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You can Download Real Numbers Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 10 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 10 Real Numbers

Real Numbers Textual Questions and Answers

Textbook Page No. 160

Hss Live Guru 9th Maths Kerala Syllabus Question 1.
Find the distance between the two points on the number line, denoted by each pair of numbers given below: .

Answer:

Real Numbers Class 9 Kerala State Board Question 2.
Find the midpoint of each pair of points in the first problem.
Answer:



Real Numbers Class 9 State Board Kerala Syllabus Question 3.
The part of the number line between the points denoted by the numbers 1/3 and 1/2 is divided into four equal parts. Find the numbers de-noting the ends of each such part.
Answer:

So the portion between 1/3 and 1/2 is divided into 4 equal parts the points are

Textbook Page No. 164

Hsslive Guru 9th Maths Kerala Syllabus Question 1.
Find those x satisfying each of the equations below:
i. |x – 1| = |x – 3|
ii. |x – 3| = |x – 4|
iii. |x + 2| = |x – 5|
iv. |x| = |x + 1|
Answer:
i. \(|x-1|=|x-3|\)
\(|x-1|\) means distance between x and 1
\(|x-3|\) means distance between x and 3. Therefore the distance from x to 1 and 3 are equal.
x is in between 1 and 3 , that is x is the midpoint of 1 and 3.
\(x=\frac{1}{2} \times(1+3)=\frac{1}{2} \times 4=2\)

ii. \(|x-3|=|x-4|\)
The distance from x to 3 and 4 are equal.
∴ x is in between 3 and 4 , that is x is the midpoint of 3 and 4.

iii. \(\begin{array}{l}{|x+2|=|x-5|} \\ {|x+2|=|x-(-2)|}\end{array}\)
The distance from x to -2 and 5 are equal.
∴ x is in between -2 and 5 , that is x is the midpoint of -2 and 5.

iv. |x| = |x + 1|

|x + l] = ]x – (-l)|
The distance from x to -1 and 0 are equal.
∴ x is in between -1 and 0, that is x is the midpoint of -1 and 0.

9th Std Kerala Syllabus Maths Solutions  Question 2.
Prove that if 1 < x < 4 and 1 < y < 4, then |x – y| < 3.
Answer:
1 < x < 4 , possible values of x = 2, 3 1 possible values of y = 2, 3
|x – y| = |2 – 2| = 0 < 3
|x – y| = |3 – 2| – 1 < 3
|x – y| = |2 – 3| = |-l| = l < 3
|x – y| = |3 – 3| = 0 < 3
From this if 1 < x < 4, 1 < y < 4, then |x – y| < 3.

Hss Live Guru Maths 9 Kerala Syllabus  Question 3.
Prove that if x < 3 and y > 7, then |x – y) > 4 .
Answer:
Since x < 3, the maximum value of x is less than 3. Since y > 7 the minimum value of y is greater than 7.
Then the difference between the minimum values of y and the maximum value of x is greater than 7 – 3 = 4.
i.e., |x – y| > 4

9th Standard Maths Notes Kerala Syllabus  Question 4.
Find two numbers x, y such that
\(|x+y|=|x|+|y|\)
Answer:
1. If x = 3, y =7, then
| x + y | =  |13 + 71| = 3 + 7 = 10
|x| + |y| = |3| + |7| = 3 + 7 = 10
|x + y| = |x| + |y|

2. If x = -6, y =-9 , then
|x + y| = |-6 + -9|
= l-15| = 15

Class 9 Maths Chapter 10 Kerala Syllabus Question 5.
Are there numbers x,y such that |x + y| < |x| + |y| ?
Answer:
x = 5, y = -3

|x+y| = |5 + -3| = |5 – 3| = |2| = 2

|x| = |5| = 5

|y| = |-3| = 3 |x| + |y| = 5 + 3 = 8

Among x, y, if one is a positive number and the other is a negative number, then
| x + y | < |x| + |y|. Question 6. Are there numbers x, y such that |x + y| > |x| + |y| ?
Answer:
No

Real Numbers 9th Standard Kerala Syllabus Question 7.
What are the numbers x, for which
|x – 2| + |x – 8| = 6
Answer:
\(|x-2|\) means distance between x and 2.
x can be to the right or left side of the number 2.
\(|x-8|\) means distance between x and 8
x can be to the right or left side of the number 8.
If we add the distance between x and 2 with x and 8 we get 6.
When x is to the left side of 2, when we add the distance of x to 2 and 8 we get a number which is greater than 6.
Distance between 2 and 8
= |2 – 8| = | -6| = 6
The difference between 2 and 8 is 6, so the position of x is between 2 and 8.
That is the value of x is in between 2 and 8 including 2 and 8.
i. e., 2 ≤ x ≤ 8

Std 9 Maths Kerala Syllabus Question 8.
What are the numbers x, for which
|x – 2| + |x – 8| = 10 ?
Taking x as different numbers, what all numbers do we get as
|x – 2| + |x – 8| ?
Answer:
|x – 2| + |x – 8| = 10, |x – 2| – |x – 8| = 10
x – 2 + x – 8 = 10
x = 10
( x – 2) – (x – 8) = 10
There is no number x which satisfies this.
– ( x- 2) + (x – 8) = 10
There is no number x which satisf¬ies this.
-(x – 2) – (x – 8) =10
-x + 2 – x + 8=10
-2x = 0
= x = 0
i.e., x = 0, 10

If x = 1
|x – 2| + |x – 8| = |1 – 2| + |1 – 8|
= | -1 | + | -7| = 1 + 7 = 8

If x = 2
|x – 2| + |x – 8| = |2 – 2| + |2 – 8|
= |0| + | -6| = 0 + 6 = 6

If x = 3
|x – 2| + |x – 8| = |3 – 2| + |3 – 8|
= | -1| + | -5| = 1 + 5 = 6

If x = 4
|x – 2| + |x – 8| = |4 – 2| + |4 – 8|
= | -2| + | -4| = 2 + 4 = 6

If x = 5
|x – 2| + |x – 8| = [5 – 2| +15 – 8|
= | -3| + | -3| = 3 + 3 = 6

If x=6
|x – 2| + |x – 8| = |6 – 2| + |6 – 8|
= | -4|+ | -2| = 4 + 2 = 6

If x = 7
|x – 2| + |x – 8| = |7 – 2| + |17 – 8|
= |-5| + |-1| = 5 + 1 = 6
i.e., |x – 2| + |x – 8| = 6 when 2 < x < 8

If x=9
|x – 2| + |x – 8| = |9 – 2| + |9 – 8|
= |7| + |1| = 7 + 1 = 8

If x= 11
|x – 2| + |x – 8| = |11 – 2| + |11 – 8|
= |9| + |3|= 9 + 3 = 12
…………. etc
Taking x as different numbers, different numbers get as |x – 2| + |x – 8|.

Real Numbers Exam Oriented Questions and Answer

Hss Live Guru Class 9 Maths Kerala Syllabus Question 1.
a. Which number indicates the point located 7 units to left of 5 on the number line?
b. Write down the numbers located 2 units apart from these points.
Answer:
a. The point which is 7 units to the left of 5 = 5 – 7 = -2
b. There is one point 2 unit apart to the left and right side of -2 .
The point which is 2 units to the left of -2 = -4
The point which is 2 units to the right of -2 = 0

9th Standard Maths Textbook Kerala Syllabus Question 2.
How many numbers are on the number line which are 11 units apart from 5. What are they ? Calculate the distance between the numbers.
Answer:
There is one point 11 unit apart to the left and right side of 5.
Point on the left side = 5 + -11 = -6
Point on the right side = 5 + 11 = 16
Distance between them = 16 – (-6) = 16 + 6 = 22 unit

Chapter 10 Maths Class 9 Kerala Syllabus Question 3.
Find the distance between the num-bers given below on the number line.
i. 4, 6
ii. 3, -2
iii. -5, -8
iv. 3/5, 5/6
Answer:
i. Distance = |6 – 4| = |2| = 2

ii. Distance = |3 – (-2)| = |3 + 2| = |5| = 5

iii. Distance = |-5 -(-8)| =|-5 + 8|
= |+3| = 3

9th Std Maths Notes Kerala Syllabus Question 4.
The endpoints of one side of an equilateral triangle are -2 and 4 on the number line. Calculate the area and perimeter of the triangle.
Answer:
The distance between -2 and 4 = 4 – (-2) = 6
One side of a equilateral triangle = 6 unit Perimeter of an equilateral triangle
= 3 × 6 = 18 unit Area of an equilateral triangle

Class 9 Maths Kerala Syllabus Question 5.
The number x is on the number line, then find x if |x – 10| = – |x – 4|.
Answer:
The distance between x to 10 and 4 are same. Therefore x is the midpoint of 4 and 10.

Real Numbers Class 10 State Syllabus Kerala Syllabus Question 6.
The quadrilateral ABCD is a square. A and B the points denote -2 and 3 on the number line.
a. Draw the square ABCD on the number line
b. Calculate the perimeter of the square ABCD.
Answer:

b. AB = |-2 – 3| = 5 unit
Perimeter = 5 × 4 = 20 unit

Question 7.
If |a + 1| = |a + 5 |, |b – 2| = |b – 6|,
|a – x| = |b -x|, then find x.
Answer:
|a + 1| = |a + 5|
Since the point, ‘a’ represented on the num¬ber line is at a the equal distance from -1 and -5.
That is ‘a’ is midpoint of -1 and -5.

|b – 2| = |b – 6|
Since the point, ‘a’ represented on the number line is at a equal distance from 2 and 6.
That is ‘a’ is midpoint of 2 and 6

|x – a| = |x – b| therefore point x is the
midpoint of a = -3 and b = 4

Question 8.
Find the value of x.
a. |x – 2| = |x – 10|
b. |x + 3| = |x – 7|
Answer:
a. The distance between x to 2 and 10 are same. Therefore x is the midpoint of 2 and 10.

b. |x + 3| = |x – (-3)|
The distance between x to -3 and 7 are same. Therefore x is the midpoint of -3 and 7.

