

Question 10.
The standard deviation of a data is 3. If each value is multiplied by 5 then the new variance is ……………. .
(1) 3
(2) 15
(3) 5
(4) 225
Answer:
(4) 225

Question 11.
A page is selected at random from a book. The probability that the digit at units place of the page number chosen is less than 7 is ……………. .
(1) $\frac { 3 }{ 10 }$
(2) $\frac { 7 }{ 10 }$
(3) $\frac { 3 }{ 9 }$
(4) $\frac { 7 }{ 9 }$
Answer:
(2) $\frac { 7 }{ 10 }$

Question 12.
The range of the relation R = {(x, x³)/x} is a prime number less than 13} is ……………. .
(1) {2,3,5,7,11}
(2) {4,9,25,49, 121}
(3) {8, 27,125, 343,1331}
(4) {1, 8, 27, 125, 343, 1331}
Answer:
(3) {8, 27,125, 343,1331}

Question 13.
If 1 + 2 + 3 + …. + n = k then 1³ + 2³ + 3³ + ……… n³ is equal to ……………. .
(1) K²
(2) K³
(3) $\frac{k(k+1)}{2}$
(4) (k + 1)K³
Answer:
(1) K²

Question 14.
Two dice are thrown simultaneously. The probability of getting a doublet is ……………. .
(1) $\frac { 1 }{ 36 }$
(2) $\frac { 1 }{ 3 }$
(3) $\frac { 1 }{ 6 }$
(4) $\frac { 2 }{ 3 }$
Answer:
(3) $\frac { 1 }{ 6 }$

PART-II

II. Answer any ten questions. Question No. 28 is compulsory. [10 × 2 = 20]

Question 15.
ARelation R is given by the set {{x,y)/y = x + 3, x ∈ {0,1,2,3,4,5}}. Determine its domain and range.
Answer:
x = {0, 1, 2, 3, 4, 5} and y = x + 3
when x = 0 ⇒ y = 0 + 3 = 3
when x = 1 ⇒ y = 1 + 3 = 4 .
when x = 2 ⇒ y = 2 + 3 = 5
when x = 3 ⇒ y = 3 + 3 = 6
whenx = 4 ⇒ y = 4 + 3 = 7
when x = 5 ⇒ y = 5 + 3 = 8
R = {(0, 3) (1, 4) (2, 5) (3, 6) (4, 7) (5, 8)}
Domain = {0, 1, 2, 3, 4, 5}
Range = {3, 4, 5, 6, 7, 8}

Question 16.
Let f = {(-1,3), (0, -1), (2, – 9)} be a linear function from Z to Z. Find f(x).
Answer:
The linear equation is f(x) = ax + b
f(-1) = 3
a(-1) + b = 3
-a + b = 3 ….(1)
f(0) = -1
a(0) + b = -1
0 + b = -1
b = -1
Substitute the value of b = -1 in (1)
– a – 1 = 3
– a = 3 + 1 ⇒ – a = 4
a = -4
∴ The linear equation is -4(x) -1 = -4x – 1 (or) – (4x + 1)

Question 17.
Solve 8x = 1 (mod 11)
Answer:
8x ≡ 1 (mod 11) can be written as 8x – 1 = 11k, for some integer k.
x = $\frac{11 k+1}{8}$
When we put k = 5, 13, 21, 29,… . then 11 k + 1 is divisible by 8.
x = $\frac{11 \times 5+1}{8}$ = 7
x = $\frac{11 \times 13+1}{8}$ = 18
Therefore, the solutions are 7,18,29,40, …….

Question 18.
How many terms of the series 1³ + 2³ + 3³ +……….. should be taken to get the sum 14400?
Answer:
1³ + 2³ + 3³ +. . . + n³ = 14400
$\left[\frac{n(n+1)}{2}\right]^{2}$ = 14400
$\frac{n(n+1)}{2}$ = √14400
$\frac{n(n+1)}{2}$ = 120 ⇒ n² + n = 240
n² + n – 240 =0

(n + 16) (n – 15) =0
(n + 16) = 0 or (n – 15) = 0
n = -16 or n = 15 (Negative will be omitted)
∴ The number of terms taken is 15

Question 19.
Given the L.C.M and G.C.D of the two polynomials are a³ – 10a² + 11a + 70 and a – 7, the polynomial p(x) is a² – 12a + 35 then find q(x).
Answer:

L.C.M. = a³ – 10a² + 11a + 70
= (a – 7) (a² – 3a – 10)

= (a – 7) (a – 5) (a + 2)
G.C.D. = (a – 7)

p(x) = a² – 12a + 35
= (a – 5) (a – 7)
q(x) = $\frac{\mathrm{LCM} \times \mathrm{GCD}}{p(x)}$
= $\frac{(a-7)(a-5)(a+2) \times(a-7)}{(a-5)(a-7)}$
q(x) = (a + 2) (a – 7)

Question 20.
Find the square root of 9x² – 24xy + 30xz – 40yz + 25z² + 16y²
Answer:
$\sqrt{9 x^{2}-24 x y+30 x z-40 y z+25 z^{2}+16 y^{2}}$
= $\sqrt{(3 x)^{2}+(4 y)^{2}+(5 z)^{2}-2(3 x)(4 y)-2(4 y)(5 z)+2(3 x)(5 z)}$
= $\sqrt{(3 x-4 y+5 z)^{2}}$ [using (a – b + c)² = a² + b² + c² – 2ab – 2bc + 2ca]
= |3x – 4y + 5z|

Question 21.
Construct a 3 × 3 matrix whose elements are a_ij = i² j²
Answer:
The general 3 × 3 matrix is given by A = $\left[ \begin{matrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{matrix} \right]$ a_ij = i² j²

a₁₁ = 1² × 1² = 1 × 1 = 1;
a₁₂ = 1² × 2² = 1 × 4 = 4;
a₁₃ = 1² × 3² = 1 × 9 = 9;

a₂₁ = 2² × 1² = 4 × 1 = 4;
a₂₂ = 2² × 2² = 4 × 4 = 16;
a₂₃ = 2² × 3² = 4 × 9 = 36;

a₃₁ = 3² × 1² = 9 × 1 = 1;
a₃₂ = 3² × 2² = 9 × 4 = 36;
a₃₃ = 3² × 3² = 9 × 9 = 81;
Hence the required matrix is A = $\left[ \begin{matrix} 1 & 4 & 9 \\ 4 & 16 & 36 \\ 9 & 36 & 81 \end{matrix} \right] $

Question 22.
An artist has created a triangular stained glass window and has one strip of small length left before completing the window. She needs to figure out the length of left out portion based on the lengths of the other sides as shown in the figure.
Answer:
Given that AE = 3 cm, EC = 4 cm, CD = 10 cm, DB = 3 cm, AF = 5 cm.
Let FB be x
Using Ceva’s theorem we have

$\frac{A E}{E C} \times \frac{C D}{D B} \times \frac{B F}{A F}=1$
$\frac{3}{4} \times \frac{10}{3} \times \frac{x}{5}=1$ $\Rightarrow \frac{2 x}{4}=1$
2x = 4 ⇒ x = $\frac{4}{2}$ = 2
The value of BF = 2

Question 23.
Find the area of the triangle whose vertices are (-3, 5), (5, 6) and (5, -2)
Answer:
Plot the points in a rough diagram and take them in counter-clockwise order. Let the vertices be A(-3,5), B(5,-2), C(5,6)
The area of ∆ABC is

= $\frac { 1 }{ 2 }$ {(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁ y₃)}
= $\frac { 1 }{ 2 }$ {(6 + 30 + 25) – (25 – 10 – 18)}
= $\frac { 1 }{ 2 }$ {(61 + 3)}
= $\frac { 1 }{ 2 }$ (64) = 32 sq. units

Question 24.
From the top of a rock 50 √3 m high, the angle of depression of a car on the ground is observed to be 30°. Find the distance of the car from the rock.
Answer:

Let the distance of the car from the rock is “x” m
In the right ∆ ABC, tan 30° = $\frac{A B}{B C}$
$\frac{1}{\sqrt{3}}=\frac{50 \sqrt{3}}{x}$
X = 50 √3 × √3 = 50 × 3
= 150 m
∴ Distance of the car from the rock =150 m

Question 25.
If n = 5, x̄ = 6, Σx² = 765, then calculate the coefficient of variation.
Answer:
Standard deviation (σ) = $\sqrt{\frac{\Sigma x^{2}}{n}-\left(\frac{\Sigma x}{n}\right)^{2}}$
$=\sqrt{\frac{765}{5}-(6)^{2}}=\sqrt{153-36}=\sqrt{117}$
σ = 10.816
Coefficient of variation = $\frac{\sigma}{\bar{x}} \times 100 \%$
= $\frac{10.816}{6} \times 100 \%$ = 180.266%
Coefficient of variation = 180. 27%

Question 26.
The roots of the equation x² + 6x – 4 = 0 are α, β. Find the quadratic equation whose roots are $\frac{2}{\alpha}$ and $\frac{2}{\beta}$.
Answer:
α and β are the roots of x² + 6x – 4 = 0
α + β = -6 ; αβ = -4
Sum of the roots = $\frac{2}{\alpha}+\frac{2}{\beta}$
$\frac{2 \beta+2 \alpha}{\alpha \beta}=\frac{2(\alpha+\beta)}{\alpha \beta}=\frac{2(-6)}{-4}$
= 3
Product of the roots = $\frac{2}{\alpha} \times \frac{2}{\beta}=\frac{4}{\alpha \beta}$
= $\frac{4}{-4}=-1$
The quadratic equation is x² – (sum of the roots) x + product of the roots = 0
∴ x² – 3x – 1 = 0 .

Question 27.
If the points A(2, 5), B(4, 6) and C(8, a) are collinear, find the value of “a” using slope concept.
Answer:
Since the three points are collinear
Slope of line = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
Slope of AB = Slope of BC
$\frac{6-5}{4-2}=\frac{a-6}{8-4}$
$\frac{1}{2}=\frac{a-6}{4}$
2a – 12 = 4 ⇒ 2a = 16
a = $\frac{16}{2}$ = 8
The value of a = 8

Question 28.
The king, Queen and Jack of clubs are removed from a deck of 52 playing cards and the remaining cards are shuffled. A card is drawn from the remaining cards, find the probability of getting (i) a card of clubs (ii) a queen of diamond
Answer:
Sample space (S) = (52 – 3) = 49
n (S) = 49
(i) Let A be the event of getting a card of clubs.
n(A) = (13 – 3) = 10
P(A) = $\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{10}{49}$
(ii) Let B be the event of getting a queen of diamond
n (B) = 1
P(B) = $\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{1}{49}$

PART-III

III. Answer any ten questions. Question No. 42 is compulsory. [10 × 5 = 50]

Question 29.
Find x if gff (x) = fgg (x), given f(x) = 3x + 1 and g(x) = x + 3.
Answer:
gff(x) = g [f {f{x)}} (This means “g of f of f of x”)
= g[f(3x + 1)] = g[3(3x + 1) + 1] = g(9x + 4)
g (9x + 4) = [ (9x + 4) + 3] = 9x + 7
fgg (x) = f[g{g (x)}] (This means “f of g of g of x”)
= f[g(x + 3)] = f[(x + 3) + 3] = f(x + 6)
f(x + 6) = [3(x + 6) + 1] = 3x + 19
These two quantities being equal, we get 9x + 7 = 3x + 19. Solving this equation we obtain x = 2.

