NCERT Exemplar Solutions For Class 6 Maths Chapter 6 Integers

Class 6 Maths Chapter 6 Integers

Exercise – 6.1

1. Write opposites of the following:

(1) Increase in weight
(2) 30 km north
(3) 80 m east
(4) Loss of? 700
(5) 1 00 m above sea level

Solution: (1) Decrease in weight

(2) 30 km south

(3) 80 m west

(4) Profit of? 700

(5) 100 m below sea level

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2. Represent the following numbers as integers with appropriate signs.

(1) An aeroplane is flying at a height, two thousand metre above the ground.
(2) A submarine is moving at a depth, eight hundred metre below the sea level.
(3) A deposit of rupees two hundred.
(4) Withdrawal of rupees seven hundred.

Solution: (1) Two thousand metre above the ground = + 2000

(2) Eight hundred metre below the sea level = -800

(3) Deposit of two hundred rupees = + 200

(4) Withdrawal of seven hundred rupees = -700

3. Represent the following numbers on a number line:

(1) +5
(2) -10
(3) + 8
(4) -1
(5) -6

Solution: (1)

Represent the following numbers on a number line -5

(2)

Represent the following numbers on a number line -10

(3)

Represent the following numbers on a number line -8

(4)

Represent the following numbers on a number line -1

(5)

Represent the following numbers on a number line -6

4. Adjacent figure is a vertical number A line, representing integers. Observe it and locate the following points:

(1) If point D is + 8, then which point is -8?
(2) Is point G a negative integer or a positive integer?
(3) Write integers for points 8 and E.
(4) Which point marked on this number line has the least value?
(5) Arrange all the points in decreasing order of value.

Adjacent figure is a vertical number line

Solution: (1) We have, point D is +8.

Therefore, 16 steps to the down from D is -8 i.e., the point F.

(2) Point G is a negative integer.

(3) Point B is four steps down from point D.

Value of point B = +8- 4 = +4

Point E is eighteen steps down from point D.

Value of point E = +8- 18 = -10

(4) Since, point E is located in the bottom.

So, point E has the least value.

(5) Decreasing order of all the points is,

D, C,B, A,0, H,G, F,E

5. Following Is the list of temperatures of five places In India on a particular day of the year.

Place                              Temperature

Siachin                           10°C below 0°C …………

Shimla                            2°C below 0°C …………

Ahmedabad                   30°C above 0°C …………

Delhi                              20°C above 0°C …………

Srinagar                          5°C below 0°C …………

(1) Write the temperatures of these places in the form of integers in the blank column.

(2) Following is the number line representing the temperature in degree Celsius

Plot the name of the city against its temperature.

(3) Which is the coolest place?

(4) Write the names of the places where temperatures are above 10°C.

Solution:

Place                           Temperature

(1) Siachin                  -10°C

Shimla                        -2°C

Ahmedabad               + 30°C

Delhi                          + 20°C

Srinagar                     -5°C

(3) Siachin is the coolest place.

(4) Ahmedabad and Delhi have temperature above 10°C.

6. In each of the following pairs, which number is to the right of the other on the number line?

(1) 2, 9
(2) -3,-8
(3) 0,-1
(4) -11,10
(5) -6,6
(6) 1,-100

Solution: (1) 9 is right to 2

(2) -3 is right to -8

(3) 0 is right to -1

(4) 10 is right to -11

(5) 6 is right to -6

(6) 1 is right to -100

7. Write all the integers between the given pairs (write them in the increasing order.)

(1) 0 and-7
(2) -4 and 4
(4) -30 and -23
(3) -8 and -15

Solution: (1) The integers between 0 and -7 are -6, -5, -4, -3, -2, -1

(2) The integers between -4 and 4 are -3, -2, -1,0, 1,2, 3

(3) The integers between-8 and -15 are -14, -13, -12, -11, -10, -9

(4) The integers between -30 and -23 are -29, -28, -27, -26, -25, -24

8. (1) Write four negative integers greater than -20.

(2) Write four integers less than- 1 0.

Solution: (1) There are 19 negative integers which are greater than -20. Four of them are -19, -18, -17, -16

(2) There are infinite integers which are less than -10. Four of them are -11, -12, -13, -14

9. For the following statements, write True (T) or False (6). If the statement is false, correct the statement.

(1) – 8 is to the right of- 1 0 on a number line.
(2) – 100 is to the right of – 50 on a number line.
(3) Smallest negative integer is -1.
(4) – 26 is greater than- 25

Solution: (1) True

(2) False

Since -100 is to the left of -50 on the number line.

(3) False

Since -1 is the greatest negative integer.

(4) False

Since -26 is less than -25.

10. Draw a number line and answer the following :

(1) Which number will we reach if we move 4 numbers to the right of- 2.
(2) Which number will we reach if we move 5 numbers to the left of 1.
(3) if wc arc at- 8 on the number line, in which direction should we move to reach – 1 3?
(4) If we are at- 6 on the number line, in which direction should we move to reach – 1 ?

Solution: (1)

we will reach 2 if we move 4 numbers to the right of -2

Thus, we will reach 2 if we move 4 numbers to the right of -2.

(2)

 

Thus, we will reach -4 if we move 5 numbers to the left of 1

Thus, we will reach -4 if we move 5 numbers to the left of 1.

(3)

Thus, we will reach -4 if we move 5 numbers to the left of 1 Thus, we should move 5 numbers to the left of-8 to reach -13

Thus, we should move 5 numbers to the left of-8 to reach -13.

(4)

Thus, we should move 5 numbers to the right of -6 to reach -1

Thus, we should move 5 numbers to the right of -6 to reach -1.

Exercise – 6.2

1. Using the number line write the integer which is:

(1) 3 more than 5
(2) 5 more than -5
(3) 6 less than 2
(4) 3 less than -2

Solution: (1)

Thus, we will reach -4 if we move 5 numbers to the left of 1

Thus, 3 more than 5 is 8.

(2)

5 more than -5

Thus, 5 more than -5 is 0.

(3)

6 less than 2

Thus, 6 less than 2 is -4.

(4)

3 less than -2

Thus, 3 less than -2 is -5.

2. Use number line and add the following integers :

(1) 9 + (-6)
(2) 5 + (-11)
(3) (-1) + (-7)
(4) (- 5) + 10
(5) (- 1 ) + (- 2) + (- 3)
(6) (- 2) + 8 + (- 4)

Solution: (1)

9 + (-6)

Thus, 9 + (-6) = 3

(2)

5 + (-11)

Thus, 5 + (-11) = -6

(3)

(-1) + (-7)

Thus, (-1) + (-7) = -8

(4)

(- 5) + 10

Thus, (-5) + 10 = 5

(5)

(- 1 ) + (- 2) + (- 3)

Thus, (-1) + (-2) + (-3) = -6

(6)

(- 2) + 8 + (- 4)

Thus, (-2) + 8 + (-4) = 2

3. Add without using number line:

(1) 1 1 + (- 7)
(2) (-13) + (+18)
(3) (-10) + (+19)
(4) (-250) + (+150)
(5) (- 380) + (- 270)
(6) (- 21 7) + (- 1 00)

Solution: (1) 11+ (-7) -11-7=4

(2) (-13) + (+18)= -13 + 18 =5

(3) (-10) + (+19)= -10 + 19 =9

(4) (-250) + (+150) = -250 + 150 = -100

(5) (-380) + (-270) = -380- 270 = -650

(6) (-217) + (-100) = -217 -100 = -317

4. Find the sum of:

(1) 137 and -354
(2) -52 and 52
(3) -31 2, 39 and 192
(4) -50, -200 and 300

Solution: (1) 137 + (-354) = 137- 354 = -217

(2) -52 + 52 = 0

(3) -312 + 39 + 192 = -312 + 231 =-81

(4) -50 + (-200) + 300 = -50 -200 + 300 = -250 + 300 = 50

5. Find the sum:

(1) (- 7) + (- 9) + 4 + 16
(2) (37) + (-2) + (-65) + (-8)

Solution: (1) (-7) + (-9) + 4 + 16

=-7-9 + 4 + 16

= -16 + 20 = 4

(2) (37) + (-2) + (-65) + (-8)

= 37-2-65-8

= 37- 75 =- 38

Exercise – 6.3

1. Find

(1) 35 -(20)
(2) 72 -(90)
(3) (-15) -(-18)
(4) (-20) -(13)
(5) 23 -(-12)
(6) (-32) -(-40)

Solution: (1) 35-20 = 15

(2) 72 -90 = -18

(3) (-15)- (-18) =- 15 + 18 = 3

(4) (-20) -(13) = -20 -13 = -33

(5) 23 -(-12) = 23 + 12 = 35

(6) (-32) -(-40) =-32 + 40 = 8

2. Fill in the blanks with >, < or = sign.

(1) (- 3) + (- 6) _________ (- 3)- (- 6)
(2) (-21) -(-10) ___________ (— 31) + (— 11)
(3) 45 – (- 11)___________ 57 + (-4)
(4) (-25) -(-42)__________ (- 42)- (- 25)

Solution: (1) < : (-3) + (-6) = -3- 6 = -9

(-3) -(-6) –3 + 6 = 3

Since, -9 < 3

(-3) + (-6) < (-3)- (-6)

(2) > : (-21)- (-10) = -21 + 10 = -11

(-31) + (-11) = -31 – 11 = -42

Since, -11 > -42

(-21) -(-10) >(-31) + (-11)

(3) > : 45- (-11) = 45 + 11 = 56

57 + (-4) = 57- 4 = 53

Since, 56 > 53

45 -(-11) >57 + (-4)

(4) > : (-25)- (-42) = -25 + 42 = 17

(-42)- (-25) = -42 + 25 = -17

Since, 17 >-17

(-25)- (-42) > (-42)- (-25)

3. Fill in the blanks

(1) (-8) +_______=0

(2) 13 +_________=0

(3) 12 + (-12) =_____

(4) (-4) +________=-12

(5) _________ -15 = -10

Solution: (1) 8 : (-8) + 8 = 0

(2) -13 : 13 + (-13) = 0

(3) 0:12 + (-12) = 0

(4) -8: (-4) + (-8) =-12

(5) 5: 5- 15 = -10

4. Find

(1) (-7)-8- (-25)
(2) (-13) + 32 -8-1
(3) (- 7) + (- 8) + (- 90)
(4) 50 -(-40) -(-2)

Solution: (1) (-7)- 8- (-25)

=-7-8 + 25

= -15 + 25 = 10

(2) (-13) +32-8-1

= -13 +32-8-1

= 32- 22 = 10

(3) (-7) + (-8) + (-90)

= -7- 8- 90 =- 105

(4) 50 -(-40) -(-2)

= 50 + 40 + 2 = 92

Section-2 NCERT Exemplar

Directions: In questions 1 to 17, only one of the four options is correct. Write the correct one.

1. Every integer less than 0 has the sign

(1) +
(2) –
(3) x
(4) ÷

Solution: (2): Every integer which is less than 0 has negative sign.

2. The integer ‘5 units to the right of 0 on the number line’ is

(1) +5
(2) -5
(3) +4
(4) -4

Solution: (1): The integer which is 5 units to the right of 0 on the number line is +5.

3. The predecessor of the integer -1 is

(1) 0
(2) 2
(3) -2
(4) 1

Solution: (3): The predecessor of the integer -1 is -2.

4. Number of integers lying between -1 and 1 is

(1) 1
(2) 2
(3) 3
(4) 0

Solution: (1): Only 1 integer lies between -1 and1 i.e., 0

5. Number of whole numbers lying between -5 and 5 is

(1) 10
(2) 3
(3) 4
(4) 5

Solution: (4): There are 5 whole numbers lying between -5 and 5 i.e., 0, 1, 2, 3 and 4.

6. The greatest integer lying between -10 and -15 is

(1) -10
(2) -11
(3) -15
(4) -14

Solution: (2): -11 is the greatest integer lying between -10 and -15

7. The least integer lying between -10 and -15 is

(1) -10
(2) -11
(3) -15
(4) -14

Solution: (4): -14 is the least integer lying between -10 and -15.

8. On the number line, the integer 5 is located

(1) to the left of 0
(2) to the right of 0
(3) to the left of 1
(4) to the left of-2

Solution: (2):

On the number line, the integer 5 is located

The above number line shows that the integer 5 is located to the right of 0

9. In which of the following pairs of integers, the first integer is not on the left of the other integer on the number line?

(1) (-1,10)
(2) (-3,-5)
(3) (-5,-3)
(4) (-6,0)

Solution: (2) :

In which of the following pairs of integers,

On observing all the options by using a number line, we get that there is only one pair (-3, -5) in which the first integer is not on the left of the other integer.

10. The integer with negative sign (-) is always less than

(1) 0
(2) -3
(3) -1
(4) -2

Solution: (1): All the negative integers are less than 0.

11. An integer with positive sign (+) is always greater than

(1) 0
(2) 1
(3) 2
(4) 3

Solution: (1): All the positive integers are greater than 0.

12. The successor of the predecessor of -50 is

(1) -48
(2) -49
(3) -50
(4) -51

Solution: (3): The predecessor of -50 is -51 and the successor of -51 is -50.

13. The additive inverse of a negative integer

(1) is always negative
(2) is always positive
(3) is the same integer
(4) zero

Solution: (2): The additive inverse of a negative integer is always positive.

14. Amulya and Amar visited two places A and B respectively in Kashmir and recorded the minimum temperatures on a particular day as -4°C at A and -1°C at 6. Which of the following statement is true?

(1) A is cooler than B
(2) Bis cooler than A
(3) There is a difference of 2°C in the temperature
(4) The temperature at A is 4°C higher than that at B.

Solution: (1) :- 4°C < -1°C [ – 4 lies on the left of-1 on the number line]

Thus, A is cooler than B.

15. When a negative integer is subtracted from another negative integer, the sign of the result

(1) is always negative
(2) is always positive
(3) is never negative
(4) depends on the numerical value of the integers

Solution: (4): When a negative integer is subtracted from another negative integer, the sign of the result depends on the numerical
value of the integers.

16. The statement “When an integer is added to itself, the sum is greater than the integer” is

(1) always true
(2) never true
(3) true only when the integer is positive
(4) true for non-negative integers

Solution: (3): When an integer is added to itself, the sum is greater than the integer only when the integer is positive.

17. Which of the following shows the maximum rise in temperature?

(1) 0°C to 1 0°C
(2) -4°C to 8°C
(3) -15°C to -8°C
(4) -7°C to 0°C

Solution: (2) : (1) Rise in temperature = (10- 0)°C = 10°C

(2) Rise in temperature = (8- (-4))°C = (8 + 4)°C = 12°C

(3) Rise in temperature = (-8- (-15))°C = (-8 + 15)°C = 7°C

(4) Rise in temperature = (0- (-7))°C = (0 + 7)°C = 7°C

Thus, option (2) has maximum rise in temperature.

Directions: In questions 18 to 39, state whether the given statements are true (T) orfalse (6).

18. The smallest natural number is zero.

Solution: False

Since,1 is the smallest natural number.

19. Zero is not an integer as it is neither positive nor negative.

Solution: False

0 is neither positive nor negative, but it is an integer.

20. The sum of all the integers between -5 and -1 is -6.

Solution: False

-4, -3 and -2 lie between -5 and -1 and their sum is (-4) + (-3) + (-2) =-4-3-2 = -9

21. The successor of the integer 1 is 0.

Solution: False

0 is the predecessor of 1.

22. Every positive integer is larger than every negative integer.

Solution: True

Since positive integers lies on the right side of 0 and negative integers lies on the left side of 0 and the integers lying on the right are
always greater.

23. The sum of any two negative integers is always greater than both the integers.

Solution: False

Since, the sum of any two negative integers is always smaller than both the integers.

24. The sum ofany two negative integers is always smaller than both the integers.

Solution: True

25. The sum of any two positive integers is greater than both the integers.

Solution: True

26. All whole numbers are integers.

Solution: True

Integers are the collection of 0, positive integers and negative integers.

Whole numbers are the collection of 0 and positive integers.

All whole numbers are integers.

27. All integers are whole numbers.

Solution: False

Integers are the collection of 0, positive integers and negative integers.

Whole numbers are the collection of 0 and positive integers.

All integers are not whole numbers.

28. Since 5 >3, therefore -5 > -3.

Solution: False

Since,5 lies on the right of 3 on the number line 5>3

And-3 lies on the right of-5 on the number line, -3 >-5.

29. Zero is less than every positive integer.

Solution: True

Since, zero lies on the left side of every positive integer on the number line. Therefore, zero is less than every positive integer.

30. Zero is larger than every negative integer.

Solution: True

Since, zero lies on the right side of every negative integer on the number line.

Therefore, zero is larger than every negative integer.

31. Zero is neither positive nor negative.

Solution: True

32. On the number line, an integer on the right of a given integer is always larger than the integer.

Solution: True

33. -2 is to the left of-5 on the number line.

Solution: False

Since, -2 lies on the right of-5 on the number line.

34. The smallest integer is 0.

Solution: False

Since, zero is greater than all the negative integers.

0 is not the smallest integer.

35. 6 and -6 are at the same distance from 0 on the number line.

Solution: True

The integer 6 is 6 units to the right of 0 and the integer -6 is 6 units to the left of 0.

Thus, 6 and -6 are at the same distance from 0 on the number line.

36. The difference between an integer and its additive inverse is always even.

Solution: True

Let a be any integer and -a is its additive inverse.

Difference = a- (-a) = a + a = 2a, which is an even number.

37. The sum of an integer and its additive inverse is always zero.

Solution: True

Let a be any integer and -a is its additive inverse.

Sum = a + (-a) =a- a = 0.

38. The sum of two negative integers is a positive
integer.

Solution: False

Since, the sum of two negative integers is always negative.

39. The sum of three different integers can never be zero.

Solution: False

Let -3, 1, 2 are three different integers.

Sum = (-3) +1 + 2 =-3 + 3 = 0

Directions: In questions 40 to 49, fill in the blanks to make the statements true.

40. On the number line, -1 5 is to the zero.

Solution: Left

41. On the number line, 10 is to the of zero.

Solution: Right

42. The additive Inverse of 14 is_.

Solution: -14: Additive inverse of an integer is obtained by changing the sign of the integer.

Additive inverse of 14 is -14.

