Plus One Maths Notes Chapter 10 Straight Lines

Kerala State Board New Syllabus Plus One Maths Notes Chapter 10 Straight Lines.

Kerala Plus One Maths Notes Chapter 10 Straight Lines

I. Slope of Line
The slope of a line is the ‘tan’ of the angle the line makes with the positive direction of the x-axis. If θ is the angle then, slope = tan θ.

The slope of the x-axis is zero and that of the y-axis is not defined.

Parallel lines have the same slope.

The product of the slopes of perpendicular lines is -1.

The slope is positive if θ < 90°. The slope is negative if θ > 90°.

The slope of a line passing through two points (x1, y1) and (x2, y2) is \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

If three points A, B, and C are collinear, then AB and BC have the same slope.

If m1 and m2 be slopes of two lines then, θ the angle between is given by tan θ = \(\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|\), 1 + m1m2 ≠ 0

II. Equation of a Line
Equation of x-axis is y = 0.

Equation of y-axis is x = 0.

The equation of a horizontal line is y = a. If ‘a’ is positive then the line is above the x-axis and if negative it will be below the x-axis.

The equation of a vertical line is x = a. If ‘a’ is positive then the line is to the right of the x-axis and if negative it will be to the left of the x-axis.

Point-slope form: y – y1 = m(x – x1), where ‘m’ is the slope and (x1, y1) is a point on the line.

Two-Point form:
y – y1 = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) (x – x1) where (x1, y1) and(x2, y2) are two point on the line.

Slope intercept form:
1. y = mx + c, where ‘m’ is the slope and ‘c’ is the y-intercept.
2. y = m(x – d), where ‘m’ is the slope and ‘d’ is the x-intercept.

Intercept form: \(\frac{x}{a}+\frac{y}{b}=1\) = 1, where ‘a’ and ‘b‘ are x and y intercept respectively.

Normal form: x cos θ + y sin θ = p, where ‘p’ is the length of the normal from the origin to the line and ‘θ’ is the angle the normal makes with the positive direction of the x-axis.

General equation of a Line: ax + by + c = 0, where a, b and c are real constants.
1. Slope of the line ax + by + c = 0 is \(-\frac{a}{b}\)

2. Parallel lines differ in constant term, i.e; a line parallel to ax + by + c = 0 is ax + by + k = 0.

3. A line perpendicular to ax + by + c = 0 is bx – ay + k = 0.

4. The equation of the family of lines passing through the intersection of the lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 is of the form a1x + b1y + c1 + k(a2x + b2y + c2) = 0.

5. The perpendicular distance of a point (x1, y1) from the line ax + by + c = 0 is \(\left|\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\right|\)

6. The distance between the parallel lines ax + by + c = 0 and ax + by + k = 0 is \(\left|\frac{c-k}{\sqrt{a^{2}+b^{2}}}\right|\)

7. Normal form of the equation ax + by + c = 0 is x cos θ + y sin θ = p;
Where cos θ = \(\pm \frac{a}{\sqrt{a^{2}+b^{2}}}\); sin θ = \(\pm \frac{b}{\sqrt{a^{2}+b^{2}}}\) and p = \(\pm \frac{c}{\sqrt{a^{2}+b^{2}}}\)

Proper choice of signs is made so that p should be positive.

III. Shifting of Origin
An equation corresponding to a set of points with reference to a system of coordinate axes by shifting the origin is shifted to a new point is called a translation of axes.

Let us take a point P (x, y) referred to the axes OX and OY. Let (h, k) be the coordinates of origin and P(X, Y) be the coordinate of P(x, y) with respect to the new axis. Then, the transformation relation between the old coordinates (x, y) and the new coordinates (X, Y) are given by X = x + h and Y = y + k.

Plus One Maths Notes Chapter 7 Permutation and Combinations

Kerala State Board New Syllabus Plus One Maths Notes Chapter 7 Permutation and Combinations.

Kerala Plus One Maths Notes Chapter 7 Permutation and Combinations

I. Fundamental Principle of Counting
If an event can occur in ‘m’ different ways, following which another event can occur in ‘n’ different ways, then the total number of occurrences of the events in the given order is m × n.

II. Permutation
A permutation is the arrangement of some or all of a number of different objects.

Factorial notation: The notation n! represents the product of first n natural numbers,
ie; n! = n(n – 1 )(n – 2) ….. 3.2.1
1. 1! = 1
2. 0! = 1

The number of permutation of ‘n’ different objects taken ‘r’ at a time, where the objects do not repeat is n(n – 1)(n – 2)……(n – r + 1) which is denoted by nPr.
Plus One Maths Notes Chapter 7 Permutation and Combinations 1
The number of permutation of ‘n’ different objects taken ‘r’ at a time, where repetition is allowed is nr.

Permutation when all the objects are not distinct.
1. The number of permutations of ‘n’ objects, where ‘p’ objects are of the same kind and rest all different = \(\frac{n !}{p !}\)

2. The number of permutations of ‘n’ objects, where ‘p1’ objects are of one kind, ‘p2’ objects are of the second kind, …….., ‘pk‘ objects are of a kth kind and rest all different = \(\frac{n !}{p_{1} ! p_{2} ! \ldots p_{k} !}\)

III. Combinations
A combination is a selection of some or all of a number of different objects (the order of selection is not important). The number of selection of ‘n’ things taken ‘r’ at a time is nCr.
Plus One Maths Notes Chapter 7 Permutation and Combinations 2

Plus One Maths Notes Chapter 5 Complex Numbers and Quadratic Equations

Kerala State Board New Syllabus Plus One Maths Notes Chapter 5 Complex Numbers and Quadratic Equations.

Kerala Plus One Maths Notes Chapter 5 Complex Numbers and Quadratic Equations

we have studied linear equations in one and two variables and quadratic equations in one variable. We have seen that the equation x2 + 1 = 0 has no real solution since the root of a negative number does not exist in a real number. So, we need to extend the real number system to a larger number system to accommodate such numbers.

I. Complex Numbers
A number of the form a + ib, where a and b are real numbers and i = √-1.
Usually, a complex number is denoted by z, a is the real part of z denoted by Re(z) and b is the imaginary part of z denoted by Im(z).

II. Algebra of Complex Numbers

Addition: Let z1 = a + ib and z2 = c + id be two complex numbers. Then the sum z1 + z2 is obtained by adding the real and imaginary parts.

1. z1 + z2 = z2 + z1, commutative.
2. z1 + (z2 + z3) = (z1 + z2) + z3, associative.
3. 0 + i0 is the identity element.
4. -z is the inverse of z.

Multiplication: Let z1 = a + ib and z2 = c + id be two complex numbers.
Then the product z1z2 is defined as follows:
z1z2 = (ac – bd) + i(ad + bc).

1. z1z2 = z2z1, commutative.
2. z1(z2z3) = (z1z2)z3, associative.
3. 1 + i0 is the identity element.
4. \(\frac{1}{z}\) is the inverse of z.
5. z1(z2 + z3) = z1z2 + z1z3, distributive law.

Power of ‘i’: i3 = -i, i4 = 1
In general i4k = 1, i4k+1 = i, i4k+2 = -1, i4k+3 = -i

Identities:
Plus One Maths Notes Chapter 5 Complex Numbers and Quadratic Equations 1
Plus One Maths Notes Chapter 5 Complex Numbers and Quadratic Equations 2

The Modulus and Conjugate of a complex number:
Consider a complex number z = a + ib . Then, the conjugate of z is denoted by \(\bar{z}\), defined as \(\bar{z}\) = a – ib and the modulus of z is denoted by |z|, defined as \(\sqrt{a^{2}+b^{2}}\).

Properties:
Plus One Maths Notes Chapter 5 Complex Numbers and Quadratic Equations 3

III. Representation of Complex Number

Argand Plane:
Plus One Maths Notes Chapter 5 Complex Numbers and Quadratic Equations 4
A complex number z = a + ib which corresponds to the ordered pair (a, b) can be represented geometrically as the unique point P(a, b) in the XY-plane, where the real part is taken along the x-axis and the imaginary part along the y-axis. Such a plane is called the Argand Plane or Complex plane.

Polar Form:
Plus One Maths Notes Chapter 5 Complex Numbers and Quadratic Equations 5
Let the point P represent the non-zero complex number z = x + iy. Let the directed line segment OP be of length ‘r’ and be the angle which OP makes with the positive direction of the x-axis. Then, P is determined by the unique ordered pair of a real number (r, θ) called polar coordinate of the point P, where x = r cos θ, y = r sin θ and therefore the polar form of z can be represented as z = r(cos θ + i sin θ).
The principle argument of z is value ‘θ’ such that -x ≤ θ ≤ π, denoted by arg z.
To find the principle argument, we find tan α = |\(\frac{y}{x}\)|, 0 ≤ α ≤ \(\frac{\pi}{2}\)

The quadrant on which ‘P’ liesarg z =
Iα
IIπ – α
IIIα – π
IV
Positive real axis0
Negative real axisπ
Positive imaginary axis\(\frac{\pi}{2}\)
Negative imaginary axis\(-\frac{\pi}{2}\)

Plus One Maths Notes Chapter 4 Principle of Mathematical Induction

Kerala State Board New Syllabus Plus One Maths Notes Chapter 4 Principle of Mathematical Induction.

Kerala Plus One Maths Notes Chapter 4 Principle of Mathematical Induction

Induction means the generalization from a particular case or facts. In contrast to deductive reasoning, inductive depends on working with each case and developing a conjecture by observing incidences till we have observed each and every case. In algebra or in another discipline of mathematics, there are certain results or statements that are formulated in terms of n, where n is a positive integer. To such statements, the well-suited principle that is based on the specific technique is known as the principle of mathematical induction.

The Principle of Mathematical Induction:
Suppose there is a statement P(n) involving the natural number n such that

1. The statement is true for n = 1, i.e; P(1) is true, and

2. If the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1, i.e; the truth of P(k) implies the truth of P(k+1). Then, P(n) is true for all natural numbers n.

Plus One Maths Notes Chapter 3 Trigonometric Functions

Kerala State Board New Syllabus Plus One Maths Notes Chapter 3 Trigonometric Functions.

Kerala Plus One Maths Notes Chapter 3 Trigonometric Functions

I. Angles
The measure of an angle is the amount of rotation performed to get the terminal side from the initial side.

1. Degree measure: If a rotation from the initial side to terminal side is \(\left(\frac{1}{360}\right)^{t h}\) of a revolution, the angle is said to have a measure of one degree, written as 1°. 1° = 60′ and f = 60″.

2. Radian measure: An angle subtended at the center by an arc of length 1 unit in a unit circle is said to be of 1 radian. Radian measure is a real number corresponding to degree measure.

180° = π radians

Radian measure = \(\frac{\pi}{180}\) × Degree measure

Degree measure = \(\frac{180}{\pi}\) × Radian measure

l = rθ, where l = arc length, r = radius of the circle and θ = angle in radian measure.

II. Trigonometric Function
Consider a unit circle with centre at the origin of the coordinate axis.
Let P (a, b) be any point on the circle which makes an angle θ° with the x-axis. Let x be the corresponding radian measure of the angle θ°, i.e; x is the arc length corresponding to θ°.

Plus One Maths Notes Chapter 3 Trigonometric Functions 1

From the ∆OMP’m the figure we get;
sin θ = sin x = \(\frac{b}{1}\) = b and cos θ = cos x = \(\frac{b}{1}\) = a
This means that for each real value of x we get corresponding unique ‘sin’ and ‘cosine’ value which is also real. Hence we can define the six trigonometric functions as follows.

1. f : R → [-1, 1] defined by f(x) = sin x
Plus One Maths Notes Chapter 3 Trigonometric Functions 2

2. f : R → [-1, 1] defined by f(x) = cos x
Plus One Maths Notes Chapter 3 Trigonometric Functions 3

3. f : R – {nπ, n ∈ Z} → R – (-1, 1) defined by f(x) = \(\frac{1}{\sin x}\) = cosec x
Plus One Maths Notes Chapter 3 Trigonometric Functions 4

4. f : R – {(2n + 1) \(\frac{\pi}{2}\)} → R – (-1, 1) defined by f(x) = \(\frac{1}{\cos x}\) = sec x
Plus One Maths Notes Chapter 3 Trigonometric Functions 5

5. f : R – {(2n + 1)π, n ∈ Z} → R defined by f(x) = \(\frac{\sin x}{\cos x}\) = tan x
Plus One Maths Notes Chapter 3 Trigonometric Functions 6

6. f : R – {nπ, n ∈ Z} → R defined by f(x) = \(\frac{\cos x}{\sin x}\) = cot x
Plus One Maths Notes Chapter 3 Trigonometric Functions 7

Sign of trigonometric functions in different quadrants;
Plus One Maths Notes Chapter 3 Trigonometric Functions 8
For odd multiple of \(\frac{\pi}{2}\) trignometric functions changes as given below.
sin → cos
cos → sin
sec → cosec
cosec → sec
tan → cot
cot → tan

The value of trigonometric functions for some specific angles;
Plus One Maths Notes Chapter 3 Trigonometric Functions 9

III. Compound Angle Formula

sin(x + y) = sin x cos y + cos x sin y

sin(x – y) = sin x cos y – cos x sin y

cos(x + y) = cos x cos y – sin x sin y

cos(x – y) = cos x cos y + sin x sin y
Plus One Maths Notes Chapter 3 Trigonometric Functions 10
sin(x + y) sin(x – y) = sin2 x – sin2 y = cos2 x – cos2 y

cos(x + y) cos(x – y) = cos2 x – sin2 y
Plus One Maths Notes Chapter 3 Trigonometric Functions 11

IV. Multiple Angle Formula

cos2x = cos2 x – sin2 x
= 1 – 2sin2 x
= 2 cos2 x – 1
= \(\frac{1-\tan ^{2} x}{1+\tan ^{2} x}\)

Plus One Maths Notes Chapter 3 Trigonometric Functions 12

V. Sub-Multiple Angle Formula
Plus One Maths Notes Chapter 3 Trigonometric Functions 13

Plus One Maths Notes Chapter 3 Trigonometric Functions 14

VI. Sum Formula
Plus One Maths Notes Chapter 3 Trigonometric Functions 16

VII. Product Formula

2 sin x cos y = sin(x + y) + sin(x – y)

2 cos x sin y = sin(x + y) – sin(x – y)

2 cos x cos y = cos(x + y) + cos(x – y)

2 sin x sin y = cos(x – y) – cos(x + y)

VIII. Solution of Trigonometric Equations

sin x = 0 gives x = nπ, where n ∈ Z

cos x = 0 gives x = (2n + 1)π, where n ∈ Z

tanx = 0 gives x = nπ, where n ∈ Z

sin x = sin y ⇒ x = nπ + (-1)n y, where n ∈ Z

cos x = cos y ⇒ x = 2nπ ± y, where n ∈ Z

tan x = tan y ⇒ x = nπ + y, where n ∈ Z

Principal solution is the solution which lies in the interval 0 ≤ x ≤ 2π.

IX. Sine and Cosine formulae

Let ABC be a triangle. By angle A we mean the angle between the sides AB and AC which lies between 0° and 180°. The angles B and C are similarly defined. The sides AB, BC, and CA opposite to the vertices C, A, and B will be denoted by c, a, and b, respectively.