Question 9.
If the difference between two numbers is 4. If one number is 8. What values will the other number have?
Answer:
Consider other number as x, then
|x – 8| = 4
i.e., x – 8 = 4 or 8 – x = 4
x = 12 or x = 4
Other number as 12 or 4.

Question 10.
Find the distance between the pairs of numbers given below.
a. 2, 8
b. -2, 8
c. -2, -8
Answer:
a. Distance between 2 and 8
= |x – y| = |2 – 8| |-6| = 6

b. Distance between -2 and 8
= |-2-8| = |-10| = 10

c. Distance between -2 and -8
= |-2 – (-8)| = |-2 + 8| = |6| = 6

Question 11.
Draw the number line. Mark the position of numbers given below.

Answer:

Question 12.
The number x is located to the right side of 3 on the number line, if
|x – 3|= 0, then
a. What is the value of x ?
b. Calculate |x – 3| + |x – 9|.
Answer:
a. |x – 3| = 0
-(x – 3) = 0 or x – 3 = 0
-x = -3 or x = 3
x=3

b. |x – 3| + |x – 9|
= |3 – 3| + |3 – 9|
= |0| + |-6 | = 6

Question 13.
If |x – 5| = 10, then find the value of x .
Answer:
The distance between x and 5 is 10.
If x is to the right side of 5 , then x = 5 + 10 = 10
If x is to the left side of 5 , then
x = 5 – 10 = -5

Question 14.
If |x + 2| = |x – 8|, then find the value of x ?
Answer:
|x + 2| = |x – (-2)|
The distance of x to -2 and 8 are equal.
Therefore position of x is between -2 and 8.

Question 15.
If |x – 1| =|x – 3|, then find the value of x.
Answer:
Distance from 1 and 3 to x are same.
Therefore position of x is between 1 and 3.
Position of x = \(\frac { 1 + 3 }{ 2 }\) = \(\frac { 4 }{ 2 }\) = 2

Question 16.
Using the common form of rational numbers, prove that the sum, difference, product and quotient of any two rational numbers is again a rational number.
Answer:
i. If a, b, c and d are natural numbers,
\(\frac { a }{ b }\), \(\frac { c }{ d }\) are rational numbers.

Sum and product of all-natural numbers are always natural numbers. So here the numerator and denominator are natural numbers.
That is if x and y are natural numbers
then this can be expressed as x/y.
Therefore the sum is rational

ii. Difference:
a, b, c and d are natural numbers

Hence it is rational.

iii. Product:
a, b, c and d are natural numbers

Hence it is rational.

iv. Quotient:
a, b, c and d are natural numbers

Hence it is rational.

Question 17.
Prove that the product of any irrational number and non – zero rational number is an irrational number.
Answer:
Let a be the irrational number and b the rational number and a × b = c, that is to prove that c is irrational.
Consider c is a rational number, a × b = c
From this we get b = c/a
That is b = \(\frac { One rational }{ other rational }\) = One rational
If we take b as irrational here we get b as a rational number. This is a wrong decision. This is due to our wrong assumption that is c is a rational number. The assumption that c is a rational number is a wrong one.
∴ c is an irrational number.
That is the product of any irrational number and non – zero rational number is an irrational number.

Question 18.
Give an example of two different irrational numbers whose product is a rational number.
Answer:
3√2 and 5√2 are irrational numbers
3√2 × 5√2 = 3 × 5 × √2 × √2 = 3 × 5 × 2 = 30 is a rational nummber. More examples
1. √3 × 7√3 = 4 × 7 × 3 = 84
2. 2√5 × 5√5 = 2 × 5 × 5 = 50
3. √20 × √5 = √20 × √5 = √100 = 10

You can Download Protectors of Biosphere Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Biology Solutions Part 1 Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Biology Solutions Chapter 1 Protectors of Biosphere

Protectors of Biosphere Textual Questions and Answers

Kerala Syllabus 9th Standard Biology Notes Chapter 1 Global Warming

Global warming is the increase of earth’s average surface temperature due to effect of greenhouse gases. Deforestation, atmospheric pollution, burning of fossil fuels, etc.
1) At least do not ruin the life of those trees, depending on whom we live”.
2) ‘Plant together let’s make the world greener.’
3) It is our duty to save environment duties.
4) Plant trees: for our future.
5) Plant trees: It is the only way to prevent Global warming.

Kerala Syllabus 9th Standard Biology Notes Question 1.
………… is the result of deforestation, atmospheric pollution, etc.?
Answer:
Global warming.

Protectors Of Biosphere Class 9 Notes Kerala Syllabus Question 2.
What is Global warming?
Answer:
Global warming is the increase of earth’s average surface temperature due to effect of greenhouse gases.

Class 9 Biology Notes Kerala Syllabus  Question 3.
What are the causes of Global warming?
Answer:
Deforestation, atmospheric pollution, burning of fossil fuels, etc.

9th Biology Notes Kerala Syllabus Question 4.
“Trees are essential for protecting our nature and life.” Prepare slogans highlighting this idea to protect our environment?
Answer:

• At least do not ruin the life of those trees, depending on whom we live”.
• ‘Plant together let’s make the world greener.’
• It is our duty to save environment duties.
• Plant trees: for our future.
• Plant trees: It is the only way to prevent Global warming.

Pigments In Leaves

9th Class Biology Notes Kerala Syllabus Question 5.
Which of the pigments given below is the main pigment that performs photosynthesis?
a) Chlorophyll a
b) Chlorophyll b
c) Xanthophyll
d) Carotene
Answer:
a) Chlorophyll a

Kerala Syllabus 9th Standard Biology Notes Malayalam Medium Chapter 1 Question 6.
Complete the equation related to photosynthesis

Answer:

Protectors Of Biosphere Class 9 Kerala Syllabus Question 7.
Complete the word relations.
a) Light reaction: Grana
Dark reaction: ……………
b) Green plants: land
……………..: Ocean
c) …………..: bluish green
Carotene: yellowish-orange
Answer:
a) Stroma
b) Algae
c) Chlorophyll a

9th Class Biology Notes Chapter 1 Kerala Syllabus Question 8.
………. is the main source of energy on the earth
Answer:
Sunlight

Kerala Syllabus 9th Standard Biology Notes Pdf Question 9.
Which group of living things can absorb solar energy directly?
Answer:
Plants

9th Class Biology Chapter 1 Kerala Syllabus Question 10.

Have you noticed Veena’s doubt’? Write down your inference?
Answer:
More number of chloroplasts are seen on the upper part of leaf than on the lower part. That is why the lower part of the leafless green in color.

9th Class Biology Chapter 1 Notes Kerala Syllabus Question 11.
Odd one out, give reason Carotene, xanthophyll, chlorophyll a
Answer:
Chlorophyll a. Others accessory pigments.

Hsslive Guru Class 9 Biology Kerala Syllabus Question 12.
………. part of leaf are seen chloroplasts more in number
Answer:
Upper part

Kerala Syllabus 9th Standard Biology Notes Malayalam Medium Question 13.
Photosynthesis in green plants taken mainly in
Answer:
leaves

Hss Live Guru 9 Biology Kerala Syllabus Question 14.
How many chloroplasts will be there in one square millimeter of a leaf?
Answer:
At an average 5 lakh chloroplasts

9th Class Biology First Chapter Kerala Syllabus Question 15.
Identify the missing parts in the picture.

Answer:
a) Grana
b) Stroma

Question 16.
The fluid part of the chloroplast is
Answer:
Stroma

Question 17.
What are the pigments seen in the grana?
Answer:
Chlorophyll a, Chlorophyll b, Carotene, and Xanthophyll are seen in the grana.

Question 18.
Describe the structure of chloroplast?
Answer:
The chloroplast is an organelle with a double-layered membrane. The fluid part of the chloroplast is the stroma. The membranous sacs arranged on above other is the grana. The membrane-bound structures joining the grana are the stroma lamellae, Pigments that can absorb sunlight are seen in the grana.

Question 19.
What is the function of accessory pigments?
Answer:
Chlorophyll b, Carotene, and Xanthophyll are accessory pigments. They are seen in the grana. They can absorb light and transfer it to chlorophyll a.

Question 20.
How do carbon dioxide and water reach the leaves for photosynthesis?

Answer:

The Chemistry Of Photosynthesis

Question 21.
Differentiate between Light reaction and Dark reaction; Prepare a table?
Answer:

Question 22.
Find out whether the statements given below are true or false?
a) Dark reaction occurs in the grana.
b) Light reaction stops when availability of light decreases.
Answer:
a) False
b) True

Question 23.
Light reaction occurs in the ………….
Answer:
grana of chloroplast

Question 24.
Dark reaction occurs in the …………..
Answer:
Stroma of chloroplast.

Question 25.
Complete the flow chart related to photosynthesis.