43. The additive inverse of-1 is

Solution: 1

44. The additive inverse of 0 is

Solution: 0

45. The number of integers lying between -5 and 5 is

Solution: 9: The integers lying between -5 and 5 are -4, -3, -2, -1, 0, 1, 2, 3, 4 i.e., 9 in number

46. (-11) + (-2) + (-1) =____________

Solution: -14: (-11) + (-2) + (-1) =-11- 2-1 = “14

47.___________ + (-11) + 111 = 130

Solution: 30

48. (-80) + 0 + (-90) =____________

Solution: -170: (-80) + 0 + (-90) =-80 + 0- 90 =-170

49. -3456 = -8910

Solution: -5454

Directions: In questions 50 to 58,fill in the blanks using <, = or >.

50. (-11) + (-15)___________ 11+15
Solution: < : (-11) + (-15) = -11-15 = -26

11 + 15 = 26 and -26 < 26

51. (-71) + (+9)___________ (-81) + (-9)
Solution: >: (-71) + (9) = -71 + 9 = -62

(-81) + (-9) = -81- 9 = -90 and -62 > -90

52. 0__________ 1

Solution: <:0<1

53. -60__________ 50

Solution: < : -60 < 50

54. -10__________ -11

Solution: >: —10 > —11

55. -101___________ -102

Solution: >: -101 >-102

56. (-2) + (-5) + (-6)__________ (-3) + (-4) + (-6)

Solution: = : (-2) + (-5) + (-6) = -2- 5- 6 = -13

(-3) + (-4) + (-6) =-3-4- 6 =-13

And -13 =-13

57. 0 __________ -2

Solution: >:0>-2

58. 1+2 + 3________ (-1 ) + (-2) + (-3)

Solution: >: 1 +2+3=6

(-1) + (-2) + (-3) =-l-2-3 =-6

And 6 > -6

59. Match the items of Column I with that of Column II:

Solution:

Match the items of Column I with that of column II

(i)–>(B), (ii)–> (E), (iii) —> (B), (iv) –> (A),(v) –> (B)

(i) The additive inverse of +2 is -2.
(ii) The greatest negative integer is -1.
(iii) The greatest negative even integer is -2.
(iv) The smallest integer 0 is greater than every negative integer.
(v) Predecessor and successor of -1 are -2 and 0 respectively.

∴ Sum = -2 + 0 = -2

60. Compute each of the following:

(1) 30 + (-25) + (-10)
(2) (-20) + (-5)
(3) 70 + (-20) + (-30)
(4) -50 + (-60) + 50
(5) 1 + (-2) + (-3) + (-4)
(6) 0 + (-5) + (-2)
(7) 0- (-6)- (+6)
(8) 0-2 -(-2)

Solution: (1) 30 + (-25) + (-10) = 30 + (-25- 10) = 30 + (-35) = 30- 35 = -5

(2) (-20) + (-5) = -20-5 = -25

(3) 70 + (-20) + (-30) = 70 + (-20- 30) = 70 + (-50) = 70- 50 = 20

(4) -50 + (-60) + 50 = (-50 -60) + 50 = -110 + 50 = -60

(5) 1 + (-2) + (-3) + (-4) =1 +(-2-3-4)=1+ (-9) =1- 9 = -8

(6) 0 + (-5) + (-2) 0 + (-5 -2) « 0 + (-7) = 0- 7 = -7

(7)0- (-6)- (+6) = 0 + 6- 6 = 6- 6 = 0

(8) 0- 2- (-2) = 0- 2 + 2 = -2 + 2 = 0

61. If we denote the height of a place above sea level by a positive integer and depth below the sea level by a negative integer, write the following using integers with the appropriate signs:

(1) 200 m above sea level
(2) 1 00 m below sea level
(3) 10m above sea level
(4) sea level

Solution: (1) 200 m above sea level = + 200

(2) 100 m below sea level =- 100

(3) 10 m above sea level = + 10

(4) Sea level = 0

62. Write the opposite of each of the following :

(1) Decrease in size
(2) Failure
(3) Profit of? 10
(4) 1000 AD
(5) Rise in water level
(6) 60 km south
(7)10m above the danger mark of river Ganga
(8) 20 m below the danger mark of the river Brahmaputra
(9) Winning by a margin of 2000 votes
(10)Depositing? 100 in the Bank account
(11) 20°C rise in temperature.

Solution: (1) Increase in size.

(2) Success.

(3) Loss of? 10

(4) 1000 BC

(5) Fall in water level.

(6) 60 km north.

(7)10 m below the danger mark of river Ganga.

(8) 20 m above the danger mark of the river Brahmaputra.

(9) Losing by a margin of 2000 votes.

(10)Withdrawing ? 100 from the Bank account.

(11) 20°C fall in temperature.

63. Temperature of a place at 12:00 noon was +5°C. Temperature increased by 3°C in first hour and decreased by 1°C in the second hour. What was the temperature at 2:00 pm?

Solution:

Given

Temperature of a place at 12:00 noon was +5°C. Temperature increased by 3°C in first hour and decreased by 1°C in the second hour.

Temperature at 12 : 00 noon = + 5°C

Temperature at 1 : 00 p.m. = 5°C + 3°C = 8°C

And temperature at 2 : 00 p.m. = 8°C- 1°C = 7°C

64. Write the digits 0, 1,2, 3, 9 in this order and insert ‘+’ or between them to get the result 3.

Solution: The digits can be written as 0-1-2-3-4-5-6+7+8+9=3

65. Write the integer which is its own additive inverse.

Solution: 0 is the integer which is its own additive inverse

66. Write six distinct integers whose sum is 7.

Solution: 1 + 2 + 3 + 6 + (-2) + (-3) = 7

The six distinct integers are 1, 2, 3, 6, -2 and -3.

67. Write the integer which is 4 more than its additive inverse.

Solution: Let x be the required integer.

According to question,

x = 4 + (-x), where (-x) is the additive inverse of x.

=> x = 4-x => x + x = 4

=> 2x = 4 => x = 2

The required integer is 2.

68. Write the integer which is 2 less than its additive inverse.

Solution: Let the required integer be x.

According to question,

x = (-x)- 2, where -x is the additive inverse of x.

x=-x-2 => x + x = -2

2x = -2 => x = —1

69. Write two integers whose sum is less than both the integers.

Solution: We can take any two negative integers, i.e., -2 and -4.

Sum = -2 + (-4) = -2- 4 = -6, which is less than both -2 and -4.

70. Write two distinct integers whose sum is equal to one of the integers.

Solution: Two distinct integers whose sum is equal to one of the integer, then one must be 0 in them.

Let us take 0 and 4.

Sum = 0 + 4 =4.

71. Using number line, how do you compare

(1) two negative integers?
(2) two positive integers?
(3) one positive and one negative integer?

Solution: Since, the integer lying on right is greater than the integer lying on left.

In all of the given cases (1), (2) and (3), we can compare by using the number line by observing which one of the given integers lie
on tine right or left.

72. Observe the following : 1 +2-3+4+5-6-7+8-9=-5 Change one’-‘sign as’+’sign to get the sum 9.

Solution: On observing the given expression, l + 2- 3 + 4 +5-6- 7 +8- 9 = -5, we noticed that (-7) should be replaced by (+7) to get a result of 9.

Thus, 1+2-3+4+5-6+7+8-9
= (l+2 + 4 + 5 + 7 + 8)-(3 + 6 + 9)
= 27-18 = 9

73. Arrange the following integers in the ascending order: -2, 1,0, -3, 4, -5

Solution: Ascending order of given integers is, -5, -3, -2, 0,1, 4

74. Arrange the following integers in the descending order: -3, 0, -1,-4, -6

Solution: Descending order of given integers is, 0, -1, -3, -4, -6

75. Write two integers whose sum is 6 and difference is also 6.

Solution: We have, 6 + 0 = 6,

6-0 = 6

The required two integers are 6 and 0.

76. Write five integers which are less than -100 but greater than -1 50.

Solution: The required five integers which are less than -100 but greater than -150 are -101,-102, -103, -104 and -105.

77. Write four pairs of integers which are at the same distance from 2 on the number line.

Solution: There are many pairs of integers which are at the same distance from 2 i.e., (1,3), (0,4), (-1, 5) and (-2, 6)

There are many pairs of integers which are at the same distance from 2

78. The sum of two integers is 30. If one of the integers is -42, then find the other.

Solution:

Given

The sum of two integers is 30. If one of the integers is -42

Let the required integer be x.

According to question,

x + (-42) = 30

=>x -42 = 30

=> x = 30 + 42 = 72

79. Sum of two integers is -80. If one of the integers is -90, then find the other.

Solution:

Given

Sum of two integers is -80. If one of the integers is -90

Let the required integer be x.

According to question,

x + (-90) = -80

=> x- 90 =- 80 => x = -80 + 90 = 10

80. If we are at 8 on the number line, in which direction should we move to reach the integer

(1) -5
(2) 11
(3) 0?

Solution: (1) If we are at 8 on the number line, then to reach the integer -5, we must move towards the left on the number line.

(2) If we are at 8 on the number line, then to reach the integer 11, we must move towards the right on the number line.

(3) If we are at 8 on the number line, then to reach the integer 0, we must move towards the left on the number line.

81. Using the number line, write the integer which is

(1) 4 more than -5
(2) 3 less than 2
(3) 2 less than -2

Solution: (1) We want to know an integer 4 more than -5.

So, we start from -5 and proceed 4 steps to the right, then we obtain -1 as shown below.

We want to know an integer 4 more than -5

Hence, 4 more than -5 is -1.

(2) We want to know an integer 3 less than 2.

So, we start from 2 and proceed 3 steps to the left, then we obtain -1 as shown below.

So, we start from 2 and proceed 3 steps to the left, then we obtain -1 as shown below

Hence, 3 less than 2 is -1.

(3) We want to know an integer 2 less than-2.

So, we start from -2 and proceed 2 steps to the left, then we obtain -4 as shown below.

So, we start from -2 and proceed 2 steps to the left, then we obtain -4 as shown below

Hence, 2 less than -2 is -4

82. Find the value of 49 -(-40) -(-3) + 69

Solution: We have,

49- (-40)- (-3) + 69

= 49 + 40 + 3 + 69 = 161

The value of 49 -(-40) -(-3) + 69= 161

83. Subtract -5308 from the sum [(-2100) + (-2001)1]

Solution: We have, [(-2100) + (-2001)]

= [-2100-2001]

= -4101

Required difference = —4101- (-5308)

= -4101 + 5308 = 1207

NCERT Exemplar Solutions For Class 6 Maths Chapter 5 Understanding Elementary Shapes

Class 6 Maths Chapter 5 Understanding Elementary Shapes

Exercise – 5.1

1. What is the disadvantage in comparing line segments by mere observation?

Solution: There may be chance of error due to improper viewing.

2. Why is it better to use a divider than a ruler, while measuring the length of a line segment?

Solution: It is better to use a divider than a ruler, because the thickness of the ruler may cause difficulties in reading of the length of a line segment. However, a divider gives accurate measurement.

3. Draw any line segment, say AB. Take any point C lying in between A and B. Measure the lengths of AB, BC and AC. Is AB=AC + CS? [Note : If A, B, C are any three points on a line such that AC + CB = AB, then we can be sure that C lies between A and B.]

Solution: AB = 4 cm, AC = 2 cm, CB = 2 cm

Hence, AC + CB = 2 cm + 2 cm = 4 cm = AB

Yes, AB = AC + CB.

Read and Learn More Class 6 Maths Exemplar Solutions

4. If A, B, C are three points on a line such that AB-5 cm, BC= 3 cm and AC=8 cm, which one of them lies between the other two?

Solution:

Given

A, B, C are three points on a line such that AB-5 cm, BC= 3 cm and AC=8 cm

Since, AC is the longest line segment.

Thus B is the point lying between A and C.

5. Verify, whether D is the mid point of AG.

Verify, whether D is the mid point of AG

Solution: AD= 3 units, DG = 3 units, AD = DG

Thus, D is Thw mid- point od AG.

6. If B is the mid point of AC and C is the mid point of BD, where A, B, C, D lie on a straight line, say why AB = CD?

Solution: B is the mid point of AC.

AB = BC …………..(1)

And C is the mid point of BD.

BC = CD ……………(2)

From (1) and (2), we get

AB = CD

7. Draw five triangles and measure their sides. Check in each case, if the sum of the lengths of any two sides is always less than the third side.

Solution: Sum of the lengths of any two sides of a triangle can never be less than length of the third side.

Draw five triangles and measure their sides

Exercise – 5.2

1. What fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from

(1) 3 to 9
(2) 4 to z7
(3) 7 to 10
(4) 12 to 9
(5) 1 to 10
(6) 6 to 3

Solution: (1)1/2 or two right angles

(2)1/4 or one right angle

(3)1/4 or one right angle

(4)3/4 or three right angles

(5)3/4 or three right angles

(6)3/4 or three right angles

2. Where will the hand of a clock stop if it

(1) starts at 12 and makes 1/2 of a revolution, clockwise?
(2) starts at 2 and makes 1/2 of a revolution, clockwise?
(3) starts at 5 and makes 1/4 of a revolution, clockwise?
(4) starts at 5 and makes 3/4 of a revolution, clockwise?

Solution: (1) At 6

(2) At 8

(3) At 8

(4) At 2

3. Which direction will you face if you start facing

(1) eastand make 1/2 ofa revolution clockwise?
(2) east and make 1×1/2 of a revolution clockwise?
(3) west and make 3/4 of a revolution anti clock wise?
(4)south and make one full revolution?
(Should we specify clockwise or anti-clockwise for this last question? Why not?)

Solution: (1) West

(2) West

(3) North

(4) South

No, it is not necessary to specify because whether we turn clockwise or anti-clockwise, one full revolution will bring us back to the
original position.

4. What part of a revolution have you turned through if you stand facing

(1) east and turn clockwise to face north?
(2) south and turn clockwise to face east?
(3) west and turn clockwise to face east7

Solution: (1) 3/4

(2) 3/4

(3) 1/2

5. Find the number of right angles turned through by the hour hand of a clock when it goes from

(1) 3 to 6
(2) 2 to 8
(3) 5 to 11
(4) 10 to 1
(5) 12 to 9
(6) 12 to 6

Solution: (1) One right angle

(2) Two right angles

(3) Two right angles

(4) One right angle

(5) Three right angles

(6) Two right angles

6. How many right angles do you make if you start facing

(1) south and turn clockwise to west?
(2) north and turn anti-clockwise to east?
(3) west and turn to west?
(4) south and turn to north?

Solution: (1) One right angle

(2) Three right angles

(3) Four right angles

(4) Two right angles

7. Where will the hour hand of a clock stop if it starts

(1) from 6 and turns through 1 right angle?
(2) from 8 and turns through 2 right angles?
(3) from 1 0 and turns through 3 right angles?
(4) from 7 and turns through 2 straight angles?

Solution: (1) At 9

(2) At 2

(3) At 7

(4) At 7

Exercise – 5.3

1. Match the following:

(i) Straight angle                (1) Less than one-fourth of a revolution

(ii) Right angle                   (2) More than half a revolution

(iii) Acute angle                 (3) Half of a revolution

(iv) Obtuse angle              (4) One-fourth of a revolution

(v) Reflex angle                 (5) Between 1/4 and 1/3 of a revolution

Solution: (i)—> (3); (ii)—> (4); (iii)—> (1);(iv)—>(5);(v)—>(2)

2. Classify each one of the following angles as right, straight, acute, obtuse or reflex :

Classify each one of the following angles

Solution: (a) Acute angle

(b) Obtuse angle

(c) Right angle

(d) Reflex angle

(e) Straight angle

(f) Acute angle

Exercise – 5.4

1. What is the measure of (i) a right angle? (ii) a straight angle?

Solution: (1) 90°

90°

(2) 180°

180°

2. Say True or False:

(1) The measure of an acute angle < 90°.
(2) The measure of an obtuse angle < 90°.
(3) The measure of a reflex angle > 1 80°.
(4) The measure of one complete revolution = 360°.
(5) If m∠A = 53° and m∠B = 35°, then m∠A > m∠B.

Solution: (1) True

(2) False

Since, measure of an obtuse angle is greater than 90° and less than 180°.

(3) True

(4) True

(5) True

3. Write down the measures of

(1) some acute angles.
(2) some obtuse angles. (give at least two examples of each).

Solution: (1) Measures of 2 acute angles are 35°, 20″

(2) Measures of 2 obtuse angles are 110°,135°

4. Measure the angles given below using the Protractor and write down the measure.

Measure the angles given below

Solution: (a) Near about 40°

(b) Near about 130°

(c) Near about 90°

(d) Near about 60°, 120°, 90°

Note: Students can measure the angles exactly with the help of protractor.

5. Which angle has a large measure ? First estimate and then measure.

Measure of Angle A =
Measure of Angle B =

Which angle has a large measure

Solution: By estimating, we observe that ZB has a large measure.

∠A = near about 40° and ∠B = near about 65°.

Note: Students can measure the angles exactly with the help of protractor.

6. From these two angles which has larger measure? Estimate and then confirm by measuring them

From these two angles which has larger masure

Solution: Second angle has larger measure.

First angle is near about .30″ and second angle is near about 70″.

Note: Students can measure the angles exactly with the help of protractor

7. Fill in the blanks with acute, obtuse, right or straight:

(1) An angle whose measure is less than that of a right angle is___________.

(2) An angle whose measure is greater than that of a right angle is____________.

(3) An angle whose measure is the sum of the measures of two right angles is__________.

(4) When the sum of the measures of two angles is that of a right angle, then each one of them is___________.

(5) When the sum of the measures of two angles is that of a straight angle and if one of them is acute then the other should be_____________.

Solution: (1) Acute

(2) Obtuse

(3) Straight

(4) Acute

(5) Obtuse

8. Find the measure of the angle shown in each figure. (First estimate with your eyes and then find the actual measure with a protractor).

Find the measure of the angle shown in each figure

Solution: (1) By estimating with our eyes, we came to know that the measure of angle is 30°.

(2) By estimating with our eyes, we came to know that the measure of angle is 120°.

(3) By estimating with our eyes, we came to know that the measure of angle is 60°.

(4) By estimating with our eyes, we came to know that the measure of angle is 150°.

Note: Students can measure the angles exactly with the help of protractor.

9. Find the angle measure between the hands of the clock in each figure :

Find the angle measure between the hands of clock

Solution: (1) 90° (Right angle)

(2) 30° (Acute angle)

(3) 180° (Straight angle)

10. Investigate In the given figure, the angle measures 30°. Look at the same figure through a magnifying glass. Does the angle becomes larger? Does the size of the angle change?

In the given figure, the angle measures 30°

Solution: No, the measure of angle will be same.

11. Measure and classify each angle:

Measure and classify each angle

Measure and classify each angle table

Solution:

Measure and classify each angle table 1

Note: Students can measure the angles exactly with the help of protractor.