Theorem 1 (sine formula): In any triangle, sides are proportional to the sines of the opposite angles. That is, in a triangle ABC
\(\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\)

Theorem 2 (Cosine formulae): Let A, B and C be angles of a triangle and a, b and c be lengths of sides opposite to angles A, B, and C, respectively, then
a2 = b2 + c2 – 2bc cos A
b2 = c2 + a2 – 2ca cos B
c2 = a2 + b2 – 2ab cos C

A convenient form of the cosine formulae, when angles are to be found are as follows:
Plus One Maths Notes Chapter 3 Trigonometric Functions 15

Plus One Maths Notes Chapter 2 Relations and Functions

Kerala State Board New Syllabus Plus One Maths Notes Chapter 2 Relations and Functions.

Kerala Plus One Maths Notes Chapter 2 Relations and Functions

I. Cartesian Product or Cross Product:
The Cartesian product between two sets A and B is denoted by A × B is the set of all ordered pairs of elements from A and B.
ie; A × B = {(a, b): a ∈ A, b ∈ B}

Properties:

  1. In general A × B ≠ B × A, but if A = B, A × B = B × A.
  2. n(A × B) = n(A) × n(B)
  3. n(A × B) = n(B × A)

II. Relations:
A relation R from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B.

Representation of a relation:

  1. Roster form
  2. Set builder form
  3. Arrow diagram.

Universal relation from A to B is A × B.

Empty relation from A to B is empty set φ.

A relation in A is a subset of A × A.

The number of relation that can be written from A to B if n(A) = p, n(B) = q is 2pq.

Domain: It is the set of all first elements of the ordered pairs in a relation.

Range: It is the set of all second elements of the ordered pairs in a relation.
If R: A → B, then R(R) ⊆ B.

Co-domain: If R: A → B, then Co-domain of R = B.

III. Functions:
A relation f from A to B (f : A → B) is said to be a function if every element of set A has one and only one image in set B.

If f : A → B is a function defined by f(x) = y.

  1. The image of x = y
  2. The pre-image of y = x
  3. Domain of f = {x ∈ A : f(x) ∈ B}
  4. Range of f = {f(x) : x ∈ D(f)}
  5. If f : A → B, then n(f) = n(B)n(A)

IV. Some Important Functions

Identity function: A function f : R → R defined by f(x) = x. Here D(f) = R, R(f) = R.
The graph of the above function is a straight line passing through the origin which makes 45 degrees with the positive direction of the x-axis.

Constant function: A function f : R → R defined by f(x) = c, where c is a constant.
Here D(f ) = R, R(f) = {c}.
The graph of the above function is a straight line parallel to the x-axis.

Polynomial function: A function f : R → R defined by
f(x) = a0 + a1x + ….. + anxn, where n is a no-negative integer and a0, a1, …., an ∈ R.

Rational function: A function f: R → R defined by \(f(x)=\frac{p(x)}{q(x)}\), where p(x), q(x) are functions of x defined in a domain, where q(x) ≠ 0

Modulus function: A function f: R → R
Plus One Maths Notes Chapter 2 Relations and Functions 1
Here D(f) = R, R(f) = [0, ∞).
The graph of the above function is ‘V’ shaped with a corner at the origin.

Signum function: A function f: R → R
Plus One Maths Notes Chapter 2 Relations and Functions 2
Here D(f) = R, R(f) = {-1, 0, 1}.
The graph of the above function has a break at x = 0.

Greatest integer function f: R → R defined by
Plus One Maths Notes Chapter 2 Relations and Functions 3
Here D(f) = R, R(f) = Z.
The graph of the above function has broken at all integral points.

V. Algebra of Functions

Let f : X → R and g : X → R be any two real functions, where X ⊂ R. Then, we define (f + g) : X → R by (f + g)(x) = f(x) + g(x) for all x ∈ X

Let f : X → R and g : X → R be any two real functions, where X ⊂ R. Then, we define (f – g) : X → R by (f – g)(x) = f(x) – g(x) for all x ∈ X

Let f : X → R be a real-valued function and k be a scalar. Then, the product kf : X → R by (kf)(x) = kf (x) for all x ∈ X

Let f : X → R and g : X → R be any two real functions, where X ⊂ R . Then, we define fg : X → R by fg(x) = f(x) × g(x) for all x ∈ X

Let f : X → R and g : X → R be any two real functions, where X ⊂ R. Then, we define \(\frac{f}{g}\) : X → R by
\(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}\) for all x ∈ X

Plus One Maths Notes Chapter 1 Sets

Kerala State Board New Syllabus Plus One Maths Notes Chapter 1 Sets.

Kerala Plus One Maths Notes Chapter 1 Sets

I. Sets
Set is a well-defined collection of distinct objects.
Examples of sets.

  • N: Set of Natural numbers.
  • Z: Set of Integers.
  • Q: Set of Rational numbers.
  • R: Set of Real numbers.
  • Z+: Set of Positive Integers numbers.
  • Q+: Set of Positive Rational numbers.
  • R+: Set of Positive Real numbers.

Representation of Sets:

  1. Roster Form: All elements are listed, are separated by commas, and closed using brackets.
  2. Set-builder Form: All elements of a set possess a single common property which is not possessed by any elements outside the set.
  3. Venn Diagram: Here sets are represented by diagrams. These diagrams consist of rectangles and closed curves usually circles. The universal et is represented by a rectangle and its subsets by circles.

II. Types of Sets:

Empty set: Set contains no element, φ or {}.

Singleton set: Set containing one element.

Finite set: Set containing a definite number of elements.

Infinite set: Set containing an infinite number of elements..

Equivalent set: Sets containing an equal number of elements.

Equal set: Sets containing identical elements.

Subset: If every element of A is an element of B, denoted by A ⊂ B. For any set A, the set A and Empty set is a subset of A. If a set A has n elements, then it has 2n subsets.

Superset: B is a superset if A is a subset of B, denoted by B ⊃ A.

Proper Subset: If A ⊂ B and A ≠ B.

Power set: The set of all subsets of a set A, denoted by P(A). If n(A) = n, then n(P(A)) = 2n

Universal set: The superset of all subsets under discussion.

Intervals as subset of R:

  1. [a, b] = {x : a ≤ x ≤ b}, closed interval.
  2. (a, b] = {x : a < x ≤ b]
  3. [a, b) = {x : a ≤ x < b}
  4. (a, b) = {x : a < x < b}, open interval.

III. Operations on Sets

Union of Sets: The union of A and B is the set which consists of all elements of A and all elements of B except the common elements. In symbol we write as A ∪ B = {x : x ∈ A or x ∈ B}.
Venn diagram representation:
Plus One Maths Notes Chapter 1 Sets 1
Properties:

  1. A ∪ B = B ∪ A, Commutative.
  2. (A ∪ B) ∪ C = A ∪ (B ∪ C), Associative
  3. A ∪ φ = A, φ is the identity.
  4. A ∪ A = A
  5. U ∪ A = U

Intersection of Sets: The intersection of A and B is the set of common elements of both A and B.
In symbol, we write as A ∩ B = {x : x ∈ A and x ∈B}.
Venn diagram representation:
Plus One Maths Notes Chapter 1 Sets 2
Properties:

  1. A ∩ B = B ∩ A, Commutative.
  2. (A ∩ B) ∩ C = A ∩ (B ∩ C), Associative
  3. A ∩φ = φ
  4. A ∩ A = A
  5. U ∩ A = A
  6. A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
  7. n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
  8. If A and B are disjoint, then n(A ∪ B) = n(A) + n(B)
  9. n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C)

Difference of Sets: The difference of the sets A and B in this order is the set of elements which belongs to A but not to B, denoted by A – B = {x : x ∈ A and x ∉ B}
Venn diagram representation:
Plus One Maths Notes Chapter 1 Sets 3
Property: A – B ≠ B – A

Complement of a Set: The complement of a set A is the set of all elements of U which are not in A, denoted by
A’ = {x : x ∈ U and x ∉ A}
Venn diagram representation:
Plus One Maths Notes Chapter 1 Sets 4
Properties:

  1. A’ ∪ A = U, Commutative.
  2. A’ ∩ A = φ, Associative
  3. (A ∩ B)’ = A’ ∪ B’
  4. (A ∪ B)’ = A’ ∩ B’
  5. U’ = φ
  6. φ’ = U
  7. (A)’ = A
  8. A – B = A ∩ B’
  9. n(A – B) = n(A ∩ B’)
  10. n(A) = n(A ∩ B’) + n(A ∩ B)
  11. n(A ∪ B) = n(A ∩ B’) + n(A’ ∩ B) + n(A ∩ B)

Plus Two Botany Chapter Wise Previous Questions Chapter 8 Environmental Issues

Kerala State Board New Syllabus Plus Two Botany Chapter Wise Previous Questions and Answers Chapter 8 Environmental Issues.

Kerala Plus Two Botany Chapter Wise Previous Questions Chapter 8 Environmental Issues

Question 1.
Destruction of forest leads to the increase of CO, in atmosphere. Recently Govt, of India instituted an award for individuals or communities from rural areas that shown extraordinary courage and dedication in protecting wildlife. (MARCH-2010)
a) Identify the award.
b) Comment on deforestation and reforestation.
Answer:
a) Amrita Devi Bishnoi award
b) Deforestation cutting down of forest trees Reforestation replanting trees which are already destroyed.

Question 2.
Industrial effluents and domestic sewage seriously affect fresh water bodies. For protecting aquatic life Govt, of India recently declared an animal as National aquatic animal. (MARCH-2010)
a) Identify the animal.
b) Distinguish biomagnification from eutrophication.
Answer:
a) Dolphin
b) Bio magnification is the increase in the amount of non biodegradable substance in successive trophic level in a food chain.
Eg: DDT Eutrophication is the overgrowth of aquatic plants as a result of accumulation of nutrients in water body.

Question 3.
In 1990s, Delhi ranked 4th among the most polluted cities of the world. But now air quality of Delhi has significantly improved mainly by switching vehicles from diesel to CNG. (MAY-2010)
a) Expand CNG.
b) CNG is betterthan diesel. Comment.
Answer:
a) Compressed Natural Gas
b) CNG is compressed natural gas. It burns more efficiently than Petrol and diesel, thus brings down the amount of pollutants from automobiles (unbumt hydrocarbons).

Question 4.
Greenhouse effect is a naturally occuring phenomenon that is responsible for heating of earth’s surface and atmosphere. (MAY-2010)
a) Explain greenhouse effect.
b) What will happen if there is absolutely no greenhouse effect over earth’s surface ?
Answer:
a) The greenhouse gases such as CO2, N20, Methane present in the atmosphere will reradiate the reflected infrared radiations. Thus raising the temperature of earth.
b) It will affect the climate of earth as the greenhouse effect causes melting of snow in the polar regions. This will in turn determine the level of water in the sea.

Question 5.
Arrange the following words into suitable categories in the given table. (MARCH-2011)
Algal bloom, Catalytic converter, Eichhornia, Electrostatic precipitator
Plus Two Botany Chapter Wise Previous Questions Chapter 8 Environmental Issues 1
Answer:
Plus Two Botany Chapter Wise Previous Questions Chapter 8 Environmental Issues 2

Question 6.
Geetha resides in a city nearby a lake. Water from this lake was used for various domestic purposes earlier. Now-a-days. this water has become turbid and is with an unpleasant odour. (MARCH-2011)
a) What can be the reason for this ?
b) Name the scientific term that explains this effect.
Answer:
a) Nutrient enrichment
b) Eutrophication or Aging of lake

Question 7.
As Head of the Vehicle Department, issue a notice to vehicle owners to observe any two measures to reduce vehicular air pollution and record the merits and demerits of CNG. (MAY-2011)
Answer:
Use of catalytic converter.
Use of lead-free petrol or diesel.
Merits of CNG
1) It burns most efficiently,
2) It is cheaper than petrol or diesel, cannot be siphoned off by thieves and adulterated like petrol or diesel.
Demerits of CNG
It is difficult of laying down pipelines to deliver CNG through distribution points/pumps and ensuring uninterrupted supply.

Question 8.
increase in green house gases. Name two green house gases. (MAY-2011)
Answer:
Green house gases – CO2 and CFC

Question 9.
The increased use of chemicals like CFCs (Chloro fluro carbons) cause adverse ecological impacts. Why CFCs are considered harmful to the environment? Mil It causes the depletion of ozone layer. (MARCH-2012)
Answer:
CFC is considered as green house gas.lt causes global warming.
The incoming UV rays cuases many diseases like skin cancer,snow blindness etc.

Question 10.
Meena an environmental activist, noticed a gradual decline in the’population of birds in the open agricultural fields near her place. She has heard of the excessive use of pesticides like DDT around that area. (MARCH-2012)
a) What might have led to the decline of bird population in that area?
b) Name the process that has caused this phenomenon.
Answer:
a) It disturbs calcium metabolism in birds due to accumulation of DDT and causes the thinning of egg shell. lt affects the premature breaking of egg.
b) Biomagnification.

Question 11.
Ammu read in the newspaper that, BOD of a water body in a nearby village was high and there is algal bloom. (MAY-2012)
a) What is BOD ?
b) What is algal bloom ?
c) Can you give possible reason for these phenomenon?
Answer:
a) Biological oxygen demand
b) Inceased algal population
c) It is due to the accumulation of inorganic nutrients like phosphate and nitrate in water body

Question 12.
Enviornmentalists usually says: There are many causes for biodiversity losses’. Illustrate four major causes of biodiversity loss. (MARCH-2013)
Answer:
i) Habitat loss and fragmentation
ii) Over-exploitation
iii) Alien species invasions
iv) Co-extinctions

Question 13.
In a study conducted, the concentration of DDT was found to increase in the successive trophic levels, The results of the study is shown below: (MAY-2013)
Plus Two Botany Chapter Wise Previous Questions Chapter 8 Environmental Issues 3
Answer:
Biomagnification
High concentrations of DDT disturb calcium metabolism in birds, which causes thinning of eggshell and their premature breaking, eventually causing decline in bird populations.

Question 14.
An article in the newspaper reports that ‘Refirigerants like Chlorofluorocarbons (CFCs) pose threat to the environment’. How CFCs are harmful to the environment? (MARCH-2014)
Answer:
CFC (CCI2F2) splits in the presence of UV and release active chlorine. This active chlorine breaks ozone molecule into O2 and (O) .It causes the thinning of stratospheric good ozone and harmful UV rays reaches earth surface.

Question 15.
Now a days many farmers are interested in organic farming. What is meant by organic farming? Can you suggest any two advantages of organic farming?  (MARCH-2014)
Answer:
It is the cyclical, sustainable and zero-waste procedure.
Advantageous
1) No need of chemical fertilizers.
2) No need of insecticides and pesticides.
3) Never kills microorganisms in the soil.

Question 16.
An aquatic ecosystem having luxurious growth of cyanobacteria (Algal bloom) leads to eutrophication.  (MARCH-2015)
a) What kind of pollutants cause algal bloom to colonize the aquatic ecosystem?
b) What are the consequences of eutrophication?
Answer:
a) Waste water contains contains large quatities of nutrients.
b) It causes ageing of lake and damage to indigenous flora and fauna.

Question 17.
During the past century, the temperature of the earth has increased by 0.6°C, most of it during the last few decades. Rise in temperature causes deleterious changes in the environment, thus leading to increased melting of polar ice caps as well as other places like Himalayan snow caps. Suggest any two control measures that will reduce global warming?  (MARCH-2015)
Answer:
1) Cutting down use of fossil fuel
2) limproving efficiency of energy usage
3) Reducing deforestation
4) Planting trees and slowing down the growth of human population.