Question 30.
Let A = {-l,l}and B = {0,2}. If the function f : A → B defined by f(x) = ax + b is an onto function? Find a and b.
Answer:
A = {-1, 1}; B = {0,2}
f(x) = ax + b
0 = -a + b
a – b = 0 …….. (1)
f(1) = a(1) + b
2 = a + b
a + b = 2 ………. (2)
Solving (1) and (2) we get

Substitute a = 1 in (1)
1 – b = 0 ⇒ -b = -1 ⇒ 6 = 1
The value of a = 1 and b = 1

Question 31.
Priya earned ₹15,000 in the first month. Thereafter her salary increases by ₹1500 per year. Her expenses are ₹13,000 during the first year and the expenses increases by ₹900 per year. How long will it take her to save ₹20,000 per month.
Answer:

Monthly savings form an A.P.
2000, 2600, 3200 ….
a = 2000 ; d = 2600 – 2000 = 600
Given t_n = 20,000
t_n = a + (n – 1) d
20000 = 2000 + (n – 1) 600
20000 = 2000 + 600n – 600
= 1400 + 600n
20000 – 1400 = 600n
1860o = 600n
n = $\frac{18600}{600}$ = 31
He will take 31 years to save ₹20,000 per month

Question 32.
Find the sum to n terms of the series 0.4 + 0.44 + 0.444 + ……… to n terms
Answer:

Question 33.
Find the GCD of 6x³ – 30x² + 60x – 48 and 3x³ – 12x² + 21x – 18.
Answer:
Let, f(x) = 6x³ – 30x² + 60x – 48 = 6(x³ – 5x² + 10x – 8) and
g(x) = 3x³ – 12x² + 21x – 18 = 3(x³ – 4x² + 7x – 6)
Now, we shall find the GCD of x³ – 5x² + 10x – 8 and x³ – 4x² + 7x – 6

Here, we get zero as remainder.
GCD of leading coeffients 3 and 6 is 3
Thus GCD
[(6x³ – 30x² + 60x – 48, 3x³ – 12x² + 21x – 18)] = 3(x – 2)

Question 34.
If the roots of the equation (c² – ab)x² – 2(a² – bc)x + b² – ac = 0 are real and equal prove that either a = 0 (or) a³ + b³ + c³ = 3abc
Answer:
(c² – ab)x² – 2(a² – bc)x + b² – ac = 0
Here a = c² – ab ; b = -2 (a² – bc) ; c = b² – ac
Since the roots are real and equal
Δ = b² – 4ac
[-2 (a² – bc)]² – 4(c² – ab) (b² – ac) = 0
4(a² – bc)² – 4[c ²b² – ac³ – ab³+ a²bc] = 0
Divided by 4 we get
(a² – bc)² [c²b² – ac³ – ab² + a²bc] = 0
a⁴ + b²c² – 2a²bc – c²b² + ac³ + ab³ – a²bc = 0
a⁴ + ab³ + ac³ – 3a²bc = 0
a(a³ +b³ + c³) = 3a²bc
a³ + b³ + c³ = $\frac{3 a^{2} b c}{a}$
a³ + b³ + c³ = 3abc
Hence it is proved

Question 35.
ABCD is a quadrilateral in which AB = AD, the bisector of ∠BAC and ∠CAD intersect the sides BC and CD at the points E and F respectively. Prove that EF || BD.
Answer:

ABCD is a quadrilateral. AB = AD.
AE and AF are the internal bisector of ∠BAC and ∠DAC.
To prove: EF || BD.
Construction: Join EF and BD
Proof: In ∆ ABC, AE is the internal bisector of ∠BAC.
By Angle bisector theorem, we have,
∴ $\frac{A B}{A C}=\frac{B E}{E C}$ …….. (1)
In ∆ ADC, AF is the internal bisector of ∠DAC
By Angle bisector theorem, we have,
$\frac{A D}{A C}=\frac{D F}{F C}$
∴ $\frac{A B}{A C}=\frac{D F}{F C}$ (AB = AD given) …….. (2)
From (1) and (2), we get
$\frac{B E}{E C}=\frac{D F}{F C}$
Hence In ∆ BCD,
BD || EF (By converse of BPT)

Question 36.
Find the equation of a straight line passing through (1,-4) and has intercepts which are in the ratio 2:5
Answer:
Let the x-intercept be 2a and the y intercept 5a .
The equation of a line is $\frac{x}{a}+\frac{y}{a}=1 \Rightarrow \frac{x}{2 a}+\frac{y}{5 a}=1$
The line passes through the point (1, -A)
$\frac{1}{2 a}+\frac{(-4)}{5 a}=1 \Rightarrow \frac{1}{2 a}-\frac{4}{5 a}=1$
Multiply by 10a
(L.C.M of 2a and 5a is 10a)
5 – 8 = 10a ⇒ -3 = 10a .
a = $\frac{-3}{10}$
The equation of the line is $\frac{x}{2(-3 / 10)}+\frac{y}{5(-3 / 10)}=1$
$\frac{x}{-3 / 5}+\frac{y}{-3 / 2}=1 \Rightarrow \frac{5 x}{-3}+\frac{2 y}{-3}=1$
$\frac{-5 x}{3}-\frac{2 y}{3}=1$
Multiply by 3
-5x – 2y = 3 ⇒ -5x – 2y – 3 = 0
5x + 2y + 3 = 0
The equation of a line is 5x + 2y + 3 = 0

Question 37.
If $\frac{\cos \theta}{1+\sin \theta}=\frac{1}{a}$, then prove that $\frac{a^{2}-1}{a^{2}+1}$ = sin θ
Answer:

Question 38.
A toy is in the shape of a cylinder surmounted by a hemisphere. The height of the toy is 25 cm. Find the total surface area of the toy if its common diameter is 12 cm.
Answer:

Let r and h be the radius and height of the cylinder respectively.
Given that, diameter d = 12 cm, radius r = 6 cm
Total height of the toy is 25 cm
Therefore, height of the cylindrical portion = 25 – 6 = 19 cm
T.S.A. of the toy = C.S.A. of the cylinder + C.S.A. of the hemisphere + Base Area of the cylinder
= 2πrh + 2πr² + πr²
= πr(2h + 3r) sq.unis
= $\frac { 22 }{ 7 }$ × 6(38 + 18)
= $\frac { 22 }{ 7 }$ × 6 × 56 = 1056
Therefore, T.S.A. of the toy is 1056 cm²

Question 39.
A coin is tossed thrice. Find the probability of getting exactly two heads or atleast one tail or two consecutive heads.
Answer:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n (S) = 8
Let A be the event of getting exactly two heads.
A = {HHT, HTH, THH}
n(A) = 3
P(A) = $\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{8}$
Let B be the event of getting atleast one tail
B = {HHT, HTH, HTT, THH, THT, TTH, TTT}
n(B) = 7
P(B) = $\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{7}{8}$
Let C be the event of getting consecutively
C = {HHH, HHT, THH}
n(C) = 3
P(C) = $\frac{n(\mathrm{C})}{n(\mathrm{S})}=\frac{3}{8}$
A ∩ B = { HHT, HTH, THH}
n(A ∩ B) = 3
p(A ∩ B) = $\frac{n(\mathrm{A} \cap \mathrm{B})}{n(\mathrm{S})}=\frac{3}{8}$
B ∩ C = {HHT, THH}
n(B ∩ C) = 2
p(B ∩ C) = $\frac{n(\mathrm{B} \cap \mathrm{C})}{n(\mathrm{S})}=\frac{2}{8}$
A ∩ C = {HHT, THH}
n(A ∩ C) = 2
P(A ∩ C) = $\frac{n(\mathrm{A} \cap \mathrm{C})}{n(\mathrm{S})}=\frac{2}{8}$
(A ∩ B ∩ C) = 2
P(A ∩ B ∩ C) = $\frac{n(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})}{n(\mathrm{S})}=\frac{2}{8}$
P(A∪B∪C) = P(A) + P(B) + P(C) -P(A ∩ B) – p(B ∩ C) – P(A ∩ C) + P(A∩B∩C)
$=\frac{3}{8}+\frac{7}{8}+\frac{3}{8}-\frac{3}{8}-\frac{2}{8}-\frac{2}{8}+\frac{2}{8}$
$=\frac{3}{8}+\frac{7}{8}-\frac{2}{8}$
$=\frac{10-2}{8}=\frac{8}{8}$ = 1
The probability is 1

Question 40.
The following table show the marks obtained by 48 students in a quiz competition in mathematics calculate the standard deviation.
Answer:

Ans.
Let us form the following table using given data

Question 41.
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4m, and slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also find the cost of canvas of the tent at the rate of ₹500 per m².
Answer:

For conical portion
Radius (r) = 2m
Slant height (l) = 2.8 m
For cylindrical portion
Radius (R) = 2m
Height (H) = 2.1 m
Area of canvas used for the tent = Curved surface area of a cone + curved area of the cylinder
= πrl + 2πRH
= π(2 × 2.8 + 2 × 2 × 2.1)
= $\frac { 22 }{ 7 }$ (5.6+ 8.4)
= $\frac { 22 }{ 7 }$ × 14 sq.m
= 44 sq. m
Cost of the canvas at the rate of ₹500 per m² = ₹ 500 × 44
= ₹ 22000

Question 42.
Seven years ago Ramkumar’s age was five time the square of Daniel’s age. Three years hence Daniel age will be two fifth of Ramkumar’s age. Find their present ages.
Answer:
Seven years ago, let Daniel’s age be x years
Seven years ago, Ramkumar age was 5x² years
Daniel’s present age = (x + 7) year
Ramkumar’s present age = 5x² + 7 years
Three years hence
Daniel’s age = (x + 7 + 3) year = (x + 10) years
Ramkumar’s age = (5x² + 7 + 3) year = (5x² + 10) years
By the given condition
x + 10 = $\frac { 2 }{ 5 }$ (5x² +10)
5x + 50 = 10x² + 20
10x² – 5x – 30 = 0
2x² – x – 6 = 0
2x² – 4x + 3x – 6 = 0
2x (x – 2) + 3(x – 2) = 0
(x – 2) (2x + 3) = 0
x – 2 = 0 or 2x + 3 = 0
x = 2 or x = $\frac { -3 }{ 2 }$
∴ x = 2 (age will not be negative)
Hence Daniel age = 9 years (2 + 7)
Ramkumar’s age = 27 years [5(2)² + 7]

PART – IV

IV. Answer all the questions. [2 × 8 = 16]

Question 43.
(a) Construct a ∆PQR in which PQ = 8 cm, ∠R = 60° and the median RG from R to PQ is 5.8 cm. Find the length of the altitude from R to PQ.
Answer:

Step 1 : Draw a line segment PQ = 8 cm.
Step 2 : At P, draw PE such that ∠QPE = 60°.
Step 3 : At P, draw PF such that ∠EPF = 90°.
Step 4 : Draw the perpendicular bisector to PQ, which intersects PF at O and PQ at G.
Step 5 : With O as centre and OP as radius draw a circle.
Step 6 : From G mark arcs of radius 5.8 cm on the circle. Mark them as R and S.
Step 7 : Join PR and RQ. Then ∆PQR is the required triangle.
Step 8 : From R draw a line RN perpendicular to LQ.
LQ meets RN at M
Step 9: The length of the altitude is RM = 3.5 cm.

[OR]

(b) Draw the two tangents from a point which is 5 cm away from the centre of a circle of diameter 6 cm. Also, measure the lengths of the tangents.
Answer:
Radius = 3cm, Distance = 5 cm.

Steps of construction:

With O as centre, draw a circle of radius 3 cm.
Draw a line OP = 5 cm.
Draw a perpendicular bisector of OP, which cuts OP at M.
With M as centre and MO as radius draw a circle which cuts previous circles at A and B.
Join AP and BP, AP and BP are the required tangents.
The length of the tangent PA = PB = 4 cm

Verification: In the right angle triangle OAP
PA² = OP² – OA²
= 5² – 3² .
= 25 – 9 = 16
PA = √l6 = 4 cm
Length of the tangent = 4 cm

Question 44.
(a) Draw the graph of y = (x – 1) (x + 3) and hence solve x² – x – 6 = 0
Answer:
y = (x – 1) (x + 3)
y = x² + 2x – 3
(i) Draw the graph of y = x² + 2x – 3 by preparing the table of values given below.

(ii) Plot the points (-4, 5), (-3, 0), (-2, -3), (-1, -4), (0, -3), (1, 0), (2, 5), (3, 12) and (4, 21) on the graph sheet using suitable scale.
(iii) To solve x² – x – 6 = 0 subtract x² – x – 6 = 0 from y = x² + 2x – 3

(iv) Draw the graph of y = 3x + 3 by preparing the table.

(v) The straight line cuts the curve at (-2, -3) and (3, 12). Draw perpendicular lines from the point to X – axis.
The line cut the X – axis at -2 and 3.
The solution set is (-2, 3)

[OR]

(b) Draw the graph of y = x² +x and hence solve x² + 1=0
Answer:
Let y = x² + x
(i) Draw the graph of y = x² + x by preparing the table.

(ii) Plot the points (-4, 12), (-3, 6), (-2, 2), (-1, 0), (0, 0), (1, 2), (2, 6), (3, 12) and (4, 20).
(iii) Join the points by a free hand to get smooth curve.
(iv) To solve x² + 1 = 0, subtract x² + 1 = 0 from x² + x we get.