Exercise 5.5

1. Which of the following arc models for perpendicular lines :

(1) The adjacent edges of a table top.
(2) The lines of a railway track.
(3) The line segments forming the letter ‘L’.
(4) The letter V.

Solution: (1) Perpendicular

(2) Not perpendicular

(3) Perpendicular

(4) Not perpendicular

2. Let PQ be the perpendicular to the line segment XY. Let PQ and XY intersect in the point A. What is the measure of ∠PAY1

Solution: ∠PAY = 90°

Angle PAY = 90°

3. There are two set-squares in your box. What are the measures of the angles that are formed at their corners? Do they have any angle measure that is common?

Solution: One set-square has angles 45°, 90°, 45° and other set-square has angles 60°, 90°, 30°.

Yes, they have angle measure 90° as common.

4. Study the diagram. The line / is perpendicular to linem

(1) Is CE = EG?

Study the diagram. The line l is perpendicular to line m

(2) Does PE bisect CGI
(3) Identify any two line segments for which PE is the perpendicular bisector.
(4) Are these true?
(i) AC>FG
(ii) CD = GH
(iii) BC < EH.

Solution: (1) Yes, both measure 2 units.

(2) Yes, because CE = EG

(3) BH and DF are two line segments for which PE is the perpendicular bisector.

(4) (i) True

Since, AC = 2 units and FG =1 unit.
AC>FG

(ii) True
Since, CD = GH= 1 unit

(iii) True
Since, BC = 1 unit, EH = 3 units

BC < EH

Exercise – 5.6

1. Name the types of following triangles:

(1) Triangle with lengths of sides 7 cm, 8 cm and 9 cm.
(2) ΔABC with AB = 8.7 cm, AC = 7 cm and BC = 6 cm.
(3) ΔPQR such that PQ = QR = PR = 5 cm.
(4) ΔDEF with m∠D = 90°
(5) ΔXYZ with m∠Y = 90° and XY = YZ.
(6) ΔLMN with m∠L= 30°, m∠M = 70° and m∠N= 80°.

Solution: (1) Scalene triangle
(2) Scalene triangle
(3) Equilateral triangle
(4) Right-angled triangle
(5) Isosceles right-angled triangle
(6) Acute-angled triangle

2. Match the following :

Measures of Triangle                                                              Type of Triangle

(i) 3 sides of equal                                                               (1) Scalene length
(ii) 2 sides of equal length                                                   (2) Isosceles right
(iii) All sides are of different length                                     (3) Obtuse angled
(iv) 3 acute angles                                                               (4) Right angled
(v) 1 right angle                                                                   (5) Equilateral
(vi) 1 obtuse angle                                                              (6) Acute angled
(vii) 1 right angle with two sides of equal length               (7) Isosceles

(i) -> (5); (ii) -» (7); (iii) -> (1); (iv) -> (6); (v) —> (4); (vi) —> (3); (vii) —> (2)

3. Name each of the following triangles in two different ways: (you may judge the nature of the angle by observation)

Name each of the following trianglesName each of the following triangles 1

Solution: (a) Acute angled triangle and Isosceles triangle

(b) Right-angled triangle and Scalene triangle

(c) Obtuse-angled triangle and Isosceles triangle

(d) Right-angled triangle and Isosceles triangle

(e) Acute angled triangle and Equilateral triangle

(f) Obtuse-angled triangle and Scalene triangle

4. Try to construct triangles using match sticks. Some are shown here. Can you make a triangle with

(1) 3 matchsticks?
(2) 4 matchsticks?
(4) 6 matchsticks?
(3) 5 matchsticks?
(Remember you have to use all the available matchsticks in each case)

Name the type of triangle in each case. If you cannot make a triangle, think of reasons for it.

Try to construct triangles using match sticks

Solution: (1) Yes, it is possible to make a triangle with 3 matchsticks because sum of lengths of two sides is greater than the length of third side.

Yes, it is possible to make a triangle

(2) No, it is not possible to make a triangle with 4 matchsticks because sum of lengths of two sides is equal to the length of third side.

(3) Yes, it is possible to make a triangle with 5 matchsticks because sum of lengths of two sides is greater than the length of third side.

Yes, it is possible to make a triangle with 5 matchsticks

(4) Yes, it is possible to make a triangle with the help of 6 matchsticks because sum of lengths of two sides is greater than the length of third side.

Yes, it is possible to make a triangle with 6 matchstics

Exercise – 5.7

1. Say True or False:

(1) Each angle of a rectangle is a right angle.
(2) The opposite sides of a rectangle are equal in length.
(3) The diagonals of a square are perpendicular to one another.
(4) All the sides of a rhombus are of equal length.
(5) All the sides of a parallelogram are of equal length.
(6) The opposite sides of a trapezium are parallel.

Solution: (1) True

(2) True

(3) True

(4) True

(5) False

Since, opposite sides of a parallelogram are of equal length.

(6) False

Since, only one pair of opposite sides of a trapezium is parallel.

2. Give reasons for the following:

(1) A square can be thought of as a special rectangle.
(2) A rectangle can be thought of as a special parallelogram.
(3) A square can be thought of as a special rhombus.
(4) Squares, rectangles, parallelograms are all quadrilaterals.
(5) Square is also a parallelogram.

Solution: (1) Because its all angles are right angle and opposite sides are equal.

(2) Because its opposite sides are equal and parallel.

(3) Because its all sides are equal and diagonals are perpendicular to each other.

(4) Because all of them have four sides’.

(5) Because its opposite sides are equal and parallel.

3. A figure is said to be regular if its sides are equal in length and angles are equal in measure. Can you identify the regular quadrilateral?

Solution:

Given

A figure is said to be regular if its sides are equal in length and angles are equal in measure.

A square is a regular quadrilateral.

Exercise – 5.8

1. Examine whether the following are polygons. If any one among them is not, say why?

Examine whether the following are polygons

Solution: (1) As it is not a closed figure, therefore, it is not a polygon.

(2) It is a polygon because it is closed by line segments.

(3) It is not a polygon because it is not made by line segments.

(4) It is not a polygon because it is not made by only line segments and also it has curved surface.

2. Name each polygon.

Name each polygon

Name each polygon 1

Make two more examples of each of these.

Solution:

(a) Quadrilateral

(b) Triangle

(c) Pentagon

(d) Octagon

Two more examples of each:

Two more examples of each

3. Draw a rough sketch of a regular hexagon. Connecting any three of its vertices, draw a triangle. Identify the type of the triangle you have drawn.

Solution:

Draw a rough sketch of a regular hexagon

ABCDEF is a regular hexagon and ΔAEF is a triangle formed by joining AE. Hence, ΔAEF is an isosceles triangle.

4. Draw a rough sketch of a regular octagon. (Use squared paper if you wish). Draw a rectangle by joining exactly four of the vertices of the octagon.

Solution:

Draw a rough sketch of a regular octagon

ABCDEFGH is a regular octagon and CDGH is a rectangle formed by joining C and H; D and G.

5. A diagonal is a line segment that joins any two vertices of the polygon and is not a side of the polygon. Draw a rough sketch of a pentagon and draw its diagonals.

Solution:

Given

A diagonal is a line segment that joins any two vertices of the polygon and is not a side of the polygon

Draw a rough sketch of a pentagon

ABODE is the required pentagon and its diagonals are AD, AC, BE, BD and CE.

Section-2 NCERT Exemplar

Directions: In each of the questions1 to 11, out of four options only one is correct. Write the correct answer.

1. Measures of the two angles between hour and minute hands of a clock at 9 O’ clock are

(1) 60°, 300°
(2) 270°, 90°
(4) 30°, 330°
(3) 75°, 285°

Solution: (2): We know that 1 minute = 6°

Measures of the two angles

The number of minutes between hour and minute hands of a clock at 9 O’ clock is 15 and 45.

The required angles are 15 * 6° = 90° and 45×6° = 270°

2. If a bicycle wheel has 48 spokes, then the angle between a pair of two consecutive spokes is

\(\text { (A) }\left(5 \frac{1}{2}\right)\)

 

\(\text { (B) }\left(7 \frac{1}{2}\right)\)

 

\(\text { (C) }\left(\frac{2}{11}\right)\)

 

\(\text { (D) }\left(\frac{2}{15}\right)\)

Solution: A bicycle wheel has 48 spokes.

The angle formed in circle is 360°.

The angle between a pair of two consecutive spokesr is 360 + 48 = \(7 \frac{24}{48}=7 \frac{1}{2}\)

3. In the given figure, ∠XYZ cannot be written as

In the given figure, ∠XYZ cannot be written as

(1)∠Y
(2) ∠ZXY
(3) ∠ZYX
(4) ∠XYP

Solution: (2): In the given figure, the name of angles formed are ∠XYP, ∠XYZ, ∠PYX and ∠ZYX

Thus, ∠ZXY is not a correct option.

4. In the given figure, if point A is shifted to point B along the ray PX such that PB = 2PA, then the measure of ∠BPY is

In the given figure, if point A is shifted to point

(1) greater than 45°
(2) Y45°
(3) less than 45°
(d) 90°

Solution: (2) : Since, the increase and decrease in the length of arms of an angle does not affect the angle made by them.
∠BPY = 45°

5. The number of obtuse angles In figure Is

The number of obtuse angles In figure Is

(1) 2
(2) 3
(3) 4
(4) 5

Solution: (3): The obtuse angles formed in the given figure are ∠AOD = 20° + 45° + 65° = 130°

∠BOD = 45° + 65° = 110°, ∠COE = 65° + 30° = 95° and ∠BOE = 45° + 65° + 30° = 140°

Thus, there are total  4 obtuse angles formed.

Thus, there are total 4 obtuse angles formed

6. The number of triangles in the given figure is

The number of triangles in the given figure is

(1) 10
(2) 12
(4) 14
(3) 13

Solution: (3): We have,

The names of triangles formed in the given figure

The names of triangles formed in the given figure are

ΔABC, ΔABD, ΔADC, ΔAFG, ΔAEG, ΔAFE, ΔFGD, ΔEGD, ΔFED, ΔFBD, ΔDEC, ΔAFD, ΔAED

7. If the sum of two angles is greater than 180°, then which of the following is not possible for the two angles?

(1) One obluso angle and one acute angle
(2) One reflex angle and one acute angle
(3) Two obtuse angles
(4) Two right angles

Solution: (4): Since, the sum of two rijÿht angles is 180°.

8. If the sum of two angles is equal to an obtuse angle, then which of the following is not possible?

(1) One obtuse angle and one acute angle
(2) One right angle and one acute angle
(3) Two acute angles
(4) Two right angles

Solution: (4): Since, the sum of two right angles is 180°, which is a straight angle, not an obtuse angle.

9. In the given figure, AB = BC and AD = BD = DC. The number of isosceles triangles in the figure is

In the given figure, AB = BC and AD = BD = DC

(1) 1
(2) 2
(3) 3
(4) 4

Solution: (3):

Given

In the given figure, AB = BC and AD = BD = DC.

We have, AB = BC and AD = BD = DC.

ΔABD, ΔBDC and ΔABC all are isosceles triangles.

There are 3 isosceles triangles formed in the given figure.

10. In the given figure, ∠BAC = 90° and AD ⊥ BC. The number of right triangles in the figure is

In the given figure, ∠BAC = 90° and AD ⊥ BC

(1) 1
(2) 2
(3) 3
(4) 4

Solution: (3):

Given

In the given figure, ∠BAC = 90° and AD ⊥ BC.

We have, ∠BAC = 90° and AD ⊥ BC

∠BDA = ∠CDA = ∠BAC = 90°

There are 3 right triangles formed in the given figure.

11. In the given figure, PQ⊥ RQ, PQ = 5 cm and QR = 5 cm. Then ΔPQR is

In the given figure, PQ ⊥ RQ, PQ = 5 cm

(1) a right triangle but not isosceles
(2) an isosceles right triangle
(3) isosceles but not a right triangle
(4) neither isosceles nor right triangle

Solution: (13) :

Given

In the given figure, PQ⊥ RQ, PQ = 5 cm and QR = 5 cm.

We have, PQ⊥ RQ, PQ = 5 cm and QR = 5 cm, which shows that in a triangle, two sides are equal and their included angle is 90°.

ΔPQR is an isosceles right triangle

Directions: In questions 12 to 17,fill in the blanks to make the statements true.

12. An angle greater than 180° and less than a complete angle is called________.

Solution: Reflex angle

13. A pair of opposite sides of a trapezium are

Solution: Parallel

14. In the given figure,

In the given figure

(1) ∠AOD is a/an __________ angle
(2) ∠COA is a/an__________ angle.
(3) ∠AOE is a/an__________ angle.

Solution: (1) Right : ∠AOD = 30° + 20° + 40° = 90°

(2) Acute : ∠COA = 20° + 30° = 50°

(3) Obtuse : ∠AOE = 40° + 40° + 20° + 30° = 130°

15. The number of triangles in figure is Their names are________.

The number of triangles in figure is

Solution: ΔAOB, ΔOC, ΔCOD, ΔCAB and ΔACD

16. Number of angles less than 180” in given figure is_______ and their names are_______.

Number of angles less than 180” in given figure

Solution: 12, ∠ABO, ∠BAO, ∠AOB, ∠BOD, ∠COD, ∠ODC, ∠OCD, ∠OCA, ∠CAO, ∠AOC, ∠BAC and ∠ACD.

17. The number of right angles in a straight angle is__________ and that in a complete angle is___________

Solution: 2, 4 : The number of right angles in a straight angle is 2 and that in a complete angle is 4.

Directions: State whether the statements given in questions 18 to 20 are true (T) or false (F).

18. A horizontal line and a vertical line always intersect at right angles.

Solution: True

19. If the arms of an angle on the paper are increased, the angle increases.

Solution: False

If the arms of an angle on the paper are increasing or decreasing, it doesn’t affect the angle made by them.

20. If the arms of an angle on the paper are decreased, the angle decreases.

Solution: False

21. Write down fifteen angles (less than 180°) involved in given figure.

Write down fifteen angles

Solution: The names of fifteen angles (less than 180°) involved in figure are :
∠AEC, ∠ADB, ∠EAD, ∠EFD, ∠EFB, ∠DFC, ∠FBC, ∠FCB, ∠BFC, ∠ABC, ∠ACB, ∠DCF, ∠FDC, ∠EBF and ∠BEF.

22. Is it possible for the same

(1) line segment to have two different lengths?
(2) angle to have two different measures?

Solution: (1) No, it is not possible for the same line segment to have two different lengths.

(2) No, it is not possible for the same angle to have two different measures.

23. Will the measure of ∠ABC and of ∠CBD make measure of ∠ABD in given figure?

Will the measure of ∠ABC and of ∠CBD

Solution: Yes, ∠ABD- ∠ABC + ∠CBD

=> ∠ABD is the sum of ∠ABC and ∠CBD.

24. Will the lengths of line segment AB and line segment BC make the length of line segment AC in given figure?

Will the lengths of line segment AB

Solution: Yes, the length of line segment AC is the sum of the lengths of line segment AB and BC.

25. Look at a given figure. Mark a point

Look at a given figure. Mark a point

(1) A which is in the interior of both∠1 and ∠2.
(2) B which is in the interior of only ∠1.
(3) Point C in the interior of ∠1.
Now, state whether points B and C lie in the interior of ∠2 also.

Solution: Yes, the given figure shows that the points B and C lie in the interior of ∠2 also.

26. In which of the following figures,

(1) perpendicular bisector is shown?
(2) bisector Is shown?
(3) only bisector is shown?
(4) only perpendicular is shown?

In which of the following figures

Solution:

(1) Figure (ii) shows the perpendicular bisector.

(2) Figure (ii) and (iii) shows the bisector.

(3) Figure (iii) shows only the bisector.

(4) Figure (i) shows only the perpendicular.

27. In given figure,

In given figure

(1) name any four angles that appear to be acute angles.
(2) name any two angles that appear to be obtuse angles.

Solution: (1) Acute angles: ∠ADE, ∠AEB,∠ABE and ∠ECD.

(2) Obtuse angles: ∠BCD and ∠BAD.

28. In given figure,

In given figure line

(1) is AC + CB = AB?
(2) is AB + AC =CB?
(3) is AB + BC = CA?

Solution: (1) Yes, AC + CB = AB

(2) No, AB- AC = CB

(3) No, AB-BC = CA

29. In given figure,

In given figure hexagon

(1) What is AE + EC?
(2) What is AC-EC?
(3) What is BD-BE?
(4) What is BD-DE?

Solution: (1) AE+EC=AC

(2) AC-EC=AE

(3) BD-BE = ED

(4) BD- DE = BE

30. Using the information given, name the right angles in each part of given figures.

(1) BA⊥BD

BA perpendicular to BD

(2) RT⊥ST

Rt perpendicular to st

(3) AC⊥BD

AC perpendicular to BD

(4) RS⊥RW

RS perpendicular to RW

(5) AC⊥BD

AC Perpendicular to BD

(6) AE⊥CE

AE Perpendicular to CE

(7) AC⊥CD

AC perpendicular to CD

(8) OP⊥AB

OP Perpendicular to AB

Solution: (1) BA ⊥ BD, The right angle is ∠ABD.

(2) RT ⊥ ST, The right angle is ∠RTS

(3) AC⊥BD, The right angles are ∠ACD and ∠ACB.

(4) RS⊥RW, The right angle is ∠SRW.

(5) AC ⊥ BD, The right angles are ∠AED, ∠AEB,∠BEC and ∠CED.

(6) AE ⊥CE, The right angle is ∠AEC.

(7) AC ⊥ CD, The right angle is ∠ACD.

(8) OP⊥AB, The right angles are ∠AKO, ∠AKP, ∠BKO and ∠BKP.

31. What conclusion can be drawn from each part of given figures, if

(1) DB is the bisector of ∠ADC?

DB is the bisector of ∠ADC

(2) BD bisects ∠ABC?

BD bisects ∠ABC

(3) DC is the bisector of ∠ADB, CA ⊥ DA and CB ⊥ DB?

DC is the bisector of ∠ADB,

Solution: (1) DB is the bisector of ∠ADC.

∠ADB =∠CDB

(2) BD bisects ∠ABC.

∠ABD = ∠CBD

(3) DC is the bisector of ∠ADB, CA ⊥ DA and CB ⊥ DB

∠ADC = ∠BDC, ∠CAD = ∠CBD =90°

32. An angle is said to be trisected, if it is divided into three equal parts. If in a given figure, ∠BAC = ∠CAD = ∠DAE, how many trisectors are there for ∠BAE?