Question 18.
Observe the diagram and answer the following:  (MAY-2015)
Plus Two Botany Chapter Wise Previous Questions Chapter 8 Environmental Issues 4
a) Suggest the reasons for the presence of DDT in the water.
b) Fish eating birds of this area have higher DDT concentration in their body. Justify.
c) What will be the impact of DDT in the birds?
OR
United Nations Framework convention on climate change, an international treaty signed by 194 countries to cooperatively discuss global climate change and its impact.
As a science student,
a) What is global warming?
b) Explain the reasons and give suggestions to control global warming?
Answer:
a) Use of pesticide in agricultural field
b) DDT is accumulated in successive trophic level and finally its concentration is very high fish eating birds.
c) DDT affect calcium metabolism, that causes thinning of eggshell and premature breaking of egg. Hence the bird population is decreased.
OR
a) Increasing the temperature of earth surface due to green house effect is called global warming
b) The incoming radiations of sunlight reaches the earth’s surface re-emits heat in the form of infrared radiation but part of this does not escape into space as atmospheric gases (e.g., carbon dioxide, methane,CFCand nitrogen oxides) absorb a major fraction of it. This cycle is repeated many a times and temperature of earth increases it leads to global warming.
It is reduced by

  1. cutting down use of fossil fuel.
  2.  improving efficiency of energy usage.
  3. reducing deforestation.
  4. planting trees and slowing down the growth of human population.

Question 19.
Increase in the concentration of toxicants at successive trophic level is called _____.  (MARCH-2016)
a) BOD
b) Biomagnification
c) Eutrophication
d) Algal Bloom
Answer:
b) Biomagnification

Question 20.
The major pollution in the environment is caused by automobiles. Expand the term CNG. Mention any two of its merits. (MARCH-2016)
Answer:
CNG – compressed natural gas It burn efficiently
It is cheaper than diesel and petrol

Question 21.
Temperature is generally increasing making the earth a hot plate. Mention any two measures to control global warming. (MARCH-2016)
Answer:
Cutting down the use of fossil fuel
Improving the efficiency of energy usage

Question 22.
Quantity of pollutants increase in successive trophic levels. Observe the flowchart regarding biomagnifications of DDT in an aquatic food chain and answerthe following: (MAY-2016)
a) What is biomagnification?
b) What are the consequences of biomagnification?
Plus Two Botany Chapter Wise Previous Questions Chapter 8 Environmental Issues 5
Answer:
a) Non bio degradable chemicals which accumulate in the body of organism and is passed on to the organisms belonging to the next trophic level, the increase in the concentration of toxicants at successive trophic level and finally concentration become high in last trophic level of the food chain.
b) High concentrations of DDT disturb calcium metabolism in birds, which causes the thinning of eggshell and their premature breaking, causing decrease in bird populations.

Question 23.
Adequate waste management is an environment issue to be considered. Discuss the advantages of Eco-san toilet. (MAY-2016)
Answer:
it is a sustainable system for handling human excreta, using dry composting toilets.
This is a practical, hygienic, efficient and cost- effective solution to human waste disposal.

Question 24.
A common cause of deforestation is slash and burn agriculture. (MARCH-2017)
a) What is the common name attributed to such type of cultivation?
b) Explain how this type of cultivation is practised?
Answer:
a) phytoplankton stage
b) Submerged plant stage
c) Submerged free floating plant Stage
d) Reed swamp stage
f) Marsh – meadow stage
g) Scrub stage
h) Forest stage

Question 25.
Particulate matter in polluted air is removed by the application of electrostatic precipitator. Explain the working principle of lectrostatic precipitator. (MARCH-2017)
Answer:
At high voltage the electrons produced in an instrument are attached to dust particles giving them a net negative charge. These charged dust particles are attracted by collecting plates. Then reducing the velocity of air between the plates which help the dust to fall. Electrostatic precipitator which can remove over 99 per cent particulate matter present in the exhaust from a thermal power plant.

Question 26.
Among the following which one Is used for reducing the emission of poisonous gases from automobiles (MAY-2017)
a) Landfills
b) Catalytic converter
c) Electrostatic precipitator
d) Earmuffs
Answer:
Catalytic converter

Question 27.
Nutrient enrichment in a fresh water lake leads to (MAY-2017)
eutrophication.
a) What happens during eutrophication?
b) How dissolved oxygen level is affected as a result of this?
Answer:
a) Due to accumulation of nitrate and phosphate into the lake, leads the rapid growth of algae, that changes the colour and quality of water.
It depletes the oxygen content of water and finally causes the death of aquatic organs.
b) Due the death and decay of algae, dissolved oxygen content is decreased in lake that causes the mortality of fish and other aquatic organs.

Plus Two Botany Chapter Wise Previous Questions Chapter 7 Ecosystem

Kerala State Board New Syllabus Plus Two Botany Chapter Wise Previous Questions and Answers Chapter 7 Ecosystem.

Kerala Plus Two Botany Chapter Wise Previous Questions Chapter 7 Ecosystem

Question 1.
During a study tour teacher showed the primary colonisers on the banks of the river ‘Nila’.(MARCH-2010)
a) Identify the succession and justify your answer.
b) List the different stages of the identified succession.
Answer:
Succession taking place in water is called Hydrarch.
b) Phytoplankton stage -> Submerged plant stage -> Submerged free floating plant stage -> Reed swamp stage -> Marsh meadow -> Scrub stage -> Forest stage

Question 2.
While learning trophic levels in class-room, teacher asked you to explain ‘standing crop’ to Raman. Explain. (MARCH-2010)
Answer:
Standing crop – The mass of living material present in each trophic level.

Question 3.
Pond is a self-sustainable unit. Some organisms related to pond ecosystem is listed below, tadpole, fish, water, plants, kingfisher. (MAY-2010)
a) Construct a food chain with the listed organisms.
b) Explain trophic level.
c) Point out trophic level of each organism in the constructed food chain.
d) Name interconnection of food chains in nature.
OR
Two students, Unni and Kannan studied inter specific interactions between different species. They made a table assigning a ‘+’ for beneficial interaction, for detrimental and ‘0’ for neutral interaction. Can you help them by naming the interaction between species in different cases ? Write one example for each interaction.

Case numberSpecies ASpecies B
1.++
2.
3.+
4.+0

Answer:
Water plants Tadpole -> Fish -> Kingfisher
b) Specific place of an organism in the food chain based on the source of nutrition or food.
c)
Plus Two Botany Chapter Wise Previous Questions Chapter 7 Ecosystem 1
d) Food web
OR
1-Mutualism Eg; Lichen
2-Competition
Eg: Competition for food (Phytoplankton) between Flamingoes and fishes
3 – Predation Eg: Lion and Deer
4 – Commensalism
Eg: Epiphyte, Vanda and Mango tree

Question 4.
Consider pond as an ecosystem showing the number of individuals in the following categories. (MARCH-2011)
Carnivores-2500, Producers-15000, Herbrivores- 5000
a) Draw the pyramid of numbers in this ecosystem.
b) Comment on the energy flow in the ecosystem
Answer:
Plus Two Botany Chapter Wise Previous Questions Chapter 7 Ecosystem 2
b) Unidirectional flow of energy from producers to consumers. It decreases in successive trophic levels

Question 5.
The gradual change in the species composition of a given area leading to the formation of climax community is called ecological succession. In a rocky area, (MARCH-2011)
a) What is the expected type of pioneer species?
b) How this pioneer species leads to the establishment of a stable climax community?(2 Scores)
Answer:
a) lichens
b) lichens -> mosses -> herbs -> shrubs ->forest.
These are different stages and leads to stable climax community.

Question 6.
Fill up the blanks with appropriate terms in the given pyramid of trophic level: (MAY-2011)
Plus Two Botany Chapter Wise Previous Questions Chapter 7 Ecosystem 3
Answer:
b – Primary consumer a- secondary consumer

Question 7.
Kalyani wrote man, hen, earthworm, mango-tree in her note book. Arrange the terms in a food chain sequence. Explain food chain and name the types of food chain. (MAY-2011)
Answer:
Earth worm—> Hen —> Man —> Mango tree
The transfer of food from the producers through a series of organisms with repeated eating and being eaten is referred to as Food chain Detritus food chain

Question 8.
In a marine ecosystem, a population of phytoplankton (150,000) supports a standing crop of fishes (40,000). (MARCH-2012)
a) Draw the pyramid of biomass and
b) The pyramid of numbers in this ecosystem,
Answer:
a) Inverted Pyramid
Plus Two Botany Chapter Wise Previous Questions Chapter 7 Ecosystem 4
b) Upright Pyramid
Plus Two Botany Chapter Wise Previous Questions Chapter 7 Ecosystem 5

Question 9.
The gradual and fairly predictable changes in the species composition in an area is called ecological succession. (MARCH-2012)
a) Name the pioneer species in the primary succession in water.
b) Give the sequence of events and climax community in the hydrarch succession.
Answer:
a) Phytoplankton
b) Phytoplankton -> submerged free floating -> Reed swamp marsh meadow -> scrub -> Forest.

Question 10.
Given number of individuals in a grassland ecosystem. (MAY-2012)
Grasshopper – 1500
Grass – 5,842,000
Wolf – 28
Birds – 215
a) Draw a pyramid of numbers showing various trophic levels.
b) Explain trophic level.
Answer:
Plus Two Botany Chapter Wise Previous Questions Chapter 7 Ecosystem 6
b) Each step in food chain is called trophic level

Question 11.
Rate of biomass production is called productivity and can be divided into GPP and NPP : (MAY-2012)
a) Define GPP and NPP.
b) How can we relate GPP and NPP?
Answer:
a) GPP – Gross primary productivity NPP – Net primary productivity
b) NPP = GPP-R

Question 12.
Final community that is in near equilibrium with environment in ecological succession is called ________. (MARCH-2013)
Answer:
Climax community

Question 13.
Natural interlinked food chains are called _______. (MARCH-2013)
Answer:
Food web

Question 14.
A list of organisms are given. Place them in different trophic levels. Grass, Man, Fishes, Birds, Lion, Grasshopper, Zooplankton, Trees. (MARCH-2013)
Answer:
First trophic level – trees, grass
Second trophic level – grasshopper, zooplankton
Third trophic level – birds, fishes
Fourth trophic level – lion, man

Question 15.
In the equation, GPP-R = NPP; If NPP = Net primary productivity. (MAY-2013)
Explain GPP – R = NPP.
Answer:
Gross primary productivity minus respiration losses (R) is the net primary productivity (NPP).
That means Net primary productivity is the available biomass for the consumption of heterotrophs (herbivores and decomposers).

Question 16.
The teacher, pointing to a forest said “ Long back, this place was a pond’’. This gradual change is an example of (MAY-2013)
1) Secondary succession
2) Xerarch succession
3) Pioneer species
4) Hydrarch succession
Answer:
Hydrarch succession

Question 17.
The species that invade a nude area are called ________ species. In a primary succession on rocks, the group that invade first are usually. (MARCH-2014)
Answer:
Pioneer species, Lichens

Question 18.
The rate of biomass production in an ecosystem is called productivity. They are of two types, gross primary productivity and net primary productivity, how these two productivities are related? (MARCH-2014)
Answer:
Gross primary productivity of an ecosystem is the rate of production of organic matterduring photosynthesis.
Gross primary productivity minus respiration losses (R), is the net primary productivity (NPP).
i.e: GPP – R = NPP

Question 19.
A list of different organisms in an ecosystem are given below. (MARCH-2014)
Arrange them in 1st, 2nd, 3rd and 4th trophic level,
i) Phytoplankton
ii) Man
iii) Fish
iv) Zooplankton
Answer:
Plus Two Botany Chapter Wise Previous Questions Chapter 7 Ecosystem 7

Question 20.
By observing the relationship of the first pair fill up the blanks: (MAY-2014)
a) Grazing food chain consists of producers and consumers whereas Detritus chain comprises dead organic matter and ______.
b) Nitrogen : Gaseous cycle
Sulphur: _______
Answer:
a) Detrivores
b) Sedimentary cycle

Question 21.
Field survey by a team of students recorded the following data related to biomass of the organisms in each tropic level of an ecosystem. Draw, name and explain the pyramid (MAY-2014)
Answer:
Plus Two Botany Chapter Wise Previous Questions Chapter 7 Ecosystem 8
Biomass increases in successive trophic levels hence the pyramid given here is inverted.

Question 22.
Gradual, sequential changes of a given area including species composition is known as ecological succession. If so name the first two stages of the succession in hydricarea. (MAY-2014)
Answer:
1) phytoplankton stage
2) submerged plant stage

Question 23.
Primary succession on rocks is known as Xerosere. Answer the following related with Xerosere. (MARCH-2015)
a) Name the pioneer community.
b) Organic acids have important roles in this succession. Justify.
Answer:
a) Lichen
b) It helps in weathering of rocks and soil formation

Question 24.
By observing the relationship of the first pair fills up the blanks. (MAY-2015)
a) Net primary productivity =
Gross primary productivity — Respiration.
Gross primary productivity is ________.
b) Carbon: Gaseous cycle
Phosphorus: ______.
Answer:
a) total organic matter
b) sedimentary cycle

Question 25.
Field survey by a team of students recorded the following data related to number of organisms in an ecosystem and plotted that into a figure shown below: (MAY-2015)
Plus Two Botany Chapter Wise Previous Questions Chapter 7 Ecosystem 9
Observe the figure and explain the pyramid.
Answer:
It is the inverted pyramid . In this number of organism increases in the successive trophic levels.

Question 26.
Hydrosere succession stages are given below. Arrange them in order. (MAY-2015)
Scrub stage — forest —submerged free floating — Marsh Meadow —Submerged stage —Reed swamp—Phytopiankton.
Answer:
Phytoplankton- submerged-submerged free floating- reed swamp -Marsh meadow -srub stage – forest.

Question 27.
Nutrients are never lost from the ecosystems and are recycled. Write about. (MARCH-2016)
a) Gaseous cycle
b) Sedimentary cycle
Answer:
a) Gaseous cycle- The reservoir of gaseous type nutrients are present in atmosphere
b) Sedimentary cycle- The reservoir of nutrients are present in earth crust

Question 28.
Ecological pyramids are usually upright. Meanwhile some, pyramid of biomass is inverted. Explain the reason. (MARCH-2016)
Answer:
Pyramid of biomass of sea is inverted because biomass of fishes far exceeds phytoplankton

Question 29.
a) Biogeochemical cycle is an important phenomenon in very ecosystem. Describe phosphorus cycle. (MAY-2016)
OR
b) The plant communities in a given area show successive changes. Mention the stages of succession in a xerosere.
Answer:
a) The cycle consists of following steps.
Plus Two Botany Chapter Wise Previous Questions Chapter 7 Ecosystem 10
OR
b) In xerach succession first plant communities appear in bare area are lichens,They secrete
carbonic acid and dissolve rocks. This process forms soil that help in the growth of mosses. Then herbs,shrubs and forest stage appears .this process takes thousands of years and form climax community.

Question 30.
Earthworms are commonly referred as farmers’ friends. Define fragmentation. (MAY-2016)
Answer:
During fragmentation large detritus (dead remains of plants & animals) is converted into into smaller particle by the detritrivores like earthworm, its surface area increases that helps in process, of decomposition.