The equation represent a straight line. Draw a line y = x – 1

Observe the graph of y = x² + 1 does not interset the parabola y = x² + x This x² + 1 has no real roots.

Samacheer Kalvi 10th English Model Question Paper 4

Leave a Comment / Class 10 / By Prasanna

Students can Download Samacheer Kalvi 10th English Model Question Paper 4 Pdf, Samacheer Kalvi 10th English Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamil Nadu Samacheer Kalvi 10th English Model Question Paper 4

General Instructions:

The question paper comprises of four parts.
You are to attempt all the sections in each part. An internal choice of questions is provided wherever applicable.
All questions of Part I, II, III and IV are to be attempted separately.
Question numbers I to 14 in Part I are Multiple Choice Questions of one mark each. These are to be answered by writing the correct answer along with the corresponding – option code.
Part II has got four sections. The questions are of two marks each. Question numbers 15 to 18 in Section I and Question numbers 19 to 22 in Section II are to be answered in about one or two sentences each. Question numbers 23 to 28 in Section III and IV are to be answered as directed.
Question numbers 29 to 45 in Part III are of five marks each and have been divided in five sections. These are to be answered as directed.
Question numbers 46 and 47 in Part IV are of eight marks each. Question number 47 has four questions of two marks each. These are to be answered as directed.

Time: 2.30 Hours
Maximum Marks: 100

Part – I

Answer all the questions. [14 x 1= 14]
Choose the most suitable answer and write the code with the corresponding answer.
Choose the appropriate synonyms for the italicised words.

Question 1.
The young seagull was alone on his ledge.
(a) sill
(b) window
(c) nest
(d) lift
Answer:
(a) sill

Question 2.
‘Burglars!’ she shouted, intuitively.
(a) intentionally
(b) thoughtfully
(c) knowingly
(d) instinctively
Answer:
(d) instinctively

Question 3.
When we realise our mistakes, we should try to rectify them.
(a) read
(b) recover
(c) recognize
(d) electrically
Answer:
(d) electrically

Choose the appropriate antonym for the italicised words.

Question 4.
The real power of women though realised earlier, is currently being projected to the world.
(a) presently
(b) formerly
(c) carefully
(d) electrically
Answer:
(b) formerly

Question 5.
Technology helps one to live in comfort.
(a) comfy
(b) discomfort
(c) misuse
(d) easiness
Answer:
(b) discomfort

Question 6.
This lesson is set in the days of France-Prussian war.
(a) avarice
(b) battle
(c) combat
(d) peace
Answer:
(d) peace

Question 7.
Choose the correct plural form of “moose“.
(a) mooses
(b) moose
(c) mooses’s
(d) moosies
Answer:
(b) moose

Question 8.
Form a derivative by adding the right suffix to the word – sad.
(a) ly
(b) ness
(c) fill
(d) ment
Answer:
(a) ly

Question 9.
Choose the correct expansion of the abbreviation IREP is ……………
(a) The Integrated Rural Energy Programmer
(b) The Integral Railway Energy Programme
(c) The Integrated Rural Energy Programme
(d) The Integral Rural Exercise President
Answer:
(c) The Integrated Rural Energy Programme

Question 10.
Complete the following sentence with the most appropriate phrasal verb given below:
Do not ……….. to temptation.
(a) give in
(b) give out
(c) give up
(d) give into
Answer:
(a) give in

Question 11.
Choose the suitable option to pair it with the word “mouth” to form a compound word.
(a) dry
(b) wash
(c) clean
(d) brush
Answer:
(b) wash

Question 12.
Fill in the blank with the most appropriate preposition given below:
The people stood ……………… the road to watch the procession go by.
(a) across
(b) along
(c) on
(d) into
Answer:
(b) along

Question 13.
Complete the following sentence using the most appropriate tense form of the verb given below:
Her teacher ………….. her the reason for her restlessness.
(a) asked
(b) will ask
(c) asking
(d) had asked
Answer:
(a) asked

Question 14.
Choose the most appropriate linker from the given four alternatives.
He completely forgot …………. he had not always been a sprightly child.
(a) that
(b) when
(c) though
(d) and
Answer:
(a) that

PART – II [10 x 2 = 20]
Section -I

Answer any THREE of the following questions in a sentence or two. [3 x 2 = 6]

Question 15.
Why did the seagull fail to fly?
Answer:
The young seagull hesitated and feared to fly. He thought that his wings would not support him. Hence the sea gull failed to fly.

Question 16.
Who was Mrs. Hudson? Why was she worried?
Answer:
Mrs. Hudson was the landlady of Mr. Holmes. She was worried because Holmes was ill and had not taken food or drink for three days continuously.

Question 17.
Who is Tara-Tarini? After whom was the sailboat named?
Answer:
Tara-Tarini is the patron deity for sailors. The sailboat was named after the famous ‘Tara- Tarini’ temple in Ganjam district of Odisha.

Question 18.
Write a few lines about the owner of the shop.
Answer:
The owner of the tea shop was over sixty, a little rustic in appearance, with his white neatly- combed hair. He looked clean. He was wearing a dhoti and a blue striped shirt that could be seen from under a green shawl.

Section – II

Read the following sets of poetic lines and answer any THREE of the following.
[3 x 2 = 6]

Question 19.
“Let me but live my life from year to year,
With forward face and unreluctant soul’”
(a) Whom does the word ‘me’ refer to?
(b) What kind of life does the poet want to lead?
Answer:
(a) ‘me’ refers to the poet, Hemy Van Dyke.
(b) The poet wants to lead a life facing every year with courage and willingness to do anything.

Question 20.
“There’s a family nobody likes to meet;
They live, it is said, on Complaining Street”
(a) Where does the family live?
(b) Why do you think the street is named as ‘Complaining Street’?
Answer:
(a) The family lives on Complaining street.
(b) I think the street is named as ‘Complaining Street’ because a family there is always complaining and no one likes to meet them.

Question 21.
“The summer of life she’s ready to see in spring.
She says, “Spring will come again, my dear
Let me care for the ones who ’re near. ”
(а) What does the word summer mean here?
(b) How does she take life?
Answer:
(a) Summer here means development.
(b) She takes life optimistically.

Question 22.
“A silly young cricket accustomed to sing .
Through the warm, sunny months of gay summer and spring.”
(а) What was the routine of the cricket?
(b) Name the seasons mentioned here.
Answer:
(a) The routine of the cricket was to sing and while away the time enjoying the spring.
(b) The seasons mentioned are summer and spring.

Section – III

Answer any THREE of the following.

Question 23.
Rewrite the following sentence to the other voice.
Answer:
When we first met, they had already offered her a job at the bank.
She had already been offered a job by them at the bank, when we first met.

Question 24.
Rewrite using indirect speech.
Answer:
“Miranda,” said Prospero, “tell me what you are looking at yonder.”
Prospero asked Miranda what she was looking at yonder.

Question 25.
Punctuate the following.
the great expanse of sea stretched down beneath and it was such a long way down miles down
Answer:
The great expanse of sea stretched down beneath, and it was such a long way down – miles down.

Question 26.
Transform the following sentence into a simple sentence.
The old man is very rich but he is a miser.
Answer:
In spite of the old man being rich, he is a miser.
[OR]
Despite being rich, the old man is a miser.

Question 27.
Rearrange the words in the correct order to make meaningful sentences:
(a) love / of others / good etiquette / and / wins the / respect
(b) when / best / they can / one is / be learnt / young
Answer:
(a) Good etiquette wins the love and respect of others.
(b) They can be learnt best when one is young.

Section – IV

Answer the following question.

Question 28.
Guide Shankar from the Railway Station to the Head Post Office. Write three instructions by way of helping him.

Answer:

Go straight from the Railway Station on the main road.
Take right from the Pandian Sweets.
The Head Post Office will be on your left on the same road.

Part – III [10 x 5 = 50]
Section-I

Answer any TWO of the following in utmost 10 lines. [2 x 5 = 10]

Question 29.
Describe the funny incident that caused the confusion in the house.
Answer:
“The Night the Ghost Got In” is a major example of the storytelling technique of James Thurber, who is far and wide considered one of the greatest comical writers that America ever produced. The story combines events that are likely to be a comic hyperbole.

The characters’ bizarre understanding of their world serves the dual purposes of entertaining readers. The story centres on a common situation where James Thurber hears a strange sound downstairs in the middle of the night when he comes out of the bath tub at 1 a.m. with a towel wrapped around him.

He assumes that it is a ghost and his mother assuming the footsteps to be of burglars, goes to the extent of throwing a shoe through a window to alert the neighbours about the burglars. The mother thinking it fancy to throw the second shoe through the neighbour’s window causes more fun and frolic. The Bodwell’s inform the police who in turn are thoroughly confused by the odd characters of the Thurber household and their way of life when they reach the suspected house.

The grandfather who is awakened by the chaos, deliberately mistakes the policemen for Meade’s men, who are deserters, illustrates the unusual characteristics of the Thurbers. By the time, it is over, one of the policemen is shot in the shoulder by the household’s disorientated grandfather, and a local news reporter is left speechless when he is told that a ghost is the cause of all the disorder, by the narrator.
“You can’t follow your heart when it is more confused than your head.”

Question 30.
Give a character sketch of M. Hamel.
Answer:
M. Hamel is a strict disciplinarian at a school in a village in the French districts of Alsace and Lorraine. M. Hamel is an experienced teacher who has been teaching in that village school for forty years. He imparts primary education in all subjects. He is a hard task master and students like Franz, who are not good learners, are in great dread of being scolded by him. He loves his profession from the bottom of his heart and holds a deep sense of respect for the mother tongue. The latest order of the Prussian conquerors upset him. He has to leave the place for ever and feels heart broken. He feels sad but exercises self-control. He has the courage to hear every lesson to the last.

This thought completely shatters a calm and composed man like M. Hamel. He tries his level best to remain calm and unruffled but breaks down at the end. His performance during the last lesson is exemplary. He is kind even to a late comer like Franz. He uses a solemn and gentle tone while addressing the students. He has a logical mind and can analyse problems and deduce the reasons responsible for it. He feels grief-stricken at the fact that people became indifferent to learning French.

He charms them to keep their language alive. Hamel is a patriot in the real sense. He regards the mother tongue to be a means of holding one’s identity and self-respect. He knows the emotional hold of a language over its users. He is a good communicator and explains everything patiently. Partings are painful and being human, M. Hamel too is no exception. He fails to say goodbye as his throat is choked.

“Good teachers are rarity. M. Hamel is an exception to rarity!”

Question 31.
What are the various innovations made in India to help the differently abled lead a normal life?
Answer:
There are various innovations made in India to help the differently abled lead a normal life.Lechal Shoes by Krispian Lawrence is an innovation for the disabled to use GPS & Blutooth. This enabled shoes to help the disabled navigate streets, based on instructions from map software on a smartphone. The App also records the route and also counts steps.

Blee Watch by industrial designers Nupura Kirloskar and Janhavi Joshi of Mumbai invented the Smart watch for the hearing impaired. It converts sound waves into vibrations and colour codes to alert users to emergency sounds and ringing doorbells. It even helps them feel the rhythm of music.

iGEST by Anil Prabhakar, IIT-M professor and cofounder of enability technologies innovated a wearable device which tracks gestures of speech-impaired people and speaks for them. I guess technology makes a differently abled person’s life easier.
“Exploration is the engine that drives Innovation.”

Question 32.
Highlight the factors responsible for the all-women Indian Navy crew to carry out their expedition.
Answer:
The support the crew members received was a major factor. When they knew that they were doing well and looked after themselves well, in spite of all apprehensions they were supportive. The crew members’ personal aim and target mattered a lot. Mostly they wanted to make sure that they complete the journey with ultimate honesty without the use of engines.

Than the destination, the journey was important. So their contention was to make sure that they go by the rules of circumnavigation without any means of repulsion and anybody else’s assistance. The presence of mind and common sense to make decisions and act quickly was an added factor. They had to quickly do an analysis of problem solving techniques. Ego should never come amidst them. Team work helped them to collaborate and work together.

Mutual understanding was important too. One would heat the water while the other would heat the gloves or even rested. Over all the confidence you had in each other than the trust and acceptance as every member of the crew to be a family was a must to carry out the expedition.