An angle is said to be trisected

Solution:

Given

An angle is said to be trisected, if it is divided into three equal parts. If in a given figure, ∠BAC = ∠CAD = ∠DAE,

We have given, ∠BAC = ∠CAD = ∠DAE

There are two trisectors namely, AC and AD.

33. Can we have two acute angles whose sum is

(1) an acute angle? Why or why not?
(2) a right angle? Why or why not?
(3) an obtuse angle? Why or why not?
(4) a straight angle? Why or why not?
(5) a reflex angle? Why or why not?

Solution: (1) Yes, v the sum of two acute angles can be the acute angle.

E.g., 30° and 40° are two acute angles and their sum = 30° + 40° = 70°, which is also an acute angle.

(2) Yes, v the sum of two acute angles be a right angle.

E.g., 30° and 60° are two acute angles and their sum = 30° + 60° = 90°, which is a right angle.

(3) Yes, the sum of two acute angles can be an obtuse angle.

E.g., 45° and 60° are two acute angles and their sum = 45° + 60° = 105°, which is an obtuse angle.

(4) No, v the sum of two acute angles is always less than 180°.

(5) No, v the sum of two acute angles is always less than 180°.

34. Can we have two obtuse angles whose sum is

(1) a reflex angle? Why or why not?
(2) a complete angle? Why or why not?

Solution: (1) Yes, v the sum of two obtuse angles is always greater than 180°.

E.g., 135° and 100° are two obtuse angles and their sum = 135° + 100° – 235°, which is greater than 180°.

(2) No, v the sum of two obtuse angles is greater than 180° but less than 360°. In the above example, we can see that the sum of 135° and 100° i.e., 235° is greater than 180° but less than 360°.

NCERT Exemplar Solutions For Class 6 Maths Chapter 4 Basic Geometrical Ideas

Class 6 Maths Chapter 4 Basic Geometrical Ideas

Directions: In each of the questions 1 to 5, out of four options only one is correct. Write the correct answer.

1. The number of lines passing through five points such that no three of them are collinear is

(1) 10
(2) 5
(3) 20
(4) 8

Solution: (1) : Since, total number of points is 5 and we need two points to form a line.

Total number of lines passing through the points is 5 x 2 = 10

2. The number of diagonals in a septagon is

(1) 21
(2) 42
(3) 7
(4) 14

Solution: (4): Since, the number of diagonals in

a polygon =n(n-3)/2

Septagon has 7 sides, i.e., n = 7

The number of diagonals in a septagon

Read and Learn More Class 6 Maths Exemplar Solutions

NCERT Exemplar Solutions For Class 6 Maths Chapter 4 Basic Geometrical Ideas

3. Number of line segments in figure is

Number of line segments in figure is

(1) 5
(2) 10
(3) 15
(4) 20

Solution: (2) : The adjacent figure shows the line segments;

The adjacent figure shows the line segment

AB, BC, CD, AC, AD, BD, AE, BE, CE, DE

Thus, there are 10 line segments.

4. The number of angles in the given figure is

The number of angles in the given figure is

(1) 3
(2) 4
(3) 5
(4) 6

Solution: (4): The names of angles formed in the given figure are ZAOB, ZAOC, ZAOD, ZBOC, ZBOD, and ZCOD.

There are a total of 6 angles formed.

There are total 6 angles formed

5. A polygon has a prime number of sides. Its number of sides is equal to the sum of the two least consecutive primes. The number of diagonals of the polygon is

(1) 4
(2) 5
(3) 7
(4) 10

Solution: (2): We have given, the number of sides of a polygon

= Sum of the two least consecutive primes

= 2 +3 =5 [ ∴ 2 and 3 are the least consecutive prime numbers]

The number of diagonals = n(n- 3)/2

where n = 5

=5(5-3)/2=5×2/2=5

Directions: In questions 6 to 14, fill in the blanks to make the statements true

6. The number of diagonals in a hexagon is

Solution: 9 : Number of diagonals = n(n-3)/2

where n = 6

=6(6-3)/2=6×3/2=9

7. In the given figure, points lying in the interior of the triangle PQR are_________, that in the exterior are__________ and that on the triangle itself are__________.

In the given figure, points lying in the interior

Solution: O and S; T and N; P, Q, R and M

8. In the given figure, points A, B, C, D and E are collinear such that AB = BC = CD = DE. Then

Given figure

(1) AD=AB +________
(2) AD=AC +_________
(3) midpoint of AE is_____________
(4) midpoint of CE is__________
(5) AE=_______x AB

Solution: (1) BD: AD=AB + BC + CD=AB + BD

(2) CD:AD = AB + BC+CD=AC + CD

(3) C: AB + BC = CD + DE

=> AC = CE :. Mid point of AE is C.

(4) D :  CD = DE

Mid point of CE is D.

(5) 4 : AB + BC + CD + DE = AE
=> AB+AB+AB + AB = AE ∴ 4AB = AE

9. The number of straight angles in given figure is____________.

The number of straight angles in given figure is

Solution: 4: The number of straight angles in the given figure is 4.

10. The number of common points in the two angles marked in the given figure is_______.

The number of common points in the two angles marked in given figure are two

Solution: Two: The two angles marked; ∠PAQ and ∠PDQ.

The number of common points are 2 and these are P and Q.

11. The number of common points in the two angles marked in a given figure is__________.

The number of common points in the two

Solution: One: The two angles marked; ∠CAB and ∠DAE.

The number of common points is 1 and that is A.

12. The number of common points in the two angles marked in given figure _________.

The number of common points in the two angles marked in given figure

Solution: Three: There are 3 common points in the two angles marked in the given figure and these are P, Q, and R.

13. The number of common points in the two angles marked in a given figure is_______.

The number of common points

Solution: Four: The number of common points in the two angles marked in the given figure is 4 and these are E, D, G and F.

14. The common part between the two angles BAC and DAB in given figure is_____.

The common part between the two angles

Solution: Ray AB

Directions: State whether the statements given in questions 15 to 21 are true (T) or false (F).

15. If line PQ|| line m, then line segment PQ || m.

Solution: True

16. Two parallel lines meet each other at some point.

Solution: False

Two lines in a plane which do not meet even when produced indefinitely in either direction, are known as parallel lines.

17. Measures of ∠ABC and ∠CBA in given figure are the same.

Measures of ABC and CBA in given figure are the same

Solution: True

ABC is same as ∠CBA.

18. Two line segments may intersect at two points.

Solution: False

The intersecting point of any two line segments is only one.

19. Many lines can pass through two given points.

Solution: False

There is only one line which passes through two given points.

20. Only one line can pass through a given point.

Solution: False

There are infinite number of lines which passes through a given point.

21. Two angles can have exactly five points in common.

Solution: False

It can have any number of points.

22. Name all the line segments in given figure.

Name all the line segments in given figure

Solution: The line segments are AB, BC, CD, DE, AC, AD, AE, BD, BE and CE

23. Name the line segments shown in given figure

Name the line segments shown in given figure

Solution: The line segments are AB, BC, CD, DE and EA

24. State the mid points of all the sides of given figure.

State the mid points of all the sides of given figure

Solution: X is a mid-point of AC,

Y is a mid-point of BC and

Z is a mid-point of AB.

25. Name the vertices and the line segments in given figure.

Name the vertices and the line segments

Solution: The vertices are : A, B, C, D and E.

The line segments are : AB, BC, CD, DE, EA, AC and AD.

26. Name the following angles of given figure,using three letters :

Name the following angles of given figure

(1) ∠1
(2) ∠2
(3) ∠3
(4) ∠1+∠2
(5) ∠2 + ∠3
(6) ∠1+∠2 + ∠3
(7) ∠CBA-∠1

Solution:(1) ∠1 = ∠CBD

(2) ∠2 = ∠DBE

(3) ∠3 = ∠EBA

(4) ∠1 +∠2 = ∠CBD + ∠DBE = ∠CBE

(5) ∠2 + ∠3 = ∠DBE +vEBA = ∠DBA

(6) ∠1 + ∠2 + ∠3 = ∠CBD + ∠DBE + ∠EBA = ∠CBA

(7)∠CBA- ∠1 = ∠CBA- ∠CBD = ∠DBA

27. Name the points and then the line segments in each of the following figures:

Name the points and then the line segments

Solution:(i) Name of the points -> A,B and C.

Name of the line segments —> AB, BC and CA.

(ii) Name of the points —>A,B,C and D.

Name of the line segments —> AB, BC, CD and DA.

(iii) Name of the points —> A, B, C, D and E.

Name of the line segments —> AB, BC, CD, DE, and EA.

(iv) Name of the points —> A, B, C, D, E and F.

Name of the line segments —> AB, CD and EF.

28. Which points in given figures, appear to be mid-points of the line segments? When you locate a mid-point, name the two equal line segments formed by it.

Which points in given figures, appear to be mid points

Solution:(i) The given figure shows there is no mid-point.

(ii) The given figure shows that O is the mid-point of AB and the name of the two equal line segments are AO and OB.

(iii) The given figure shows that D is the mid-point of BC and the name of the two equal line segments are BD and DC.

29. Find out the incorrect statement, if any, in the following : An angle is formed when we have

(1) two rays with a common end-point
(2) two line segments with a common endpoint
(3) a ray and a line segment with a common end-point

Solution: All the three statements (1), (2) and (3) are incorrect.

The common initial point of two rays forms an angle.

30. What is common in the following figures (i) and (ii)? Is figure (i) that of triangle? if not, why?

common in the following figures

Solution: Both the figures (i) and (ii) have 3 line segments.

No, Fig. (i) is not a triangle since the three line segments does not form a closed figure.

31. If two rays intersect, will their point of intersection be the vertex ofan angle of which the rays are the two sides?

Solution: Yes

32. How many points are marked in given figure?

1 line segment

Solution: Two points A and B are marked

33. How many line segments are there in given figure?

1 line segment

Solution: Only one line segment, AB is there.

34. In given figure, how many points are marked? Name them.

3 points

Solution: From the given figure, Three points A, B, and C are marked.

35. How many line segments are there in the given figure? Name them.

3 line segments

Solution: From the given figure, Three line segments, namely AB, BC, and AC are there.

36. In the given figure, how many points are marked? Name them.

4 Points

Solution: From the given figure, Four points A, B, C, and D are marked.

37. In the given figure how many line segments are there? Name them.

4 line segments

Solution: Six line segments, namely AB, AC, AD, BC, BD, and CD.

38. In the given figure, how many points are marked? Name them

5 line segments

Solution: From the given figure, Five points are marked, namely A, B, D, E and C.

39. In given figure how many line segments are there? Name them

10 line segments

Solution: From the given figure, Ten line segments, namely AB, AD, AE, AC, BD, BE, BC, DE, DC and EC.

40. In given figure,O is the centre of the circle.

In the given figure of the circle

(1) Name all chords of the circle.
(2) Name all radii of the circle.
(3) Name a chord, which is not the diameter of the circle.
(4) Shade sectors OAC and OPB.
(5) Shade the smaller segment of the circle formed by CP.

Solution:(1) Name of chords : PC and BA.

(2) Name of radii : PO, OC, OB and OA.

(3) PC is a chord which is not the diameter of the circle

(4) Shade sectors OAC and OPB

(5) Shade the smaller segment of the circle formes by cp

41. Write the name of

(1) vertices
(2) edges, and
(3) faces of the prism shown In given figure.

PRISM

Solution: (1) Vertices: A, B, C, D, E and F.

(2) Edges: AB, BC, AC, DF, FC, BD. EF, ED and AE.

(3) Faces: EACF, EDBA, ABC, DEF and DBCF.

42. How many edges, faces and vertices are there in a sphere?

Solution: In a sphere, edges – 0, faces – 0 and vertices- 0.

43. Draw all the diagonals of a pentagon ABCDE and name them.

Solution: The diagonals of a pentagon ABCDE are AC, AD, BE, BD and EC.

all the diagonals of a pentagon ABCDE

NCERT Exemplar Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Class 6 Maths Chapter 3 Playing With Numbers

Directions: In questions 1 to 14, out of the four options, only one is correct. Write the correct answer.

1 Sum of the number of primes between 16 to 80 and 90 to 100 is

(1) 20
(2) 18
(4) 16
(3) 17

Solution: (3): Prime numbers between 16 to 80 are 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73 and 79.

So, there are 16 prime numbers between 16 to 80.

Also, 97 is the only prime number between 90 to 100.

So, there is only a prime number between 90 to 100.

Required sum = 16 +1 = 17

Read and Learn More Class 6 Maths Exemplar Solutions

NCERT Exemplar Solutions for Class 6 Maths Chapter 3 Playing With Numbers

 

2. Which of the following statements is not true?

(1) The HCF of two distinct prime numbers is 1
(2) The HCF of two coprime numbers is 1
(3) The HCF oftwo consecutive even numbers is 2
(4) The HCF of an even and an odd number is even

Solution: (4): The HCF of an even and an odd number is always an odd number.

3. The number of distinct prime factors of the largest 4-digit number is

(1) 2
(3) 5
(2) 3
(4) 11

Solution: (2): The largest 4-digit number is 9999.

Now,

The largest 4-digit number is 9999

9999 = 3 x 3×11 x101

So, distinct prime factors of 9999 are 3, 11 and 101, i.e., 3 distinct prime factors.

4. The number of distinct prime factors of the smallest 5-digit number is

(1) 2
(2) 4
(4) 8
(3) 6

Solution: (1): The smallest 5-digit number is 10000.

Now,

the smallest 5 digit number

10000 = 2x2x2x2x5x5x5x5

So, distinct prime factors of10000 are 2 and 5, i.e., 2 distinct prime factors.

5. If the number 7254*98 is divisible by 22, the digit at* is

(1) 1
(2) 2
(3) 6
(4) 0

Solution: (3): 7254*98 is divisible by 22, if it is divisible by both 2 and 11.

Given number is even, therefore it is divisible by 2.

7254*98 is divisible by 11, if (7 + 5 + * + 8) – (2 + 4 + 9) or (20 + *)- 15 or 5 + * is divisible by 11.

The digit at * should be filled by 6.

6. The largest number which always divides the sum of any pair of consecutive odd numbers is

(1) 2
(2) 4
(3) 6
(4) 8

Solution:(2): The sum of any pair of consecutive odd numbers comesin the form ofmultiple of 4.

The required largest number is 4.

7. A number is divisible by 5 and 6.It may not be divisible by

(1) 10
(3) 30
(2) 15
(4) 60

Solution: (4): The LCM of 5 and 6 is 30.

And 30 is divisible by 10, 15 and 30 but not by 60.

8. The sum of the prime factors of 1 729 is

(1) 13
(2) 19
(4) 39
(3) 32

Solution: (4): We have,

prime factors of 1729

The prime factors of 1729 = 7 x 13 x 19.

The sum of the prime factors of 1729 = 7+ 13 +19 = 39

9. The greatest number which always divides the product of the predecessor and successor of an odd natural number other than 1, is

(1) 6
(2) 4
(3) 16
(4) 8

Solution: (2): Since the odd natural numbers other than1 are 3, 5, 7, 9 and so on.

Now, the predecessor and successor of 3 are 2 and 4 respectively, and their product is 2×4 = 8

Similarly, the predecessor and successor of 5 are 4 and 6 respectively and their product is 4 x 6 = 24 and so on.

Thus, the above shows that the greatest number which always divides the product of the predecessor and successor of an odd natural other than 1 is 4.

10. The number of common prime factors of 75, 60,105 is

(1) 2
(2) 3
(3) 4
(4) 5

Solution: (1): We have,

The number of common prime factors of 75, 60.105

75 = 3 x 5 x 5

60 = 2 x 2 x 3 x 5

105 = 3 x 5 x 7

The common prime factors of 75, 60 and 105 are 3 and 5 i.e., 2 in number.

11. Which of the following pairs is not coprime?

(1) 8,10
(3) 1,3
(2) 11, 12
(4) 31,33

Solution: (1): 8 and 10 are not co-prime numbers.

Their common factor other than 1 is 2

12. Which of the following numbers is divisible by 11?

(1) 1011011
(2) 1111111
(3) 22222222
(4) 3333333

Solution: (3): The difference of the sum of digits of 22222222 at even and odd places is 0. It must be divisible by 11.

13. LCM of10, 15 and 20 is

(1) 30
(2) 60
(4) 180
(3) 90

Solution: (2): We have,

lcm 10 15,20

The LCM of 10, 15 and 20 is 2x2x3x5=60

14. LCM of two numbers is 1 80. Then which of the following is not the HCF of the numbers?

(1) 45
(2) 60
(3) 75
(4) 90

Solution:(3): We have,

hcf of 180

The factors of 180 = 2 x 2 x 3 x 3 x 5

75 does not divide 180

75 can not be the HCF of the numbers whose LCM is 180.

Directions: In questions 15 to 32 state whether the given statements are true (T) orfalse (6).

15. Sum of two consecutive odd numbers is always divisible by 4.

Solution: True

16. If a number divides three numbers exactly, it must divide their sum exactly.

Solution: True

17. If a number exactly divides the sum of three numbers, it must exactly divide the numbers separately.

Solution: False

18. If a number is divisible both by 2 and 3, then it is divisible by 12.

Solution: False

Since, a number is divisible by both 2 and 3 implies that it is also divisible by 6.

19. A number with three or more digits is divisible by 6, if the number formed by its last two digits (i.e., ones and tens) is divisible by 6.

Solution: False

A number is divisible by 6 if the number is divisible by both 2 and 3.

20. A number with 4 or more digits is divisible by 8, if the number formed by the last three digits is divisible by 8.

Solution: True

21. If the sum of the digits of a number is divisible by 3, then the number itself is divisible by 9.

Solution: False

If the sum of the digits of a number is divisible by 3, then the number it self is divisible by 3.

22. All numbers which are divisible by 4 may not be divisible by 8.

Solution: True

Since, 12 is divisible by 4 but not by 8.

23. The Highest Common Factor of two or more numbers is greater than their Lowest Common Multiple.

Solution: False

HCF of two or more numbers is smaller than their LCM.

24. LCM of two or more numbers is divisible by their HCF.

Solution: True

25. LCM of two numbers is 28 and their HCF is 8.

Solution: False

The LCM of two numbers i.e., 28 is not divisible by their HCF, i.e., 8.

Their HCF cannot be 8.

26. LCM of two or more numbers may be one of the numbers.

Solution: True

27. HCF of two or more numbers may be one of the numbers.

Solution: True

28. Any two consecutive numbers are coprime.

Solution: True

The common factor of any two consecutive numbers is 1.