Answer:
a) Light is required
b) Glucose is produced
c) Takes place in the grana

Question 26.
During light reaction, light energy is converted to ……… and stored as ………..
Answer:
Chemical energy, ATP molecules

Question 27.
Water splits into & in the light reaction
Answer:
Hydrogen & Oxygen

Question 28.
Define Dark reaction?
Answer:
Sunlight is not utilized in this phase that occurs in the stroma of chloroplast. In this process, hydrogen is added to C02 to form glucose using the energy of ATP molecules.

Question 29.
Illustrate an experiment to prove that plants released oxygen during the time of photosynthesis.
Answer:

Question 30.
Dark reaction is also known as …………
Answer:
Calvin cycle

Question 31.
……….. is known as the energy currencies of the cell.
Answer:
ATP

Question 32.
Expansion of ATP =
Answer:
Adenosine Tri Phosphate

Question 33.
About 70-80% of oxygen in the atmospheric air is contributed by ………..
Answer:
algae in the ocean

Question 34.
What is the contribution of Melvin Calvin?
Answer:
Melvin Calvin explained the various stages of the formation of glucose during photosynthesis.

Question 35.
Prepare the equation showing the process of photosynthesis?
Answer:

After Photosynthesis

Question 36.

Isn’t doubt genuine? What happens to the glucose formed as a result of photosynthesis? Explain your opinion?
Answer:
Glucose is easily soluble in water. It can’t be stored in plant body. Therefore plants store glucose in the form of insoluble starch in leaves. Plants utilize starch as a source of energy for life activities and to prepare substances required for growth. Starch is converted to sucrose and it transported through phloem to various plant parts and to store in different forms such as starch, protein, fat, fructose and sucrose.

Question 37.
Prepare a flow chart showing the chemical changes of glucose after its formation
Answer:

Question 38.
In which form is starch transported to various parts of the plant body for storage?
Answer:
Sucrose

Question 39.
Complete the table?

Answer:

Question 40.
Plants store glucose in the form of ………….. in leaves.
Answer:
Insoluble starch.

Question 41.
Write the examples of economically important plant products.
Answer:
Coffee, Rubber, Pepper, Cocoa

Ocean At Par With Land

Question 42.
‘Ocean – an ecosystem’ – Prepare a flow chart.
Answer:
Sunlight + CO₂ → Algae & other aquatic Organisms → Photosynthesis
Fish and other aquatic organisms → Birds & Terrestrial organisms

Question 43.
How does pollution in the ocean affect the organisms?
Answer:
Ocean is polluted due to sewage, toxic chemicals from industries, land runoff, large scale oil spills, ocean mining, Littering, etc. As a result of ocean pollution, there are several negative impacts such as
1) Effect of toxic wastes on marine animals.
2) Disruption to the cycle of coral reefs
3) Depletes oxygen content in water
4) Failure in the reproductive system of sea animals.
5) Effects on food chain.
6) It affects human health.
In this way ocean pollution adversely affects the organisms.

Plants – Earth’S Wealth

Question 44.
Do we obtain food only from producers?
Answer:
No plants are the cheapest and effective natural mechanism for the purification of air. It offers great service to the biological world by absorbing CO₂ from the
atmosphere and releasing oxygen.

Question 45.
The service rendered by plants for the sustenance of the living world is unique comment on this statement.
Answer:
Plants serve as the cheapest, effective and natural means for the purification of air. Plants also have a . major role in the mitigation of natural disasters. Mangrove forests help in controlling Tsunami to some extend. Bamboo forests, reed, vetiver, lemongrass, etc protect the river banks from collapsing during flood. Trees and bushes in mountains and hills prevent soil erosion and landslides.

Protectors of Biosphere Additional Question and Answers

Question 46.
“Trees are essential for protecting our nature and life.” Prepare slogans highlighting this idea to protect our environment?
Answer:

• At least do not ruin the life of those trees, depending on whom we live”.
• Plant together let’s make the world greener.’
• It is our duty to save environment duties.
• Plant trees: for our future.
• Plant trees: It is the only way to prevent Global warming.

Question 47
What is photosynthesis?
Answer:
Photosynthesis is a process used by plants and other organisms to convert light energy into chemical energy that can be later released to fuel the organisms’ activities.

Question 48.
What do you mean by greenhouse gases? Give examples?
Answer:
Greenhouse gases are gases that contribute to the greenhouse effect by absorbing infrared radiation. Carbon dioxide and chloroform carbons are examples of greenhouse gases.

Question 49.
“They kill Good Trees To put out Bad Newspapers” -comment.
Answer:
There are wide range of cutting trees all over the world. Many such instances are for making newspapers. In the same newspapers, we proclaim that trees should not be cut down. It is really a controversy. Similarly, bad news and wrong messages are passed onto people & generations by spoiling the life-supporting trees. Therefore
protection of trees is more important.

Question 50.
“Sale of AC’ has increased in Kerala due to severe heat” – Comment on this pews paper statement?
Answer:
The temperature in Kerala has gone up to the record figure this year. This has happened due to the practice of cutting down trees and pollution. To escape from the heat people started purchasing AC instead of planting trees. The excessive use of AC will again increase heat in the atmosphere by increased emission of chloroform carbon. This will finally intensify the magnitude of Global warming.

Question 51.
Plants give out approximately ………. of oxygen when they use one tonne CO₂
Answer:
118 kg

Question 52.
Why the plants are called the lungs of the earth? Explain.
Answer:
Plants are the cheapest and effective natural mechanism for the purification of air. They absorb carbon dioxide from the atmosphere & release oxygen. It is estimated that plants give out approximately 118 kilograms of oxygen when they use one tonne CO₂. As the plant cover on the earth decreases, this recycling mechanism stops and air pollution becomes severe. That is why the plants are called the lungs of the earth.

You can Download Reflection of Light in Spherical Mirrors Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 18 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 18 Reflection of Light in Spherical Mirrors

The phenomena of light is always a surprising one. From long time ago man has tried to study about light. This chapter explains to draw the figure of the images formed by mirrors, their uses, magnification etc.

Reflection Of Light In Spherical Mirrors Class 8 Notes Spherical mirrors

Images are formed not only in-plane mirrors but also in smooth curved surfaces. Spherical mirrors are mirrors in which the reflecting surface is a part of the sphere. The center of a sphere of which the mirror is a part is the center of curvature. Any I line drawn from the center of curvature to the mirror is normal to the mirror. Radius of curvature of a mirror is the radius of the sphere of which is a part. The reflecting surface of the mirror is called the aperture of a mirror. The midpoint of the reflecting surface is called the pole. The straight line connecting the pole and center of curvature of a mirror is the principal axis of the mirror.

The angle of incident and angle of reflection are equal in spherical mirrors. Rays of light incident on a concave mirror, parallel to the principal axis, passes through a fixed point on the principal axis after reflection. This point is the principal focus of the concave mirror. Rays of light incident on a convex mirror parallel to the principal axis appear to come from a fixed point on the other side of the mirror. The point is the principal axis of the convex mirror.

Focal length of a mirror is the distance from pole of the mirror and the principal’s focus of the mirror. Rays of light coming from infinity get focused on a plane perpendicular to the principal axis. This plane is the focus plane. The focus plane passes through the focus plane.

Reflection Of Light In Spherical Mirrors Class 8 Images formed by spherical mirrors

The image of an object placed different positions in front of a mirror is.formed in different positions. Rays of light incident on a concave mirror, parallel to the principal axis, passes through a fixed point on the principal axis after reflection Rays incident on an a concave mirror through the principal focus are reflected parallel to the principal axis.

Rays incident through the center of curvature reflected through the same path. Rays incident on the pole makes an angle with the principal axis. The ray diagrams of images formed by spherical mirrors are drawn according to the above rules. An object placed between F and C in front of a concave mirror, the image will be real and inverted. If the object is at F the image is formed at infinity. The paths of reflected rays are parallel to each other. If the object is in between F and P the image is formed behind the mirror. The image is erect and virtual.

Reflection Of Light In Spherical Mirrors 8th Magnification

Magnification is the ratio of the height of the image to the height of the object.

Reflection Of Light In Spherical Mirrors Class 8 Notes Pdf Uses of spherical mirrors

Spherical mirrors are used in lighthouses and reflectors.

Reflection of Light in Spherical Mirrors Textbook Questions and Answers

Class 8 Physics Notes Kerala Syllabus Questions 1.
Classify the following statements as to those related to concave mirrors and convex mirrors and tabulate them accordingly.
a. to view the face
b. as makeup mirror
c. as rearview mirrors in vehicles
d. in solar concentrators
e. in periscopes
f. as shaving mirror
Answer:
Concave mirror:

• In solar concentrators
• Makeup mirror
• Shaving mirror

Convex mirror:

• In rearview mirrors of vehicles
• In Searchlights

Plane mirror:

• In periscopes
• To see face

8th Class Physics Notes Pdf Question 2.
Calculate the radius of curvature of a convex mirror of focal length 12 cm.
Answer:
focal length of the mirror = 12 cm
\(\mathrm{f}=\frac{\mathrm{R}}{2} \quad 12=\frac{\mathrm{R}}{2}\)
R = 2 × 12 = 24 cm

Reflection Of Light At Curved Surfaces Questions And Answers 8th Question 3.
A ray of light is made to fall on the pole of a concave mirror making an angle 30° with the principal axis.
a. What is the angle of reflection?
b. Justify your answer.
c. Draw the ray diagram.
Answer:
a. Angle of reflection is 30°
b. angle of incidence and angle of reflection are equal
c. the figure showing angle of incidence and angle of reflection is 210°

Chemistry Class 8 Kerala Syllabus Question 4.
Which type of mirror always gives an erect and diminished image?
Answer:
Convex mirror

Kerala Syllabus 8th Standard Chemistry Notes Question 5.
A ray of light incident on a spherical mirror gets reflected along the same path. If so, show the light incident on the mirror.
Answer:

8th Class Biology Notes Pdf Question 6.
OA is a ray of light incident on a concave mirror.
a. Draw the path of the reflected ray

b. On what basis did you mark the reflected ray?
Answer:

b. In mirrors angle of incidence and angle of reflection are same. The normal to the point of incidence is passed through the center of curva¬ture. The angle between ray of reflection and the normal is the same as angle of incidence.