Question 31.
The different stages of primary succession in water are represented below. Fill the gaps that are unfilled. (MARCH-2017)
a) Phytoplankton
b) _______
c) Submerged free floating plant stage
d) _______
e)________
f) Shrub stage
g) _______
Answer:
a) phytoplankton stage.
b) Submerged plant stage.
c) Submerged free floating plant Stage
d) Reed swamp stage.
f) Marsh – meadow stage
g) Scrub stage.
h) Forest stage

Question 32.
An ecosystem consist of the following population: (MARCH-2017)
Phytoplankton
Man
Fish
Zooplankton
Draw a food chain denoting each trophic level
Answer:
Plus Two Botany Chapter Wise Previous Questions Chapter 7 Ecosystem 11

Question 33.
The natural reservoir of phosphorous is rock where it is present in the form of phosphates. How this phosphorous is cycled in ecosystem? (MAY-2017)
Answer:
The reservoir of phosphorus is rock. During weathering process orthophosphates reaches the soil solution. It is taken by plants, that passes through food chain and reaches animals. After the death and decay, the organic form of phosphorus is coming into the soil solution as orthophosphates by the activity of phosphate solubilising bacteria.

Question 34.
Birds represent members in a food chain. (MAY-2017)
Draw a food chain representing each of the above in different tropic levels.
Answer:
Plus Two Botany Chapter Wise Previous Questions Chapter 7 Ecosystem 12

Plus Two Botany Chapter Wise Previous Questions Chapter 6 Organisms and Populations

Kerala State Board New Syllabus Plus Two Botany Chapter Wise Previous Questions and Answers Chapter 6 Organisms and Populations.

Kerala Plus Two Botany Chapter Wise Previous Questions Chapter 6 Organisms and Populations

Question 1.
Given below is a table which shows the interspecific interaction.’+’ sign indicates beneficial,sign indicates detrimental and ‘0’ indicates neutral. (MARCH-2010)
a) Fill in the blanks.

Species ASpecies BInteraction
Competiton
0…………….
+0……………

b) Name the interactions where one species is benefited and the other is detrimental.
Answer:
Amensalism
Commensalism

Question 2.
Small bottle labelled with rDNA insulin. (MARCH-2010)
a) Does it a natural insulin ?
b) Identify the major steps involved in this rDNA insulin production.
Answer:
No.
1. Isolation of desired genes.
2. Insertion of desired Genes into plasmids of E. coli.
3. Introduction of plasmids into E coli cells.
4. Culture of E coli cells.
5. After this, a polypeptide chains A and B are separated and connected together by disulphide linkages.
Thus, genetically engineered insulin is prepared
Plus Two Botany Chapter Wise Previous Questions Chapter 6 Organisms and Populations 1

Question 3.
Given below a schematic representation with circles and squares, which shows four factors/processes influence the population density. (MARCH-2010)
Write the positive factors in circles and negative factors in squares.
Plus Two Botany Chapter Wise Previous Questions Chapter 6 Organisms and Populations 2
Answer:
Plus Two Botany Chapter Wise Previous Questions Chapter 6 Organisms and Populations 3

Question 4.
Snakes change their body temperature with changes in external temperature, but human beings not. Organism may be classed according to above character with explanation. (MAY-2010)
Answer:
Temperature has a significant role in the kinetics of enzymes and thus influence the metabotic activities and physiological functions. Accordingly organisms can be classified into eurythermal (tolerate wide range of temperature) and Stenothermal (restricted to narrow range of temperature)

Question 5.
Density of a population in a given habitat during a given period, fluctuates due to changes in 4 basic processes – Natality, Mortality, Immigration & Emigration. (MAY-2010)
a) Differentiate Natality and Mortality.
b) Differentiate Immigration and Emigration.
Answer:
a) Mortality is the number of death of a population at a given period and Natality is the number of birth during a given period.
b) Immigration is the number of individuals of the same species that have come into the habitat from elsewhere during the time period. Emigration is the number of individuals of the population who left the habitat and gone elsewhere during the time period.

Question 6.
Plus Two Botany Chapter Wise Previous Questions Chapter 6 Organisms and Populations 4
Given above is the bar diagram showing age structure of three different populations. Observe the diagram carefully and answer the following questions. (MARCH-2011)
a) Select the stable population.
b) Compare the nature of population growth in A,B, and C

Answer:
a) ‘B’Stable population
b)

ABC
Expanding Positive growthStable Zero growthDeclining negative growth

Question 7.
By observing the relationship of the first, fill in the blanks. (MARCH-2011)
a) Unisexual male flower-Staminate ______
Unisexual female flower ______
b) Organisms tolerating a wide range of temperature – eurythermal
Organisms tolerating a narrow range of temperature ______
Answer:
a) pistillate
b) stenothermal

Question 8.
Population interactions: (MAY-2011)

CaseSpecies xSpecies ySpecies z
1++0
20+
3+

Where’+’ beneficial interaction detrimental interaction ‘0’ neutral interaction.
Observe the interactions of populations of 3 species as shown in the table. Name the interactions
a) Species x and species y in case 1.
b) Species y and species z in case 2.
c) Species x and species z in case 3.
d) Species y and species z in case 1
Answer:
a) Mutualism
b) Predation/Parasitism
c) Competition
d) Commensalism

Question 9.
Mohammed and his family left to Dubai from Kozhikode on March, 2009. In Kozhikode they are referred as after 2009. How it affects Kozhikode population? (MAY-2011)
Answer:
Emigrants, Decrease the size of population

Question 10.
Inter specific interaction from the interaction of populations of two different species. If we assign + for beneficial, — for detrimental and 0 for neutral interactions, copy and complete the following chart. (MARCH-2012)

Species ASpecies BName of interaction
……………………………….Mutualism
…………..
…………….………………….Commensalism
…………….…………………….Amensalism
+………………

Answer:

Species ASpecies BName of interaction
++
Competition
+0
0
Parasitism /predation

Question 11.
Prakash parked his car in bright sunlight for a few hours, with glass windows fully raised. After sometime inside of the car was very hot. (MAY-2012)
a) Name the phenomenon.
b) How can you correlate this phenomenon with global warming?
Answer:
a) Green house effect
b) The green house effect is due to various green house gases, of which the percentage of carbon dioxide is very high. This causes the increase of temperature on earth called Global warming.

Question 12.
Students involved in nature club activity found some interspecific interactions between organisms in a garden area. They made a table of interaction giving’+’ for beneficial interaction,for detrimental and ‘O’ for neutral interaction. (MAY-2012)

SnoSpecies ASpecies B
i.++
ii.
iii.+0
Iv0

a) Give name of interaction in each case.
b) Explain how parasitism differ from predation.
c) Give the significance of species interaction.
Answer:
i – Mutualism
ii – competition
iii – commensalism
iv- amensalism
b) In parasitism, parasite absorb nutrients from the living host In predation, predator kills and eat the prey.
c) Species interaction is beneficial, detrimental or neutral (neither harm nor benefit) to one of the species or both.

Question 13.
Read the statements below and identify the mode of interaction between the species. (MARCH-2013)
a) Tiger eating deer
b) Butterfly feeding pollen
c) Human liver fluke feed on snail
d) Lice on humans
e) Orchid attached to a tree
f) Mycorrhizal association of fungi and roots of higher plants.
g) Sparrow eating seed
h) Egrets foraging close to cattle
Answer:
a) predation
b) mutualism
c) predation or parasitism
d) parasitism
e) commensalism
f) mutualism
g) predation
h) commensalism

Question 14.
In summer we use air conditioners and in winter we use heaters. Here homeostasis is accomplished by artificial means. Explain four ways by which other living organisms cope with the situation. (MARCH-2013)
Answer:
Hibernation – winter sleeping
Aestivation – Summer sleeping
Migration – Moving into more suitable area
Diapause – inactive in adverse condition

Question 15.
Many desert plant have adaptations to prevent loss of water from their body. Mention any two adaptations to minimise water loss from plant body. (MAY-2013)
Answer:
a) Many desert plants have a thick cuticle on their leaf surfaces and have their stomata arranged in deep pits to minimise water loss through transpiration.
b) They also have a special photosynthetic pathway (CAM) that enables their stomata to remain closed during daytime.

Question 16.
The size of a population is not static. Which of the following leads to decrease in population? (MAY-2013)
1) Natality and Mortality
2) Mortality and Emigration
3) Mortality and immigration
4) Natality and Immigration
Answer:
Mortality and Emigration

Question 17.
Some type of Orchids live on the branches of Mango trees. The relationship between mango tree and Orchid is an example of. (MAY-2013)
1) Mutualism
2) Predation
3) Commensalism
4) Parasitism
Answer:
Commensalism

Question 18.
The density of population in a given habitat increase or decrease due to different reasons. Name two factors responsible for increase in population in a given area. (MARCH-2014)
Answer:
Natality and Immigration

Question 19.
Observe the diagram:(MAY-2014)
Plus Two Botany Chapter Wise Previous Questions Chapter 6 Organisms and Populations 5
Define the following terms:
a) Natality
b) Mortality
c) Emigration
d) Immigration
Answer:
a) Birth rate or total number of live births per 1,000 of a population in a year.
b) Death rate or total number of death per 1,000 of a population in a year.
c) Emigration- movement of individuals out of the population
d) Immigration- movement of individuals into the population

Question 20.
Response of organisms to abiotic stress involves different methods. Explain any two such responses with suitable examples. (MAY-2014)
Answer:
i) Regulate: Organisms are able to maintain constant body temperature and constant osmotic Concentration. Eg- birds and mammals
ii) Conform: Organisms cannot maintain a constant internal environment.
Eg- Majority (99 percent) of animals and all plants.

Question 21.
Suckerfish and shark live in close association, is a classic example of commensalism. What is commensalism? (MARCH-2015)
Answer:
Commensalism -This is the interaction in which one species benefits and the other is neither harmed nor benefited.

Question 22.
Desert plants like Opuntia are able to grow in extreme conditions. Suggest any two adaptations of this plant. (MARCH-2015)
Answer:
1) They have a thick cuticle on their leaf surfaces
and have their stomata arranged in deep pits to minimise water loss through transpiration.
2) They also have a special photosynthetic pathway (CAM) that enables their stomata to remain closed during day time
3) The photosynthetic function is carried out by the flattened stems.

Question 23.
With regard to population growth rate, when responses are limiting the plit is logistic. Verhulst-Pearl Ligstic growth is represented by the equation. (MARCH-2015)
\(\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=\mathrm{rN} \frac{(\mathrm{K}-\mathrm{N})}{\mathrm{K}}\) what,are
a) r
b) K
Answer:
a) r-lntrinsic rate of natural increase or( b-d)
b) k-Carrying capacity

Question 24.
Observe the equation (MAY-2015)
\(\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=\mathrm{rN} \frac{(\mathrm{K}-\mathrm{N})}{\mathrm{K}}\)
a) Which type of growth curve does it represents?
b) What do the following notations represent:
a) N b) r c) K
Answer:
a) Logistic growth
b) N-population of size
b) r- intrinsic rate of natural increase
c) K-Carrying capacity

Question 25.
On earth, life exists even in extreme and harsh conditions. Mention any two major biomes in India. (MARCH-2016)
Answer:
Tropical deciduous forest
Rain forest

Question 26.
a) Population interactions may be beneficial or not. Write any three interactions in detail. (MARCH-2016)
OR
b) Organism are influenced by biotic and abiotic factors. Write an account of any three abiotic environmental factors.
Answer:
a) Mutualism- in this both partners are benefitted eg lichen (+,+)
Commensalism- In this one partner is benefitted other partner is neither benefitted nor harmed (-, +) Competition- In this both partners have detrimental effect or negative effect (-, -)
OR
b) Temperature- it affect the enzyme kinetics of reaction. Enzyme works at optimum temperature
Water- it affect productivity and distribution of plants in aquatic ecosystem.
Light- It influence the photoperiodic flowering of plants

Question 27.
Population growth may be exponential or logistic. Differentiate between them. (MAY-2016)
Answer:
When the resources in the habitat are unlimited, each species has the ability to grow in number. Here the population grows in an exponential or geometric fashion. dN/dt = rN
Limited resources leads to competition between individuals and the ‘fittest’ individual will survive and reproduce. This is called logistic growth
\(\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=\mathrm{rN} \frac{(\mathrm{K}-\mathrm{N})}{\mathrm{K}}\)

Question 28.
Plants are adapted to grow in different habitats. Name any four adaptations of plants in desert habitat. (MAY-2016)
Answer:
Desert plants have a thick cuticle on their leaf surfaces and stomata arranged in deep pits to minimise water loss through transpiration.
They also have CAM pathway in which they open stomata during night and closed during day time.

Question 29.
In a given habitat, the maximum number possible for a species is called _________ of that species in that habitat. (MARCH-2017)
Answer:
Carrying capacity (K)

Question 30.
Different types of population interaction has been observed in a population. (MARCH-2017)
Write the types of interaction observed among the following species:

Species ASpecies BType of interaction
Orchid ophrysBees——————
CattleCattle egret——————
Sea anemoneClown fish——————-
TicksDogs——————–
CuscutaHedge plant——————-
TigerDeer——————–

OR
B) Organisms other than human beings manage or adapt to stressful conditions by adopting different mechanisms. Explain any three mechanisms adopted by them to maintain the internal environment.
Answer:
A) Mutualism or pseudocopulation Commensalism Commensalism Parasitism Parasitism Predation
OR
B) 1) Conform : About 99% of animals and all plants cannot maintain a constant internal environment according to the external environment. They change their body temperature and osmotic concentration of body fluid when external environment changes.
2) Migrate : Some organisms move away temporarily from the stressful habitat to a more hospitable area and return when stressful period is over.
3) Suspend : Some organisms like bacteria, fungi and lower plants produce thick walled spores to tide over unfavourable conditions.
Some organism avoid the stress by escaping in time by method of hibernation during winter (eg. polar bear) or aestivation to avoid summer related problem (eg: snails and shells).

Question 31.
There are four mechanism by which living organisms other than human beings maintain the constancy of internal environment. Name these processes. (MAY-2017)
Answer:
Organisms maintain internal environment as constant by sweating shivering deposition of fat layer below skin and hairy covering on body surface.

Question 32.
Adaptations are the attributes of the organism that enables it to survive and reproduce in its habitat. Give the adaptations of  (MARCH-2014)
a) Cactus plant in desert
b) Kangaroo rat in desert
c) Seals in polar region.
Answer:
a) Desert plants have a thick cuticle on their leaf surfaces and have theirstomata arranged in deep pits to minimise water loss through transpiration
b) kangaroo rat in North American deserts is capable of meeting all its water requirements through its internal fat oxidation
c) In the polar seas aquatic mammals like seals have a thick layer of fat (blubber) below their skin that acts as an insulator and reduces loss of body heat.

Plus Two Botany Chapter Wise Previous Questions Chapter 5 Biotechnology and its Applications

Kerala State Board New Syllabus Plus Two Botany Chapter Wise Previous Questions and Answers Chapter 5 Biotechnology and its Applications.

Kerala Plus Two Botany Chapter Wise Previous Questions Chapter 5 Biotechnology and its Applications

Question 1.
Teacher asked Balan to write the principle of RNAi technology. Help him by describing the method.(MARCH-2010)
Answer:
The silencing of mRNA by a double stranded RNA (dsRNA) is called RNA interference.

Question 2.
Bt. Cotton is a well known example of application of Biotechnology in Agriculture. Bt. Cotton reduces use of pesticides. Explain. (MAY-2010)
Answer:
Bt. Cotton is a transgenic plant which contains the cry gene of bacillus thuringiensis. The cry gene produces cry protein which kills larvae of lepedopteran, diptern insects which attack cotton plants. This protein is inactive in bacteria and becomes active in the intestine of insects. When the cry gene is transferred to the plant it becomes resistant to the insects.