“It’s about women helping women and women doing things
together and supporting each other ”

Section – II

Answer any TWO of the following in utmost 10 lines. [2 x 5 = 10]

Question 33.
Describe the journey of life as depicted in the poem by Henry Van Dyke.
Answer:
Henry Van Dyke, one of the greatest American short story writers and poets, has surpassed the act of writing skillfully. ‘Life’ is no doubt one of his priced literary pieces. This poem is a pinnacle of expressive embarkment on the quest of self-revival from the glum beats of monotony. It has a very deep and farsighted meaning held within it and this is evident from the very beginning of the poem. The poem is the poet’s own reflection on his life and tells his point of view on the more important things in life. The poet advises the readers from his life experiences.

We feel that he is now an older man reflecting on his younger days. Through his words he is explaining to us what he is taking away as most important to live is the best life possible. Life is too short to get caught up in the moment or worry about the past. On the other hand, it suggests that we look forward to what the future holds. We sometimes find ourselves brooding on the bad times and we forget about how much good there is in the future. The poet is making us understand this concept and be more aware of reality.
“Life is short and if we enjoy every moment of everyday
We will be happy no matter what happens or changes our way]”

Question 34.
Elucidate – We are nothing more than children of your brain.
Answer:
“The Secret of the Machines” looks back to the poems rejoicing modem technology which had pre-occupied Kipling during the mid-1990s. This poem by Rudyard Kipling, a famous British poet, is about the great status of machines in the age of the industrial progress. The poet qualifies machines to define the situation from their point of view! In many of those poems it is the machines themselves who speak, taking on human characteristics and feelings.

In “The Secret of the Machines” the anthropomorphism is choral, with the machines conveying their message in a collective chant, informing the reader, posing rhetorical questions, offering a stem warning about their potential strength and finally submitting to their masters – the human brain. This is yet another example of Kipling’s constant, though often tilted is the denial of religion. The machines have the ability to change the environment of all created things except

The Gods, the creator of human-beings. The machines assert that they have produced a kind of smoke-screen which is momentarily covering the fact that it is you, the human beings, who are the true gods, with the machines nothing more than children of your brain! If the machines are to function to their full capacity they must be dutifully controlled by the Gods who have created them. That is why it is so important for man to understand the Law that machines ‘are not built to comprehend a lie.’
“ Your beliefs become your thoughts,
Your thoughts-your words

Question 35.
Read the following stanza and answer the questions given below.
“It sat alone.
What happened there is still today unknown.
It is a very mysterious place,
And inside you can tell it has a ton of space,
But at the same time it is bare to the bone.”
(i) Which adjective is used to describe the house?
(ii) Give the rhyming word for place and bone.
(iii) Write the rhyme scheme of the above stanza.
(iv) Identify the figure of speech employed in the fifth line of the given stanza.
Answer:
(i) The adjective ‘mysterious’ is used to describe the house.
(ii) The rhyming word for place is space and for bone, it is unknown and alone.
(iii) The rhyme scheme is aabba.
(iv) The figure of speech is alliteration in the last line.

Question 36.
Paraphrase the following stanza.
Answer:
Remember, no men are strange, no countries foreign Beneath all uniforms, a single body breathes Like ours: the land our brothers walk upon Is earth like this, in which we all shall lie.Emphasising the value of universal brotherhood, the poet draws our attention to the absence of any differences amongst the people of different countries.

He asks us never to forget that people living in other countries are not strange or unfamiliar. The uniforms worn by people in different parts of the world may be different, but the bodies beneath them are the same. All human bodies live and breathe in a similar fashion. We are all brothers because we walk upon the same earth that we have divided into countries. Also, we shall all meet this same earth when we die and be buried in it.

Section – III

Answer any ONE of the following. [1 x 5 = 5]

Question 37.
Rearrange the following sentences in coherent order.
(i) In the army, Mulan proved to be a brave soldier.
(ii) With Mulan, they won every battle and so some soldiers accepted her as the General.
(iii) She went to her father and said that she would go in his place.
(iv) Mulan, a teenage girl of China, heard the news when she was washing clothes.
(v) After a few years Mulan was given the top job – General of the entire army.
Answer:
Rearranged number sequence :(iv), (iii), (i) ,(v),(ii)
(iv) Mulan, a teenage girl of China, heard the news when she was washing clothes.
(iii) She went to her father and said that she would go in his place.
(iv) In the army, Mulan proved to be a brave soldier.
(v) After a few years Mulan was given the top job – General of the entire army.
(ii) With Mulan, they won every battle and so some soldiers accepted her as the General.

Question 38.
Read the following passage and answer the questions that follow.
Answer:
Francis Bennett went on into the reporters’ room. His fifteen hundred reporters, placed before an equal number of telephones, were passing on to subscribers the news which had come in during the night from the four quarters of the earth. In addition to his telephone, each reporter has in front of him a series of commutators, which allow him to get into communication with this or that telephotic line.

Thus the subscribers have not only the story but the sight of these events. Francis Bennett questioned one of the ten astronomical reporters – a service which was growing because of the recent discoveries in the stellar world. ‘Well, Cash, what have you got?’ ‘Phototelegrams from Mercury, Venus and Mars, Sir.’ ‘Interesting! And Jupiter?’ ‘Nothing so far! We haven’t been able to understand the signals the Jovians make. Perhaps ours haven’t reached them? ….’ ‘Aren’t you getting some result from the moon, at any rate?’ ‘Not yet, Mr Bennett.’

‘Well, this time, you can’t blame optical science! The moon is six hundred times nearer than Mars, and yet our correspondence service is in regular operation with Mars. It can’t be telescopes we need…’ ‘No, it’s the inhabitants,’ Corley replied. ‘You dare tell me that the moon is uninhabited?’ ‘On the face it turns towards us, at any rate, Mr Bennett. Who knows whether on the other side…’ ‘Well, there’s a very simple method of finding out.’

(i) Who is Cash?
(ii) What were the fifteen hundred reporters doing in the room?
(iii) Why can’t we blame optical science to get some result from the moon?
(iv) How do the spectators enjoy the visual effect of the story told by the reporters?
(v) From where did Cash get Phototelegrams?
Answer:
(i) Cash is one of the ten astronomical reporters who was questioned by Francis Bennett.
(ii) The fifteen hundred reporters were passing on to subscribers the news which had come in during the night from the four quarters of the earth from the reporters room.
(iii) We can’t blame optical science because the moon is six hundred times nearer than mars and their correspondence service is in regular operation with Mars.
(iv) When Francis Bennett’s fifteen hundred reporters, placed before an equal number of telephones, were passing on to subscribers the news, each reporter has in front of him a series of commutators, which help the spectators to enjoy the visual effect of the story told by the reporters.
(v) Cash got Phototelegrams from Mercury, Venus and Mars.

Section – IV

Answer any FOUR of the following. [4 x 5 = 20]

Question 39.
Prepare an attractive advertisement using the hints given below.
Crispy and Tasty – to tingle the taste buds – Good variety of non-vegetarian food items.

Question 40.
Write a letter to the Collector of your District expressing your desire to contribute to the welfare of the flood affected victims in your area.
Answer:
16th December, 2020
From
XXX
YYY
To
The District Collector YYY
Respected Sir,
Sub : Contribution for Flood Affected Victims I am extremely devastated to see Thiruvannamalai district submerged in water. The plight of the flood victims pains my heart. I remember the days I did my schooling at St. Paul’s Matriculation Higher Secondary School. The Principal and teachers imbibed the true service to mankind among us during the Community Development Period that we had once a week.

The seeds sown by my mentors now urges me to contribute towards the welfare of the flood victims affected in our area. Now that I am placed well in my career as a the CEO of a private company, I wish to contribute a meagre sum of one lakh towards the same. I believe it will reach the flood victims at the earliest and help them in a small way to brave the storms and overcome the situation.
Thank you,
Yours sincerely,
XXX
Address on the envelope
The District Collector,
YYY

Question 41.
You are Priyaraj/Priya, the cultural secretary of St. Joseph’s Matriculation Higher Secondary School Coonoor. You have been asked to inform students of class IX and XI about the auditions for an inter-school Dramatics Competition. Draft a notice in not more than 50 words for the students’ Notice board. Invent all the necessary details.
Answer:

St. Joseph’s Mat., Higher Secondary School Coonoor.
NOTICE
30 July 2016
Inter – School Dramatics Competition – Audition

An audition for the Inter-school Dramatics Competition will be at the School Auditorium on 23rd August, 2020. Selected students will be part of a Shakespearean drama to represent the school team. Those interested kindly give your names to the undersigned. The details of the audition are given below:

Date: 23-08-2020
Time: 12.30p.m.
Venue: School Auditorium
Eligibility: Students of classes IX and XI
Last date for Registration: 25-08-2020

Priyaraj / Priya,
Cultural Secretary

Question 42.
Look at the following picture and express your views on it in about five sentences

A Happy Family

This picture denotes a real life situation on a weekend. This picture could depict a joint family too as we see five children in the hall. The father is seen reading the newspaper while the mother brings him a cup of coffee. The children are reading books. One boy is trying to fix a picture on the wall. A baby is sleeping in the cradle and near it we can see a woman folding clothes.

Question 43.
Make notes or write a summary of the following passage.
Answer:
We all know what we mean by a “good” man. The ideal good man does not drink or smoke, avoids bad language, converses in the presence of men exactly as he would if there were ladies present, attends church services regularly and holds the correct opinion on all subjects.

He has a wholesome horror of wrong-doing and realizes that it is our painful duty to reprimand sin. He has a still greater horror of wrong thinking, and considers it the business of the authorities to safeguard the young against those who question the wisdom of the views generally accepted by middle-aged successful citizens.

Apart from his professional duties, at which he is diligent, he spends much time in good works: he may encourage patriotism and military training; he may promote industry, seriousness and virtue among wage earners and their children by seeing to it that failures in these respects receive due punishment; he may be a trustee of a university and prevent an ill-judged respect for learning from allowing the employment of professors with rebellious ideas. Above all, of course, his “morals” in the narrow sense must be blameless.
Notes
Title: Characteristics of a Good Person
Good man – intensely religious, avoid bad habits.
Decent language, correct opinion on all subjects Diligent in professional duties Encourages patriotism and military training
Promotes industry, seriousness and virtue among wage earners and their children Prevents the employment of professors with rebellious ideas Morals – blameless
Title: Characteristics of a Good Person

Rough Draft
The author opines that an ideally good man is intensely religious and avoids unacceptable behavioural habit? such as moking and drinking. He does not use obscene or indecent language and has a fear to do wrongThings and think negatively. He is professionally dutiful and takes active interest in promoting virtues such as patriotism, industry and seriousness amidst workers and their children. Whenever possible he even prevehts the-ertigloyment of professors with rebellious ideas.

Fair draft
Title: Characteristics Of A Good Person
The author opines a good man to be intensely religious avoiding bad habits viz. smoking and drinking. He does not use obscene or indecent language and has a fear of wrong doing and wrong thinking. He is professionally dutiful and takes active interest in promoting virtues such as patriotism, industry and seriousness among workers and their children. Whenever possible he prevents the employment of professors with rebellious ideas. No. of words written in the summary: 68

Question 44.
Identify and correct the errors in the following sentences.
(a) Divide these mangoes among Reema and Seema.
(b) I do not know who of the new trainees should be confirmed.
(c) She has not completed her course, isn’t it?
(d) The manager visits the office regularly, isn’t it?
(e) David and me are like brothers but from different mothers.
Answer:
(a) Divide these mangoes between Reema and Seema.
(b) I do not know which of the new trainees should be confirmed.
(c) She has not completed her course, hasn’t she?
(d) The manager visits the office regularly, doesn’t he?
(e) David and I are like brothers but from different mothers.

Section – V

Quote from memory. [1 x 5 = 5]

Question 45.
The summer of life ………….. faith and beliefs.
Answer:
The summer of life she s ready to see in spring.
She says, “Spring will come again, my dear.
Let me care for the ones who ’re near. ”
She’s The Woman – she has no fear!
Strong is she in her faith and beliefs.

Part – IV

Write a paragraph of about 150 words by developing the following hints. [2 x 8 = 16]

Question 46.
(a) Foot of the mountain – poor farmer – Shining – aged mother – tyrannical leader – rare announcement – farmer filled with remorse – unwhitened rice – gourd filled – mother on his back – summit of Obatsayuma – helpless mother noticed perilous path – unconditional love – face the verdict together.
Answer:
“The Story of the Aged Mother” by Matsuo Basho is a story in which a son and his mother’s pain in their battle against the decree of putting aged people to death is portrayed. The story happened at the foot of the mountain where a poor farmer lived in a country called Shining with his aged mother whom he loved. Their place was governed by a tyrannical leader who sent the announcement of killing all aged people which was not common then.