29. If the HCF of two numbers is one of the numbers, then their LCM is the other number.

Solution: True

The product of two numbers =HCF * LCM.

30. The HCF of two numbers is smaller than the smaller ofthe numbers.

Solution: False

Since, HCF of two numbers is less than or equal to the smaller of the numbers.

31. The LCM of two numbers is greater than the larger ofthe numbers.

Solution: False

Since, LCM of two numbers is greater than or equal to the larger of the numbers.

32. The LCM of two coprime numbers is equal to the product ofthe numbers.

Solution: True

33. A number is a _________ of each ofits factor.

Solution: Multiple

34.__________ is a factor of every number.

Solution:1

35. The number of factors of a prime number is_______

Solution: 2

36. A number for which the sum of all its factors is equal to twice the number is called a___________ number.

Solution: Perfect

37. The numbers having more than two factors are called___________

Solution: Composite

38. 2 is the only__________

Solution: Prime

39. Two numbers having only 1 as a common factor are called___________numbers.

Solution: Co-prime

40. Number of primes between 1 to 100 is_______.

Solution: 25 : Prime numbers between I to 100,are 2. 3, 5. 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

So, there are 25 primes between 1 to 100.

41. If a number has ________in ones place, then it is divisible by 10.

Solution: 0

42. A number is divisible by 5, if it has____________ or in its ones place.

Solution: 0, 5

43. A number is divisible by________if it has any of the digits 0, 2, 4, 6, or 8 in its ones place. 1,1,1

Solution: 2

44. If the sum of the digits in a number is a __________ of 3, then the number is divisible by

Solution: Multiple

45. If the difference between the sum of digits at odd places (from the right) and the sum of digits at even places (from the right) of a number is either 0 or divisible by_____________, then the number is divisible by 11.

Solution: 11

46. The LCM of two or more given numbers is the lowest of their common________.

Solution: Multiple

47. The HCF of two or more given numbers is the highest of their common____________.

Solution: Factors

48. Find the LCM of 80, 96, 125, 160.

Solution: We have,

The LCM of 80, 96, 125 and 160

The LCM of 80, 96, 125 and 160 is = 2x2x2x2x2x3x5x5x5 = 12000

49. Find the LCM of 160, 170 and 90.

Solution: We have,

The LCM of 160, 170 and 190

The LCM of 160, 170 and 190

=2x2x2x2x2x3x3x5x17

= 24480

50. Determine the sum of the four numbers as given below:

(1) successor of 32
(2) predecessor of 49
(3) predecessor of the predecessor of 56
(4) successor of the successor of 67

Solution:

Since, successor of32is 33, predecessor of 49 is 48, predecessor of the predecessor of 56 is 54 and successor of the successor of 67 is 69.

The required sum = 33 + 48 + 54 + 69 = 204

51. Determine the least number which when divided by 3, 4 and 5 leaves remainder 2 in each case.

Solution: We have,

lcm of 3 4 5

Since, the LCM of 3, 4 and 5 is 2 x 2 x 3 x 5 = 60.

The required number is 60 + 2 = 62

So, 62 is the least number which when divided by 3, 4 and 5 leaves remainder 2 in each case.

52. A merchant has 120 litres of oil of one kind, 180 litres of another kind and 240 litres of a third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin?

Solution:

Given:

merchant has 120 litres of oil of one kind, 180 litres of another kind and 240 litres of a third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity.

A merchant has 3 kinds of oil with different quantities like 120 litres, 180 litres and 240 litres

Since, he wants lo soli the oil by filling the three kinds of oil in tins of equal capacity, so the greatest capacity of such a tin is the HCF of 120, 180 and 240.

hcf of 180,240,120

Now, 120 = 2x2x2x3x5

180 = 2x2x3x3x5

240 = 2x2x2x2x3x5

The required greatest capacity of a tin

= (2 x 2 x 3 x 5) litres

= 60 litres

The greatest capacity of a tin = 60 litres

53. Find a 4-digit odd number using each of the digits 1, 2, 4 and 5 only once such that when the first and the last digits are interchanged, it is divisible by 4.

Solution: By using the digits 1, 2, 4 and 5 only once, we get a 4 digit odd number 4521.

When we interchanged its first and last digits we get a new number, i.e., 1524 which is divisible by 4.

Thus, the required number is 4521.

54. Using each of the digits 1, 2, 3 and 4 only once, determine the smallest 4-digit number divisible by 4.

Solution: By using the digits 1, 2, 3 and 4 only once, the smallest 4-digit number which is divisible by 4 is 1324.

55. Fatima wants to mail three parcels to three village schools. She finds that the postal charges are 20, 28 and 36, respectively. If she wants to buy stamps only of one denomination, what is the greatest denomination of stamps she must buy to mail the three parcels?

Solution:

Given

Fatima wants to mail three parcels to three village schools. She finds that the postal charges are 20, 28 and 36, respectively.

The postal charges to mail three parcels are 20, 28 and 36 respectively.

Also, Fatima wants to buy stamps only of one denomination.

So, to find the greatest denomination of stamps, we find the HCF of 20, 28 and 36.

 

hcf of 20, 28,36

Now, 20 = 2 x 2 x 5

28 = 2 x 2 x 7

36 = 2 x 2 x 3 x 3

The HCF of 20, 28 and 36 is 2×2 = 4

So, ? 4 is the greatest denomination of stamps, she must buy to mail the three parcels.

56. Three brands A, B and C of biscuits are available in packets of 12, 15 and 21 biscuits respectively. If a shopkeeper wants to buy an equal number of biscuits, of each brand, what is the minimum number of packets of each brand, he should buy?

Solution:

Given

Three brands A, B and C of biscuits are available in packets of 12, 15 and 21 biscuits respectively. If a shopkeeper wants to buy an equal number of biscuits, of each brand

A shopkeeper has three brands A, B and C of biscuits are available in packets of 12, 15 and 21 biscuits respectively.
Also, a shopkeeper wants to buy equal number of biscuits of each brand for that we need to find the LCM of 12, 15 and 21.

 

lcm of 12,15,21

The LCM of 12, 15 and 21 = 2x2x3x5x7 = 420

Thus, the required number of packets of

brand A =420/12= 35,

brand B =420/15 =28 and

brand C =420/21 = 20

57. The floor of a room is 8 m 96 cm long and 6 m 72 cm broad. Find the minimum number of square tiles of the same size needed to cover the entire floor.

Solution:

Given

The floor of a room is 8 m 96 cm long and 6 m 72 cm broad.

Length of floor of a room = 8 m 96 cm = 896 cm

Breadth of floor of the room = 6 m 72 cm = 672 cm

To find the minimum number of square tiles of same size needed to cover the entire floor, we find the LCM of 896 cm and 672 cm.

lcm of 672 and 896

LCM of 672 and 896 is 2 X 2 X 2 x 2 X 2 X 2 X 2 X 3 X 7 = 2688

Number of square tiles =( 2688/672×2688/896)

= 4×3 = 12

58. In a school library, there are 780 books of English and 364 books of Science. Ms. Yakang, the librarian of the school wants to store these books in shelves such that each shelf should have the same number of books of each subject. What should be the minimum number of books in each shelf?

Solution:

Given

In a school library, there are 780 books of English and 364 books of Science. Ms. Yakang, the librarian of the school wants to store these books in shelves such that each shelf should have the same number of books of each subject.

Number of books of English = 780

Number of books of Science = 364

 

hcf of 780 and 364

780 = 2x2x3x5x13

364 = 2 x 2 x 7 x 13

The HCF of 780 and 364 = 2 x 2 x 13 = 52

Thus, the minimum number of books in each shelf = 52

59. In a colony of 100 blocks of flats numbering 1 to 1 00, a school van stops at every sixth block while a school bus stops at every tenth block. On which stops will both of them stop if they start from the entrance of the colony?

Solution:

Given

In a colony of 100 blocks of flats numbering 1 to 1 00, a school van stops at every sixth block while a school bus stops at every tenth block.

There are 100 blocks in a colony numbering 1 to 100.

A school van stops at every sixth block and a school bus stops at every tenth block.

We have to find the common stops at which they both stop if they start from a same entrance.

We need to find the LCM of 6 and 10.

lcm of 6 and 10

The LCM of 6 and 10 is 2 * 3 * 5 = 30. i.e.

Firstly both will stop at 30th block, then at 60th block and lastly at 90th block.

60. Test the divisibility of following numbers by 11

(1) 5335
(2) 9020814

Solution: (1) We have the difference between the sum of digits at odd places (from the right) and the sum of digits at even places (from the right) of a number 5335 is (5 + 3) = (3 + 5) = 8- 8 = 0, which is divisible by 11.

5335 is divisible by 11.

(2) We have the difference between the sum of digits at odd places (from the right) and the sum of digits at even places (from
the right) of the number 9020814 is (4 + 8 + 2 + 9) -(1 + 0 + 0) = 23-1 = 22, which is divisible by 11.

9020814 is divisible by 11.

61. Using divisibility tests, determine which of the following numbers are divisible by 4?

(1) 4096
(2) 21084
(3) 31795012

Solution: (1) We have, 4096

Since, the last two digits 96 is divisible by 4.

4096 must be divisible by 4.

(2) We have, 21084

Since, the last two digits 84 is divisible by 4.

21084 must be divisible by 4.

(3) We have, 31795012

Since, the last two digits 12 is divisible by 4.

31795012 must be divisible by 4.

62. Using divisibility test determine which of the following numbers are divisible by 9?

(1) 672
(2) 5652

Solution: (1) We have, 672

Since, the sum of all the digits of 672 is 15, which is not divisible by 9.

672 is not divisible by 9.

(2) We have, 5652

Since, the sum of all the digits of 5652 is 18, which is divisible by 9.

5652 must be divisible by 9.

 

 

NCERT Exemplar Solutions For Class 6 Maths Chapter 2 Whole Numbers

Class 6 Maths Chapter 2 Whole Numbers

Directions: In questions 1 to 11, out of the four options, only one is correct. Write the correct answer.

1. The number of whole numbers between 38 and 68 is

(1) 31
(2) 30
(3) 29
(4) 28

Solution (3): There are 29 whole numbers between 38 and 68.

2. The product of the successor and predecessor of 999 is

(1) 999000
(2) 998000
(3) 989000
(4) 1998

Solution (2): Successor of 999 = 999 +1 = 1000

Predecessor of 999 = 999-1 = 998

Now, their product = 998 x 1000 = 998000

Read and Learn More Class 6 Maths Exemplar Solutions

NCERT Exemplar Solutions For Class 6 Maths Chapter 2 Whole Numbers

3. The product of a non-zero whole number and its successor is always

(1) an even number
(2) an odd number
(3) a prime number
(4) divisible by 3

Solution (1): The product of a non-zero whole number (even/odd) and its successor (odd/ even) is always an even number.

4. A whole number is added to 25 and the same number is subtracted from 25. The sum of the resulting numbers is

(1) 0
(2) 25
(3) 50
(4) 75

Solution(3): Let the whole number be x.

According to question,

Required sum = (x + 25) + (25- x)

= x + 25 + 25- x = 50

5. Which of the following is not true?

(1) (7 + 8) + 9 = 7 + (8 + 9)
(2) (7 x 8) x 9 = 7 x (8 x 9)
(3) 7 + 8 x 9 = (7 + 8) x (7 + 9)
(4) 7 x (8 + 9) = (7 x 8) + (7 x 9)

Solution (3): 7 + 8 x 9 = 7 + 72 = 79,

(7 + 8) x (7 + 9)- 15 × 16 = 240 and 79 x 240

6. By using dot (•) patterns, which ofthe following numbers can be arranged in all the three ways namely a line, a triangle and a rectangle?

(1) 9
(2) 10
(3) 11
(4) 12

Solution (2): As we know that every number can be arranged as a line.

The number 10 can be shown as

Also, the number 10 can be shown as a triangle as given below :

And, the number 10 can also be shown as a rectangle, as given below :

7. Which of the following statements is not true?

(1) Both addition and multiplication are associative for whole numbers.
(2) Zero is the identity for multiplication of whole numbers.
(3) Addition and multiplication both are commutative for whole numbers.
(4) Multiplication is distributive over addition for whole numbers.

Solution (2): Zero is the identity for addition of whole numbers.

8. Which of the following statements is not true?

(1) 0 + 0 = 0
(3) 0x0 = 0
(2) 0-0 = 0
(4) 0-0 = 0

Solution (4): 0 + 0 is not defined.

9. The predecessor of 1 lakh is

(1) 99000
(2) 99999
(4) 100001
(3) 999999

Solution (2): 1 lakh = 100000

Predecessor of 100000 = 100000-1 = 99999

10. The successor of 1 million is

(1) 2 millions
(2) 1000001
(3) 100001
(4) 10001

Solution(2): 1 million = 1000000

Successor of 1000000 = 1000000 +1 = 1000001

11. Number of even numbers between 58 and 80 is

(1) 10
(2) 11
(3) 12
(4) 13

Solution (1): Even numbers between 58 and 80 are 60, 62, 64, 66, 68, 70, 72, 74, 76, 78.

So, here are 10 even numbers between 58 and 80.

Directions: In questions 12 to .37 stole whether the given statements are true (T) or false (F)

12. Successor of a one digit number is always a one digit number.

Solution: False

The successor of a one digit number 9 is a two digit number, i.c., 10.

13. Successor of a 3-digit number is always a 3-digit number.

Solution: False

The successor of a 3-digit number 999 is a 4-digit number, i.e., 1000.

14. Predecessor of a two digit number is always a two digit number.

Solution: False

The predecessor of a two digit number 10 is a one digit number, i.e., 9.

15. Every whole number has its successor.

Solution: True

16. Every whole number has its predecessor.

Solution: False

0 is a whole number and it does not have any predecessor.

17. Between any two natural numbers, there is one natural number.

Solution: False

Since, 1 and 2 are two natural numbers and there is no natural number between1 and 2.

18. The smallest 4-digit number is the successor of the largest 3-digit number.

Solution: True

The smallest 4-digit number is 1000, i.e., 999 +1 and the largest 3-digit number is 999.

19. Of the given two natural numbers, the one having more digits is greater.

Solution: True

20. Natural numbers are closed under addition.

Solution: True

The sum of two natural numbers is also a natural number.

21. Natural numbers are not closed under multiplication.

Solution: False

The multiplication of two natural numbers is also a natural number.

22. Natural numbers are closed under subtraction.

Solution: False

2 and 5 are natural numbers, but their subtraction, 2-5 =-3 isnot anatural number.

23. Addition is commutative for natural numbers.

Solution: True

Since, a + b = b + a, where ‘a’ and ‘V are natural numbers.

24. 1 is the identity for addition of whole numbers.

Solution: False

0 is the identity for addition of whole numbers.

As 0 + a = a + 0 = a, where a is any whole number.

25. 1 is the identity for multiplication of whole numbers.

Solution: True

a×1 = 1 x a = a, where ‘a’ is any whole number.

26. There is a whole number which when added to a whole number, gives the number itself.

Solution: True

27. There is a natural number which when added to a natural number, gives the number itself.

Solution: False

28. If a whole number is divided by another whole number, which is greater than the first one, the quotient is not equal to zero.

Solution: True

29. Any non-zero whole number divided by itself gives the quotient 1.

Solution: True

30. The product of two whole numbers need not be a whole number.

Solution: False

The product of any two whole numbers will always be a whole number.

31. A whole number divided by another whole number greater than 1 never gives the quotient equal to the former.

Solution : True

32. Every multiple of a number is greater than or equal to the number.

Solution: True

33. The number of multiples of a given number is finite.

Solution: False

Since, the number of multiples of a given number is infinite.

34. Every number is a multiple of itself.

Solution: True

35. Every whole number is the successor of another whole number.

Solution: False

0 is a whole number but it is not a successor of any whole number.

36. Sum of two whole numbers is always less than their product.

Solution: False

Product of two whole numbers may or may not be greater than their sum.

37. If the sum of two distinct whole numbers is odd, then their difference also must be odd.

Solution: True

Directions:In questions 38 to 59,fill in the blanks to make the statements true.

38. The smallest whole number is_________.

Solution 0 : 0 is the smallest whole number.

39. Successor of 1061 59 is_________.

Solution 106160: Successor of 106159 is 106159 +1, i.e., 106160

40. Predecessor of 1 00000 is_________.

Solution 99999: Predecessor of 100000 is 100000-l, i.e., 99999

41. 400 is the predecessor of__________.

Solution 401: 400 is the predecessor of 400 + 1, i.e., 401

42.________ is the successor of the largest 3 digit number.

Solution 1000: Largest 3 digit number = 999

Successor of 999 is 999 + 1, i.e., 1000

43. If 0 is subtracted from a whole number, then the result is the_________ itself.

Solution: Number

44. Whole numbers are closed under and under________.

Solution: Addition, multiplication

45. Natural numbers are closed under and under_______.

Solution: Addition, multiplication

46. Division of a whole number by ___________ is not defined.

Solution: 0

47. Multiplication is distributive over__________ for whole numbers.

Solution: Addition

48. 2395 x_________ = 6195×2395

Solution 6195: Since, multiplication is commutative for whole numbers

49. 1001 x 2002 = 1001 x (1001 +_________)

Solution: 1001

1001 x 2002 = 1001 x (1001 +1001)

50. 10001 x 0 = __________

Solution: 0.

10001 x 0 = 0

51. 2916 x_________ = 0

Solution: 0.

2916 x 0 = 0

52. 9128 x_______= 9128

Solution 1 : Since,1 is the multiplicative identity for whole numbers.

53. 125 + (68 + 17) = (125 +_______)+ 17

Solution 68: Since, addition is associative for whole numbers.

54. 8925x 1 =_____

Solution: 8925

8925x 1= 8925

55. 19 x 12 + 19 = 19 x (12 +_______)

Solution 1: Since multiplication is distributive over addition for whole numbers.

56. 24 x 35 = 24 x 1 8 + 24 X_______

Solution: 17

24 x 35 = 24 x 1 8 + 24 X 17

57. 32x(27×19) = (32x__________)x19

Solution 27: Since, multiplication is associative for whole numbers.

58. 786×3 + 786×7 =_______

Solution: 7860 : 786 x 3 + 786 x 7 = 786 * (3 + 7)

= 786 x 10 = 7860

786×3 + 786×7 =7860

59. 24 x 25 = 24 X □÷4= 600

Solution: 100

60. Given below are two columns – Column I and Column II. Match each item of Column I with the corresponding item of Column II.