Question 3.
Adenosine deaminase (ADA) deficiency is a hereditary disease, where ADA, which is crucial for functioning of immune system is absent. Explain how ADA deficiency can be treated. (MAY-2010)
Answer:
Lymphocytes are taken from patient’s blood. It is grown invitro culture. By using retroviral vector functional ADA c DNA (from WBC of normal person) is introduced into invitro cultured lymphocytes of patients. These genetically engineered lymphocytes (corrected functional lymphocytes) are re injected to patients. If the ADA producing genes are introduce into cells at early embryonic stage it could be a permanent cure.

Question 4.
Bt toxin is produced by Bacillus thurungiensis that can kill certain insects. (MARCH-2011)
a) Name the bacterical gene that is producing this toxin.
b) Why the toxin produced by the bacterium is nontoxic to it?
Answer:
a) Cry-gene
b) Bt. toxin protein exist as inactive ie. protoxin in bacteria

Question 5.
Raju is a diabetic parient who takes insulin injections regulary. The insulin used by such patients is producted by genetically engineered organisms . Write the different steps involved in the production of insulin by genetic engineering. (MARCH-2011)
Answer:
1) Preparation of DNA sequences corresponding to A&B chain of human insulin.
2) Introduce them in plasmid of E.coli.
3) Products of A&B Chains are separated.
4) Combine A&B chains by creating disulphide bonds

Question 6.
Expand the common short forms used in biotechnology. (MAY-2011)
i) GEAC
ii) GMO
iii) PCR
iv) ELISA
OR
Bt toxins are not toxic to bacillus and Bt cotton plant but toxic to insects. Explain.
Answer:
i) GEAC – Genetic Engineering Approval Committee
ii) GMO – Genetically Modified Organisms
iii) PCR -Polymerase Chain reaction
iv) ELISA – Enzyme Linked Immuno Sorbent Assay
OR
The protein crystals are solubilised in the presence of alkaline pH of insect gut, hence becomes toxic to insects.

Question 7.
In human beings, certain diseases are caused due to genetic disorders. (MARCH-2012)
a) Name the method that allows the correction of a gene defect that has been diagnosed in a child or embryo.
b) How this method has been used for treating ADA (Adenosine deaminase) deficiency?
Answer:
a) Gene therapy
b) Introduction of functionalADAcDNAintothe lymphocyte of defective person.

Question 8.
Infection by nematodes cause threat to cultivation and yield loss of tobacco plants. A strategy has been developed at RNA level to control this infestation. (MARCH-2012)
a) Name the process.
b) Explain how this process works at the molecular level.
Answer:
a) RNA interference
b) In this process specific mRNA of the nematode is silenced due to a complementary ds RNA, that prevents the translation of mRNA

Question 9.
Using genetically modified crops, farmers can minimize use of insecticides and pesticides during cultivation. (MAY-2012)
a) Give name of one such genetically modified pest resistant crop.
b) Which gene is used for its production?
c) Name the source of pest resistant gene.
d) Write about its mode of action.
Answer:
a) Bt cotton
b) cry gene
c) Bacillus thuringiencis
d) The cry gene produce the insecticidal protein which solubilised in the alkaline PH of insect gut and make pores in the epithelial cells.This causes the death of insect.

Question 10.
Nita found that her Grandma used to inject human insulin that is genetically engineered. She wants to know how such insulin can be produced. Give her an idea about structure of insulin and production of genetically engineered insulin. (MAY-2012)
Answer:
It consists of two short polypeptide chains: chain A and chain B, that are linked together by disulphide bridges.
It is necessary to prepare two DNA sequences corresponding to A and B, chains of human insulin and inserted in plasmids of E. coli to produce insulin chains. Chains A and B are produced separately, extracted and combined by creating disulfide bonds to form human insulin.

Question 11.
A novel strategy to prevent nematode infestation is (MARCH-2013)
based on ‘RNA interference’
a) Explain RNA interference.
b) Can you suggest, how it can be used for producing nematode resistant plant.
Answer:
a) it involves silencing of a specific mRNA of nematode Here the complementary dsRNA molecule that binds to and prevents translation of the mRNA (silencing).
b) Afterthe insertion of nematode-specific genes by Agrobacterium vectors into the host plant, it produce both sense and anti-sense RNA in the host cells. These double stranded RNA (dsRNA) that initiated RNAi and silenced the specific mRNA of the nematode.

Question 12.
Sophie was born with a genetic disorder – ADA deficiency. (MARCH-2013)
a) What is ADA deficiency?
b) Can you suggest methods to treat this ADA deficiency?
Answer:
a) Adenosine deaminase (ADA) deficiency
b) Bone marrow transplantation or gene therapy (Lymphocytes from the blood of the patient are cultured and functional ADA cDNA is introduced in it.Then, these cells are return back to the patient) .

Question 13.
Gene therapy aims in correcting diseases caused by defective genes. A Child is suffering from a disease due to deficiency of ADA enzyme. ADA gene which normally produce the enzyme is missing in the patient. Recommend any two methods to treat the child. (MAY-2013)
Answer:
Enzyme replacement therapy & Genetic engineering.

Question 14.
Expand the short forms used in Biotechnology. (MAY-2013)
1) PCR
2) ELISA
3) GEAC
4) GMO
Answer:
PCR – Polymerase Chain reaction
ELISA – Enzyme Linked Immuno Sorbent Assay GEAC – Genetic Engineering Approval Committee GMO – Genetically Modified Organisms

Question 15.
Bt Cotton is regarded as an important achievement of genetic engineering. What does Bt stands for? (MARCH-2014)
Answer:
Bacillus thuringiensis

Question 16.
Animals that had their DNA manipulated to possess and express foreign DNA are called transgenic animals. Write briefly any three benefits of such transgenic animals to human beings. (MARCH-2014)
Answer:
a) Transgenic mice are used to test the safety of the polio vaccine.
b) Transgenic cow, Rosie, produced human protein enriched milk (2.4 grams per litre).The milk contained the human alpha-lactalbumin It is the balanced product for human babies than natural cow- milk.
c) Transgenic animals that carry genes which make them more sensitive to toxic substances than non- transgenic animals. Toxicity testing in such animals get results in less time.

Question 17.
Consider you are appointed as biotechnologist in a National Institute: What are the basic steps to be designed to produce a genetically modified organism? (Hint 3 points) (MAY-2014)
Answer:
i) Identification of DNA with desirable genes.
ii) Introduction of the identified DNA into the host.
iii) Maintenance of introduced DNA in the host and transfer of the DNA to its progeny.

Question 18.
Pharmaceutical companies are producing large quantities of human insulin by genetic engineering. Briefly explain the process. (MAY-2014)
Answer:
1. Prepare two DNA sequences corresponding to A and B, chains of human insulin.
2. Introduced them in plasmids of E. coli to produce insulin chains.
3. Chains A and B were produced separately, extracted and combined by creating disulfide bonds to form human insulin.

Question 19.
in the 2012 children’s science congress one of the speaker summarized like this – if we are not vigilant, countries or individuals encash our resources as their right. (MAY-2014)
Explain this with an example
Answer:
Basmati rice is distinct for its aroma and flavour and 27 documented varieties of Basmati are grown in India. There is reference to Basmati in ancient texts, folklore and poetry, as it has been grown for centuries.
In 1997, an American company got patent rights on Basmati rice through the US Patent and Trademark Office. This allowed the company to sell a ‘new’ variety of Basmati, in the US and abroad.

Question 20.
In a class room seminar on cloning vectors, your friend asked to explain the steps to introduce the plasmid DNA to a bacteria cell, Microinjection and bilestics. Answer his questions. (MAY-2014)
Answer:
For this, bacterial cells must first be made ‘competent’ to take up DNA. This is done by treating them with a specific concentration of a divalent cation, such as calcium, Recombinant DNA can then be forced into such cells by incubating the cells with recombinant DNA on ice, followed by placing them briefly at 42°C (heat shock), and then putting them back on ice. This helps to introduce the plasmid DNA into bacterial cell.
Microinjection – In this recombinant DNA is directly injected into the nucleus of an animal cell. Biolistics – In this cells are bombarded with high velocity micro-particles of gold or tungsten coated with DNA

Question 21.
Genetically Modified Organism (GMO) is always a debatable topic among scientists,’academicians and public. State any four usefulness of GMOs. (MARCH-2015)
Answer:
i) More tolerant to abiotic stresses (cold, drought, salt, heat).
ii) Reduced reliance on chemical pesticides (pest- resistant crops).
iii) Helped to reduce post harvest losses.
iv) Increased efficiency of mineral usage by plants
v) Enhanced nutritional value of food, e.g., Vitamin ‘A’enriched rice.

Question 22.
Biotechnology in agriculture will lead to pest resistant plants, which could decrease the amount of pesticides used. For example Bt cotton. Expand the letter ‘B’ and ‘t’. (MARCH-2015)
Answer:
B-Bacillus t-thuringiensis

Question 23.
In 1997, an American company got patent rights on Basmati rice through the U.S. Patent and Trademark Office. Variety of Basmati had actually been derived from Indian farmer’s varieties. If so, what is Biopiracy? (MARCH-2015)
Answer:
It is the use of bio-resources by multinational companies and other organisations without proper authorisation from the countries and people concerned without compensatory payment.

Question 24.
One of the speaker in the National Children’s Science Congress delivered a talk about Transgenic animals. Explore any 2 benefits of Transgenic animals. (MAY-2015)
Answer:
i) Study of disease: Transgenic animals can be used to know, how genes contribute to the development of disease.
ii) Biological products: Transgenic animals that produce useful biological products. In 1997, the first transgenic cow, Rosie, produced human protein-enriched milk contained the human alpha- lactalbumin. It is nutritionally a more balanced product for human babies than natural cow-milk.

Question 25.
The recombinant DNA technological process have made immense impact in the area of healthcare. How Eli Lilly produced Insulin? (MARCH-2016)
Answer:
Eli lily prepared two DNA sequences corresponding to A and B chains by using recombinant DNA technology , it is introduced into plasmid of E coli and produced polypeptide chains A and B.
These chains are separated and connected by using disulphide linkage to form human insulin

Question 26.
A farmer approached an Agriculture officer to tell his grievance i.e.. reduction in tobacco yield due to root damage by nematodes, (MARCH-2016)
a) What is your suggestion to prevent this infestation?
b) Briefly explain the process.
Answer:
a) RNA interference (RNAi)
b) The best method used to prevent the attack of nematode is RNA interference (RNAi). It involves silencing of a specific mRNA of nematode. Here the complementary dsRNA molecule that binds to and prevents translation of the mRNA (silencing).

Question 27.
RNA can suppress the activity of a gene. Explain it with suitable example. (MAY-2016)
Answer:
In this process nematode specific genes were introduced into the host plant through agrobacterium,it produce both sense and antisense RNA, since this two RNAs are complementary to each other, form a dsRNA RNAi is operated and the nematode can’t produce proteins.

Question 28.
Match the following: (MARCH-2017)
Plus Two Botany Chapter Wise Previous Questions Chapter 5 Biotechnology and its Applications 1

Answer:
Plus Two Botany Chapter Wise Previous Questions Chapter 5 Biotechnology and its Applications 2

Question 29.
Insulin getting assembled into a mature form was the major challenge in commercial insulin production by rDNA technology. How did Eli Nilly found a solution to this problem? (MARCH-2017)
Answer:
An American company Eli Lilly in 1983 prepared two DNA sequences corresponding to A and B, chains of human insulin and inserted in plasmids of E, coli to produce insulin chains. Chains A and B were produced separately, extracted and combined by creating disulfide bonds to form human insulin.

Question 30.
A) Bt cotton is an example of genetically engineered cotton. (MAY-2017)
a) What does Bt stands for?
b) Name the gene responsible for Bt toxin production.
c) How does the toxin kill the insect?
OR
B) Gene therapy is a corrective therapy for a hereditary disease.
a) Name the disease which was successfully corrected by gene therapy for the first time.
b) How gene therapy is practiced for a permanent cure of the disease?
Answer:
A) a) Bacillus thuringiensis
b) Cry gene
c) Bt toxin protein exist as inactive protoxins it is converted into an active form in the presence of the alkaline pH of insect gut. The activated toxin binds to the surface of midgut epithelial cells and create pores that cause cell swelling and lysis and results in the death of insect.
OR
B) a) Adenosine deaminase deficiency (ADA)
b) In this, functional ADA cDNA is introduced into embryonic stage. It is permanent cure forthe disease.

Question 31.
Antigen-antibody reaction is the basis of the technique called ____. (MAY-2017)
a) ELISA
b) PCR
c) RNA interference
d) Gene therapy
Answer:
ELISA

Plus Two Botany Chapter Wise Previous Questions Chapter 4 Biotechnology: Principles and Processes

Kerala State Board New Syllabus Plus Two Botany Chapter Wise Previous Questions and Answers Chapter 4 Biotechnology: Principles and Processes.

Kerala Plus Two Botany Chapter Wise Previous Questions Chapter 4 Biotechnology: Principles and Processes

Question 1.
Rinku with a circular DNA contains sequence (MARCH-2010)
5′-> GGAATTCC -> 3′
3′ -> CCTTAAGG -> 5′
She wishes to add a new segment of DNA into it.
a) Identify the technology she planned.
b) Suggest the specific enzyme to make a cut in the DNA with above sequence.
c) Name the category of enzyme you suggested.
d) How this enzyme identifies the sequence ?
e) Draw the cut ends of the DNA with sequence.
OR
Rashid isolated a natural plasmid from a bacterium and planning to facilitate cloning.
a) What are the minimum requirements for considering the isolate & plasmid as a vector ?
b) How he identifies whether a foreign DNA is inserted or not after cloning ?
Answer:
a) Recombinant DNA technology
b) Eco- R-1
c) Restriction endonuclease
d) Palindromic sequence
e)
Plus Two Botany Chapter Wise Previous Questions Chapter 4 Biotechnology Principles and Processes 1
OR
a) It should have origin of replication (ori), selectable marker for identifiying transformants, cloning sites.
b) Selectable marker which helps in identifying and eliminating non transforms and selectively permitting the growth of transformants. The markers commonly unsed are the gene encoding reistance to antibiotics such as tetracycline, ampicillin etc.

Question 2.
Diagram shows a typical agarose gel showing migration fragments. (MAY-2010)
Plus Two Botany Chapter Wise Previous Questions Chapter 4 Biotechnology Principles and Processes 2
a) Which of the bands has largest and smallest DNA fragments?
b) How can you make fragments of DNA for electrophoresis ?
c) Explain separation of DNA fragments using electrophoresis.
d) Point out a method to visualize the separated DNA fragments after electrophoresis.
Answer:
a) a is the largest DNA fragment and d is the smallest DNA fragment
b) DNA fragments can be made by cutting the DNA by restriction endonuclease.
c) Separation of DNA fragments takes place through gel electrophoresis where the cut DNA fragments
matrix. The separation takes place according to the size of the DNA.
d) The DNA separated by electrophoresis is visualized by staining it with Ethidium Bromide dye and viewing it under UV light. The fragments of DNA appears as orange coloured bands.