8th Class Biology Notes Pdf Malayalam Medium Question 7.
Write down the type of mirrors that should be used for getting the following type of images.
a. real and magnified
b. virtual and magnified
c. virtual and diminished
d. real and diminished
Answer:
a. concave mirror
b. concave mirror
c. convex mirror
d. concave mirror

Kerala Syllabus 8th Standard Chemistry Notes Malayalam Medium Question 8.
The height of an object kept 12 cm away from a concave mirror is 1 cm. Calculate the magnifica¬tion if an image of height 2.5 cm is formed in front of the mirror.
Answer:

8th Standard Chemistry Textbook Question 9.
Which type of mirror always gives a virtual and erect-image, b. Is this image magnified or diminished?
Answer:
a. convex mirror
b. Diminished

Reflection of Light in Spherical Mirrors Additional Questions and Answers

8th Class Biology Notes Pdf Kerala Syllabus Question 1.
Complete the table

Answer:
a. 30°
b. 40°
c. 50°
d. 60°

Kerala Syllabus 8th Standard Physics Notes Question 2.
If the radius of curvature of a concave mirror is 24 cm what is its focal length?
Answer:
R = 24 cm

8th Standard Chemistry Textbook Kerala Syllabus Question 3.
Find the radius of curvature of a convex mirror of focal length 0.6 m
Answer:

Basic Science Class 8 Solutions Question 4.
Write the characteristics of the image formed by an object plac¬ed at the center of curvature of a concave mirror?
Answer:
Position: At the center of curvature at the same side
Size : Same as the size of the object Nature: Real, Inverted

Reflection Of Light By Spherical Mirrors 8th Question 5.
Complete the figure

Answer:

Hsslive Guru Physics 8th Standard Question 6.
Complete the table

Answer:
a. Passes through principal focus
b. Seem to come from the principal focus
c, d. Returns parallel to the principal axis
e, f. Returns through the same path
g, h. Reflects in the same angle of incident ray.

Question 7.
Write three differences of real image and virtual image which is made by spherical mirrors
Answer:
Real image:

1. Inverted
2. can be shown on the screen
3. can measure the length of the image and distance to the image

Virtual image:

1. Cannot show on the screen
2. Cannot measure
3. Erect

Question 8.
Write uses of concave mirrors
Answer:

•  As shaving mirror
• As makeup mirror
• As head mirrors used by doctors
• In film projectors

Question 9.
Examine the position off the object given in the figure and table the following peculiarities

a. position of the image
b. size of the image
c. nature of the image
Answer:
a. Behind the mirror.
b. Larger than the object.
c. Erect and virtual

Question 10.
When an object of height 4 cm is placed in front of a concave mirror an image of height 8cm is formed. Find magnification.
Answer:
h_i = 4cm h_o= -8 cm
Magnification = \(\frac{h_{i}}{h_{0}}\) = \(\frac{h_{-8}}{h_{4}}\) = -2

Question 11.

a. Examine the figure and find the magnification
b. What is the height of the object if height of the image is 4cm when the object is placed on the same position in front of the mirror.
Answer:

Question 12.
Write the uses of convex mirror
Answer:

• As reflector in the street light
• As rearview mirror
• In searchlights

You can Download Statistics Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 13 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 13 Statistics

Statistics Textual Questions and Answers

Textbook Page No. 193

Statistics Class 9 Kerala Syllabus Question 1.
The weight of 6 players in a volleyball team are all different and the average weight is 60 kilograms.
i. Prove that the team has at least one player weighing more than 60 kilograms.
ii. Prove that the team has at least one player weighing less than 60 kilograms.
Answer:
i. Total weight of 6 players = 60 × 6 = 360 kg
The team contains players having weights 60, less than 60 or greater than 60. If the weights of all the players are less than 60, the average will also be less than 60. This is not possible. Therefore there will be at least one player having weight greater than 60.

ii. If the weight of all the players are greater than 60, the average will also be greater than 60. Therefore there will be at least one player having weight less than 60.

Hss Live Guru 9th Maths Kerala Syllabus Question 2.
Find two sets of 6 numbers with average 60, satisfying each of the conditions below:
i. 4 of the numbers are less than 60 and 2 of them greater than 60.
ii. 4 of the numbers are greater than 60 and 2 of them less than 60.
Answer:
Total sum = 60 × 6 = 360
i. 20, 30, 40, 50, 100, 120

ii. 5, 15, 70, 80, 90, 100
Other ways are also possible.

Kerala Syllabus 9th Standard Maths Notes Question 3.
The table shows the children in a class, sorted according to the marks they got for a math test.

Calculate the average marks of the class.
Answer:
Total number of children is 45. Repeated addition can be written as multiplication.

Average mark = \(\frac { Total }{ Number }\) = \(\frac { 297 }{ 45 }\) = 6.6

Class 9 Maths Solutions Kerala Syllabus Question 4.
The table below shows the days in a month sorted according to the amount of rainfall in a locality

What is the average rainfall per day during this month?
Answer:

The average of the rain fall per day during that month = \(\frac { Total rain fall }{ Number of days }\)
= \(\frac { 1545 }{ 30 }\) = 51.5mm

Hsslive Guru 9th Maths Kerala Syllabus Question 5.
The details of rubber sheets a farmer got during a month are given below.

i. How many kilograms of rubber did he get a day on average in this month?
ii. The price of rubber is 120 rupees per kilogram. What is his average income per day this month from selling rubber?
Answer:


i. Average Quantity of rubber per day = \(\frac { 381 }{ 30 }\) = 12.77kg
ii. . If the price is Rs. 120 per kg, then average incomeperday = 12.7 × 120 = Rs.1524

Textbook Page No. 197

Kerala Syllabus 9th Standard Maths Solutions Question 1.
Find different sets of 6 different numbers between 10 and 30 with each number given below as mean:
i. 20
ii. 15
iii. 25
Answer:
i. The mean of 6 numbers is 20.
ie; sum = 6 x 20= 120 (Write 3 pairs with sum 40)
i.e., 15, 25, 18, 22, 19, 21

ii. The mean of 6 numbers is 15
i.e; sum =6 x 15=90
(Write 30 pairs with sum 3)
12, 18, 13, 17, 14, 16

iii. Mean is 25
sum = 25 x 6 = 150
(Write 50 pairs with sum 3)
22, 28, 23, 27, 24, 26

Hss Live Guru Class 9 Maths Kerala Syllabus Question 2.
The table below shows the children in a class, sorted according to their heights.

What is the mean height of a child in this class?
Answer:

Mean height = \(\frac { Total height }{ No of children }\)
= \(\frac { 7892 }{ 50 }\) = 157.84 cm

Hss Live Class 9 Maths Kerala Syllabus Question 3.
The teachers in a university are sorted according to their ages, as shown below.

What is the mean age of a teacher in this university?
Answer:

Mean age = \(\frac { Total age }{ No of persons }\)
= \(\frac { 2775 }{ 70 }\) = 39.64

Kerala Syllabus 9th Standard Maths Question 4.
The table below shows children in a class sorted according to their weights.

The mean weight is calculated as 26 kilograms. How many children have weights between 23 and 25 kilograms?
Answer:
Let’s prepare the table for finding the mean by considering the number of children in the group 23 to 25, as ‘x’.

Mean weight = 26 kg

i.e; the number of children having weight between 23 to 25 is 7.

Statistics Exam oriented Questions and Answers

Kerala Syllabus Std 9 Maths Solutions Question 1.
The details of rubber sheets got for a month by a farmer are given in the table.

During this month he got an average of 10 sheets per day. If so in how many days did he get 11kg per day ?
Answer:

Let ‘x’ be the number of days in which he got 11 kg rubber sheet.

∴ He got 11 kgs of sheets for 5 days.

Kerala Syllabus 9th Standard Maths Notes Malayalam Medium Question 2.
In a factory, there are workers belonging to four categories. The average income and the number of workers in each category are given. What is the mean income when all the workers in the four categories are combined?

Answer:

Mean income = \(\frac { 320000 }{ 40 }\)
= Rs. 8000

Kerala Syllabus 9th Standard Maths Solution Question 3.
The daily wages of 10 workers in a factory are given below.
400, 350, 450, 500, 400, 500, 350, 500, 350, 450
If one more person is joined, the mean becomes Rs. 450. What is the daily wage of the new person?
Answer:
Total wages of 10 workers = 4250 Total wages of 11 workers= 11 × 450 = Rs. 4950
Wage of the 11th person = 4950 – 4250

Kerala State Class 9 Maths Solutions Question 4.
Find 10 different numbers between 10 and 30 whose mean is 20.
Answer:
Given mean is 20
Sum 20 × 10 = 200.
We have to find 10 different numbers whose sum is 200 (for this find 5 pairs of sum 40)
(15, 25) (16, 24) (17, 23) (18, 22) (19, 21)
The numbers are 15, 16, 17, 18, 19, 21, 22, 23, 24, 25

9th Maths Notes Kerala Syllabus Question 5.
A table categorizing the workers in an office on the basis of their salary is given below.