Match the following

 

 

NCERT Exemplar Solutions For Class 6 Maths Chapter 1 Knowing Our Numbers

Class 6 Maths Chapter 1 Knowing Our Numbers

Directions: In questions 1 to 13, out of the four options, only one is correct. Write the correct answer.

1. The product of the place values of two 2’s in 428721 is

(1) 4
(2) 40000
(3) 400000
(4) 40000000

Solution : (3) Place values of 2’s in 428721 are 20000 and 20

The required product = 20000 x 20 = 400000

NCERT Exemplar Solutions For Class 6 Maths Chapter 1 Knowing Our Numbers

2. 3x 1 0000 + 7x 1 000 + 9×1 00 + 0x10 + 4 is the same as

(1) 3794
(2) 37940
(3) 37904
(4) 379409

Read and Learn More Class 6 Maths Exemplar Solutions

Solution (3) : 3 x 10000 + 7 x 1000 + 9 x 100 + 0 x 10 + 4

= 30000 + 7000 + 900 + 4 = 37904

3. If 1 is added to the greatest 7-digit number, it will be equal to

(1) 10 thousand
(2) 1 lakh
(3) 10 lakh
(4)1,00,00,000

Solution (4): The greatest 7-digit number = 99,99,999

Now, 99,99,999 +1 = 1,00,00,000

4. The expanded form of the number 9578 is

(1) 9×10000 + 5×1000 + 7×10 + 8×1
(2) 9x 1000 + 5x 100 + 7x 10 + 8×1
(3) 9x 1000 + 57x 10 + 8×1
(4) 9x 100 + 5x 100 + 7x 10 + 8×1

Solution (2) : Expanded form of 9578 = 9 x 1000 + 5 x 100 + 7 x 10 + 8×1

5. When rounded off to the nearest thousands, the number 85642 is

(1) 85600
(2) 85700
(3) 85000
(4) 86000

Solution (4): Round off 85642 to the nearest thousands = 86000

6. The largest 4-digit number, using any one digit twice, from digits 5, 9, 2 and 6 is

(1) 9652
(2) 9562
(3) 9659
(4) 9965

Solution: (4): The largest 4-digitnumber formed by 5, 9, 2 and 6 using digit ‘9’ twice = 9965

7. In Indian System of Numeration, the number 58695376 is written as

(1) 58,69,53,76
(2) 58,695,376
(3) 5,86,95,376
(4) 586,95,376

Solution: (3): In Indian System of Numeration, 58695376 can be written as 5,86,95,376

8. One million is equal to

(1) 1 lakh
(3) 1 crore
(2) 10 lakh
(4) 10 crore

Solution: (2): One million = 1,000,000

Also, 10,00,000 = 10 lakh

9. The greatest number which on rounding off to nearest thousands gives 5000, is

(1) 5001
(2) 5559
(3) 5999
(4) 5499

Solution: (4) : (1) Rounding off 5001 to nearest thousands = 5000

(2) Rounding off 5559 to nearest thousands = 6000

(3) Rounding off 5999 to nearest thousands = 6000

(4) Rounding off 5499 to nearest thousands = 5000

And 5499 > 5001

10. Keeping the place of 6 in the number 6350947 same, the smallest number obtained by rearranging other digits is

(1) 6975430
(2) 6043579
(3) 6034579
(4) 6034759

Solution (3): The new number formed = 6034579

11. Which of the following numbers in Roman numerals is incorrect?

(1) LXXX
(2) LXX
(3) LX
(4) LLX

Solution: (4) : LLX is incorrect because L can be used exactly once in Roman numerals.

12. The largest 5-digit number having three different digits is

(1) 98978
(2) 99897
(3) 99987
(4) 98799

Solution: (3) : The largest 5-digit number with three different digits is 99987

13. The smallest 4-digit number having three different digits is

(1) 1102
(2) 1012
(3) 1020
(4) 1002

Solution: (4) : The smallest 4-digit number with three different digits is 1002.

Directions: In questions 14 to 29 state whether the given statements are true (T) or false (F)

14. In Roman numeration, a symbol is not repeated more than three times.

Solution: True

15. In Roman numeration, if a symbol is repeated, its value is multiplied as many times as it occurs.

Solution: False

If a symbol is repeated, then its value is added as many times as it occurs in Roman numeration

16. 5555 = 5×1000 + 5×100 + 5×10 + 5×1

Solution: True

17. 39746 = 3 × 10000 + 9 × 1000 + 7 × 100 + 4×10 + 6

Solution: True

18. 82546 = 8 x 1000 + 2 × 1000 + 5 × 100 + 4 × 10 + 6

Solution: False

Since, 82546 = 8 × 10000 + 2× 1000 + 5×100 + 4×10 + 6

19. 532235 = 5 X 100000 + 3 x 10000 + 2 x 1000 + 2X 100 + 3X 10 + 5

Solution: True

20. XXIX = 31

Solution: False

Since, XXIX = 29

21. LXXIV = 74

Solution: True

22. The number LIV is greater than LVI.

Solution: False

Since, LIV = 54 and LVI = 56

LVI is greater than LTV

23. The numbers 4578, 4587, 5478, 5487 are in descending order.

Solution: False

Since, the numbers 4578, 4587, 5478, 5487 are in ascending order.

24. The number 85764 rounded off to nearest hundreds is written as 85700.

Solution: False

The number 85764 rounded off to nearest hundreds is 85800.

25. Estimated sum of 7826 and 1 2469 rounded off to hundreds is 20,000.

Solution: False

Round off 7826 to hundreds is 7800.

Round off 12469 to hundreds is 12500.

Required sum = 7800 + 12500 = 20300

26. The largest six digit telephone number that can be formed by using digits 5, 3, 4, 7, 0, 8 only once is 875403.

Solution: False

The largest six digit telephone number that can be formed by using digits 5, 3, 4, 7, 0, 8 only once is 875430.

27. The number 81652318 will be read as eighty one crore six lakh fifty two thousand three hundred eighteen.

Solution: False

The number 81652318 will be read as eight crore sixteen lakh fifty two thousand three hundred eighteen.

28. The largest 4-diglt number formed by the digits 6, 7, 0, 9 using each digit only once is 9760.

Solution: True

29. Among kilo, milli and centi, the smallest is centi.

Solution: False

Among kilo, milli and centi, the smallest is milli.

Directions: In questions 30 to 44, fill in the blanks to make the statements true

30. (1) 10 million =__________ crore

Solution:1

(2) 10 lakh =_______ million.

Solution: 1

31. (1) 1 metre = ____________ millimetres

Solution: 1000

(2) 1 centimetre =___________ millimetres

Solution: 10

(3) 1 kilometre =__________millimetres

Solution: 10,00,000

33. 100 thousands =________ lakh.

Solution: 1

34. Height of a person is 1 m 65 cm. His height in millimetres is_________.

Solution: 1650 : 1 m 65 cm = (1000 + 650) mm = 1650mm

35. Length of river ‘Narmada’ is about 1290 km. Its length in metres is___________.

Solution: 1290000 : 1290 km = (1290 x 1000) m = 1290000m

36. The distance between Srinagar and Leh is 422 km. The same distance in metres is___________

Solution: 422000: 422 km= (422 x 1000) m = 422000 m

37. Writing of numbers from the greatest to the smallest is called an arrangement in__________ order.

Solution: Descending

38. By reversing the order of digits of the greatest number made by five different non zero digits, the new number Is the________ number of five digits.

Solution: Smallest*

By reversing the order of digits of the greatest number made by five different non zero digits, the new number is the smallest number of these digits.

39. By adding 1 to greatest_______ digt number, we get ten lakh.

Solution: 6: Greatest 6-digit number = 999999 By adding1 to 999999, we get 1000000.

40. The number five crore twenty three lakh seventy eight thousand four hundred one can be written, using commas, in the Indian
System of Numeration as_______.

Solution: 5,23,78,401

41. In Roman Numeration, the symbol X can be subtracted from ________, M and C only.

Solution: L

42. The number 66 in Roman numerals is__________.

Solution: LXVI: 66 =LXVI

43. The population of Pune was 2,538,473 in 2001. Rounded off to nearest thousands, the population was____________.

Solution: 2,538,000

44. The smallest 6 digit natural number ending in 5 is__________

Solution: 100005

45. Arrange the following numbersin descending order: 8435, 4835, 13584, 5348, 25843

Solution: Descending order of given numbers is, 25843, 13584, 8435, 5348, 4835

46. Of the following numbers which is the greatest? Which is the smallest? 38051425, 30040700, 67205602

Solution: The greatest number is 67205602 and the smallest number is 30040700.

47. Write In expanded form :

(1) 74836
(2) 574021
(3) 8907010

Solution:(1) 74836- 7 x10000) x 4 x 1000 + 8 x 100 +3x 10+6×1

(2) 574021 =5x 1 00000 + 7x 1 0000 + 4x 1000 + 0x 100 + 2x 10+ 1 x1

(3) 8907010 = 8 x 1000000 + 9x 1 00000 + 0 x 10000 + 7 x 1000 + 0 x 100 + 1 x 10 + 0 x 1

48. As per the census of 2001, the population of four states are given below. Arrange the states in ascending and descending order of their population.

(1) Maharashtra 96878627
(2) Andhra Pradesh 76210007
(3) Bihar 82998509
(4) Uttar Pradesh 166197921

Solution: Ascending order -» (2), (3), (1), (4)

Descending order —» (4), (1), (3), (2)

49. The diameter of Jupiter is 142800000 metres. Insert commas suitably and write the diameter according to International System of Numeration.

Solution:

Given

The diameter of Jupiter is 142800000 metres.

The diameter of Jupiter is 142,800,000 metres.

50. India’s population has been steadily increasing from 439 millions in 1961 to 1028 millions in 2001. Find the total increase in population from 1961 to 2001. Write the increase in population in Indian System of Numeration, using commas suitably.

Solution:

Given

India’s population has been steadily increasing from 439 millions in 1961 to 1028 millions in 2001. Find the total increase in population from 1961 to 2001.

Total increaseinpopulation from 1961 to 2001= 1028 millions- 439 millions = 589 millions

According toIndian System of Numeration, the increase in population = 58,90,00,000

51. Radius of the Earth is 6400 km and that of Mars is 4300000 m. Whose radius is bigger and by how much?

Solution:

Given

Radius of the Earth is 6400 km and that of Mars is 4300000 m.

Radius of the Earth = 6400 km = 6400 x 1000 m = 6400000 m

And radius of the Mars = 4300000 m

Radius of the Earth is greater than the radius of the Mars by (6400(X)0- 4300000) m = 2100000 m.

52. In 2001, the populations of Tripura and Meghalaya were 3,199,203 and 2,318,822, respectively. Write the populations of these two states in words.

Solution:

Given

In 2001, the populations of Tripura and Meghalaya were 3,199,203 and 2,318,822, respectively.

The population of Tripura = 3,199,203 i.e., Three million one hundred ninety-nine thousand two hundred three.

And the population of Meghalaya = 2,318,822, i.e., Two million three hundred eighteen thousand eight hundred twenty two.

53. In a city, polio drops were given to 2,12,583 children on Sunday in March 2008 and to 2,16,813 children in the next month. Find the difference of the number of children getting polio drops in the two months.

Solution:

Given

In a city, polio drops were given to 2,12,583 children on Sunday in March 2008 and to 2,16,813 children in the next month.

Number of children getting polio drops in March 2008 = 2,12,583

Number of children getting polio drops in April 2008 = 2,16,813

The required difference of the number of children getting polio drops in the two months = 2,16,813- 2,12,583 = 4,230

54. A person had ? 1000000 with him. He purchased a colour T.V. for ? 16580, a motor cycle for ? 45890 and a flat for ? 870000. How much money was left with him?

Solution:

Given :

A person had 1000000 with him. He purchased a colour T.V. for 16580, a motor cycle for  45890 and a flat for  870000.

Total amount a personhad = 1000000

The amounthe spent on a colour T.V. = 16580

The amount he spent on a motorcycle = 45890

The amount he spent on a flat =870000

Total amount he spent = (16580 + 45890 + 870000) = 932470

Thus, the amount left with him = 1000000- 932470 = 67530

55. Out of 1 80000 tablets of Vitamin A, 1 8734 are distributed among the students in a district. Find the number of the remaining vitamin tablets.

Solution:

Given:

Out of 1 80000 tablets of Vitamin A, 1 8734 are distributed among the students in a district.

Total tablets of Vitamin A= 180000

Number of tablets distributed among the students in a district = 18734

The number of remaining vitamin tablets -180000-18734-161266

56. Chinmay had ? 610000. He gave 87500 to Jyoti, 126380 to Javed and 350000 to John. How much money was left with him?

Solution:

Chinmay had total amount = 610000

The amount he gave to Jyoti = 87500

The amount he gave to Javed = 126380

The amount he gave to John = 350000

Total amount given by Chinmay = (87500 + 126380+ 350000) = 563880

Thus, the amount left with him = 610000- 563880 = 46120

57. Find the difference between the largest number of seven digits and the smallest number of eight digits.

Solution: The smallest number of eight digits = 10000000

The largest number of seven digits = 9999999

The required difference = 10000000-9999999 =1

58. A mobile number consists of ten digits. The first four digits of the number are 9, 9, 8 and 7. The last three digits are 3, 5 and 5. The remaining digits are distinct and make the mobile number, the greatest possible number. What are these digits?

Solution:

Given

A mobile number consists of ten digits. The first four digits of the number are 9, 9, 8 and 7. The last three digits are 3, 5 and 5. The remaining digits are distinct and make the mobile number, the greatest possible number.

A mobile number consists of 10 digits.

If the first four digits of the number are 9, 9, 8 and 7 and the last three digits of the number are 3, 5 and 5.

Thus, for the greatest possible number, the remaining distinct digits are 6, 4 and 2.

59. A mobile number consists of ten digits. First four digits are 9, 9, 7 and 9. Make the smallest mobile number by using only one digit twice from 8, 3, 5, 6, 0.

Solution:

Given

A mobile number consists of ten digits. First four digits are 9, 9, 7 and 9.

A mobile number consists of 10 digits.

If the first four digits are 9, 9, 7 and 9.

Thus, the smallest mobile number by using only one digit twice from 8, 3, 5, 6, 0 is 9979003568.

60. In a five digit number, digit at ten’s place is 4, digit at unit’s place is one fourth of ten’s place digit, digit at hundred’s place Is 0, digit at thousand’s place is 5 times of the digit at unit’s place and ten thousand’s place digit is double the digit at ten’s place. Write the number.

Solution:

Given

In a five digit number, digit at ten’s place is 4, digit at unit’s place is one fourth of ten’s place digit, digit at hundred’s place Is 0, digit at thousand’s place is 5 times of the digit at unit’s place and ten thousand’s place digit is double the digit at ten’s place.

A number consists of 5 digits.

Now, the digit at ten’s place = 4,

the digit at unit’s place = 1/4×4=1,

tine digit at hundred’s place = 0,

the digit at thousand’s place = 5 x1 = 5

the digit at ten thousand’s place = 2×4 =8

Therefore, the number is 85041.

61. Find the sum of the greatest and the least six digit numbers formed by the digits 2, 0, 4, 7,6, 5 using each digit only once.

Solution: By using the digits 2, 0, 4, 7, 6, 5

The greatest number formed = 765420,

and the least number formed = 204567

The required sum= 765420 + 204567= 969987

62. A factory has a container filled with 35874 litres of cold drink. In how many bottles of 200 ml capacity each can it be filled?

Solution:

Given:

A factory has a container filled with 35874 litres of cold drink.

Quantity of cold drink in a container = 35874 litres = 35874 x 1000 ml = 35874000 ml

The capacity of one bottle = 200 ml

The required number of bottles = 35874000 + 200 = 179370

Therefore, 179370 bottles can be filledby cold drink.

63. The population of a town is 450772. In a survey, it was reported that one out of every 14 persons is illiterate. In all how many illiterate persons are there in the town?

Solution:

Given:

The population of a town is 450772. In a survey, it was reported that one out of every 14 persons is illiterate.

Total population of a town = 450772

Since, one out of every 14 personsis illiterate.

The number of illiterate persons in the town = 450772 + 14 = 32198

The number of illiterate persons in the town

Therefore, 32198 persons are illiterate in the town.

64. Make the greatest and the smallest 5-digit numbers using different digits in which 5 appears at ten’s place.

Solution: By using the digit 5 at ten’s place, the greatest 5-digit number is 98756, and the smallest 5-digitnumber is 10253

65. How many grams should be added to 2 kg 300 g to make it 5 kg 68 g?

Solution: 5 kg 68 g = (5 x 1000 + 68) g = 5068 g and 2 kg 300 g = (2 x 1000 + 300) g = 2300 g

The required number of grams should be added = 5068 g- 2300 g = 2768 g or 2kg 768 g

2768 grams should be added to 2 kg 300 g to make it 5 kg 68 g

66. A box contains 50 packets of biscuits each weighing 1 20 g. How many such boxes can be loaded in a van which cannot carry beyond 900 kg?

Solution:

Given

A box contains 50 packets of biscuits each weighing 1 20 g.

The total weight of a box containing 50 packets ofbiscuits each weighing 120 g = 50 x 120 g = 6000 g

The capacity of a van = 900 kg = 900 x 1000 g = 900000 g

The required number of boxes = 900000/6000 = 150

Therefore, 150 boxes can be loaded in the van.

67. How many lakhs make five billions?

Solution: 50000 lakhs make 5 billions.

68. How many millions make 3 crores?

Solution: 30 millions make 3 crores.