Question 3.
The picture given below shows the technique used forgenerating multiple copies of the gene of interest. (MARCH-2011)
Plus Two Botany Chapter Wise Previous Questions Chapter 4 Biotechnology Principles and Processes 3
a) What is the technique called?
b) Name the reactions at Step I, Step II, Step III.
c) Explain the principle underlying this technique of DNA amplification.
Plus Two Botany Chapter Wise Previous Questions Chapter 4 Biotechnology Principles and Processes 4
The above picture shows cloning vector pBR 322.
a) What is Ori? Give its importance.
b) How does the insertion of foreign DNA at Bam Hi site selected ? What is ampR?
c) How many cloning sites are depicted in this vector as shown in the figure?
Answer:
a) Polymerase chain reaction
b) Step I denaturation
Step II annealing
Stepl II extention/amplification
c) Multiple copies of gene of interest is synthesized invitro by using the set of primer,/Taq polymerase and deoxy nucleotides
OR
a) Origin of replication. Controlling the copy no.of linked DNA
b) Due to insertion of Foreign DNA at Bam H1 site, recombinant plasmid will lose tetracycline resistance, thus recombinants will not grow in the medium containing tetracycline
c) 5 cloning sites-Eco R1, Bam H1, PuvII, Pst1, Pvu1

Question 4.
In a class room seminar on cloning vectors, your friend asked to explain the steps to introduce the plasmid DNA to a bacteria cell, Micro injection and biolistics. Answer his questions. (MAY-2011)
Answer:
For this, bacterial cells must first be made ‘competent’ to take up DNA. This is done by treating them with a specific concentration of a divalent cation, such as calcium, Recombinant DNA can then be forced into such cells by incubating the cells with recombinant DNA on ice, followed by placing them briefly at 420C (heat shock), and then putting them back on ice. This helps to introduce the plasmid DNA into bacterial cell.
Micro injection – In this recombinant DNA is directly injected into the nucleus of an animal cell.
Biolistics – In this cells are bombarded with high velocity micro-particles of gold or tungsten coated with DNA.

Question 5.
Observe the relation in the first pair and fill up the blank in the second. (MAY-2011)
Cry I Ac: Control cotton bollworm
……………: Control corn borer
Answer:
Cry I Ab ………. control corn borer.

Question 6.
Restriction endonucleases are the enzymes used to cut the DNA molecules. (MARCH-2012)
a) Give the general term of the specific sequences where these enzymes cut the DNA.
b) Name the enzyme that joints the foreign DNA and vector DNA.
c) Give any two procedures to introduce the recombinant DNA into the host cell.
OR
During genetic engineering Vector with foreign DNA is transferred into a host bacterium. The next target will be the selection of transformants from non- ‘ transformants.
How antibiotic resistance and insertional inactivation is exploited for this purpose?
Answer:
a) Palindromic sequence or recognition sequence
b) DNAIigase
c) Direct gene transfer by gene gun (biolistics) Direct gene transfer by micro injection
OR
Antibiotic gene is used us selectable marker to identify the recombinants from non recombinant. Insertional inactivation is used to identify the transformants as white coloursed and non transformants as blue coloured

Question 7.
While studying nucleotide sequence. Raj found the following sequence which can be recognized by some enzymes : (MAY-2012)
5′- GAATTC – 3′
3′- CTTAAG – 5′
a) Give salient features of this sequence.
b) Write name of enzymes which recognize such sequences.
c) Elaborate importance of this enzyme in Genetic engineering.
OR
A group of students came to know about recombinant DNA technology. They want to know how scientists can produce a new desired product using rDNAtechnology. Can you give them an idea about the important steps that are involved in this process?
Answer:
a) Palindromic sequence
b) Restriction endonuclease enzyme
c) The enzyme cut at specific nucleotide sequence and get sticky ends.The same restriction enzyme is used to cut both foreign DNA and cloning vector.
OR
Recombinant DNA technology involves several steps.
They are
1) Isolation of the Genetic Material (DNA)
2) Cutting of DNA at Specific Locations
3) Amplification of Gene of Interest using PCR
4) Insertion of Recombinant DNA into the Host Cell Organism
5) Obtaining the Foreign Gene Product
6) Down stream processing

Question 8.
Jaya read in a Biotechnology book that alien DNA can be introduced into host ceil by micro injection and biolistics. Explain these methods. (MARCH-2013)
Answer:
1) Micro-injection- The recombinant DNA is directly injected into the nucleus of an animal cell.
2) Biolistics or gene gun- The cells are bombarded with high velocity micro-particles of gold or tungsten coated with DNA. It is suitable for plants.

Question 9.
Genetic Engineering include creation of recombinant DNA with the help of restriction enzymes. (MARCH-2013)
a) Explain recombinant DNA technology.
b) What are restriction enzymes? Name a restriction enzyme.
Answer:
a) Recombinant DNA technology involves several steps.
They are
1) Isolation of the Genetic Material (DNA)
2) Cutting of DNA at Specific Locations
3) Amplification of Gene of Interest using PCR
4) Insertion of Recombinant DNA into the Host Cell /Organism
5) Obtaining the Foreign Gene Product
6) Downstream Processing
b) Restriction enzyme is used to cut DNA at specific nucleotide sequence Examples are EcoR1 ,Hind111 etc.

Question 10.
Gel electrophoresis is a technique to separate fragments of DNA from a mixture. Some of the events of electrophoresis are given below. Arrange the events in order: (MAY-2013)
1) Cutout DNA bands
2) Expose to UV
3) Force DNA to move through gel
4) Stain DNA with ethidium bromide.
Answer:
Force DNA to move through the gel, stain DNA with ethidium bromide expose to UV ,cut out DNA bands.

Question 11.
Identify palindrome sequence from the following. (MAY-2013)
1) 5′-GAATTC-3′
3′-CTTAAG-5′
2) 5′-ATCG-3′
3′-TACG-5′
3) 5′ – AAAAA – 5′
5′ – TTTTT – 3′
4) 5′-CCCCC-3′
3′-GGGGG-5′
Answer:
5′ ___GAATTC ____ 3′
3′ ____ CTTAAG ____ 5′

Question 12.
_________ are the enzymes used for cutting the DNA molecule into fragments. An example for this type of enzyme is Eco Ri. What does Eco, R and I stand for? (MARCH-2014)
Answer:
UB Restriction Endonuclease,
Eco- Escherichia coli, R- letter ‘R’ is derived from the name of strain, I— It indicate the order in which the enzymes were isolated from that strain of bacteria.

Question 13.
Use of a thermostable DNA polymerase from the bacterium, Thermus aquaticus made it possible to generate billion copies of DNA in a very short time using a process. (MARCH-2014)
a) Name the process.
b) Name the three important steps involved in this process.
Answer:
a) Polymerase chain reaction
b) Denaturation, Anealing, Extension

Question 14.
There are many features required to facilitate successful cloning in to a vector. Write shortly any two such features required by a vector. (MAY-2014)
Answer:
Origin of replication (ori): This is a sequence from where replication starts and any piece of DNA when linked to this sequence can be made to replicate within the host cells.
Selectable market: Genes encoding resistance to antibiotics are considered as useful selectable markers which helps in identifying and eliminating non transformants and selectively permitting the growth of the transformants.

Question 15.
Recombinant DNA technology can be accomplished only if we have the following key tools, ie. Restriction enzymes, Polymerase enzyme, Ligases and Vectors. (MARCH-2015)
State the functions of
a) Ligases
b) Restriction Enzymes
Answer:
a) Ligase- Enzyme which is used to join the DNA fragments
b) Restriction Enzymes- It is used to cut DNA at specific locations

Question 16.
Figure representing the reactions associated with Polymer Chain Reaction (PCR). Name the steps A, B, C in the process (MARCH-2015)
Plus Two Botany Chapter Wise Previous Questions Chapter 4 Biotechnology Principles and Processes 5
Answer:
A — Denaturation
B — Annealing
C — Extension/Elongation

Question 17.
Observe the cloning vector and explain the following: (MAY-2015)
Plus Two Botany Chapter Wise Previous Questions Chapter 4 Biotechnology Principles and Processes 6
a) Ori
b) Bam HI
Answer:
a) ori-origin of replication
b) Bam HI- restriction enzyme

Question 18.
A multinational company successfully cloned a gene of interest and also optimized the conditions to induce the expression of target protein. (MAY-2015)
a) Name the apparatus for large scale production of such proteins.
b) Briefly explain the apparatus.
Answer:
a) Bioreactor
b) It is the large culture vessel (100-1000 litres) used for the production of large quantities of recombinant protein, enzymes, etc. it,provides optimum growth conditions (temperature, pH, substrate, salts, vitamins and oxygen). It consist of agitator system, an oxygen delivery system and a foam control system, a temperature control system, pH control system and sampling ports.

Question 19.
Observe the sketch of stirred-tank bioreactor and label the parts A, B. C and D. (MARCH-2016)
Plus Two Botany Chapter Wise Previous Questions Chapter 4 Biotechnology Principles and Processes 7
Answer:
A – motor
B – Foam breaker
C- Flat bladed impeller
D – Acid or base for PH Control

Question 20.
Manipulating with nucleic acid is a trend in Biotechnology. (MARCH-2016)
a) Name any one organism used as vector.
b) What are DNA polymerase?
Answer:
a) Ecoli
b) It helps to add nucleotide one by one on template srtand (polymerization of deoxy ribonucleotides)

Question 21.
Electrophoresis is a method commonly used in Biotechnology. Write briefly about GelElectrophoresis. (MAY-2016)
Answer:
In this method DNA fragments are separated accord-ing to their charge and size.
DNA is negatively charged molecules they move through agarose gel towards positively charged anode, DNA fragments are stained by Ethidium bromide, separated fragments can be observed as orange coloured bands under UV light.

Question 22.
Genetic engineering is a promising branch recently developed in biological science. (MAY-2016)
a) Expand PCR and name three steps in each cycle.
b) What is a plasmid? Name three features required for cloning vectors.
Answer:
a)PCR or polymerase chain reaction involves 4 steps
1) Denaturation – It involves the separating of DNA strands.
2) Annealing – The double strands are synthesised from free nucleotides by the action of DNA polymerase.
3) Extension – The length of the strands are increased as a result of addition of more and more nucleotides. The process of replication is repeated many times & billions of copies of DNA is synthesised.
OR
b)Plasmids are circular double stranded DNA molecules occurring in extra chromosomal state
1) Ori (Origin of replication) it is the sequence from where replication starts.
2) Cloning sites: For linking the alien DNA into the vector, there must be preferably single recognition sites because more than one recognition sites within the vector results several fragments.
3) Selectable markers are employed in rDNA technology for selecting the recombinants from non recombinants.

Question 23.
The following photograph shows the result of a technique showing the separation of DNA. (MARCH-2017)
Plus Two Botany Chapter Wise Previous Questions Chapter 4 Biotechnology Principles and Processes 8
a) Name the technique.
b) How the separated DNA is visualized?
c) DNA fragments of size 500bp and 2000 bp are separated by this process. Which fragment will migrate fast. Why?
OR
B) Different methods have been suggested to introduce alien DNA into host cells. Given and explain any three methods adopted for this purpose.
Answer:
A) a) Gel electrophoresis
b) The separated fragments of DNA can be visualized only after staining with ethidium bromide followed by exposure to uv radiation. DNA will be visualized as bright orange coloured bands on the gel.
c) 500 bp migrate fast because it is the smaller fragment found nearer to anode than other fragments on the gel.
OR
B) 1) Micro-injection-recombinant DNA is directly injected into the nucleus of an animal cell.
2) Biolistics or gene gun – cells are bombarded with high velocity micro-particles of gold or tungsten coated with DNA. It is suitable for plants.
3) Disarmed pathogen – vectors when allowed to infect the cell, transfer the recombinant DNA into the host.

Question 24.
Sequences of base pairs in DNA that reads the same on both the strands when the orientation of reading is kept the same are called ______ sequences. (MARCH-2017)
Answer:
Palindromic nucleotides

Question 25.
Origin of replication and selectable markers are the two important features required for a cloning vector. Explain their role in facilitating cloning. (MAY-2017)
Answer:
Up Origin of replication – It is the start of replication required for making many copies of the desired gene. Selectable markers – They are antibiotic-resistant genes that helpful in identifying recombinants from non-recombinants.

Question 26.
Rhizome, bulbil, offset and the bulb is different methods of vegetative reproduction in plants. Of these, the vegetative reproductive structures of Agave and Ginger are and respectively. (MAY-2017)
Answer:
Bulbil, Rhizome

Question 27.
Denaturation, Annealing and Extension are three steps of a process used for gene amplification: (MAY-2017)
a) Name the process.
b) Name the organism from which the DNA polymerase for this process is extracted.
Answer:
a) Polymerase Chain Reaction (PCR)
b) Thermusaquaticus

Plus Two Botany Chapter Wise Previous Questions Chapter 3 Strategies for Enhancement in Food Production

Kerala State Board New Syllabus Plus Two Botany Chapter Wise Previous Questions and Answers Chapter 3 Strategies for Enhancement in Food Production.

Kerala Plus Two Botany Chapter Wise Previous Questions Chapter 3 Strategies for Enhancement in Food Production

Question 1.
Johny, a Plus Two student is from a tribal colony with lower level of vitamins, minerals and protein deficiency. He wishes to be a plant breeder to help the public by producing new crops with high levels of vitamins, minerals, protein etc. Identify the phenomenon. (MARCH-2010)
Answer:
Biofortification

Question 2.
In mid 1960’s as a result of various plant breeding techniques, there was a drastic increase in food production in our country. This phase is often referred as _______. (MAY-2010)
Answer:
Green revolution

Question 3.
Tissue culture is a fast and efficient system for crop improvement. Scientists in a research institution wants to produce a hybrid of potato and tomato. Is it possible to make such a hybrid ? If possible, explain how. (MAY-2010)
Answer:
Somatic hybridisation – The isolated cells from to-mato and potato are converted into naked protoplasts by digesting the cell walls. Isolated protoplasts of the two plants are fused to get hybrid protoplasts which is further grown to form a new plant. Somatic hybrid of tomato and potato is called Pomato.

Question 4.
Ram Singh is a conventional wheat breeder, One the promising wheat varieties is found to be susceptible to leaf rust.
What breeding steps he will adopt to make his original promising variety resistant to leaf rust? (MARCH-2011)
Answer:
Breeding steps
a) screening germ plasm
b) hybridization of selected parents
c) selection and evaluation of hybrid
d) testing and release of new varieties

Question 5.
The method of producing thousands of plantlets through tissue culture is called ______.  (MARCH-2011)
These plantlets which are genetically identical to each other are called _______.
Answer:
a) Micropropagation
b) Somaclones

Question 6.
Being a member of the Committee of People’s Planning Programme of your Panchayat, suggest two common fresh water fishes to grow in the fresh water fishery project undertaken by your Panchayat.  (MAY-2011)
Answer:
Catla and Rohu

Question 7.
If the tomato plants of your village are virus affected, which part of the plant would you recommend to culture for virus free plants?  (MAY-2011)
Answer:
Meristem /Shoot tip

Question 8.
The regeneration of whole plants from any part of the plant grown under sterile conditions is called tissue culture.
a) The general term for the part of the plant taken out for tissue culture is _______.  (MARCH-2012)
b) The capacity to generate a whole plant from any plant cell is ______.
Answer:
a) Explant
b) Toti potency

Question 9.
Raju went to a Rice Research station on his study tour. There he noticed a scientist working on rice plants using scissors and forceps. To his surprise he saw the scientist covering the inflorescences with paper bags. (MARCH-2012)
a) Name the techniques the scientist was doing.
b) Give the purpose of these techniques.
Answer:
a) Emasculation & Bagging
b) It helps to prevent the contamination of unwanted pollen

Question 10.
A newspaper report read like this. (MARCH-2012)
“Conventional agricultural products like cereals, pulses and other seeds may not be able to meet the demand of food according to the increase in population. So focus has to be shifted to alternate food sources like SCP’s.
a) What are SCP’s?
b) Give one example of SCP’s.
c) What are the advantages of SCP’s?
Answer:
a) Single cell protein. It is alternative food source human consumption
b) Spirulina
c) They are protein rich food besides carbohydrate, fat and minerals.