Find the mean of salary.
Answer:

Mean income = \(\frac { 349500 }{ 15 }\)
= R.s 23300

Kerala Syllabus 9th Standard Notes Maths Question 6.
i. Find the mean of natural numbers from 1 to 100.
ii. What is the mean of even numbers from 1 to 100? What is the mean of odd numbers?
iii. What is the difference between the means of the first 100 even numbers and odd numbers?
iv. What is the difference between the means of the first 200 even numbers and 200 odd numbers?
Answer:
Sum of the natrural numbers from 1 to n = \(\frac n{ n + 1 }{ 2 }\)


Sum of the first 100 odd numbers
= 1002 = 100 × 100

In general, the difference between the means of n even numbers and n odd numbers is always 1.

Question 7.
A table tabulating the players in a cricket team on the basis of their age is given below.

Calculate the mean age of the players?
Answer:

Mean age of players = 306/12 = 25.5

You can Download अकाल में सारस Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Hindi Solutions Unit 4 Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Hindi Solutions Unit 4 Chapter 1 अकाल में सारस (कविता)

अकाल में सारस Textual Questions and Answers

अकाल में सारस विश्लेषणात्मक प्रश्न

अकाल में सारस कविता का आशय Kerala Syllabus 9th प्रश्ना 1.
धान की सूखी पत्तियों की गंध’ –से किसका आभास होता है?

उत्तर:
धान के पौधे गर्मियों में सूखते हैं। धान की सूखी पत्तियों की गंध से गर्मियों का आभास मिलता है।

Akal Mem Saras Hindi Poem Kerala Syllabus 9th प्रश्ना 2.
सारसों ने जल भर कटोरे को क्यों न देखा होगा?

उत्तर:
सारसों के लिए कटोरे भर का जल पर्याप्त नहीं है। वे ताल-तलैयों की तलाश कर रहे थे। उनके लिए प्राकृतिक जल स्रोतों की ज़रूरत थी। इसलिए वे जल भर कटोरे को न देखा होगा।

अकाल में सारस Summary Kerala Syllabus 9th प्रश्ना 3.
सारसों ने जाते-जाते शहर की ओर क्यों मुड़कर देखा होगा?

उत्तर:
शायद सारसों को वह जगह पसंदीदा थी। लेकिन उनकी आशा के विरुद्ध वहाँ कोई जलस्रोत न देखा। वे निराश लौटने लगे। वहाँ के लोगों के प्रति उनके मन में दया या घृणा की भावना थी। दया इसलिए कि ये लोग कितने अभागे हैं। घृणा शायद इसलिए कि इन्होंने प्राकृतिक जलस्रोतों का ह्रास किया है।

अकाल में सारस Additional Questions and Answers

अकाल में सारस आशयग्रहण के प्रश्न

Akaal Mein Saras Summary In Hindi Kerala Syllabus 9th प्रश्ना 1.
वे देर तक करते रहे शहर की परिक्रमा’ -सारस किसकी तलाश में शहर की परिक्रमा करते होंगे?

उत्तर:
सारस देश-देशांतर में जानेवाले पक्षी हैं। वे ऋतुओं के बदलने के अनुसार एक देश से दूसरे में चले जाते हैं। यहाँ सारस पानी की तलाश कर रहे हैं।

अकाल में सारस कविता की व्याख्या Kerala Syllabus 9th प्रश्ना 2.
बुढ़िया ने क्या सोचकर पानी लाकर रखा होगा?

उत्तर:
सारसों को देखकर बुढ़िया ने सोचा कि वे पानी की तलाश में हैं। इसलिए बुढ़िया ने कटोरे में पानी लाकर रखा।

अकाल में सारस Grammar

अकाल में सारस व्याकरण के प्रश्न

अकाल में सारस Summary In Hindi Kerala Syllabus 9th प्रश्ना 1.
तुलना करें :

i. तीन बजे दिन में / आ गए वे

ii. वे दिन में तीन बजे आ गए।

iii. कवितांश की संरचना और वाक्य की संरचना का अंतर पहचानें।

iv. कविता को गद्य में बदलने पर शब्दों के क्रम में क्या परिवर्तन आया है?

उत्तर:
वाक्य की संरचना कर्ता, कर्म और क्रिया का पालन किया है। लेकिन कविता की संरचना में यह क्रम नहीं है। कविता की सुंदरता को बढ़ाने के लिए कभी-कुभी शब्दों के क्रम में आवश्यक परिवर्तन करते हैं।

Hss Live Guru 9th Hindi Kerala Syllabus प्रश्ना 2.
इन पंक्तियों को गद्य की संरचना में बदलकर लिखें।

उत्तर:

9th Class Hindi Notes Kerala Syllabus प्रश्ना 3.
लेख लिखें :

1. प्रकृति के जलस्रोतों का नाश होता जा रहा है।
2. साथ ही पर्यावरण का संतुलन भी बिगड़ रहा है।

उत्तर:
अकाल में सारस’ कविता मनुष्यों के बुरे व्यवहार की संकेत करनेवाली कविता है। प्राकृतिक संसाधन केवल मनुष्यों के लिए नहीं है। पशु-पक्षी, मछलियाँ और दूसरे प्राणी भी इनके हकदार है। दरअसल ऋतु बदलने पर सारस पक्षी पानी आदि की तलाश में एक देश से दूसरे देश में चले जाते हैं। लेकिन यहाँ कहीं पानी नहीं है। सारसों को तो पता तक नहीं था कि लोग उन्हें सारस कहते हैं। यहाँ तो पानी और धान की पत्तियाँ सूख गई हैं। उनकी गंध हवा में हैं। यही गंध सारसों के डैनों से नीचे झर रही है। एक बुढ़िया अपने आँगन में जलभरा कटोरा रख देती है।

लेकिन सारसों ने न तो बुढ़िया को देखा। क्योंकि वे जलस्रोतों की तलाश में आए थे। उनकी निगाहों में दया थी या घृणा, हम समझ नहीं पाते। दया शायद इसलिए कि सारस तो उड़कर कहीं और चले जाएँगे, जहाँ पानी होगा। किंतु ये दयनीय लोग जो अपना पानी तक नहीं बचा पाए, अपने खेत-खलिहान और घर-बार छोड़कर, कहाँ और कैसे जाएँगे? और घृणा इसलिए कि अपना पानी तो नष्ट किया ही हमारा भी नष्ट कर दिया। क्योंकि पर्यावरण नष्ट करने के ज़िम्मेदार तो मनुष्य ही हैं।

अकाल में सारस Summary in Malayalam and Translation

Akal प्रश्ना 1.
बरसों बीते
बादलों को इधर
बरसे नहीं।
ये पंक्तियाँ किस हालत की ओर ।
संकेत करती है?

उत्तर:
पानी की कमी एक बड़ी समस्था हो.
पर्यावरण में हुए बदलाव से बारिश
की मात्रा कम होती जा रही है। इस
हालत की ओर ये पंक्तियाँ संकेत देती है 

अकाल में सारस शब्दार्थ

You can Download Unravelling Genetic Mysteries Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Biology Solutions Chapter 6 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Biology Solutions Chapter 6 Unravelling Genetic Mysteries

Unraveling Genetic Mysteries Text Book Questions and Answers

Emergence Of Genetics

Genetics is the branch of science emerged at the beginning of the 20th century. It influences almost all areas of life like diagnosis, therapeutic and food production. Gregor Johann Mendel’s hybridization experiments in pea plants have led to the foundation of genetics. So he is considered as the father of Genetics.

Unravelling Genetic Mysteries Kerala Syllabus 10th Chapter 6 Question 1.
What are the traits that were experimented by Mendel?
Answer:
The traits that were experimented by Mendel

• Height of the plant (Tall – Dwarf)
• Position of the flower (Terminal – Axial)
• Shape of the seed (Round – Wrinkled)
• Colour of the seed (Green-Yellow)
• Colour of the flower (Purple, White)
• Shape of the pod (Inflated – Constricted)
• Colour of the fruit (Yellow-Green)

Experiments Of Mendel

Sslc Biology Chapter 6 Kerala Syllabus Chapter 6 Question 2.
Which trait of the pea plant was considered in these experiments?
Answer:
Height of the pea plant was considered as the trait in this experiment.

Biology Chapter 6 Class 10 Kerala Syllabus Chapter 6 Question 3.
What variant forms of the trait are considered here?
Answer:
Tall and dwarf are the variant forms of the trait considered here.

Unravelling Genetic Mysteries Notes Kerala Syllabus 10th Chapter 6 Question 4.
Which forms of the trait was expressed in the first generation?
Answer:
Tall is the trait was expressed in the first generation.