When the poor farmer heard the decree he felt sorrow in his heart for his aged mother. Just as his labour ended, he prepared food from a portion of unwhitened rice, covered it in a square cloth and knotted it in his neck along with a gourd filled with cool, sweet water. He then carried his mother on his back and started walking towards the summit of Obatsayuma, the mountain of the“abandoning of aged”.

On their way to the top, the helpless mother noticed the perilous path and thought that her son didn’t know the paths and might pass through danger. So she reached for twigs, snapped them to pieces and quietly dropped them until they reached the summit.

The poor farmer freed his mother and made a comfortable cushion with pine needles and bid farewell. Before he left her, she told him about the trail she made with the piles of twigs. His mother’s words clearly portrayed the unconditional love that the mother had for her son, an unselfish love that thumped the heart of the poor child and made him cry. Now, the farmer realized that they need to face the verdict together. So, he again carried his mother on his back and started to walk back to their humble hut in the valley.

[OR]

(b) Uncle Philip – leaves a letter – the message – precious stones – have expectations – remember uncle – charity – interesting mechanism – explosion – never doubt me – blown to atoms – believe never fulfilled.
Answer:
A Dilemma by Silas Weir Mitchell is indeed a story with a twist. The letter was addressed to Tom. It said that the box contained a large number of very fine pigeon-blood rubies and a fair lot of diamonds. One was blue and a beauty indeed. There were hundreds of pearls—one the famous green pearl and a necklace of blue pearls, for which any woman would sell her soul— or her affections.

He wanted his nephew to continue to have expectations and continuously remember his dear uncle. He had stated that he would have left those stones to some charity, but he hated the poor. It stated that the box contained an interesting mechanism, which would act with certainty as he unlocked it, and explode ten ounces of an improved, supersensitive dynamite. He then made it clear that to be accurate, there were only nine and a half ounces.

He said that if he doubted his uncle, on opening it, he would be blown to atoms. On the contrary if he believed him, he would continue to nourish expectations which would never be fulfilled. He also added at the end that since he was a considerate man, he counselled extreme care in handling the box and never forget his affectionate uncle.

Question 47.
Read the following passage and answer the questions given below:
The little boy of twelve will remember this day forever. It is not every day you get to see a world championship at Hyatt Regency on November 19,2013. The visit of the Children was the result of an initiative by the organizers of the tournament and the School education Department to get school children be a part of the world championship.

Many children from schools gathered to witness the ongoing world Championship and there was an overwhelming response. The entry being free, children found it a rare and unforgettable opportunity’. Student visitors were given a chance to spend an hour in the chess puzzles contest. No wonder the visitors would be hailed heroes in their schools after their visit to the spot.

Children enthusiastically tried to move the big chess pieces at the hotel lobby, thereby founding that the city of Chennai would produce many masters of chess dazzling with intelligence and talent.

Questions:
(a) What made the day memorable for children?
(b) Mention the purpose for which the visit of the children had been organized.
(c) Which places were the student visitors permitted into?
(d) Who were responsible for such an exciting initiative?
Answer:
(a) The day was made memorable for the children because they got to see a world championship at Hyatt Regency.
(b) The purpose for which the visit of the Children had been organised was to get school children be a part of the world championship thereby making way for many masters of chess to be produced in the future.
(c) The student visitors were permitted into the Hotel’s lobby.
(d) The Organisers of the tournament and the school education department were responsible for such an exciting initiative.

[OR]

Read the following poem and answer the questions given below:
The Leader
Patient and steady with all he must bear,
Ready to meet every challenge with care,
Easy in manner, yet solid as steel,
Strong in his faith, refreshingly real,
Isn’t afraid to propose what is bold,
Doesn’t conform to the usual mould,
Eyes that have foresight, for hindsight won’t do,
Never backs down when he sees what is true,
Tells it all straight, and means it all too.
Going forward and knowing he’s right,
Even when doubted for why he would fight,
Again and again, he makes the case far clear
Wants to reach those who don’t hear
Growing in strength, he won’t be unnerved,
Ever assuring he’ll stand by his word.
Wanting the world to join his firm stand,
Bracing for war, but praying for peace,
Using his power so evil will cease:
A trustworthy person is a leader so far,
He is a person who knows what he must act upon.

Questions:
(a) What is the poem about?
(b) Explain the line, ‘Doesn’t conform to the usual mould’.
(c) Mention any two qualities of a true leader portrayed in this poem.
(d) What does a real leader stand for according to the poem?
Answer:
(a) The poem is about a leader and the leadership qualities one needs to possess.
(b) ‘Doesn’t conform to the usual mould’, suggests that the leader who is described has qualities that are totally different.
(c) Intelligence, courageous, strong in faith, Patience, steadfast (any two)
(d) According to the poem, a real leader stands for truth.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction

Samacheer Kalvi 12th Bio Zoology Human Reproduction Text Book Back Questions and Answers

Question 1.
The mature sperms are stored in the ________
(a) Seminiferous tubules
(b) Vas deferens
(c) Epididymis
(d) Seminal vesicle
Answer:
(c) Epididymis

Question 2.
The male sex hormone testosterone is secreted from ________
(a) Sertoli cells
(b) Leydig cell
(c) Epididymis
(d) Prostate gland
Answer:
(b) Leydig cell

Question 3.
The glandular accessory organ which produces the largest proportion of semen is ________
(a) Seminal vesicle
(b) Bulbourethral gland
(c) Prostate gland
(d) Mucous gland
Answer:
(a) Seminal vesicle

Question 4.
The male homologue of the female clitoris is ________
(a) Scrotum
(b) Penis
(c) Urethra
(d) Testis
Answer:
(b) Penis

Question 5.
The site of embryo implantation is the ________
(a) Uterus
(b) Peritoneal cavity
(c) Vagina
(d) Fallopian tube
Answer:
(a) Uterus

Question 6.
The foetal membrane that forms the basis of the umbilical cord is ________
(a) Allantois
(b) Amnion
(c) Chorion
(d) Yolk sac
Answer:
(a) Allantois

Question 7.
The most important hormone in initiating and maintaining lactation after birth is ________
(a) Oestrogen
(b) FSH
(c) Prolactin
(d) Oxytocin
Answer:
(c) Prolactin

Question 8.
Mammalian egg is ________
(a) Mesolecithal and non-cleidoic
(b) Microlecithal and non-cleidoic
(c) Alecithal and non-cleidoic
(d) Alecithal and cleidoic
Answer:
(c) Alecithal and non-cleidoic

Question 9.
The process which the sperm undergoes before penetrating the ovum is ________
(a) Spermiation
(b) Cortical reaction
(c) Spermiogenesis
(d) Capacitation
Answer:
(d) Capacitation

Question 10.
The milk secreted by the mammary glands soon after child birth is called ________
(a) Mucous
(b) Colostrum
(c) Lactose
(d) Lactose
Answer:
(b) Colostrum

Question 11.
Colostrum is rich in ________
(a) IgE
(b) IgA
(c) IgD
(d) Ig M
Answer:
(b) IgA

Question 12.
The Androgen Binding Protein (ABP) is produced by ________
(a) Leydig cells
(b) Hypothalamus
(c) Sertoli cells
(d) Pituitary gland
Answer:
(c) Sertoli cells

Question 13.
Which one of the following menstrual irregularities is correctly matched?
(a) Menorrhagia – excessive menstruation
(b) Amenorrhoea – absence of menstruation
(c) Dysmenorrhoea – irregularity of menstruation
(d) Oligomenorrhoea – painful menstruation
Answer:
(b) Amenorrhoea – absence of menstruation

Question 14.
Find the wrongly matched pair:
(a) Bleeding phase – fall in oestrogen and progesterone
(b) Follicular phase – rise in oestrogen
(c) Luteal phase – rise in FSH level
(d) Ovulatory phase – LH surge
Answer:
(c) Luteal phase – rise in FSH level

Question 15.
A – In human male, testes are extra abdominal and lie in scrotal sacs.
R – Scrotum acts as thermoregulator and keeps temperature lower by 2°C for normal sperm production.
Assertion (A) and Reason (R)
(a) A and R are true, R is the correct explanation of A
(b) A and R are true, R is not the correct explanation of A
(c) A is true, R is false
(d) Both A and R are false
Answer:
(a) A and R are true, R is the correct explanation of A

Question 16.
A – Ovulation is the release of ovum from the Graafian follicle.
R – It occurs during the follicular phase of the menstrual cycle.
Assertion (A) and Reason (R)
(a) A and R are true, R is the correct explanation of A
(b) A and R are true, R is not the correct explanation of A
(c) A is true, R is false
(d) Both A and R are false
Answer:
(c) A is true, R is false

Question 17.
A – Head of the sperm consists of acrosome and mitochondria.
R – Acrosome contains spiral rows of mitochondria.
(d) Both A and R are false
Assertion (A) and Reason (R)
(a) A and R are true, R is the correct explanation of A
(b) A and R are true, R is not the correct explanation of A
(c) A is true, R is false
(d) Both A and R are false
Answer:
(d) Both A and R are false

Question 18.
Mention the differences between spermiogenesis and spermatogenesis
Answer:
Spermiogenesis: Transformation of spermatids into mature sperm.

Spermatogenesis: Spermatogenesis is the sequence of events in the seminiferous tubules of testes that produces male gametes, the sperms.

Question 19.
At what stage of development are the gametes formed in newborn male and females?
Answer:
In males, at puberty, the spermatogonia (sperm mother cells) begin to undergo meiotic division and produces sperms throughout life, whereas in females during the stage of foetal development, the germinal epithelial cells undergo mitosis and produce oogonia (egg mother cells) and they further enter prophase-I of meiosis-I forming primary oocytes and get arrested. No more oogonia is formed further. At puberty, out of million eggs (prime oocytes) produced at birth only 300-400 will ovulate till menopause.

Question 20.
Expand the acronyms

Answer:

FSH – Follicle Stimulating Hormone
LH – Luteinizing Hormone
hCG – human Chorionic Gonadotropin
hPL – human Placental Lactogen

Question 21.
How is polyspermy avoided in humans?
Answer:
Once fertilization is accomplished, cortical granules from the cytoplasm of the ovum form a barrier called the fertilization membrane around the ovum preventing further penetration of other sperms. Thus polyspermy is prevented.

Question 22.
What is colostrum? Write its significance.
Answer:
The mammary glands secrete a yellowish fluid called colostrum during the initial few days after parturition. It has less lactose than milk and almost no fat, but it contains more proteins, vitamin A and minerals. Colostrum is also rich in IgA antibodies. This helps to protect the infant’s digestive tract against bacterial infection.

Question 23.
Placenta is an endocrine tissue. Justify.
Answer:
During pregnancy, the placenta acts as a temporary endocrine gland and produces large quantities of human Chorionic Gonadotropin (hCG), human Chorionic Somatomammotropin (hCS) or human Placental Lactogen (hPL), oestrogens and progesterone which are essential For a normal pregnancy. A hormone called relaxin is also secreted during the later phase of pregnancy which helps in relaxation of the pelvic ligaments at the time of parturition.

Question 24.
Draw a labeled sketch of a spermatozoan.
Answer:

Question 25.
What is inhibin? State its functions.
Answer:
inhibin is a hormone secreted by Sertoli cells of the testes which is involved in the negative feedback control of sperm production.

Question 26.
Mention the importance of the position of the testes in humans.
Answer:
The testes are positioned in such a way hanging out from the body in the scrotal sac that provides optimal temperature 2°C to 3°C lower than internal body temperature for effective sperm production.

Question 27.
What is the composition of semen?
Answer:
Semen or seminal fluid is a milky white fluid which contains sperms and seminal plasma, which is secreted from the seminal vesicles, prostate gland and bulbourethral glands.

Question 28.
Name the hormones produced from the placenta during pregnancy, human Chorionic Gonadotropin (hCG)
Answer:

human Placental Lactogen (hPL)
Relaxin.

Question 29.
Define gametogenesis.
Answer:
Gametogenesis is the process of formation of gametes i.e., sperms and ovaries from the primary sex organs in all sexually reproducing organisms. Meiosis plays the most significant role in the process of gametogenesis.