Question 11.
Continued inbreeding, usually reduces fertility and causes non productivity. This is called ________. (MAY-2012)
Answer:
Inbreeding depression

Question 12.
Bee keeping requires some specialized knowledge for success. (MAY-2012)
a) What is the alternate name for Bee Keeping?
b) Give your suggestions for successful bee
Answer:
a) apiculture
b) i) Knowledge of the nature and habits of bees
ii) Selection of suitable location for keeping the beehives
iii) Catching and hiving of swarms (group of bees)
iv) Management of beehives during different seasons, and
Handling and collection of honey and beewax.

Question 13.
Plant breeding programmes are carried out in a systematic way in research organizations. Explain main steps in breeding to produce a new genetic variety. (MARCH-2013)
Answer:
i) Collection of variability
ii) Evaluation and selection of parents:
iii) Cross hybridisation among the selected parents:
iv) Selection and testing of superior recombinants.
v) Testing, release and commercialisation of new cultivars:

Question 14.
MOET is a programme for herd improvement. Expand MOET. (MARCH-2013)
Answer:
MOET- Multiple Ovulation Embryo Transfer Technology

Question 15.
It is observed that continuous inbreeding of animals for 4 – 6 generations produce progeny with reduced fertility and productivity. What measures can be taken to improve fertility and productivity of progeny? (MAY-2013)
Answer:
The selected animals of the breeding population should be mated with unrelated superior animals of the same breed. This is usually helps to restore fertility and yield. The offspring of such a mating is known as an out-cross.
A single out cross helps to overcome inbreeding depression.

Question 16.
crop are completely used up and hence genetic variations are to be created for crop improvement. Suggest any one method for creating genetic variation. (MAY-2013)
Answer:
Mutation breeding

Question 17.
Vidya got a plant which was affected with a viral disease. Her objective is to raise a disease free plant from this infected plant through tissue culture. (MARCH-2014)
a) Which part of the plant should be selected as the explant?
b) State the reason for the selection of this part as the explant.
Answer:
a) Shoot tip
b) Conducting tissues are not present

Question 18.
Plant breeding involves techniques for manipulating plants in order to create the desired plant types. State the steps involved in the production of a new genetic variety of a crop. (MARCH-2014)
Answer:
The main steps in breeding a new genetic variety of a crop are:
i) Collection of variability
ii) Evaluation and selection of parents:
iii) Cross hybridisation among the selected parents:
iv) Selection and testing of superior recombinants Testing, release and commercialisation of new cultivars.

Question 19.
The local people in a village wanted to produce a crop with improved nutritional qualities. What are the major objectives to be included to improve the nutritional qualities? (MAY-2014)
Answer:
i) Protein content and quality
ii) Oil content and quality
iii) Vitamin content; and
iv) Micro nutrient and mineral content

Question 20.
Observe the relation in the first pair and fill up the blank in the second. (MAY-2014)
a)

CropVarietyResistance to disease
ChilliPusa

Sadabahar

Chilly or tobacco

Mosaic virus

Brassica…………………White rust

b)

CropVarietyInsect pest
Flat beanPusa

sawani

Jassids ,fruit borer and aphids
Okra………………….Shoot and fruit borer

Answer:
a) Pusaswarnim
b) PusaSawani

Question 21.
In a debate one of the speaker reported like this. (MARCH-2015)
“Continuous inbreeding leads to inbreeding depression.” If so, define the following:
a) Outcross
b) Crossbreeding
Answer:
a) It is the practice of mating of animals within the same breed, but having no common ancestors on either side of their pedigree up to 4-6 generations. The offspring of such a mating is known as an out-cross.
b) In this superior males of one breed are mated with superior females of another breed. Hisardale is a new breed of sheep developed in Punjab by crossing Bikaneri ewes and Marino rams.

Question 22.
250 Kg. cow produces 200g of protein/day. In the same period 250 g of Methylophilus methylotrophus produce 25 tonnes of protein. Then what is single cell protein?  (MARCH-2015)
Answer:
One of the alternate sources of proteins for animal and human nutrition is Single Cell Protein (SCP)
Eg- Spirulina

Question 23.
In a Grama Panchayat, Members wanted to start a Bee-keeping industry. What are your suggestions for successful bee keeping ?  (MAY-2015)
Answer:
For successful bee-keeping it requires
i) Knowledge of the nature and habits of bees.
ii) Selection of suitable location for keeping the beehives.
iii) Catching and hiving of swarms (group of bees).
iv) Management of beehives during different seasons.

Question 24.
Observe the relation in the first pair and fill up the blank in the second.  (MAY-2015)
a)

CropVarietyResistance to disease
BrassicaPusa swarnimWheat rust
Chilli………………..Chilly mosaic virus

b)

CropVarietyInsect pest
OkraPusa sawaniShoot and fruit borer
Flat bean…………..Juassids ,fruit borer and aphids

Answer:
a) Pusa sadabahar
b) Pusa sem 2, Pusa sem 3

Question 25.
Resistance is the ability to prevent the pathogen from causing disease.  (MARCH-2016)
1) Elucidate the steps in breeding for disease resistance.
2) Cite two examples for virus resistant plants.
OR
Tissue culture is an achievement in plant breeding. What is a somaclone ? Describe the production of somatic hybrid.
Answer:
1) a) screening of germ plasm for resistance
b) Hybridization of selected plants
c) Selection and evaluation of hybrids
d) Testing and release of new varieties
2) pusa sadabahar, parbhani kranti
OR
Morphologically and genetically similar off springs are produced through tissue culture called somaclones. Isolation of somatic cells from two different varieties Digestion of cell wall using enzymes.
Fusion of protoplast of two different varieties forming somatic hybrid protoplast.
Culture of protoplast hybrid to produce somatic hybrid.

Question 26.
a) Describe the major steps followed for the production of new genetic variety starting from the collection of germplasm upto elucidating the cultivars.  (MAY-2016)
b) A plant breeder has a rare variety of cultivar with him but unfortunately it has become infected with vims. Suggest a suitable technique to produce many viable number of progenies with a short note.
Answer:
a) Collection of variability
Evaluation and selection of parents Cross hybridisation among the selected parents Testing, release and commercialisation of new cultivars. Selection and testing of superior recombinants,
b) Meristem culture In this method virus free plants are developed because it lack conducting tissues

Question 27.
Match the following varieties with their respective crops:  (MARCH-2017)

VarietyCrop
a)  Pusa Swarnim

b) Pusa Snowball

c)  Pusa Swani

d)  Pusa Sadabahar

i)    Chilly

ii)   Bhindi

iii)  Cauliflower

iv)  Brassica

Answer:

VarietyCrop
a)  Pusa swarrnim

b)  Pusashubhra

c)  Pusasawani

d)  Pusa Sadabahar

i)    Brassica

ii)   Cauliflower

iii)  Bhindi

iv)  Chilli

Question 28.
Breeding crops with the objective of increased nutritional quality is called _______.  (MARCH-2017)
Answer:
Biofortification

Question 29.
Out crossing and cross breeding are two different aspects of out breeding in animals. How out crossing is different from cross breeding?  (MAY-2017)
Answer:
Out-crossing
It is mating of animals within the same breed, but having no common ancestors on either side of their pedigree up to 4-6 generations.
Cross-breeding
It is the method of mating superior males of one breed with superior females of another breed.

Question 30.
The practice of maintenance of honeybees for the production is called _______. (MAY-2017)
Answer:
Bee keeping (Apiculture)

Plus Two Botany Chapter Wise Previous Questions Chapter 2 Sexual Reproduction in Flowering Plants

Kerala State Board New Syllabus Plus Two Botany Chapter Wise Previous Questions and Answers Chapter 2 Sexual Reproduction in Flowering Plants.

Kerala Plus Two Botany Chapter Wise Previous Questions Chapter 2 Sexual Reproduction in Flowering Plants

Question 1.
Given below in the diagram showing the transfer of pollen grains. (MARCH-2010)
i) Identify a & b with technical terms.
ii) Critically evaluate a & b.
Plus Two Botany Chapter Wise Previous Questions Chapter 2 Sexual Reproduction in Flowering Plants 1
Answer:
i) (a) Autogamy & (b) Geitonogamy
ii) (a) Transfer of pollen from antherto the stigma of the same flower is called autogamy.
b) Transfer of pollen from anther to the stigma of a different flower on the same plant.

Question 2.
Microsporangium is generally surrounded by four wall layers. Name the layer which nourishes developing pollen grains.   (MAY-2010)
Answer:
Tapetum

Question 3.
Match the following:  (MAY-2010)
Plus Two Botany Chapter Wise Previous Questions Chapter 2 Sexual Reproduction in Flowering Plants 2
Answer:
Plus Two Botany Chapter Wise Previous Questions Chapter 2 Sexual Reproduction in Flowering Plants 3

Question 4.
Plus Two Botany Chapter Wise Previous Questions Chapter 2 Sexual Reproduction in Flowering Plants 4
Copy the picture given above and mark the following:  (MARCH-2011)
a) Connective tissue
b) Endothecium
c) Tapetum
d) Sporogenous tissue
Answer:
Plus Two Botany Chapter Wise Previous Questions Chapter 2 Sexual Reproduction in Flowering Plants 5

Question 5.
Transfer of pollengrains from the anther to the sitgma of a flower is called pollination. Grass plants generally have small, inconspicuous flowers while plants belonging to many angiosperm families bear conspicuous coloured flowers. (MARCH-2011)
a) Comment on the type of pollination taking place in these two groups.
b) What are the salient features present in these two groups for effective pollination?
Answer:
a) 1) Wind pollination
2) Pollination by biotic agents

b) Floral features of wind pollination :-

  1. Light pollengrain
  2. Dry, smooth & large quantities of pollengrains
  3. Exposed stigma
  4. Lack of scent
  5. Lack of nectar

c) Floral features of animal pollination :-

  1. Large flower
  2. Colourful and fragrant with nectar
  3. Sticky pollengrain

Question 6.
After fertilization in flowering plants, seeds bearing embryos are found inside the fruits. If seeds are developed from ovules. (MARCH-2011)
a) Name the parts that given rise to embryo and fruits.
b) What is the thick wall of the fruit that is protective in function called?
Answer:
a) Zygote -> Embryo
Ovary -> Fruit
b) Pericarp

Question 7.
Teacher wrote the steps of a crop improvement programme on the blackboard as follows: (MAY-2011)
Bagging
Emasculation
Arrange the steps in correct order, explain them and name the process of crop imdorvement programme with the given steps.
Answer:
Emasculation
Bagging
If the female parent bears bisexual flowers, removal of anthers from the flower bud before the anther dehisces This step is called emasculation. Emasculated flowers have to be covered with a bag of suitable size prevent the contamination of its stigma with unwanted pollen. This process is called bagging.
The above two process are coming under Artificial Hybridisation

Question 8.
+2 students of a school at Kasargod district on their study tour collected flowers showing the following character
1) Flowers are with light pollen grains. (MAY-2011)
2) Colourful flowers.
3) Nectar producing flowers.
4) Flowers with feathery stigma.
a) Arrange the characters underdifferent pollination groups in the given table.
b) Write the name of 2 flowers pollinating through water.
Plus Two Botany Chapter Wise Previous Questions Chapter 2 Sexual Reproduction in Flowering Plants 6
Answer:
a)

Entamophilous flowersAnemophilous flowers
Colourful flowers .Flowers are with pollen grains.
Nectar producing flowers.Flowers are with feathery stigma.

b) Hydrilla, vallisneria

Question 9.
a) Hilum (MARCH-2012)
b) Funicle
c) Micropylarpole
d) Nucellus
e) Chalazalpole
d) Embryosac
Answer:
Plus Two Botany Chapter Wise Previous Questions Chapter 2 Sexual Reproduction in Flowering Plants 7

Question 10.
Innermost wall layer of microsporangium which nourishes the developing pollen grain is called ________. (MAY-2012)
Answer:
Tapetum

Question 11.
In large number of plants, pollination is carried out by insects. List four characters of flowers that helps insect pollination. (MAY-2012)
Answer:
Characters of entamophilous flowers.
1) Flowers are large, colourful, fragrant and rich in nectar.
2) small flowers are clustered into an inflorescence to make them conspicuous.

Question 12.
In Papaya, male and female flowers are present in separate plants. They are said to be ______. (MAY-2012)
Answer:
Dioceous

Question 13.
In artificial hybridization, it is important to make sure that sigma is protected from unwanted pollen. This is achieved by emasculation and bagging techniques. Can you explain, how emasculation and bagging techniques are performed? (MAY-2012)
Answer:
Anthers are removed before the dehiscence of anther of female parent that bears bisexual flowers. This step is called as emasculation.
It is covered with a bag of suitable size, to prevent contamination of its stigma with unwanted pollen. This process is called bagging.

Question 14.
After syngamy and triple fusion in embryosac, embryo will be diploid and endosperm will be ________. (MARCH-2013)
Answer:
Triploid

Question 15.
Flowering plants evolved an array of adaptations to achieve pollination.(MARCH-2013)
a) Explain pollination.
b) Point out adaptations found in flowers for insect pollination and wind pollination.
c) Illustrate pollination in Vallisnaria.
(OR)
Artificial hybridization is one of the major approaches for crop improvement programme. In such crosses it is important to avoid unwanted pollen.
a) Explain how can we protect stigma from unwanted pollen.
b) How artificial pollination can be performed?
Answer:
a) It is the transfer of pollen grains to the stigma of a pistil.
b) Characters of entamophilous flowers
1) Flowers are large, colourful, fragrant and rich in nectar.
2) small flowers are clustered into an inflorescence to make them conspicuous.
Characters of anemophilous flowers
1) pollen grains are light and non-sticky.
2) They possess well-exposed stamens and feathery stigma.
3) The flowers have a single ovule in each ovary and numerous flowers packed into an inflorescence.
c) In Vallisneria, the female flower reach the surface of water by the long stalk and the male flowers or pollen grains are released on to the surface of water. The anthers eventually reach the female flowers and the stigma.
(OR)
a) It is covered with a bag of suitable size, to prevent contamination of its stigma with unwanted pollen. This process is called bagging.
b) 1) Anthers are removed before the dehiscence of anther of female parent that bears bisexual flowers. This step is called as emasculation.
2) It is covered with a bag of suitable size, to prevent contamination of its stigma with unwanted pollen. This process is called bagging.
3) When the stigma of bagged flower attains receptivity, mature pollen grains collected from anthers of the male parent are dusted on the stigma, and the flowers are rebagged, and the fruits allowed to develop.

Question 16.
In many grasses seeds are formed only after fertilization. There are reports that in some grasses, seeds are formed without fertilization. Explain the phenomenon. (MAY-2013)
Answer:
The phenomenon of formation of seeds without fertilization is called Apomixis.
Apomixis is a form of asexual reproduction that mimics sexual reproduction. In this phenomenon , the diploid egg cell is formed without reduction division and develops into the embryo without fertilization.