69.Estimate each of the following by rounding off each number to nearest hundreds:

(1) 874 + 478
(2) 793 + 397
(3) 11244 + 3507
(4) 17677 + 13589

Solution:(1) 874 rounded off to The nearest hundreds-900

478 rounded off to the nearest hundreds -500

Estimated sum = 900 + 500 – 1400

(2) 793 rounded off to the nearest hundreds = 800

397 rounded off to the nearest hundreds = 400

Estimated sum = 800 + 400 = 1200

(3) 11244 rounded off to the nearest hundreds= 11200

3507 rounded off to the nearest hundreds = 3500

Estimated sum = 11200 + 3500 = 14700

(4) 17677roundedofftothenearesthundreds = 17700

13589 rounded off to the nearest hundreds = 13600

Estimated sum = 17700 + 13600 = 31300

70. Estimate each of the following by rounding off each number to nearest tens:

(1) 11963-9369
(2) 76877-7783
(3) 10732-4354
(4) 78203-16407

Solution: (1) 11963 rounded off to the nearest tens- 11960

9369 rounded off to the nearest tens = 9370

Estimated difference=11960- 9370= 2590

(2) 76877 rounded off to the nearest tens = 76880

7783 rounded off to the nearest tens = 7780

Estimated difference= 76880- 7780= 69100

(3) 10732 rounded off to the nearest tens = 10730

4354 rounded off to the nearest tens = 4350

Estimated difference = 10730- 4350 = 6380

(4) 78203 rounded off to the nearest tens = 78200

16407 rounded off to the nearest tens = 16410

Estimated difference = 78200 – 16410 = 61790

71. Estimate each of the following products by rounding off each number to nearest tens:

(1) 87×32
(2) 311 × 113
(3) 3239 × 28
(4) 1385×789

Solution: (1) 87 rounded off to the nearest tens = 90

32 rounded off to the nearest tens = 30

Estimated product = 90 * 30 = 2700

(2) 311 rounded off to the nearest tens = 310

113 rounded off to the nearest tens = 110

Estimated product = 310 * 110 = 34100

(3) 3239 rounded off to the nearest tens= 3240

28 rounded off to the nearest tens = 30

Estimated product = 3240 * 30 = 97200

(4) 1385 rounded off to the nearest tens=1390

789 rounded off to the nearest tens = 790

Estimated product= 1390 * 790= 1098100

72. The population of a town was 78787 in the year 1991 and 95833 in the year 2001. Estimate the increase in population by
rounding off each population to nearest hundreds.

Solution:

Given

The population of a town was 78787 in the year 1991 and 95833 in the year 2001.

78787 rounded off to the nearest hundreds = 78800

95833 rounded off to the nearest hundreds = 95800

The estimated increase in population = 95800-78800 = 17000

73. Estimate the product 758 x 6784 using the general rule.

Solution: 758 can be rounded off to 800 and 6784 can be rounded off to 7000 Estimated product= 800 × 7000 = 5600000

The product 758 x 6784 using the general rule = 5600000

74. A garment factory produced 216315 shirts, 182736 trousers and 58704 jackets in a year. What is the total production of all the three items in that year?

Solution:

Given

A garment factory produced 216315 shirts, 182736 trousers and 58704 jackets in a year.

Number of shirts produced by the factory = 216315

Number of trousers produced by the factory = 182736

Number of jackets produced by the factory = 58704

Total production of the factory = 216315 + 182736 + 58704 = 457755

Total production of the factory = 457755

75. A vessel has 13 litres 200 mL of fruit juice. In how many glasses each of capacity 60 mL can it be filled?

Solution:

Given:

A vessel has 13 litres 200 mL of fruit juice.

Quantity of fruit juice in a vessel = 13 L 200 mL

= (13 x 1000 + 200) mL = 13200 mL

Capacity of one glass = 60 mL

The required number of glasses = 13200 ÷ 60 = 220

Therefore, 220 glasses can be filled by fruit juice.

76. A loading tempo can carry 482 boxes of biscuits weighing 15 kg each, whereas a van can carry 518 boxes each of the same weight. Find the total weight that can be carried by both the vehicles.

Solution:

Given

A loading tempo can carry 482 boxes of biscuits weighing 15 kg each, whereas a van can carry 518 boxes each of the same weight.

Total weight can be carried by a tempo= (482×15) kg = 7230kg

and the total weight can be carried by a van = (518×15) kg = 7770 kg

Thus, the total weight that can be carried by both the vehicles = (7230 + 7770) kg = 15000 kg

The total weight that can be carried by both the vehicles = 15000 kg

77. In the marriage of her daughter, Lecla spent 216766 on food and decoration,122322 on jewellery, 88234 on furniture and 26780 on kitchen items. Find the total amount spent by her on the above items.

Solution:

Given :

In the marriage of her daughter, Lecla spent 216766 on food and decoration,122322 on jewellery, 88234 on furniture and 26780 on kitchen items.

Amount spent by Leela on food and decoration = 216766

Amount spent by her on jewellery = 122322

Amount spent by her on furniture = 88234

Amount spent by her on kitchen items = 26780

Total amount spent by her = (216766 + 122322 + 88234 + 26780) = 454102

Total amount spent by her = 454102

78. A box contains 5 strips each having 12 capsules of 500 mg medicine in each capsule. Find the total weight in grams of medicine in 32 such boxes.

Solution:

Given :

A box contains 5 strips each having 12 capsules of 500 mg medicine in each capsule.

Quantity of medicine in one capsule = 500 mg

Quantity ofmedicinein 12 capsules or 1 strip = (500 x 12) mg = 6000 mg- 6 g

Quantity of medicine in 5 strips or 1 box = (6 x 5) g = 30 g

Quantity of medicinein 32 boxes = (30 x 32)g = 960g

The total weight in grams of medicine in 32 such boxes = 960g

 

Class 10 Science Notes For Heredity And Common Genetic Diseases

Heredity (How are the characteristics of an organism transmitted from one generation to another ?)

Heredity and Variation :

The resemblance of offspring to their parents is termed heredity. Members of a family share many similarities .in appearance, such as height, eye colour, hair colour. There are
differences in the manner how characteristics are inherited to- offsprings.

Different types of eye colour Different types of natural hair colour

Offsprings do not look exactly like their parents. So, the offspring resemble their parents in major respect but also differ from the parents in. many minor respects.

An Austrian monk, Gregor Johan Mendel performed’ a series of simple experiments and discovered how heredity worked.

Definition: Heredity is the transmission of characteristics or traits through genes from one generation to another by reproduction.

The term ‘genetics’ was coined by Bateson (1905). It is derived from the Greek word ‘gen’, means “to grow into”, whereas the term ‘gene’ was first proposed by Johannsen (1909).

The science of heredity and the principles governing the inheritance of characters from parent to the progeny (i.e. next generation offsprings) is called genetics.

Mutation—Sometimes there may be some change in chromosome structure or chromosome number or alteration in composition of DNA of a gene. This change in genetic constitution of an organism is called mutation.

Heridity parents and child

Mutation is the cause of variation among organisms. Mutation gives rise to origin of new characteristics or adaptive changes that may result into evolution in the long run. That s why it is sai,”Mutation is the raw material of evolution”.

Any mutation or variation may have some good effects or bad effects.

Read and Learn More Class 10 Science

Variation—The permanent change in chromosome or DNA may result in expression of characters, called as variation. So. variation signifies any difference between cells or individual organisms or groups of organisms of any species.

This is caused either by genetic difference (genotypic variation) or by the effect of environmental factors on the external manifestation of the organisms (phenotypic variation).

Variation shows deviation in characters in an individual from the group to which it belongs or deviation in characters of offspring from those of its parents. Genotypic variation may result Into phenotypic variation.

So, mutation is the cause variation whereas variation is the effect of mutation.

Examples of variation in man :

Free and attached ear lobe —In man, if ear lobes hang free, it is called free or detached ear lobe. On the contrary, if the ear lobes are connected directly to the sides of head, they are attached.

Ear lobe in man (Genetically Inherited)

This variation in different persons Is controlled by genes. Free ear lobe is dominant trait and that of attached ear lobe is recessive trait.

Roller and normal tongue—Some people can curl up (fold) the sides of their tongue to form a tube shape called tongue rolling, while others can not. It has been proposed tongue rolling Is due to genetic variation (but sometimes it might be a learning process also (environmental factor)].

Tongue rolling (Genetically inherited character)

Roller tongue Is dominant variation whereas normal unfolded tongue is recessive variation.

Key terms associated with heredity:

Characteristics: Definition—All special features of an organism that are transmitted through genes from one generation to next are called characteristics, e.g. Eye colour, height
of a person, gender (sex), blood groups, skin colour etc.

Traits: Definition—The phenotypic expression of a particular characteristic is called trait e.g. Eye colour is the characteristic whereas a trait would be blue eyes or brown eyes etc.

Allele or Allelomorph: Definition—The contrasting pair of genes located in the same locus of the homologous chromosome and control same characteristic are called allele
or allelomorph, e.g.

In a hybrid tall plant, T (Tall plant) and t (dwarf plant) genes are alleles. Similarly gene for brown eye colour (B) and blue eye colour (b) are also alleles.

Locus: Definition—The point or place on a chromosome where a particular gene is present is known, as locus of that gene. e.g. a chromosome may contain many loci linearly
present on a chromosome.

Each gene is present at a definite locus on the chromosome.

Unit of inheritance (Factor/Gene): Definition—Gene is a part of DMA, located on a particular point (locus) of a chromosome, functions as hereditary unit (responsible for inheritance of characteristics from parents to offsprings), also controls biological functions.

G. j. Mendel, father of genetics, called these genes as ‘’factors’ which control all the characteristics of an organism, e.g. Gene for tallness and gene for dwarfness in pea plant; gene for black hair colour and white hair colour in guineapig etc.

Monohybrid cross: Definition—The genetic experiment of cross breeding where only one pair of contrasting character (i.e. two alternate traits of a particular character) is considered
is known as monohybrid cross, e.g.

In Mendel’s experiment, the cross between pure tall and pure dwarf pea plant; in guineapig, the cross between pure black hair and pure white hair etc.

Dihybrid cross: Definition—The genetic experiment of cross breeding where two pairs of contrasting characters (i.e. two alternate forms of two different characters) considered is known as dihybrid cross, e.g.

In Mendel’s dihybrid cross with pea plant, he considered shape of seed (Round or Wrinkled) as well as colour of seed (Yellow or Green); in Guineapig, hair colour (Black or White) as well as texture of hair (Rough and Smooth).

Homozygous organisms: Definition—Whan the pair of genes present in the corresponding locus of the homologous chromosome are exactly same, the organism is called as
homozygous organism, e.g. homozygous tall pea plant (IT), homozygous dwarf pea plant (tt) etc.

Heterozygous organism: Definition-When the pair of genes (alleles) present in the corresponding locus of the homologous chromosome are of contrasting type, the organism
is called as heterozygous organism e.g. heterozygous tali pea plant (Tt); heterozygous black guineapig (Bb) etc.

Hybridization: Definition—it is the process of interbreeding between two genetically different (divergent) individuals. They may be of same species (intraspecific hybridization) or
different species (interspecific hybridization), e.g.

In Mendel’s monohybrid cross, there cross breeding or cross pollination between pure tali plant and pure dwarf plant.”

Pure and hybrid: Definition—Homozygous organisms are known as pure organisms and hetero-zygous organisms are called hybrid organisms, e.g, homozygous tall pea plant isTT, which Is also known as pure tall, whereas heterozygous pea plant is Tt, which is known as hybrid tall.

Parental generation: Deflnition-The first -set of parents used in the cross of a genetic experiment is called parental generation. So the generation from where a genetic experiment starts is known as parental generation and is represented by P. e.g.

In Mendel s monohybrid cross, parental generation (P) Consists of the pure tall and pure dwarf plants that were cross-pollinated.

Filial generation (F1, F2): Definition—In a genetic experiment, all the offsprings produced from the parental generation (P) are together known as filial generation. It is
represented by F.

The first generation offspring of the parental generation is known- as First filial generation (F1) and the offspring of F1 is known as second filial generation (F2) and so on. e.g.

In Mendel’s monohybrid cross in F1 generation, all plants were hybrid tall. In F2 generation, there were 3/4 tall plants and 1/4 dwarf plants.

Dominant characteristics: Definition—Out of two contrasting alleles in a hybrid organism, one is expressed and other is suppressed—the expressed phenotype is known as
dominant characteristic, the expressed gene is called dominant gene and the phenomenon is known as dominance, e.g.

In Mendel’s experiment, gene (factor) for tallness of plant (T) is dominant over the gene for dwarfness of plant (t); in Guineapig, black hair colour (B) is dominant over white hair colour (b).

Recessive characteristics: Definition—In a hybrid organism, out of two contrasting alleles, one is expressed and other is suppressed—the suppressed phenotype is known as recessive
characteristic, the suppressed gene is -called as recessive gene and the phenomenon is known as recessiveness. e.g.

In Mendel’s monohybrid cross, gene (factor) for dwarfness (t) is recessive to gene for tallness (T); in Guineapig, white hair coiour (b) is recessive to black hair colour (B).

Phenotype and Genotype: Definition—The externally visible (or measurable) characteristics of an organism are called as phenotype, e.g. Tall plant, dwarf plant, red flower, .white flower, black eye, blue eye etc.

Definition—The genetic constitution of an organism is called genotype. It is represented by some symbols to explain the composition of genes of a particular character, e.g. gene for
tallness is represented by (T), that of dwarfness by (t); the genotype of a pure tall plant is TT, that of hybrid tall plant is (Tt) etc.

In Mendel’s monohybrid cross, phenotype and genotype can be explained as follows :

In Mendel's monohybrid cross, phenotype and genotype can be explained as follows

In the above example, phenotype of the P generation plants were pure tall and pure dwarf and their corresponding genotype were (TT) and (tt) respectively. In the first filial generation
(F1) all the offspring plants were hybrid tall.

Here ‘hybrid tall’ is the phenotype and the genotype is (Tt). Usually dominant gene is represented by capital letter and recessive gene by small letter to explain genotype.

Differences between Homozygous (pure) and Heterozygous (hybrid) :

Differences between Homozygous (pure) and Heterozygous (hybrid)

Differences between Dominant and Recessive characters :

Differences between Dominant and Recessive characters

Differences between Phenotype and Genotype :

Differences between Phenotype and Genotype

Mendel’s work on Pea plant :

Gregor Johann Mendel (1822—1884) is known as “Father of genetics” because he was the first person to demonstrate the mechanism of inheritance of characters from parents to offsprings.

Mendel studied Botany and Physics in the University of Vienna. He was a priest in St. Thomas Church in Austria.

In 1856, Mendel observed the varieties of characteristics of pea plants in his monastery. He performed varieties of experiments (1856-1866) of self pollination and cross pollination with the pea plants.

Mendel’s research work was not recognised by the contemporary scientists for various reasons. Mendel died in 1884 without any recognition.

Gregor Johann Mendel

Self and Cross-pollination with pea plants: Mendel selected Garden pea (edible pea) plant (Pisum sativum; 2n = 14) for his experiments. Mendel performed his experiment at a time when there was no concept of chromosome, DNA, gene, meiosis and so on.

For Self-pollination, Mendel covered the bisexual flowers of pea plant by paper bags so that no foreign pollen grain could contaminate the flowers during experiment.

For cross-pollination, Mendel followed three steps :

  1. Emasculation—From the selected bisexual flowers, the stamens were cut off by a scissor as if the masculine part of the flower was removed.
  2. Dusting—With the help of brush, the pollen grains of one flower were transferred to the stigma of another flower and vice versa.
  3. Bagging – After cross pollination oil the flowers were covered with paper bags to avoid any Contamination of undesired pollen grains.

Finally all fir fertilization, the ovary of the flower was modified into fruit (pea pod) and ovules wore modified Into seeds. When the seeds were matured, they were collected and put Into the soil, then the seeds germinated into new saplings (F1 generation).

Reasons behind mendel’s success:

Mendel performed his experiments when biologists were unaware of the chromosome, DNA, genes etc and also the process of cell division (mitosis or meiosis). Even the process of
union between gametes during sexual reproduction was not very clear.

Simply on the basis of his breeding (hybridization) experiments, Mendel came to several conclusions (known as Mendel’s law) which constituted the foundation of modern science of genetics.

The reasons behind Mendel’s success can be explained as. follows :

Mendel repeated his experiments many times to accumulate a good deal of data for statistical analysis.

He analysed the data by applying principles of Mathematics and Statistics confirmed his conclusions or laws.

Mendel conducted his experiment with extreme care and considered one or two pairs of contrasting character for monohybrid and dihybrid cross respectively.

He considered only distinctive contrasting pairs of characters for his hybridization experiment. So, there was no confusion on the results e.g. Tall plant was 6-7′ whereas dwarf plant is 1-1×1/2.

Hence there is no possibility of any overlapping result of tall and dwarf plant.

By repeated Inbreeding (self-pollination) over the generations, Mendel selected only pure breeding varieties of pea plant for his experiment.

Mendel took enough care to avoid contamination of foreign pollen grain by covering the flowers with paper bags.

Mendel did not select those characters that were present on same chromosome—thus he avoided the complexity of linkage, crossing over etc.

Mendel’s chosen contrasting characteristics in pea plant:

Mendel’s experiment and Mendel’s laws: (How did Mendel derive the laws of heredity from his experiments ?)

Mendel performed monohybrid and dihybrid crosses with seven pairs of contrasting characters and from these experiments, he derived laws of inheritance.

Seven pairs of contrasting traits in pea plant as studied by mandel

Monohybrid Cross :

Definition: The genetic experiment of cross breeding between two varieties of same species bearing one pair of contrasting traits is known as monohybrid cross.

Experiment and Analysis of results of Monohybrid cross in Pea Plant: Mendel performed several monohybrid experiments taking each of the seven pairs of contrasting
characters separately and obtained similar result.

However, his monohybrid cross on length of pea plant (Tall and Dwarf variety) is very famous.

Experiment :

Mendel first selected pure tall pea plant (6′-7′) and pure dwarf pea plant (1′-1×1/2′) by repeated inbreeding (self pollination) generation after generation. Thus he selected parental generation (P) of pure tall arid pure dwarf plants.

In P generation,Mendel cross pollinated pure tall plants with pure dwarf plants by the process of emasculation, dusting and bagging. In first filial generation (F1) all the plants appeared tall and there was no dwarf plant at all.

All the plants were self pollinated by bagging. In second filial generation (F2); it was observed that out of total number of 1064 pea plants, 787 were tall plants and 277 were dwarf plants.

So, almost 75% (3/4th) of the plants were tall and 25% (1/4th) were dwarf. Therefore, it comes approximately to a ratio of 3 (Tall): 1 (dwarf) which is called as phenotypic ratio.

Phenotypic ratio of F2 generation in a monohybrid cross is also known as monohybrid ratio.

Diagrammatic representation of Mendel's monohybrid cross.

Analysis of results of Mendel’s monohybrid

According to Mendel, tall and dwarf characters in pea plants are transmitted from one generation to the next by the hereditary units known as factors. (Now-a-days, Mendelian
‘factors’ are known as genes).

He postulated that at least two ‘factors’ are responsible for a particular character. (Presently we know that in higher organisms, chromosomes are present in homologous pair— one paternal and one maternal. So, genes are also present in pair.

Hence, at least two genes (one paternal gene and one maternal gene) are responsible for a particular character).

 

Analysis of results of Mendel's monohybrid

He considered T for tall trait and ‘t’ for dwarf trait. So, in tall plants, the factors will be (TT) and that of dwarf plant (tt).