Sslc Biology Chapter 6 Notes Kerala Syllabus Question 5.
The following illustration showing hybridization experiment in pea plants using symbols for the factors that control traits.
Answer:

Peculiarities of offsprings in the second generation

Mendel self-pollinated the plants obtained in the first generation and produced the second generation. Among the 1064 plants obtained in the second generation, 787 plants were tall and 277 plants were dwarf. The ratio of the result obtained is about 3:1

Statistics In Mendel’S Experiment

10th Class Biology 6th Chapter Kerala Syllabus Question 6.
Complete the table

Answer:
a) Axial
b)3:1
c) Round
d) 3:1

Inferences of Mendel:

• A trait is controlled by the combination of two traits.
• One character is expressed (dominant character) and the other character remains hidden (recessive character) in the offspring first generation.
• The character which remains hidden in the first generation appears in the second generation.
• The ratio of the dominant character and recessive character in the second generation is 3:1

Gene – Allele:
The gene present in the chromosome of the nucleus determines the character. A gene that controls a trait has different forms. The different forms of a gene are called alleles. Generally, a gene has two alleles. When.we illustrate hybridization experiment, the allele that controls the dominant character that is expressed in the first generation is indicated by a capital letter and the allele that controls recessive character is indicated by a small letter.

Sslc Biology Chapter 6 Malayalam Medium Question 7.
Which are the alleles of a tall plant?
Answer:
‘TT are the alleles of a tall plant.

Hsslive Guru 10th Biology Kerala Syllabus Chapter 6 Question 8.
Which are the alleles of the dwarf planets?
Answer:
‘tt’ are the alleles of the dwarf planet.

Sslc Biology Focus Area Questions And Answers Kerala Syllabus Chapter 6 Question 9.
How do the allele combination of the first generation differ from parental plants?
Answer:
The allele combination ‘Tt’ is responsible for the character height in first-generation instead of TT. One character is expressed and the other character remains hidden in the offsprings.

Kerala Genetics Question 10.
How can you differentiate alleles that control the dominant character and recessive character?
Answer:
The allele that controls dominant character generally indicated by capital letter and the allele that controls recessive character is indicated by a small letter.

Biology Class 10 Kerala Syllabus Kerala Syllabus Chapter 6 Question 11.
Observe the illustration showing the hybridization experiment conducted by Mendel on two traits namely height and color of flowers. Complete the illustration and table suitably based on the indicators, analyze illustration and write down inferences.

Answer:

Tall, red flower TtRr
Self-pollination first-generation TtRr x TtRr

Hss Live Guru 10th Biology Kerala Syllabus Chapter 6 Question 12.
What are the characters expressed in the offsprings of the first generation? Which are the recessive ones?
Answer:
Expressed characters – Tall and red flower
Recessive characters – Dwarf and white flower.

Kerala Syllabus 10th Standard Biology Notes Pdf Chapter 6 Question 13.
Are there new combination of characters different from parents appeared in the second generation? Which are they?
Answer:
Yes. New combination of characters different form parents appeared in the second generation. The appearance of new combination of characters in offsprings is due to the independent assortment of each character.
New combination of characters appeared in second generation

• Dwarf, red flower plants
• Tall, white flower plant

DNA (Deoxyribonucleic Acid)

Sslc Biology Chapter Wise Questions And Answers Kerala Syllabus Chapter 6  Question 14.
What are the peculiarities of double-helical model of DNA presented by Watson and Crick?
Answer:
James Watson and Francis Crick presented the double-helical model of DNA in 1953. As per the double-helical model, DNA contains two strands. Two strands of DNA are made of phosphate and pentose sugar and steps with nitrogen bases. The nitrogen bases are adenine, thymine, guanine, and cytosine.

Structure of DNA molecule

DNA molecule contains two long strands with deoxyribose sugar, phosphate, and steps with nitrogen bases. Nitrogen bases are molecules that contain nitrogen and are alkaline in nature. The nitrogen bases are of four types. In DNA the bases adenine pairs with thymine and guanine pairs with cytosine. One deoxyribose sugar molecule one phosphate molecule and one nitrogenous base join together to form a nucleotide. Nucleotides are the basic unit of DNA. Since DNA has four kinds of nitrogen bases, DNA has four kinds of nucleotides.

Kerala Syllabus 10th Standard Biology Pdf Chapter 6 Question 15.
Observe the illustration and complete its second strand.

Answer:

Terminal Axial Question 16.
Compare the structure of DNA and RNA and complete the table.

Answer:

How Do Genes Act?

Question 17.
Observe the following illustration and write down the inferences in the science diary.

Answer:
DNA does not participate directly in protein synthesis. RNA is the molecule that carries information form DNA to ribosomes and controls protein synthesis. This RNA is messenger of DNA, it is called messenger RNA or mRNA. Besides mRNA, there are tRNA (Transfer RNA) that carry amino acids to the ribosomes and rRNA (Ribosomal RNA) that are seen associated with ribosomes. Protein molecule is synthesized by the combined activities of all these molecules.

Question 18.
Prepare a flow chart of Protein synthesize.
Answer:

There are 46 chromosomes in human beings. Of these 44 are somatic chromosomes and two are sex chromosomes. A somatic chromosome pair contains two identical chromosomes. Thus in human beings, there are 22 pairs of somatic chromosomes. Sex chromosomes are two types. They are called x chromosomes and y chromosomes. Females have 44 + xx and genetics of variation that of male 44 + XY.

Genetics Of Variation

Crossing over in chromosomes: A source of variation:

During the initial phase of meiosis, chromosomes pair and exchange their parts. This process is called crossing over. As a result of this, part of a DNA crosses over to become the part of another DNA. This causes a difference in the distribution of genes. When these chromosomes are transferred to the next generation, it causes the expression of new characters in offsprings.

Combination of Allele during fertilization:
When gametes undergo fusion, the combination of allele changes. This causes the expression of characteristics in offsprings that are different from parents. Thus fertilization causes variations int the next generation.

Mutation

Question 19.
What is mutation?
Answer:
Mutation is a sudden heritable change in the genetic constitution of an organism.

Question 20.
What are the causes of mutations?
Answer:
The defects in the duplication of DNA, certain chemicals and radiations.

Question 21.
What is the importance of mutations?
Answer:
Certain mutations are harmful and some are helpful for survival of the organisms. Mutations lead to variations in characters. Mutation has great relevance in evolution.

Is The Child – Male Or Female

Observe illustration and write down the inferences in the science diary.

Question 22.
Is there any difference in the number of chromosomes in male and female.
Answer:
No. In male and female, there is not any difference in the number of chromosomes. In men and women 46 chromosomes are present.

Question 23.
Which chromosome is different in male and female?
Answer:
Sex chromosome is different. The sex chromosome of female is are XX and in male are XY.

Question 24.
What is the possibility for the birth of a male or a female child? Discuss.
Answer:
The possibility for the birth of a male or a female child is more or less equal. If the X chromosomes in male unite with X chromosome in female the offspring will be female and if with the Y chromosome In female the offspring will be male.

Question 25.
Is it fair to criticize mothers who deliver only female- children. Substantiate your opinion scientifically.
Answer:
The XY chromosomes of the father determine whether the child is male or female. Child with XX sex chromosomes is female arid with XY sex chromosomes is made. So male sex chromosomes have greater importance than female sex chromosomes in sex determination. So the view of her husband and relatives is wrong.

Difference In Colour

Melanin, a pigment-protein imparts color to the skin. The difference in gene function is the reason for the color difference of skin. This is simply an adaption to live under sun.

Let Us Assess

Question 1.
The nitrogen base absent in RNA.
a) Admin
b) Thymine
c) Uracil
d) Cytosine
Answer:
b) Thymine

Question 2.
Arrange the stages of protein synthesis in the form of a flow chart.
1. Combines amino acid.
2. mRNA reaches ribosomes
3. mRNA is formed
4. Amino acids are carried to the ribosomes.
Answer:
mRNA is formed → mRNA reaches ribosomes → amino acids are carried to the ribosomes → combines amino acids.

Question 3.
Observe the hybridization experiment given below

a) Prepare an illustration of this hybridization experiment using symbols
b) Prepare an illustration for the second generation.
Answer:

b) Parental plants (Self-pollination of first-generation) F2 → Gg × Gg

Extended Activities

Question 1.
‘ Prepare an excerpt including information on scientist who made contributions in the progress of genetics. (Hints – Gregor Johann Mendel, Walter, S. Sutton, Boveri, Friedrich Meischer, Johannsen, Avery, James Watson, Francis Crick, Marshall, Nirenberg, Har Gobind Khorana)
Answer:
Milestones in the History of Genetics

Question 2.
Prepare models of DNA and RNA using locally available materials and present them in a science exhibition.

Unraveling Genetic Mysteries More Questions And Answers

Question 1.
The father of genetics?
Answer:
Gregor Johann Mendel

Question 2.
Mendel conducted the process of hybridization using one pair of contrasting characters. In all his experiments only one character was expressed. So which was the method he adopted to find out recessive character?
Answer:
Mendel self-pollinated the plants obtained as the first generation.

Question 3.
When Mendel conducted experiment using one pair of contrasting characters, the plants obtained in the F2 generation is was in ………. ratio.
Answer:
3:1

Question 4.
Find out the dominant characters in plants having TTRR, TTRr, TtRR, TtRr traits.
Answer:
All plants are tall with red flowers.

Question 5.
Hereditary factors which Mendel had described are now known as …………….
Answer:
Genes

Question 6.
What are genes?
Answer:
Genes are the specific units of DNA that control metabolic activities and responsible for the specific characters of any organism.

Question 7.
“Offsprings of the same parents show differences among themselves” Give reasons for this?
Answer:
During fertilization alleles from the chromosomes of gametes segregate and causes change in the allele combination. This change causes variations in the offsprings. So offsprings of the same parents also show differences.