Question 17.
The diagramatic view of a typical anatropous ovule is show below. Copy the diagram and label the unlabelled parts. (MAY-2013)
Plus Two Botany Chapter Wise Previous Questions Chapter 2 Sexual Reproduction in Flowering Plants 8
Answer:
Plus Two Botany Chapter Wise Previous Questions Chapter 2 Sexual Reproduction in Flowering Plants 9

Question 18.
In flowering plants, double fertilization occurs during sexual reproduction. One of the events of double fer-tilization is triple fusion. Name the other event. (MAY-2013)
Answer:
Syngamy. (Fusion of egg cell with male gamete that leads the formation of zygote)

Question 19.
In flowering plants during double fertilization two events take place in the embryosac namely ______ and ______. (MARCH-2014)
Answer:
Syngamy and Triple fusion

Question 20.
From the following, select the two having haploid Chromosome number. (MARCH-2014)
a) Egg
b) endosperm
c) Zygote
d) Pollen
Answer:
Egg, pollen

Question 21.
Sunflower is pollinated by insects while rice is pollinated by wind.  (MARCH-2014)
a) How these plants are adapted to their respective type of pollination method? (Hint-any 4 points)
b) Plants can be self or cross pollinated. Write any two mechanisms existing in nature to promote cross pollination.
(OR)
a) The diagram given below shows the transverse section of a young anther. Identify the parts a, b, c and d.
Plus Two Botany Chapter Wise Previous Questions Chapter 2 Sexual Reproduction in Flowering Plants 10
The developmental stages of male gametes in plants consist of microsporogenesis and male ga- metophyte. Arrange the following terms in their correct developmental sequence.
Pollen grain
Sporogenous tissue
anther
microspore tetrad
pollen mother cell
male gamete.
Answer:
Adaptation for wind pollination
1) The pollen grains are light and non-sticky
2) They possess well-exposed stamens and feathery stigma
Adaptation for insect pollination
1) Insect-pollinated flowers are large, colourful, fragrant and rich in nectar.
2) The flowers are small, a number of flowers are clustered into an inflorescence
b) 1) pollen release and stigma receptivity are not at the same time.
2) Anther and stigma are placed at different positions so that the pollen cannot come in contact with the stigma of the same flower. Both these devices prevent autogamy.
(OR)
a) a-connective
b- epidermis
c – sporogenous tissue
d-Tapetum
b) Anther -> sporogenous tissue —> pollen mother cell —> microspore tetrad —> pollen grain —> malegamete

Question 22.
Most of the plants produce single type of flowers but Viola, Commelina and Oxalis produce two type of flowers. Explain. (MAY-2014)
Answer:
Chasm ogamous
Flowers with exposed anthers and stigma.
Cleistogamous
Flowers which do not open at all. In such flowers, the anthers and stigma lie close to each other. These flowers are example for autogamous flowers (self pollination)

Question 23.
Egg cell formation in angiosperms involves me- gasporogenesis and female gametophyte development. (MAY-2014)
a) Briefly write the various steps involved in female gametophyte development.
b) Mature angiosperm embryosac at maturity, though 8 nucleated is 7 celled.
What is your explanation related to this statement? Explain.
Answer:
a) Single megaspore mother cell (MMC) in the micropylar region of the nucellus undergoes meiotic division results in the production of four megaspores. In a majority of flowering plants, one of the megaspores is functional while the other three degenerate. Only one functional megaspore develops into the female gametophyte (embryo sac).

b) 2 polar nuclei are situated below the egg apparatus in the large central cell.
Three cells are grouped together at the micropylar end and constitute the egg apparatus. The egg apparatus consists of two synergids and one egg cell.
The synergids have special cellular thickenings at the micropylar tip called filiform apparatus, which play an important role in guiding the pollen tubes into the synergid. Three cells are at the chalazal end and are called the antipodals.

Question 24.
Development of fruit without fertilization and are seedless known as ______. (MARCH-2015)
a) Polyembryony
b) Apomixix
c) Parthenocarpy
d) Parthenogenesis
Answer:
Parthenocarpy

Question 25.
Given below are the components related to simplified model of mineral cycling in a terrestrial ecosystem. Construct a flow chart. (MARCH-2015)
(Hint: Weathering of rock)
Plus Two Botany Chapter Wise Previous Questions Chapter 2 Sexual Reproduction in Flowering Plants 11
Answer:
Producers consumers -> detritus -> soil solution -> Minerals in rock

Question 26.
a) Reeja a science student observed the structure of mature embryosac comprising antipodals, central cells and egg apparatus. Explain each one of them. (MARCH-2015)
(OR)
b) Three different flowers are given to you in the practical class.
i) Maize
ii) Vallisneria
iii) Rose
You are asked to group them based on pollinating agents. Describe the adaptations of each flower related with the agents of pollination.
Answer:
The typical female gametophyte or embryo sac is 8-nucleate and 7-celled.
2 polar nuclei are situated below the egg apparatus in the large centraftfell.
Three cells are grouped together at the micropylar end and constitute the egg apparatus. The egg apparatus consists of two synergids and one egg cell.
The synergids have special cellular thickenings at the micropylar tip called filiform apparatus, which play an important role in guiding the pollen tubes into the synergid.
Three cells are at the chalazal end and are called the antipodals.
(OR)
1) Maize wind pollination
a. The pollen grains are light and non-sticky
b. They possess well-exposed stamens and feathery stigma
2) Vallisneria water pollination
a. Water pollinated flowers are not very colourful and do not produce nectar.
b. Pollen grains are protected from wetting by a mucilaginous covering
3) Rose Insect polination
a. Insect-pollinated flowers are large, colourful, fragrant and rich in nectar.
b. The flowers are small, a number of flowersare clustered into an inflorescence.

Question 27.
You are supplied with three different flowers such as Maize, Vallisneria and Rose and they have different pollinating agents also. (MAY-2015)
a) Differentiate the type of pollination.
b) Write their various adaptability in the plants suited to pollination.
Answer:
a) Maize -> wind pollination Vallisneria-> water pollination Rose -> insect pollination
b) In maize pollen grains are produced in large quantity, small flowers are packed into inflorescence, and stamen and stigma are exposed.
In vallisneria pollen grains are non sticky and surrounded by mucilaginous layer to prevent from wetting. They have not produced nectar and scent. In rose pollen grains are sticky and produced strong scent, the flowers are brightly coloured. it attract insect for pollination.

Question 28.
In some seeds the nucellus may be persistent. Such nucellus is called _______. (MARCH-2016)
a) Endosperm
b) Scutellum
c) Plumule
d) Perisperm
Answer:
d) Perisperm

Question 29.
What is a false fruit ? Cite an example. (MARCH-2016)
Answer:
Fruit is developed from parts of flower other than ovary.
Eg. apple and strawberry

Question 30.
Many of the flowering plants have developed some devices for discouraging in breeding. Write any two of them. (MARCH-2016)
Answer:
Self incompatibility
Production of unisexual flowers

Question 31.
The development of pollen grains in Angiosperms is called _______. (MAY-2016)
a) Microsporogenesis
b) Embryogenesis
c) Megasporogenesis
d) Gametogenesis
Answer:
Microsporogenesis

Question 32.
Which of the following part in a flower is haploid? (MAY-2016)
a) Antherwall
b) Pollen mother cell
c) Synergid
d) Secondary nucleus
Answer:
c) Synergid

Question 33.
Observe the following diagram and label A, B, C and D. (MAY-2016)
Plus Two Botany Chapter Wise Previous Questions Chapter 2 Sexual Reproduction in Flowering Plants 12
Answer:
A) Epidermis
B) Endothecium
C) Middle layers and
D) Tapetum

Question 34.
In aquatic plants like water hyacinth and water Lily the pollinating agent is ________. (MAY-2016)
a) Wind and insect
b) Water
c) Birds and butterflies
d) Aquatic organisms
Answer:
a) Wind and insect

Question 35.
A date palm seed discovered during archeological investigation retained viability even after 10000 years. The retention of viability is due to the state of inactivity of embryo called ________. (MARCH-2017)
Answer:
Dormancy

Question 36.
When the pollen is transferred from anther to the stigma of the same flower, the pollination is called autogamy.
a) Cleistogamous flowers are invariably autogamous. Explain. (MARCH-2017)
b) Geitonogamy is functionally cross pollination, but genetically similar to autogamy. Justify the statement.
Answer:
a) In this, flowers are not open pollen falls to the stigma of the same flower and seed setting takesplace without the influence of external agency.
b) In geitonogamy transfer of pollen grains from the anther to the stigma of another flower of the same plant.
Functionally geitonogamy is a type of cross pollination but it is genetically similar to autogamy since the pollen grains come from the same plant.

Question 37.
The thick protective covering of the fruit is known as ______. (MARCH-2017)
Answer:
Pericarp

Question 38.
Nature has mechanisms to promote outbreeding in plants. Explain any two mechanisms existing in plants to promote outbreeding. (MARCH-2017)
Answer:
1) Anther and stigma are placed at different positions so that the pollen cannot come in contact with the stigma of the same flower.
2) Self-incompatibility is the genetic mechanism in which pollen cannot germinate on the stigma of the same flower or other flowers of the same plant by inhibiting pollen germination or pollen tube growth in the pistil.

Question 39.
Rose is a flower pollinated by insect while in paddy pollination is by wind. Give any three adaptations existing in these plants to facilitate their respective mode of pollination. (MAY-2017)
(OR)
(B) Double fertilization and triple fusion are the two terms associated with angiosperm fertilization.
a) What is double fertilization?
b) Explain triple fusion.
c) Give the ploidy level of
i) endosperm
ii) zygote
Answer:
(A) Pollination in Rose
1. Flowers are large and colourful
2. They are fragrant and rich in nectar
3. Pollen grain are sticky
Pollination in Paddv
1. Pollen grains are light weight and non-sticky.
2. They possess well-exposed stamens
3. They possess feathery stigma.
(OR)
(B) a) It is the fertilization take place in two times. It involves syngamy and triple fusion.
b) It is the fusion of two polar nuclei and male gametes and results primary endosperm nucleus (PEN)
c) i) endosperm – 3n
ii) zygote-2n

Question 40
Identify the following parts of a dicot embryo. (MAY-2017)
Plus Two Botany Chapter Wise Previous Questions Chapter 2 Sexual Reproduction in Flowering Plants 13
Answer:
a- plumule
b – cotyledons
c – hypocotyle
d – radicle

Plus Two Botany Chapter Wise Previous Questions Chapter 1 Reproduction in Organisms

Kerala State Board New Syllabus Plus Two Botany Chapter Wise Previous Questions and Answers Chapter 1 Reproduction in Organisms.

Kerala Plus Two Botany Chapter Wise Previous Questions Chapter 1 Reproduction in Organisms

Question 1.
Fill up the blanks after reading the statement: (MARCH-2010)
The postfertilisation events in angiosperms.
Zygote : Embryo
Ovule : ______
Ovary : _______
Answer:
Seed
Fruit

Question 2.
Given below are 3 gametes a, b and c. a and b gametes undergoes fusion. (MARCH-2010)
a) Identify the fusion.
Plus Two Botany Chapter Wise Previous Questions Chapter 1 Reproduction in Organisms 1
b) Give the explanation for the identification.
Answer:
a) Isogamy
b) Fusion of two morphologically similar gametes (male & female) is called Isogamy

Question 3.
Raman is learning the post-fertilization changes of an angiosperm embryo sac with the help of slides. He identified the egg nucleus and polar nuclei with the help of his teacher. (MARCH-2010)
a) Name the other nuclei present in the embryo sac.
b) Help Raman by giving the changes that takes place with egg nucleus and polar nuclei after fertilization.
Answer:
a) Antipodals and synergids
b) Egg nucleus + Sperm -> Sygote (2n)
Polar nuclei + Sperm -> PEN (3n) primary endosperm nucleus.

Question 4.
Find out which of the statements are true. (MAY-2010)
a) Ovary develops into fruit.
b) In flowering plants, zygote is formed outside the ovule.
c) Ovules develops into fruit.
d) Zygote develops into embryo.
Answer:
Ovary develops into fruit.

Question 5.
Farmers are propagating plants using vegetative structures. Can you mention the names of any two such structures. (MAY-2010)
Answer:
Rhizome – Ginger
Bulb – Onion

Question 6.
Prefertilization events of sexual reproduction in all organisms are gametogenesis and gamete transfer. What are the post fertilization events? (MAY-2011)
Answer:
Formation of zygote
Embryogenesis

Question 7.
You are asked to plant ginger in your home. Which part of ginger you will plant? Name the type of reproduction. (MAY-2011)
Answer:
Underground stem
Asexual reproduction

Question 8.
In asexual reproduction, off springs are produced by a single parent with or without the involvement of gamete formation.(MARCH-2012)
Name the asexual reproductive structures (a & b) given below.
Answer:
a) Bud
b) conidia

Question 9.
Zoospores are common asexual reproductive structures in plants and animals with relatively simple organization. Name two other asexual reproductive structures seen in the group. (MAY-2012)
Answer:
Gemmules in sponges, conidia in penicillium

Question 10.
In honey bees and some lizards,female gamete undergoes development to form new organisms without fertilization. This phenomenon is called ________. (MARCH-2013)
Answer:
Parthenogenesis

Question 11.
Morphologically and genetically similar individuals are called ______ . (MARCH-2013)
Answer:
Clones

Question 12.
Match the following terms with regard to vegetative reproduction in plants. (MAY-2013)
Plus Two Botany Chapter Wise Previous Questions Chapter 1 Reproduction in Organisms 2
Answer:
a-c
b-d
c-a
d-b

Question 13.
a) Amoeba asexually multiplies by binary fission whereas Sponge by _______ (MAY-2014)
b) Water hyacinth vegetatively multiples by offset agave by ______
Answer:
a) Gemmule
b) Bulbil

Question 14.
Match the Column A with B: (MARCH-2015)
Plus Two Botany Chapter Wise Previous Questions Chapter 1 Reproduction in Organisms 3
Answer:
Plus Two Botany Chapter Wise Previous Questions Chapter 1 Reproduction in Organisms 4

Question 15.
a) Yeast asexually multiples by budding whereas Penicillium by. (MAY-2015)
b) Bryophyllum vegetatively multiples by adventitious buds water hyacinth by
Answer:
a) Conidia
b) offset/stem

Question 16.
The chromosome number of onion is 16 (2n). Find the chromosome number in the following cells with reasons. (MAY-2015)
a) Endosperm cell
b) Zygote
Answer:
a) 24
b) 16

Question 17.
When a gamete without any fusion develop into a new organism the phenomenon is called ______. (MARCH-2016)
a) Syngamy
b) External fertilization
c) Parthenogenesis
d) Parthenocarpy
Answer:
c) Parthenogenesis

Question 18.
A unisexual flower having no androecium is called _____. (MARCH-2016)
a) Dithecous
b) Dioecious
c) Monoecious
d) Pistillate
Answer:
d) Pistillate

Question 19.
Select the one which is not helping vegetative propagation. (MAY-2016)
a) Bulb
b) Clone
c) Adventitious
d) Eyes of the potato
Answer:
b) Clone

Question 20.
The plant in which adventitious buds along the mar¬gin of leaves give rise to new plants is (MARCH-2017)
a) Water Hyacinth
b) Agave
c) Bryophyllum
d) Dahlia
Answer:
Bryophyllum

Question 21.
In flowering plants male flower is called ______ flower and female flower is known as _______ flower. (MAY-2017)
Answer:
Staminate
Pistillate