Question 30.
Describe the structure of the human ovum with a neat labelled diagram.
Answer:

Human ovum is non-cleidoic, alecithal and microscopic in nature. Its cytoplasm called ooplasm contains a large nucleus called the germinal vesicle. The ovum is surrounded by three coverings namely an inner thin transparent vitelline membrane, middle thick zona pellucida and outer thick coat of follicular cells called corona radiata. Between the vitelline membrane and zona pellucida is a narrow perivitelline space.

Question 31.
Give a schematic representation of spermatogenesis and oogenesis in humans
Answer:

Question 32.
Explain the various phases of the menstrual cycle.
Answer:
Menstrual cycle: The menstrual or ovarian cycle occurs approximately once in every 28/29 days during the reproductive life of the female from menarche (puberty) to menopause except during pregnancy. The cycle of events starting from one menstrual period till the next one is called the menstrual cycle during which cyclic changes occurs in the endometrium every month. Cyclic menstruation is an indicator of normal reproductive phase.

Menstrual cycle comprises of the following phases

Menstrual phase
Follicular or proliferative phase
Ovulatory phase
Luteal or secretory phase

1. Menstrual phase: The cycle starts with the menstrual phase when menstrual flow occurs and lasts for 3-5 days. Menstrual flow is due to the breakdown of endometrial lining of the uterus, and its blood vessels due to decline in the level of progesterone and oestrogen. Menstruation occurs only if the released ovum is not fertilized. Absence of menstruation may be an indicator of pregnancy. However it could also be due to stress, hormonal disorder and anaemia.

2. Follicular or proliferative phase: The follicular phase extends from the 5th day of the cycle until the time of ovulation. During this phase, the primary follicle in the ovary grows to become a fully mature Graafian follicle and simultaneously, the endometrium regenerates through proliferation. These changes in the ovary and the uterus are induced by the secretion of gonadotropins like FSH and LH, which increase gradually during the follicular phase. It stimulates follicular development and secretion of oestrogen by the follicle cells.

3. Ovulatory phase: Both LH and FSH attain peak level in the middle of the cycle (about the 14th day). Maximum secretion of LH during the mid cycle called LH surge induces the rupture of the Graafian follicle and the release of the ovum (secondary oocyte) from the ovary wall into the peritoneal cavity. This process is called as ovulation.

4. Luteal or secretory phase: During luteal phase, the remaining part of the Graafian follicle is transformed into a transitory endocrine gland called corpus luteum. The corpus luteum secretes large amount of progesterone which is essential for the maintenance of the endometrium. If fertilization takes place, it paves way for the implantation of the fertilized ovum. The uterine wall secretes nutritious fluid in the uterus for the foetus. So, this phase is also called as a secretory phase. During pregnancy, all events of the menstrual cycle stop and there is no menstruation.
In the absence of fertilization, the corpus luteum degenerates completely and leaves a scar tissue called corpus Albicans. It also initiates the disintegration of the endometrium leading to menstruation, marking the next cycle.

Question 33.
Explain the role of oxytocin and relaxin in parturition and lactation.
Answer:
Relaxin is the hormone secreted by the placenta that causes the contraction of pelvic joints and promotes parturition (childbirth).
Oxytocin causes the Let-down reflex – the actual ejection of milk from the alveoli of mammary glands. Oxytocin also stimulates the uterus to regain its pre-pregnancy size after childbirth.

Question 34.
Identify the given image and label its parts marked as a, b, c, and d.

Answer:
The given image is the diagram of the human egg cell or ovum
a – vitelline membrane
b – Nucleus
c – Zona pellucida
d – Corona Radiata

Question 35.
The following is the illustration of the sequence of ovarian events (a-i) in a human female.

(a) Identify the figure that illustrates ovulation and mention the stage of oogenesis it represents.
(b) Name the ovarian hormone and the pituitary hormone that have caused the above- mentioned events.
(c) Explain the changes that occurs in the uterus simultaneously in anticipation.
(d) Write the difference between C and H.
Answer:
(a) name of types:
A- Primordial follicle
B- Primary follicle
C- Secondary follicle
D-Tertiary follicle
E- Mature graafian follicle
F- Ovulation (release of egg)
G- Empty Graafian follicle
H- Corpus luteum
I – Corpus Albicans.

(b) Pituitary hormones: Follicle Stimulating Hormones (FSH) and Lutenizing Hormone (LH). Ovarian hormones: Estrogen and Progesterone.

(c) At the start of menstrual cycle, the endometrium of uterus starts regenerating through proliferation of cells induced by FSH and CH. After ovulation, the progesterone secreted by corpus luteum prepares the endometrium (uterine wall) to receive the egg if it is fertilized.

(d) C- Secondary follicle H – Corpus luteum During development of ovum, the primary follicle gets surrounded by many layers of granular Cells and forms a new layer called secondary follicle. Corpus luteum is the empty graafian follicle that remains after ovulation. It acts as a transitory endocrine gland secreting progesterone to maintain pregnancy.

Samacheer Kalvi 12th Bio Zoology Human Reproduction Additional Questions and Answers

1 – Mark Questions

Question 1.
The developing spermatozoa are nourished by _________
(a) Leydig cells
(b) Sertoli cells
(c) Follicular cells
(d) Epididymis
Answer:
(b) Sertoli cells

Question 2.
Identify the correct sequence of reproductive events in human beings.
(a) Insemination, Implantation, Fertilization, Parturition and Placentation.
(b) Implantation, Fertilization, Insemination, Placentation and Parturition.
(c) Implantation, Insemination, Fertilization, Parturition and Placentation.
(d) Insemination, Fertilization, Implantation, Placentation and Parturition.
Answer:
(d) Insemination, Fertilization, Implantation, Placentation and Parturition.

Question 3.
Expulsion of baby from the mother’s womb is referred as _________
Answer:
(a) Ejection
(b) Relaxation
(c) Parturition
(d) Implantation
Answer:
(c) Parturition

Question 4.
Match the Column I with Column II

Answer:
(a) a – ii, b – i, c – iv, d – iii

Question 5.
Which of the following statement is not correct?
(i) Interstitial cells are seen surrounding the seminiferous tubule.
(ii) Nurse cells secrete inhibin.
(iii)Males have single prostate gland which encircles the urethra.
(iv) Insemination, Fertilization, Implantation, Placentation and Parturition.
(a) i and ii
(b) iii only
(c) iii and iv
(d) iv only
(d) iv only
Answer:
(d) iv only

Question 6.
Assertion (A): In scrotum, the temperature is maintained 2 – 3°C lower than body temperature.
Reason (R): Reduced temperature results in efficient sperm production.
(a) R explains A.
(b) A is right R is wrong.
(c) A and R are right. R does not explain A.
(d) Both A and R are wrong.
Answer:
(a) R explains A.

Question 7.
Assertion (A): The acrosome of the sperm cell contains sperm lysin.
Reason (R): Sperm lysin destroys the deformed sperm cells.
(a) R explains A.
(b) A is right, R is wrong.
(c) A and R are right. R does not explains A.
(d) Both A and R are wrong.
Answer:
(A) A is right, R is wrong.

Question 8.
Assertion (A): Human ovum is non – cieidoic
Reason (R): Human does not contain yolk.
(a) R explains A.
(b) A is right, R is wrong.
(c) A and R are right. R does not explains A.
(d) Both A and R are wrong.
Answer:
(c) A and R are right. R does not explains A.

Question 9.
Assertion (A): Menopause refers to the absence of menstruation during pregnancy.
Reason (R): Ovulation occurs during menstrual phase.
(a) R explains A.
(b) A is right, R is wrong.
(c) A and R are right. R does not explains A.
(d) Both A and R are wrong.
Answer:
(d) Both A and R are wrong.

Question 10.
Assertion (A): Cervix is common site of ectopic pregnancies
Reason (R): Implantation of fertilized ovum outside uterus.
(a) A is wrong, R is right.
(b) A is right, R is wrong.
(c) A and R are right. R does not explains A.
(d) Both A and R are wrong.
Answer:
(a) A is wrong, R is right.

Question 11.
Which of the following contributes to the seminal plasma?
(i) Cowper’s gland
(ii) Seminal vesicles
(iii) Prostate gland
(iv) Bulbourethral gland
(a) ii, iii and ii
(b) i, ii, and iii
(c) i, iii and iv
(d) all the above
Answer:
(d) all the above

Question 12.
Organ of copulation in human female is _________
(a) Cevix
(b) Fundus
(c) Vagina
(d) Uterus
Answer:
(c) Vagina

Question 13.
Identify the gland which is homologous to the Cowper’s glands of male.
(a) Bartholin’s gland
(b) Bulbourethral gland
(c) Prostate gland
(d) Skene’s gland
Answer:
(a) Bartholin’s gland

Question 14.
Find out the proper sequence representing the parts of female reproductive system.
(a) Vagina → Ovary → Uterus → Cervix → Infundibulum → Oviduct
(b) Vagina → Ovary → Oviduct → Infundibulum → Cervix → Uterus
(c) Ovary → Infundibulum → Oviduct → Uterus → Cervix → Vagina
(d) Oviduct → Ovary → Uterus → Infundibulum Vagina → Cervix
Answer:
(c) Ovary- Infundibulum → Oviduct → Uterus → Cervix → Vagina

Question 15.
Spermatid → spermatozoa. What does ‘A’ stands for?
(a) Spermatogenesis
(b) Spermiation
(c) Spermiogenesis
(d) Gametogenesis
Answer:
(c) Spermiogenesis

Question 16.
An adult male produces an average of…………….. sperms per day
(a) 200 million
(b) 300 million
(c) 300 billion
(d) 120 million
Answer:
(a) 200 million

Question 17.
Statement (1): During spermiation, the sperms are released into the cavity of I seminiferous tubule.
Statement (2): During spermiogenesis, the spermatids get mature into sperms.
(a) Statement 1 is correct; statement 2 is incorrect.
(b) Statement 1 is incorrect; statement 2 is correct.
(c) Both statements 1 and 2 are correct.
(d) Both statements 1 and 2 are incorrect.
Answer:
(c) Both statements 1 and 2 are correct.

Question 18.
Statement (1): Siamese twins are conjoined twins who are joined during birth.
Statement (2): Dizygotic twins will be of same-sex.
(a) Statement 1 is correct; statement 2 is incorrect.
(b) Statement 1 is incorrect; statement 2 is correct.
(c) Both statements 1 and 2 are correct.
(d) Both statements 1 and 2 are incorrect.
Answer:
(a) Statment 1 is correct; statement 2 is incorrect.

Question 19.
Statement (1): The endometrium acts as a transitory endocrine gland secreting
progesterone
Statement (2): Progesterone maintains pregnancy
(а) Statement 1 is correct; statement 2 is incorrect.
(b) Statement 1 is incorrect; statement 2 is correct.
(c) Both statements 1 and 2 are correct.
(d) Both statements 1 and 2 are incorrect.
Answer:
(b) Statement 1 is incorrect; statement 2 is correct

Question 20.
Statement (1): Human pregnancy lasts for 35 weeks.
Statement (2): During gestation, the embryo’s heat develops during the 12th week.
(a) Statement 1 is correct; statement 2 is incorrect.
(b) Statement 1 is incorrect; statement 2 is correct.
(c) Both statements 1 and 2 are correct.
(d) Both statements 1 and 2 are incorrect.
Answer:
(d) Both statements 1 and 2 are incorrect.

Question 21.
Statement (1): Menstrual cycle occurs once in every 29 days.
Statement (2): The average age of menopause is 45-50 years.
(a) Statement 1 is correct; statement 2 is incorrect.
(b) Statement 1 is incorrect; statement 2 is correct.
(c) Both statements 1 and 2 are correct.
(d) Both statements 1 and 2 are incorrect.
Answer:
(c) Both statements 1 and 2 are correct.

Question 22.
The first ejaculation of the semen in male is called as ___________
Answer:
Spermarche

Question 23.
Identify the mismatched pair.
(a) Castration – Orchidectomy
(b) Spermiogenesis – Release of sperms into the cavity of seminiferous tubule
(c) Ovulation – Release of egg from ovary
(d) Capacitation – Process enabling the sperm to penetrate the egg.
Answer:
(b) Spermiogenesis – Release of sperms into the cavity of seminiferous tuble.

Question 24.
Given below are the extraembryonic membranes of which identify the outermost membrane
(a) Amnion
(b) Chorion
(c) Yolk sac
(d) Allantois
Answer:
(b) Chorion

Question 25.
Identify the given figure and select the correct option representing X, Y and Z.