Question 8.
When a woman gave birth to girl children in her consecutive deliveries, her husband and relatives blamed her. Evaluate this social situation and write your opinion.
Answer:
The XY chromosomes of the father determine whether the child is male or female. Child with XX sex chromosomes is female and with XY sex chromosomes is male. So male sex hormones have greater importance than female sex chromosomes in sex determination. So the view of her husband and relatives is wrong.

Question 9.
What is the possible ratio for the birth of a male or a female child? What is the reason for this?
Answer:
The possibility for the birth of a male or a female child is more or less equal. If the X chromosomes in male unite with X chromosome in female the offspring will be female and if with the Y chromosome in female the offspring will be male.

Question 10.
Which are the two types of nucleic acids?
Answer:
DNA, RNA

Question 11.
What is the figure shows?
Answer:
a) a DNA molecule
b) a RNA molecule
c) a nucleotide
d) a chromosome
Answer:
a) nucleotide

Question 12.
Identify the picture? From where is it seen?

Answer:
DNA molecule. DNA molecules are seen in chromosome.

Question 13.
Identify the nucleotides presented in DNA and RNA. Which factor help you to identify this?
Answer:

B and C are nucleotides seen in DNA because the adenine and thymine are seen in DNA. A and C are nucleotide seen in RNA because Adenine and Uracil are seen in RNA.

Question 14.
Give examples for different kinds of RNA seen in the cell?
Answer:
mRNA(messenger RNA), tRNA (transfer RNA), rRNA (ribosomal RNA).

Question 15.
What is the following illustration represents?
Answer:

Answer:
Protein synthesis of DNA.

Question 16.
Explain the role of mRNA in protein synthesis.
Answer:
DNA does not participate directly in protein synthesis. It unwinds and mRNA is synthesized which carries the information from DNA to ribosomes. Based on the information in mRNA protein is synthesized by proper adding amino acids.

Question 17.
Observe the diagram and find out the names of the labeled parts a. b and c.
Answer:

Answer:
a) mRNA
b) Ribosomes
c) Protein molecule

Question 18.
Observe the figures and identify the process given below.

Answer:
Crossing over

Question 19.
‘ While some mutations are harmful, some of them are helpful.” Analyze the statement.
Answer:
This statement is correct. Certain mutations are harmful and some are helpful for survival of the organism. Mutations also lead to evolution.

Question 20.
‘‘Some specific processes during Meiosis helps to create variation in characters among organisms Analyse this statement and explain the process.
Answer:
Homologous chromosomes from father and mother pair exchange chromosomal material during meiosis. Tbi3 is “Crossing Over” in meiosis.

Question 21.
Diagrammatically represent with symbols the First generation of progenies of Tall and Dwarf pea plants when cross-pollinated as in Mendel’s first stage of Experiment (March 2015)
Answer:

Question 22.
Observe the flow chart given below. (Model 2015)
Answer:
i) RNA is synthesized
↓
ii) DNA unwinds
↓
iii) RNA combines with ribosome
↓
iv) RNA comes out through nuclear membrane.
↓
v) Protein molecule is formed.
↓
vi) Different amino acids are formed.
↓
a) Arrange the flow orderly.
b) Identify the process.
Answer:
a) DNA unwinds
↓
RNA Is synthesized
↓
RNA comes out through nuclear membrane.
↓
RNA combines with ribosome
↓
Different amino acids are formed.
↓
Protein molecule is formed
b) Protein synthesis or gene action

Question 23.

(Model 2014)
Observe the diagram Identify the process which occurs in Chromosomes during meiosis.
Answer:
Crossing over

Question 24.
A pea plant with genetic constitution TtRr (Tall plant which produces red flower) was subjected to self-pollination. The genetic constitution of some of the progenies obtained are given below.
Write the expressed character of the given progenies based on their genetic constitution. (Model 2014)
(a) TTRr
(b) ttrr
(c) ttRr
(d) Ttrr
Answer
a) TTRr – tall plant with red flower
b) ttrr – dwarf plant with white flower
c) ttRr – dwarf plant with red flower
d) Ttrr – tall plant with white flower

Question 25.
Mr. Rajan decided to divorce his wife by arguing that she is incapable to give birth to a boy child.
a) Can you agree with Mr. Rajan?
b) Give scientific explanation, to this problem with the help of an illustration showing the role of sex chromosomes in determining sex in human beings.
Answer:
a) I cannot agree with Rajan

Y chromosome from male decide the sex of child

Question 26.
Find the odd one. Write down the common feature of the others. (March 2014)
a) Bypass surgery, ECG, EEG, Pacemaker
b) Adenine, Cytosine, Thymine, Uracil
Answer:
a) EEG: Related with the treatment of heart diseases
b) Uracil: Nitrogen bases in DNA/Thymine: Nitrogen bases in RNA.

Question 27.
Gopalettan is trying to develop new varieties of pea plants in his garden. Given below is the illustration of the experiment which he conducted. Observe this and answer the questions that follow. (Model 2013)

a) Which law of hereditary can be used to explain the reason of violet flowered plants in the F1 generation?
b) Among these, which are the dominant and recessive characters?
c) What will be the characters appearing in the F2 generation if the plants in the F1 generation are self-pollinated? In what ratio?
Answer:
a) Law of dominance. (When a pair of contrasting characters combines, only one character is expressed, while the other remains hidden.)
b) Dominant is violet and recessive is white.
c) Yellow and green seeds in 3:1 ratio.

Question 28.
Fill up the blanks in the table showing gene action in protein synthesis. (Model 2013)

Answer:
a) RNA is being synthesized
b) RNA comes out through nuclear membrane
c) RNA1 combines with ribosomes

Question 29.
Rearrange B and C according to the data given in A. (Model 2013)

Answer:

Question 30.
A portion of DNA molecule is shown below. Find out the missing nitrogen base pair from those given below. (Model 2012)

Answer:
(b) C – G

Question 31.
The fusing of gametes during self-pollination of F1 in pea plants with two separate characters combined together were tabulated; (Model 2012)

A) Fill up the blanks.
B) How many white-flowered plants will be formed if 16 pea plants in F₂ generation were produced?
Answer;
A) (a) TtRR
(b) Ttrr
(c) ttRR
(d) ttrr
B) 4

Question 32.
Given below is the illustration showing how sex determination is taking place in man. (March 2012)

a) Observe the illustration and examine the 4 types of possible offsprings. Specify their sex chromosomes.
b) What are the inferences you arrive at from this illustration?
c) What is the probable ratio of formation of male child and female children in man ? Illustrate your answer.
d) Which are the sex-determining chromosomes in man?
Answer:
a) (i) and (iii) are female children having XX chromosomes
(ii) and (iv) are male children having XY chromosomes
b) The sex chromosomes of male determine the sex of a child. The probability of formation of male and female children is almost equal (1:1)
c) 1: 1
d) Male XY and female XX.

Question 33.
Using given indicators, construct RNA nucleotide and anyone DNA nucleotide. (March 2012)

Answer:

Unraveling Genetic Mysteries Questions & Answers

Question 1.
Find the word pair relationship and fill in the blanks appropriately. (Question Pool – 2017)
a) DNA: Thymine
RNA:……………..
b) Adenine: Thymine
Guanine:…………………
c) The character which is expressed: dominant
The character which remains hidden:……………
Answer:
a) Uracil
b) Cytosine
c) Recessive character

Question 2.
Given below are certain indicators exhibited by Anu in her slide presentation while conducting seminar on the topic ” Emergence of Genetics”. What explanations would you give for these indicators?
a) Heredity
b) Variation
c) Genetics
d) The father of Genetics
Answer:
a) Transmission of features of parent to offspring.
b) Features seen in offspring that are different from their parents.
c) The branch of science that deals with heredity and variation.
d) Gregor John Mendel

Question 3.
The note prepared by Shahana on Mendel’s inferences during the classroom analysis of Mendel’s hybridization experiment in pea plants, based on a single trait is given below. Analyze the statements in the note and correct those that are wrong ones.
a) A trait is controlled by a specific factor.
b) A character is expressed and the other remains hidden in the first generation.
c) The character that remains hidden, in the first generation does not appear in the second generation.
d) The ratio of characters in the second generation is 3: 1
Answer:
a) One trait is controlled by the combination of two factors.
b) The characters that remain hidden in the first generation appears in the second generation.

Question 4.
Fill in the blanks in the illustration given below. (Question Pool -2017)

Answer:
A) tt
B) t
C) Tt
D) dwarf

Question 5.
Complete the flowchart illustrating the location of gene by using the information given in the box: (Question Pool – 2017)
nucleus, gene, DNA, cell, chromosome

Answer:
A – Cell
B – Nucleus
C – Chromosome
D – DNA

Question 6.
Analyze the article and answer the questions. The experiments performed by Gregor John Mendel in pea plants led to the emergence of a new branch of science that has today grown and expanded to a great extent. This branch of science has untravelled several mysteries regarding the similarities and variations found in the characters of organisms. (Question Pool – 2017)
a) Which branch of science does the article refer to?
b) List out any 4 traits selected by Mendel for performing hybridization in pea plants.
Answer:
a) Genetics
b) Height, colour of the seed, colour of the flower, shape of the seed, color of the fruit, shape of the pod

Question 7.
Observe the illustration given below and answer the questions. (Question Pool-2017)

a) Identify the dominant character
b) How does the parental plant with green colored seed and the plant in the first generation differ in their alleles.
c) Describe alleles.
Answer:
a) Green
b) Alleles in parental plant – G, G Allele in the first generation – G, g
c) Different forms of a gene

Question 8.
Fill in the blanks in the illustration related to chromosomes in man. (Question Pool-2017)

Answer:
A) Autosomes
B) 2
C) XY

Question 9.
Identify the word pair relationship and fill in the blanks: (Question Pool -2017)
Female : 44 + XX
Male: …………….
Answer:
44 + XY

Question 10.
Fill in the blanks in the illustration. (Question Pool-2017)

Answer:
A. ggww
B. GW
C. Green colored round seed

Question 11.
The indicators given below are about the plant in the first generation formed as a result of the hybridization between a tall plant with red flowers and a dwarf plant with white flowers. (Question Pool-2017)

a) Identify the alleles in the plant related to the trait ’tallness’.
b) Identify the gametes formed from this plant.
Answer:
a) T, t

Question 12.
Complete the illustration of the second generation obtained from the hybridization in which two traits of a plant are considered. (Question Pool-2017)
Indicators:
Dominant character – Tallness, red color of flower Recessive character – Dwarfness, white color of flower

Answer:
A. TTRr
B. TtRr
C.TTRr
D. TtRr
E. TtRr
F. ttRR
G. ttRr
H. TtRr

Question 13.

Given below are some of the offspring obtained by self-pollinating the above plant. Analyze the offspring and answer the questions.
a) ttRr
b) ttRR
c) TTrr
d) ttrr
i) Identify the dominant characters in each of the offspring?
ii) What explanation would you give for the expression of characters in the offspring which were hidden in the parental plant?
Answer:
i. a) dwarf, red flower
b) dwarf, red flower
c) tall, white flower
d) dwarf, white flower
ii. The expression of characters in the offsprings which were hidden in the parental plant is due to the independent assortment of each character.

Question 14.
Analyze the illustration of a nucleotide molecule and answer the questions. (Question Pool-2017)

a) Identify A and B in the illustration.
b) ‘Nucleotides are found in DNA alone’. What is your opinion regarding these statements? Substantiate.
Answer:
a) A – Phosphate
B – Sugar
b) 1. does not agree
2. Like DNA, RNA is also made up of nucleotides

Question 15.
Genes which are the specific units of DNA control the metabolic activities and are also responsible for specific characters. They control the process of protein synthesis. Binu has a doubt on the above note. (Question Pool – 2017)
‘Does the RNA have no role in protein synthesis?’ What explanation would you give to Binu’s doubt? Substantiate.
Answer:

• RNA has role
• DNA is not directly involved in protein synthesis
• mRNA is formed from DNA.
• mRNA that carries information from DNA, controls the protein synthesis
• tRNA carries amino acids to ribosomes
• rRNA associated with ribosomes also have a role in protein synthesis.

Question 16.
The components and features of nucleic acid are given below. Analyze them and complete the table. (Question Pool-2017)
a) ribose sugar
b) double helical shape
c) Uracil
d) one strand
e) deoxyribose sugar
f) thymine

Answer:

Question 17.
Observe the nucleotide strands given below and answer the questions. (Question Pool-2017)

a) Identify the strand that is found in DNA only.
b) Identify the strand that can be found in both DNA and RNA.
c) What is a nucleotide?
Answer:
a) B
b) A
c) A unit of sugar, phosphate and nitrogen base / Component of nucleic acid

Question 18.
The stages in the process of protein synthesis are given below. Prepare a flowchart using the stages. (Question Pool-2017)
a) tRNA carries different kinds of amino acids to the ribosome.
b) mRNA reaches outside the nucleus.
c) mRNA forms from DNA
d) Amino acids are added based on the information in mRNA
e) mRNA reaches ribosome.
f) Proteins are synthesized.
Answer:
c → b → e → a → d → f

Question 19.
A part of the article. Variations in ourself is given below: The features seen in offspring that are different form their parents are called variations. Certain processes taking place in the initial phase of meiosis are responsible for such variations. (Question Pool-2017)
a) Which process, as mentioned in the article, is responsible for variations?
b) How does this process bring about variations?
Answer:
a) Crossing over of chromosomes
b) 1. Part of a DNA crosses over to become the part of another DNA.
2. This causes difference in the distributions of genes
3. When these chromosomes are transferred to the next generation, new characters are expressed.

Question 20.
The process of crossing over of chromosomes that takes place in the initial phase of meiosis is illustrated below. Analyse-it and answer the questions. (Question Pool – 2017)

a) Arrange the stages appropriately.
b) This process brings about variations in offspring. How?
Answer:
a) C, A, B
b) 1. Part of a DNA crosses Over to become the part of another DNA.
2. This causes difference in the distributions of genes.
3. When these chromosomes are transferred to the next generation, new characters are expressed.

Question 21.
Given below is an illustration regarding sex determination Observe the illustration and answer the questions. (Question Pool-2017)

a) Complete A, B, C in the illustration.
b) What is the possibility of the formation of a male child or a female child? Explain.
Answer:
a) A) 44 + XY
B) 22 + X
C) 22+ Y
b) Equal chance. The number of male gametes with X chromosome and those with Y chromosome are equal.
Egg with the X chromosome has equal chance to combine with sperm having Y chromosome and those having X chromosome.

Question 22.
The practice of blaming those mothers who give birth to girl children exists even today. (Question Pool – 2017)
a) As a science student, how will you respond to this situation? Substantiate.
Answer:

• No, it is wrong to blame the mother
• The possibility of the birth of male or female child is equal.
• The gender of the child is determined by the XY chromosome of the father.

Question 23.
The chromosomes from the father determine whether the child is male or female. Evaluate this statement on a scientific basis. (Question Pool-2017)
Answer:
Males have 2 types of sex chromosomes. (X, Y) Females have only one type of sex chromosome (X, X) Sex determination is based on the type of male gamete that fuses with the egg. If the male gamete with Y chromosome fuses with the egg, then male child is born, if the male gamete with X chromosome fuses with the egg, then female child is born.

Question 24.
Given below is a placard exhibited in a school rally organized ‘Against Racism’. It is not racial difference that makes the skin color different;
This is an adaptation to live under the sun.
a) How ; will you. explain the difference in skin color of people living in different parts of the world?
b) What attitude should be adopted by a scientifically enlightened society towards the idea in the placard? Substantiate.
Answer:
a) 1. Melanin, a pigment-protein imparts color to skin.
2. It is due to the difference in gene function,
b) 1. Skin color is an adaptation to live under the sun.
2. Races among making are only cultural.
3. Scientifically, all men are of the same race.
4. Consider all men as equal, without any racial difference.

Question 25.
Vipin wrote the following as situations that create variations in organisms. Choose the right ones. (Question Pool – 2017)
a) Mutation
b) Formation of mRNA
c) Crossing over of chromosomes
d) Action of rRNA
Answer:
a) Mutation
c) Crossing over of chromosomes

Question 26.
The components of nucleic acids are given below. Answer the questions through illustrations using these components: (Question Pool – 2017)

a) Illustrate the nucleotide which is found only in RNA.
b) Illustrate the nucleotide which is found only in DNA.
Answer:

Question 27.
Analyze the nitrogen bases given below and write the nitrogen base pairs found in DNA. (Question Pool-2017)

Answer:
Thymine – Adenine.
Guanine – Cytosine

Question 28.
‘Gene itself is allele; allele itself is gene. (Question Pool – 2017)
Answer:
Statement is party correct.
Each character is controlled by pair factors called genes.
Different forms of a gene are called alleles. Generally, a gene has two alleles. ,
Alleles can be of the same type (TT) or different types (Tt).
If the alleles are of different types, only one trait represented by anyone of the alleles get expressed.

Question 29.
Offsprings may vary in characters from their parents.
a) What are reasons of this variation in the light of genetics?
b) How does the changes take place during meiosis cause variations in next generation?
c) How does the chemical substances and the radiations cause variation in characters?
Answer:
a) Crossing over and mutations.
b) When a part of a particular DNA become the part of another DNA, the sequencing of nucleotides in the DNA become differs and hence variation may occur in the offsprings.
c) Mutation may occur due to chemicals and radiations.

Question 30.
Observe the illustration related to Mendel’s experiment based on two contrasting characters.
a) Complete the illustration appropriately

b) What are the characters observed in the second generation?
Answer:

b) Tall with red flowers = 9
Tall with white flower = 3
Dwarf with red flowers = 3
Dwarf with white flowers = 1

Question 31.
The different stages of protein synthesis given below. Rearrange them appropriately. (Orukkam – 2017)
a) tRNA carries different types of amino acids.
b) mRNA come out from the nucleus.
c) mRNA is formed from DNA.
d) Amino acids are joined together based on the messages in mRNA.
e) mRNA reaches in ribosome.
f) Protein is synthesized.
Answer:
c → b → e → a → d → f

Question 32.
Observe the figure and answer the questions given below. (Orukkam – 2017)

a) Name the process shown in the figure.
b) Write the importance of this process.
Answer:
a) Crossing over
b) Causes variation

Question 33.
Write the role of mRNA and tRNA in protein synthesis. (Orukkam – 2017)
Answer:
mRNA carries messages for protein synthesis from DNA to the ribosomes.
tRNA carries amino acids to ribosomes according to the message in the mRNA.

Question 34.
What is mutation? Write the reasons? (Orukkam – 2017)
Answer:
Mutation is a sudden heritable change occur in the genetic material (chromosomes)
Reasons: – Radiations, chemicals, changes in the replication of DNA.