Answer:
(b) Acrosome, Nucleus and Mitochondria

Question 26.
The entire process of spermatogenesis takes about ________ days
(a) 60 days
(b) 44 days
(c) 64 days
(d) 50 days
Answer:
(c) 64 days

Question 27.
Observe the diagram and select the correct option denoting the proper sequence of parts.

Answer:
(b) Fimbriae, Infundibulum, Uterus and Isthmus

Question 28.
Pick out the incorrect statements.
(a) The upper rounded portion of uterus is fundus.
(b) Uterus open into vagina through narrow cervix.
(c) Cervix is the organ of copulation in female.
(d) Vagina extends from the cervix and opens to exterior.
Answer:
(c) Cervix is the organ of copulation in female.

Question 29.
What is the role of fimbriae?
(a) Secretion of oestrogen and prolactin.
(b) Helps in the collection of the ovum after ovulation.
(c) Attaches the ovary to the abdominal cavity.
(d) Connects oviduct with ovary.
Answer:
(b) Helps in the collection of the ovum after ovulation.

Question 30.
Name the enzyme found in the acrosomal tip of sperm cell.
Answer:
Hyaluronidase

Question 31.
Which is not a correct statement regarding Oogenesis?
Answer:
(i) During foetal development, cells in germinal epithelium of foetal ovary undergo, mitosis and produce oogonia.
(ii) Oogonial cell divide and enter into prophase I of meiosis I and from primary oocytes.
(iii)Primary oocytes later develop into primary follicles.
(iv) No oogonia is formed or added after the foetal birth.
(a) Only i
(b) ii and iii
(c) iv only
(d) None of the above
Answer:
(d) None of the above

Question 32.
In embryo development of human beings, how long does it takes for a zygote to convert into morula?
(a) 24hrs
(b) 36hrs
(c)48hrs
(d) 72hrs
Answer:
(d) 72 hrs

Question 33.
Identify the hormone which is produced only during the time of pregnancy
(a) Relaxin
(b) Oxytocin
(c) Progesterone
(d) Cortisol
Answer:
(a) Relaxin

Question 34.
The type of antibodies present in colostrum.
(a) IgE
(b) IgM
(c) IgA
(d) IgB
Answer:
(c) IgA

2 – Mark Questions

Question 1.
Enumerate the functions of the reproductive system.
Answer:
The reproductive system has four main functions namely,

to produce the gametes namely sperms and ova
to transport and sustain these gametes
to nurture the developing offspring
to produce hormones

Question 2.
Define the terms

Insemination
Fertilization.

Answer:

Insemination: Transfer of sperms by the male into the female genital tract.
Fertilization: Fusion of male and female gametes to form zygote, called fertilization.

Question 3.
What are seminiferous tubules? Mention its role.
Answer:
Seminiferous tubules are highly coiled tubules seen in the lobules of testis. They occupy 80% – of testicular substance. They are the site for sperm production.

Question 4.
Name the cells noticed in the epithelial layer of seminiferous tubule.
Answer:

Sertoli cells or Nurse cells
Spermatogonic cells or male germ cells.

Question 5.
Mention the role epididymis.
Answer:

Epididymis is a temporary store house for sperms.
Sperms undergo physiological maturation, increased motility and fertilizing capacity inside epididymis.

Question 6.
Seminal plasma is acidic or alkaline. Write its composition.
Answer:

Seminal plasma is alkaline in nature.
It contains fructose, ascorbic acid, prostaglandins and a coagulating enzyme called vesiculase.

Question 7.
Define Semen.
Answer:
Semen or seminal fluid is a milky white fluid which contains sperms and seminal plasma, which is secreted from the seminal vesicles, prostate gland, and bulbourethral glands.

Question 8.
Why do males have Penis?
Answer:

Penis is the male external genitalia.
It functions as both excretory and copulatory organ.
It is made of special tissue that erects the penis to facilitate insemination.

Question 9.
Point out the female accessory organs.
Answer:
Fallopian tubes, Uterus and Vagina.

Question 10.
Define the nature of uterus.
Answer:
The uterus or womb is a hollow, thick-walled, muscular, highly vascular and inverted pear shaped structure lying in the pelvic cavity between the urinary bladder and rectum.

Question 11.
What are the components that make up external genitalia female?
Answer:
Labia Majora, Labia Minora, Hymen and Clitoris.

Question 12.
Name the accessory reproductive glands in female which are homologous to (a) Cowper’s gland and (b) Prostate gland.
Answer:
In female, Bartholin’s gland is homologous to Cowper’s gland and Skene’s gland is homologous to prostate gland.

Question 13.
Define Gametogenesis.
Answer:
Gametogenesis is the process of formation of gametes i.e., sperms and ovary from the primary ,sex organs in all sexually reproducing organisms. Meiosis plays the most significant role in the process of gametogenesis.

Question 14.
Define the terms

Spermiogenesis
Spermination

Answer:

Spermiogenesis: Transformation of spermatids into mature sperm.
Spermiation: Release of mature sperm into the lumen of the seminiferous tubule.

Question 15.
What do you mean by ‘Sperm lysin’? Mention its function.
Answer:

Sperm lysin is a proteolytic enzyme secreted in the acrosome of sperm.
It helps to penetrate the ovum during fertilization.
It is also called hyaluronidase.

Question 16.
Name the four phases of the menstrual cycle.
Answer:

Menstrual phase
Follicular or proliferative phase
Ovulatory phase
Luteal or secretory phase

Question 17.
What is corpus Albicans?
Answer:
In the absence of fertilization, the corpus luteum degenerates completely and leaves a scar tissue called corpus Albicans. It also initiates the disintegration of the endometrium leading to menstruation, marking the next cycle.

Question 18.
Define menopause.
Answer:
Menopause is the phase in a women’s life when ovulation and menstruation stops. The average age of menopause is 45-50 years. It indicates the permanent cessation of the primary functions of the ovaries.

Question 19.
When does capacitation occur? Define it.
Answer:
The sperms deposited in the female reproductive tract undergo capacitation. It is a biochemical event that makes the sperm to penetrate and fertilize the egg.

Question 20.
Write a brief note on ectopic pregnancy?
Answer:
If the fertilized ovum is implanted outside the uterus it results in ectopic pregnancy. About 95 percent of ectopic pregnancies occur in the fallopian tube. The growth of the embryo may cause internal bleeding, infection, and in some cases even death due to rupture of the fallopian tube.

Question 21.
Point out the extraembryonic membranes of the human embryo.
Answer:

amnion
Chorion
allantois
Yolk nac

Question 22.
What is the placenta?
Answer:
Placenta is a temporary endocrine organ formed during pregnancy and it connects the foetus to the uterine wall through the umbilical cord. It is the organ by which the nutritive, respiratory and excretory functions are fulfilled.

Question 23.
Name the organs developed from embryonic ectoderm.
Answer:
Brain and spinal cord (CNS), peripheral nervous system (PNS), epidermis and its derivatives, and mammary glands.

Question 24.
Mention the hormones secreted by the placenta during pregnancy.
Answer:

human Chorionic Gonadotropin (hCG)
human Chorionic Somatomammotropin (hCS)
human Placental Lactogen (hPL)
Oestrogen and progesterone and relaxin.

Question 25.
Name the hormones that are secreted in human only during pregnancy.
Answer:

human Chorionic Gonadotropin (hCG)
human Chorionic Somatomammotropin (hCS)
relaxin

Question 26.
State the role of relaxin.
Answer:
Relaxin is an hormone secreted by the placenta during the later phase of pregnancy. It helps in the relaxation of the pelvis during childbirth.

Question 27.
Define parturition and labour.
Answer:
Parturition is the completion of pregnancy and giving birth to the baby. The series of events that expels the infant from the uterus is collectively called “labour”.

Question 28.
What do you mean by ‘false labour’?
Answer:
Throughout pregnancy the uterus undergoes periodic episodes of weak and strong contractions, fhese contractions called Braxter-Hick’s contractions lead to false labour.

Question 29.
Explain the term C-section.
Answer:
When normal vaginal delivery is not possible due to factors like position of the baby and nature of the placenta, the baby is delivered through a surgical incision in the woman’s abdomen and uterus. It is also termed as abdominal delivery or Caesarean Section or ‘C’ Section.

3 – Mark Questions

Question 1.
Compare gametogenesis with organogenesis.
Answer:
Gametogenesis:

Formation of gametes i.e., Sperm and Ova.
It takes place in gonads.

Organogenesis:

Formation of tissues, organs and organ system.
It takes place in the embryonic germ layers.

Question 2.
What are primary reproductive organs? What role does they play in organisms?
Answer:
The primary reproductive organs namely the ovary and testis are responsible for producing the ova and sperms respectively. Hormones secreted by the pituitary gland and the gonads help in the development of the secondary sexual characteristics, maturation of the reproductive system and regulation of normal functioning of the reproductive system.

Question 3.
Scrotum acts as a thermoregulator – Justify.
Answer:
The scrotum is a sac of skin that hangs outside the abdominal cavity. Since viable sperms cannot be produced at normal body temperature, the scrotum is placed outside the abdominal cavity to provide a temperature 2-3°C lower than the normal internal body temperature. Thus, the scrotum acts as a thermoregulator for spermatogenesis.

Question 4.
Write any three statements on Sertoli cells.
Answer:

Sertoli cells are elongated and pyramidal cells.
They provide nourishment to sperm till maturation.
They secrete a hormone called inhibin which is involved in negative feedback control of sperm production.

Question 5.
Give a brief account on leydig cells.
Answer:

Interstitial cells of Leydig are seen embedded in the soft connective tissue surrounding the seminiferous tubules.
These cells are endocrine natured and produce testosterone (androgen).
These cells are characteristic to mammalian testes.

Question 6.
Name the accessory glands of male reproductive system.
Answer:

A pair of seminal vesicles.
A pair of bulbourethral gland (Cowper’s gland).
A single prostate gland.

Question 7.
State the location and secretion of prostate gland.
Answer:
The prostate encircles the urethra and is just below the urinary bladder and secretes a slightly acidic fluid that contains citrate, several enzymes and prostate specific antigens.

Question 8.
Write a note on hymen.
Answer:
The external opening of the vagina is partially closed by a thin ring of tissue called the hymen. The hymen is often tom during the first coitus (physical union). However in some women it remains intact. It can be stretched or tom due to a sudden fall or jolt and also during strenuous physical activities such as cycling and horseback riding, etc., and therefore cannot be considered as an indicator of a woman’s virginity.

Question 9.
“Role of hormones in spermatogenesis” – comment on the statement.
Answer:
Spermatogenesis starts at the age of puberty and is initiated due to the increase in the release of Gonadotropin Releasing Hormone (GnRH) by the hypothalamus. GnRH acts on the anterior pituitary gland and stimulates the secretion of two gonadotropins namely Follicle Stimulating Hormone (FSH) and Lutenizing Hormone (LH). FSH stimulates testicular growth and enhances the production of Androgen Binding Protein (ABP) by the Sertoli cells and helps in the process of spermiogenesis. LH acts on the Leydig cells and stimulates the synthesis of testosterone which in turn stimulates the process of spermatogenesis.

Question 10.
Define menstrual cycle.
Answer:
The menstrual or ovarian cycle occurs approximately once in every 28/29 days during the reproductive life of the female from menarche (puberty) to menopause except during pregnancy. The cycle of events starting from one menstrual period till the next one is called the menstrual cycle during which cyclic changes occurs in the endometrium every month. Cyclic menstruation is an indicator of normal reproductive phase.

Question 11.
Luteal phase of menstrual cycle is also called as secretory phase. Why?
Answer:
After ovulation, the graafian follicle turns into corpus luteum (a transistory endocrine gland) which secrets progesterone. Progesterone maintain endometrium for implantation of fertilized ovum, the endometrium of uterus also secretes nutritious fluid for the foetus. Hence this phase is also referred as secretory phase.

Question 12.
Menstrual hygiene is essential for women. Why?
Answer:
Menstrual hygiene is vital for good health, well-being, dignity, empowerment and productivity of women. The impact of poor menstrual hygiene on girls is increased stress levels, fear and embarrassment during menstruation. This can keep girls inactive during such periods leading to absenteeism from school.

Question 13.
Name the absorbents or materials used to manage menstruation.
Answer:
Clean and safe absorbable clothing materials, sanitary napkins, pads, tampons and menstrual cups have been identified as materials used to manage menstruation.

Question 14.
Explain acrosomal reaction.
Answer:
The follicular cells of egg are held together by an adhesive cementing substance called hyaluronic acid. The acrosomal membrane disintegrates releasing the proteolytic enzyme, hyaluronidase during sperm entry through the corona radiata and zona pellucida. This is called
acrosomal reaction.

Question 15.
Differentiate between monozygotic and Dizygotic twins.
Answer:
Monozygotic twins:

Monozygotic (Identical) twins are produced when a single fertilized egg splits into two during the first cleavage.
They are of the same sex, look alike and share the same genes.

Dizygotic twins:

Dizygotic (Fraternal) twins are produced when two separate eggs are fertilized by two separate sperms.
The twins may be of the same sex or different sex and are non-identical.

Question 16.
What is morula?
Answer:
The first cleavage in zygote produces two identical cells called blastomeres. These produce 4 cells, then 8 and so on. After 72 hours of fertilization, a loose collection of cells forms a berry shaped cluster of 16 or more cells called the morula.

Question 17.
Explain gastrulation.
Answer:
The inner cell mass in the blastula is differentiated into epiblast and hypoblast immediately after implantation. The hypoblast is the embryonic endoderm and the epiblast is the ectoderm. The cells remaining in between the epiblast and the endoderm form the mesoderm. Thus the transformation of the blastocyst into a gastrula with the primary germ layers by the movement of the blastomeres is called gastrulation.

Question 18.
Name the three primary germ layers of embryo. Also mention any three organs or ortgan systems developing from each layer.
Answer:

Question 19.
Define lactation and explain its hormonal background.
Answer:
Lactation is the production of milk by mammary glands. The mammary glands show changes during every menstrual cycle, during pregnancy and lactation. Increased level of oestrogens, progesterone and human Placental Lactogen (hPL) towards the end of pregnancy stimulate the hypothalamus towards prolactin-releasing factors. The anterior pituitary responds by secreting prolactin which plays a major role in lactogenesis.

Question 20.
What is “let-down reflex”?
Answer:
Oxytocin causes the “Let-Down” reflex the actual ejection of milk from the alveoli of the mammary glands. During lactation, oxytocin also stimulates the recently emptied uterus to contract, helping it to return to pre-pregnancy size.

Question 21.
Explain Foetal-ejection reflex.
Answer:
As the pregnancy progresses, increase in the oestrogen concentration promotes uterine contractions. These uterine contractions facilitate moulding of the foetus and downward movement of the foetus. The descent of the foetus causes dilation of cervix of the uterus and vaginal canal resulting in a neurohumoral reflex called Foetal ejection reflex or Ferguson reflex. This initiates the secretion of oxytocin from the neurohypophysis which in turn brings about the powerful contraction of the uterine muscles and leads to the expulsion of the baby through the birth canal.

Question 22.
Give an account on uterine wall layers.
Answer:
The wall of the uterus has three layers of tissues. The outermost thin membranous serous layer called the perimetrium, the middle thick muscular layer called myometrium and the inner glandular layer called endometrium. The endometrium undergoes cyclic changes during the menstrual cycle while myometrium exhibits strong contractions during parturition

5 – Mark Questions

Question 1.
Describe the structure of human ovary.
Answer:

Ovaries are the primary female sex organ producing eggs (ovum).
They are located one on each side of lower abdomen (pelvis).
The ovary is attached in pelvic wall uterus by an ovarian ligament called mesovarium.
It is an elliptical structure of 2-4 cm long
Each ovary is covered by thin cuboidal germinal epithelium encloses ovarian stroma.
Below germinal epithelium is a dense connective tissue called tunica albuginea.
The stroma is differentiated into outer cortex and inner medulla.
The cortex is dense and granular due to follicular cells at varying development stages.
The medulla is a loose connective tissue with blood vessels, lymph vessels and nerve

Question 2.
Explain the structure and function of mammary glands.
Answer:
The mammary glands are modified sweat glands present in both sexes. It is rudimentary in the males and functional in the females. A pair of mammary glands is located in the thoracic region. It contains glandular tissue and variable quantities of fat with a median nipple surrounded by a pigmented area called the areola. Several sebaceous glands called the areolar glands are found on the surface and they reduce cracking of the skin of the nipple. Internally each mammary gland consists of 2-25 lobes, separated by fat and connective tissues.

Each lobe is made up of lobules which contain acini or alveoli lined by epithelial cells. Cells of the alveoli secrete milk. The alveoli open into mammary tubules. The tubules of each lobe join to form a mammary duct. Several mammary ducts join to form a wider mammary ampulla which is connected to the lactiferous duct in the nipple. Under the nipple, each lactiferous duct expands to form the lactiferous sinus which serves as a reservoir of milk. Each lactiferous duct opens separately by a minute pore on the surface of the nipple.

Normal development of the breast begins at puberty and progresses with changes during each menstrual cycle. In non-pregnant women, the glandular structure is largely underdeveloped and the breast size is largely due to the amount of fat deposits. The size of the breast does not have an influence on the efficiency of lactation.

Question 3.
Describe the spermatogenesis with a diagram.
Answer:
Spermatogenesis is the sequence of events in the seminiferous tubules of the testes that produce the male gametes, the sperms.
During development, the primordial germ cells migrate into the testes and become immature germ cells called sperm mother cells or spermatogonia in the inner surfaces of the seminiferous tubules. The spermatogonia begin to undergo mitotic division at puberty and continues throughout life.

In the first stage of spermatogenesis, the spermatogonia migrate among Sertoli cells towards the central lumen of the seminiferous tubule and become modified and enlarged to form primary spermatocytes which are diploid with 23 pairs i.e., 46 chromosomes. Some of the primary spermatocytes undergo the first meiotic division to form two secondary spermatocytes which are haploid with 23 chromosomes each.

The secondary spermatocytes undergo second meiotic division to produce four haploid spermatids. The spermatids are transformed into mature spermatozoa (sperms) by the process called spermiogenesis. Sperms are finally released into the cavity of seminiferous tubules by a process called spermiation The whole process of spermatogenesis takes about 64 days. At any given time, different regions of the seminiferous tubules contain spermatocytes in different stages of development.

The sperm production remains nearly constant at a rate of about 200 million sperms per day. Spermatogenesis starts at the age of puberty and is initiated due to the increase in the – release of Gonadotropin-Releasing Hormone (GnRH) by the hypothalamus. GnRH acts on the anterior pituitary gland and stimulates the secretion of two gonadotropins namely Follicle Stimulating Hormone (FSH) and Lutenizing Hormone (LH). FSH stimulates testicular growth and enhances the production of Androgen Binding Protein (ABP) by the Sertoli cells and helps in the process of spermiogenesis. LH acts on the Leydig cells and stimulates the synthesis of testosterone which in turn stimulates the process of spermatogenesis.

Question 4.
Describe the structure of the human spermatozoan.
Answer:
The human sperm is a microscopic, flagellated and motile gamete. The whole body of the sperm is enveloped by P plasma membrane and is composed of a head, neck and tail. The head comprises of two parts namely the acrosome and nucleus. Acrosome is a small cap like pointed structure present at the tip of the nucleus and is formed mainly from the Golgi body of the spermatid. It contains hyaluronidase, a proteolytic enzyme, popularly known as sperm lysin which I helps to penetrate the ovum during fertilization. The nucleus is flat and oval. The neck is very short and is present between the head and the middle piece.

It contains the proximal centriole towards the nucleus which plays a role in the first division of the zygote and the distal centriole gives rise to I the axial filament of the sperm. The middle piece possesses mitochondria spirally twisted around the axial filament called mitochondrial spiral or nebenkem. It produces energy in the form of ATP molecules for the movement of sperms. The tail is the longest part of the sperm and is slender and tapering. It is formed of a central axial filament or axoneme and an outer protoplasmic sheath. The lashing movements of the tail, push the sperm forward.

Question 5.
Explain the process of oogenesis.
Answer:
Oogenesis is the process of development of the female gamete or ovum or egg in the ovaries. During foetal development, certain cells in the germinal epithelium of the foetal ovary divide by mitosis and produce millions of egg mother cells or oogonia. No more oogonia are formed or added after birth.

The oogonial cells start dividing and enter into Prophase-I of meiotic division-I to form the primary oocytes which are temporarily arrested at this stage. The primary oocytes then get surrounded by a single layer of granulosa cells to form the primordial or primary follicles. A large number of follicles degenerate during the period from birth to puberty, so at puberty, only 60,000 to 80,000 follicles are left in each ovary.

The primary follicle gets surrounded by many layers of granulosa cells and a new theca layer to form the secondary follicle. A fluid filled Oogenesis space, the antrum develops in the follicle and gets transformed into a tertiary follicle. The theca layer gets organized into an inner theca interna and an outer theca externa. At this time, the primary oocyte within the tertiary follicle grows in size and completes its first meiotic division and forms the secondary oocyte.

It is an unequal division resulting in the formation of a large haploid secondary oocyte and a first polar body. The first polar body disintegrates. During fertilization, the secondary oocyte undergoes second meiotic division and produces a large cell, the ovum and a second polar body. The second polar body also degenerates. The tertiary follicle eventually becomes a mature follicle or Graafian follicle. If fertilisation does not take place, second meiotic division is never completed and the egg disintegrates. At the end of gametogenesis in females, each primary oocyte gives rise to only one haploid ovum.

Question 6.
Write a note on embryonic membranes
Answer:
The extra embryonic membranes include amnion, yolk sac, allantois and chorion. They protect the embryo from dessication, mechanical shock, absorption of nutrients, gaseous exchange and placental formatio
Answer:

Question 7.
Name the three primary germ layers of embryo. Also mention any three organs or organ systems developing from each layer.
Answer:

Higher Order Thinking Skills (HOTs) Questions

Questio 1.
Give the alternate terminologies for

Spermatogonia
Embryonic ectoderm

Answer:

Spermatogonia = Sperm mother cell
Embryonic ectoderm = hypoblast

Question 2.
Mention the production site and active site of the following hormones.
(a) GnRH
(b) Relaxin
Answer:

Question 3.
May 28th is celebrated as the annual Menstrual Hygiene Day (MHD). State its importance.
Answer:
MHgD aims to create awareness of the importance for women and girls to hygienically manage their menstruation. Menstrual hygiene is vital for good health, well-being, dignity, empowerment, and productivity of women.

Question 4.
Suggest a few hygiene tips for face healthy and happy menses.
Answer:

Change the napkins periodically for 4 to 6 hours.
Wash your genitals properly using clean lukewarm water.
Avoid using soaps and vaginal hygiene products.
Discard the sanitary napkins by incinerating them.
Wear clean and comfortable underwear. Avoid tight clothing.
Have a healthy diet rich in Iron content and vitamins.

Search Results for:

Tamilnadu Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Samacheer Kalvi 11th Accountancy Bank Reconciliation Statement Text Book Back Questions and Answers

Tamilnadu Samacheer Kalvi 12th Computer Applications Solutions Chapter 5 PHP Function and Array

Samacheer Kalvi 12th Computer Applications PHP Function and Array Text Book Back Questions and Answers

Samacheer Kalvi 12th Computer Applications Solutions PHP Function and Array Additional Questions and Answers

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation

Samacheer Kalvi 12th Bio Zoology Principles of Inheritance and Variation Text Book Back Questions and Answers

Samacheer Kalvi 12th Bio Zoology Principles of Inheritance and Variation Additional Questions and Answers

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 3 Vegetative Morphology

Samacheer Kalvi 11th Bio Botany Vegetative Morphology Text Book Back Questions and Answers

Samacheer Kalvi 11th Bio Botany Vegetative Morphology Additional Questions and Answers

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 2 Plant Kingdom

Samacheer Kalvi 11th Bio Botany Plant Kingdom Text Book Back Questions and Answers

Samacheer Kalvi 11th Bio Botany Plant Kingdom Additional Questions and Answers

Tamil Nadu Samacheer Kalvi 10th Science Model Question Paper 4 English Medium

Tamilnadu Samacheer Kalvi 10th Maths Model Question Paper 3 English Medium

Tamil Nadu Samacheer Kalvi 10th English Model Question Paper 4

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction

Samacheer Kalvi 12th Bio Zoology Human Reproduction Text Book Back Questions and Answers

Samacheer Kalvi 12th Bio Zoology Human Reproduction Additional Questions and Answers