Mendel also assumed that the ‘factors’ are segregated (separated) during gamete formation (since he had no idea about meiosis). (Now, it. is very dear that the gametes are formed by meiosis which is reduction division where the chromosome number is reduced

to half—so number of genes will be aiso reduced to half]. Thus out of two ‘factors’ (genes),only one factor (gene) would be present in each gamete. He also calculated that during
cross pollination, along with transfer of gametes, there were transfer of ‘factors’ between tall and dwarf plants. So gametes are the bearer of factors.

Hence, from tall plant (TT), the gamete would bear T factor (gene) and that of dwarf plant (tt), the gamete would bear ‘t’ factor.

After cross pollination, T and t unite together to form zygote (Tt). So, the genotype of all F1 plants was (Tt) i.e. hybrid tall plant. In spite of presence of t, only T was expressed, so tall trait (T) is dominant over dwarf (t) and dwarf (t) is recessive to tall (T).

During gametogenesis, the F1 hybrid tall plants produce two types of gametes—(T) and (t) by segregation (meiotic separation). So, two types of male gametes from pollen grains and two types of eggs in ovule were formed.

During self pollination of F1 plants, any male gamete could unite with any female gamete at random that would give rise to F2 generation plants which could be shown by checker board (Punnett Square).

Mendel’s conclusion from the results of monohybrid cross :

Concept of two factors—At least two ‘factors’ are responsible for a particular character.

Existence of unit character—’Two- gametes (T) and (t) of two varieties of P generation plants formed the zygote (Tt) from where F1. generation hybrid plants developed. In the hybrid
plants, T and Y ‘factors’ were present side by side and the two ‘factors’ remained perfectly independent from each other without loosing their identities.

So, the factors maintain their purity generation after generation.

Dominance-Out of two contrasting factors—tall (T) and dwarf (t), one unit factor (T) has expressed called dominant and other factor (t) remained hidden called recessive.

Principle of segregation-Dwarf factor (t) was present in P, suppressed in F1 reexpressed in F2. The zygotes of F1 hybrid plant contains two unit character which lie side by side without
mixing.

In F2 generation, the plants produced both parental type (Tall and dwarf) as well as hybrid type. The phenomenon when the hybrid type separates into pure parental and
were hybrid types is called segregation,

Experiment and Analysis of results of Monohybrid cross in Animal (Guineapig):

Many geneticists performed monohybrid cross with animals and obtained resuit like that of Mendel.

Experiment:

When pure (homozygous) black haired guineapigs were crossed artificially with pure (homozygous) white haired guineapigs, all the F1 hybrid guineapigs appeared black haired.

The F2 guineapigs (male and female) were intercrossed.

In F2 generation, the ratio of the offsprings were 3 (Black) : 1 (White).

 

Experiment and Analysis of results of Monohybrid cross in Animal (Guineapig)

Analysis of the result :

In P generation, pure black haired guineapig was (BB) and that of white haired (bb).

During gametogenesis, pure black guineapig and pure white guineapig produced (B) and (b) gamete respectively.

In F1 generation, the genotype of hybrid black guineapig, was (Bb). In spite of presence of two contrasting genes of ‘B’ and ‘b’, the F1 animals appeared black which signified black (B) is dominant over white (b).

Diagrammatic representation of Monohybrid cross in guineapig

F1 hybrids produced two types of gametes— (B) and (b)

After intercrossing, the ratio of F2 progeny was 3 (Black) : 1 (White).

Hence, it shows that the inheritance of hair colour in animal like guineapig follows similar method of segregation like that of Mendel.

Analysis of the result

Concept of ‘Dominance’ from Mendel’s experiment :

Concept of 'Dominance' from Mendel's experiment

Phenotypic ratio = 3 (Black) : 1 (White)
Genotypic ratio = 1 (Pure tall) TT : 2 (Hybrid tall) Tt : 1 (Pure dwarf) tt
= 1 (TT) : 2 (Tt) : 1 (tt)

Concept of dominance :

In P generation, pure dwarf plant (1 -1×1/2) was present.

In F1 generation, all plants were hybrid tall (Tt) and there was no trace of any draft plant. .

In F2 generation, dwarf reappeared ( 3 tall : 1 dwarf).

The dwarf plants of P generation and F2 generation were exactly same.

In fact the dwarf gene (t) was suppressed in F1 generation by the Tall gene (T). So, T is dominant over ‘t’ gene. Hence T is expressed and ‘t’ is suppressed.

However, in hybrid tall plants (Tt) of F1 generation, t gene was suppressed by ‘V gene but was not destroyed. Moreover, the T and t genes were not mixed up or blended in F1 hybrids but they maintained their individual purity or identity.

That’s why, tt (pure dwarf) plant reappeared in F2 generation. This signifies that dominant gene can suppress the recessive gene but can not destroy the recessive gene.

The genes maintain their purity or potentiality generation after generation (unless it is ‘mutated’). It does not matter whether the gene is expressed or suppressed. A particular genemay be suppressed over the generations but its potentiality remains unchanged.

Dihybrid Cross :

Definition: The genetic experiment of cross breeding between two varieties of same species bearing two pairs of contrasting traits is known as dihybrid cross

diagrammatic representation of mendels dihybrid cross

Experiment and Analysis of results of Dihybrld cross in Pea plant : For dihybrid experiment, Mendel selected two characters-shape of seed and colour of seed (cotyledon).
contrasting traits of shape of seed were round and wrinkled whereas two contrasting traits of colour of seed were yellow and green.

Experiment :

Mendel selected pure round yellow seeded plants and pure wrinkled green seeded Plants by repeated Inbreeding (self pollination) for few generations. Thus he selected parenta
generation (P) of pure round yellow and pure wrinkled green plants.

In P generation, Mendel cross pollinated pure round yellow seeded plants with pure wrinkled green seeded plants by emasculation, dusting and bagging technique.

In F1 generation, all pea plants developed round yellow seeds.

All F1 plants were allowed self pollination.

In F2 generation, it was observed that out of 556 plants produced, 315 were Round yellow, 108 Round green, 101 Wrinkled yellow and 32 were Wrinkled green i.e. 9/16 Round
yellow,3/16 Round green, 3/16 Wrinkled yellow and 1/16 Wrinkled green.

So, the phenotypic ratio was 9:3:3: 1. Phenotypic ratio of F2 progeny in a dihybrid cross is also known as dihybrid ratio.

 

dihybrid ratio

Analysis Of results of Mendel’s Dihybrid Cross :

In P generation pure round yellow seeded plants were crossed with pure wrinkled green seeded plants. In F1 generation, all offsprings were round yellow which signifies seed is dominant trait over wrinkled seed and yellow colour of cotyledon is dominant green colour of cotyledon.

If we consider, R as round gene (dominant) and r as wrinkled gene (recessive) Y as yellow gene (dominant) and y as green gene (recessive), then in P generation, the genotype
of pure round yellow plant will be RRYY and the genotype of green wrinkled plant, will be rryy.

From pure round yellow (RRYY) plant, the gamete was (RY) and that of pure wrinkled green (rryy) plant, the gamete was (ry).

After fertilization of (RY) and )ry) gametes, the hybrid rouna yellow plants were formed with the genotype RrYy in F1 generation.

The F1 hybrid round yellow plants (RrYy) produced four types of gametes—RY, Ry, rY, ry.

During self pollinadon of F1 plnts,these dour types of male gemetesmunited with form types of female gametes at random. The possible combination of those gemetes to form
zygotes is shown in the checker board or punnett square.

 

Analysis of results of Mendel's Dihybrid Cross

Round yellow = 9, Round green = 3, Wrinkled yellow = 3 and Wrinkled green = 1 Therefore the phenotypic ratio = 9 : 3 : 3 : 1

Mendel’s conclusion from the results of dihybrid

Mendel concluded that the ‘factors’ are separated as well as assorted randomly and independently of each other during gamete formation.

He also concluded that during gametic union (fertilization) any gamete can unite with any other gamete irrespective to its dominance or recessiveness. Since the gametes are
the “bearer of factors” so random union of gametes means random assortment (union) of ‘factors’ or genes.

Thus during gametic union any gene/factor (whether dominant or recessive) can combine independently with any other gene/factor.

Experiment and Analysis of results of Dihybrid cross in Animal (Guineapig).

Experiment :

For dihybrid cross in guineapig, two pairs of contrasting characters were were— hair colour (black or white) and texture of hair (rough or smooth).

In P generation, pure black colour and rough hair variety of guineapig was cross pure white colour and smooth hair variety of guineapig.

In F1 generation, all offsprings appeared with black rough hair.

F1 male and female guineapigs were intercrossed.

Diagrammatic representation of dihybrid cross with Guineapig

In F2 generation, the offspring were in the ratio of 9 (black rough) : 3 (block smooth):3(white rough): 1 (white smooth).

 

 

the offspring ratio

Analysis of results of Dihybrid Cross In Guineapig :

If we consider gene for black hair ‘B’, white hair ‘b’ rough hair ‘R’ and smooth hair r, then in P1 generation, the genotype of pure black rough hair guineapig will be BBRR and that of white smooth hair will be bbrr.

P generation pure Black rough hair guineapig (BBRR) produces’gamete (Bf)) whereas pure white smooth hair guineapig (bbrr) produces gamete (br). These gametes fertilize to give birth to F1 hybrids!

In F1 generation, the hybrid black rough hair guineapig had the genotype BbRr. In spite of presence of white and smooth gene, all offsprings appeared black rough hair which signified black (B) is dominant over white (b) and rough (R) is dominant over smooth (r).

Analysis of results of Dihybrid Cross In Guineapig

The F1 hybrids (BbRr) produced gametes when Bb and Rr alleles were segregated and assorted independently to produce four types of gametes—BR, Br, bR and br. Thus four types
of male gametes and same four types of female gametes are produced.

These gametes fertilise at random to produce 16 different types of zygotes.

The phenotypic ratio of these zygotes were 9 : 3 : 3 : 1 which can be represented in checker board as follows:

Phenotypic ratio : 9 (Black Rough) : 3 (Black Smooth) : 3 (White Rough) : 1 (White Smooth).

Mendel’s Law :

From monohybrid cross, Mendel formulated his first law known as Law of Segregation.

1st Law of Mendel: Law of Segregation—Factors for contrasting character in a zygote do not blend or contaminate with each other but segregate and pass into different gametes and offsprings randomly.

From dihybrid cross, Mendel formulated his second law known as Law of Independent Assortment.

2nd Law of Mendel: Law of Independent Assortment— Two or more contrasting pairs of factors (characters) are assorted independently and may recombine randomly in all possible combinations governed by chance alone.

Deviation of Mendel’s Laws of Heredity :

In Mendel’s monohybrid cross, tall (T) trait was completely dominant over dwarf (t) trait in pea plant. That’s why, in F1 hybrid tail plants (Tt), T was fully expressed and ‘t’ was totally
suppressed. This is known as complete dominance.

After Mendel, long time passed away. Many scientist made their research work with various plants and animals. Lot of new genetic phenomenon have been discovered. One such
interesting discovery is incomplete dominance.

Definition: In a heterozygous organism, out of contrasting dominant and recessive allele, if the dominant gene cannot express its dominant phenotype completely but an intermediate
or mixed phenotype between dominant and recessive gene is expressed, the phenomenon is known as incomplete dominance or partial dominance or semidominance.

Experiment:

Correns reported incomplete dominance in case of flowers of Four O’ clock plant (Mirabilis jalapa).

In P generation, a pure red flowered plant was cross pollinated with pure white flowered plant.

In F1 generation, all hybrid plants produced pink flowers.

Diagrammatic representation of imcomplete dominance in 4'0 clock plant

All F1 plants were self pollinated.

In F1 generation, 25% plants produced red flowers, 50% pink flowers and 25% white flowers. Phenotypic ratio = 1 (Red) : 2 (Pink): 1 (White).

Analysis of result :

In P generation, Pure red flowered plant is (RR) and pure white flowered plant was (rr).

During gametogenesis by meiosis, the plant produced (R) and (r) gametes respectively.

After cross pollination of P plants, (R) and (r) unite together to form (Rr) zygote that developed into F1 plants producing pinkflowers.

When the F2 hybrid pink flowered plants were self pollinated, it produced F2 plants in the ratio of 1 (red) : 2 ( pink) : 1 (white).

The genotypic ratio was 1 (RR): 2 (Rr): 1 (rr)

Incomplete dominance (Flower colour of Mirabilis jalapa)

Phenotypic end genotypic ratio are samo In F2 generation,

In F1( hybrids, rod gene (R) did not destroy white gene (r), That’s why, they reappeared In F2 Ronomtion as pure red (RR) and pure white (rr),

In fact, In this example of Mlrabllls Jalapa, red gone Is Incompletely dominant over white gene. So, both red and white genes give rise to an Intermediate pink coloured flowers
in hybrid condition.

Phenotypic end genotypic ratio are samo

Sex determination in man :

Definition: The mechanism by which the male and female individuals of a species are

differentiated is known as sex determination. In man, mechanism of sex determination is known as XX-XY mechanism.

In human {Homo sapiens), the chromosome number is 2n = 46. There are 22 pairs autosome and 1 pair sex chromosome.

Sex chromosomes determine sex. In human female, the sex chromosomes (allosome) are homologous (XX) but in human male, the sex chromosomes are nonhomologous (XY).

Sex determination in man

Female (44A + XX) Is homogametic and produces only one type of ovum by meiosis (during oogenesis) (22A + X). (ÿ stands for autosome and ’X for allosome).

Male (44A + XY) is heterogametic and produces two different types of sperm by meiosis (during spermatogenesis) – (22A + X) and (22A + Y). (‘A’ for autosome and X, Y designate
allosome). X containing sperm is known as gynosperm and Y containing sperm is known as androsperm.

During fertilization between sperm and ovum, any type of sperm may fertilize the ovum. Whether gynosperm (22A + X) or androsperm (22A + Y) would fertilize the ovum-is purely
a matter of chance. Nothing can be predicted earlier.

The principle of sex determination in man is that presence of Y in the zygote indicates maleness whereas absence of Y in zygote indicates femaleness.

If gynosperm (22A + X) fertilizes the ovum (22A + X), the zygote would be (44A + XX) which gives birth to a female baby. On the contrary, if androsperm (22A + Y) fertilizes the
ovum (22A + X), the zygote would be (44A + XY) which gives birth to a male baby.

So, whether a male baby (son) or a female baby (daughter) will be born (Sex determination of the baby), depends completely on father and never on the mother (in no way).

Sex determination of the baby

Gender determination by checker board in man

Some Common Genetic Diseases (What are the causes of genetic diseases?)

Common genetic diseases in population:

Many genetic diseases in man have been discovered. These are mainly because of various mutant genes that are located either in the autosome (autosomal gene) or in the allosome
/sex chromosome (sex linked gene).

If the sex-linked gene is present in the X-chrornosome or Y-chromosome, it may be called as X-linked gene or Y-Iinked gene respectively

Thalassemia : it is a type of haemolytic anaemia caused by abnormal synthesis of faulty haemoglobin. It was discovered by American doctor Cooley (1925).

Cause :

  1. Thalassemia is a genetic disorder caused by autosomal recessive gene.
  2. The defects occur due to mutation of genes located in the chromosome 16 and chromosome 11.
  3. The mutations cause defective synthesis of a globin chain or (3 globin chain of haemoglobin.
  4. Thalassemia is generally of three types—Alpha, Beta and Delta. (Beta thalassemia is called cooley’s anaemia)

Symptoms :

  1. Synthesis of less haemoglobin or formation of abnormal haemoglobin.
  2. Premature destruction of RBC and development of anaemia.
  3. Expansion of bone narrow and heart problem.
  4. Enlargement of liver—hepatomegaly.
  5. Enlargement of spleen—Splenomegaly and increased risk of infection.
  6. Skeletal deformation and iron overload.
  7. Retardation of body growth.
  8. Affected fetus may die inside mother’s uterus or the baby may die soon after birth (jaundice and erythroblastosis).

Haemophilia: It is a clinical condition in which the ability of blood clotting of a person is severely reduced, causing the sufferer to bleed severely even from a slight injury. It is also known as bleeder’s disease. This was first discovered by an American doctor John Otto (1803).

Cause :

Haemophilia is caused due to presence of a recessive sex-linked gene h, located on X-chromosome.

There are two types of haemophilia—Haemophilia A and Haemophilia B, caused by two mutant genes present on the X-chromosome in man with a distinct gap.

Haemophilia A (Royal disease) is caused due to deficiency of a protein for blood clotting known as “Factor 8” or “Antihaemophilic factor” (AHF). About 80% Haemophilia are of this
type.

Haemophilia B (Christmas disease) is caused due to deficiency of another blood clotting protein known as “Factor 9” or Plasma thromboplastin. About 20% Haemophilia are of this
type. It is called Christmas disease according to the name of first patient—Stephen Christmas.

The disease (Haemophilia A) was first discovered in Queen Victoria. Probably the gene was inherited to her from parents or the mutation was developed in herself.

Symptoms :

For normal blood clotting, both factor 8 and 9 are needed. Defect or deficiency of anyone factor will impair blood clotting. So blood does not clot within 3-8 minutes but it takes hours.
As a result, there will be continuous bleeding and the patient may die.

There is no permanent cure of this genetic disease but it may be partly managed by blood transfusion.

Females may be ‘carrier’ with one dominant normal gene and one recessive ‘h’ gene of haemophilia.

Haemophilia is found more in males than in females.

Colour blindness: It is the defective colour vision when someone can not distinguish between certain colours, usually between green and red, and occasionally blue due to absence
or reduced amount of visual pigments.

First scientific paper about colour blindness was wtitten by John Dalton in 1793. Dalton himself was red-green colour blind. That’s why red-green colour blindness is sometimes called as Daltonism.

Cause :

The gene for normal vision is dominant. Red-green colour blindness is caused by recessive gene located on X-chromosome (sex-linked gene) in man.

The mutant gene for red blindness is protan and the mutant gene for green blindness is deutan. They are located at different locus of X-chromosome

Symptoms :

  1. Colour blindness is inheritable disease of X-linked recessive gene.
  2. The cone cells of eye in these persons fail to distinguish red and.green colour
  3. Besides the problem of colour vision, these colourblind persons will have otherwise normal vision for reading, writing etc,
  4. Colour blindness Is more common In males than In females because males have one X and other Y sex chromosome whereas In human female there are two X-chromosomes.

Thalassemia as an autosomal chromosomal disorder and genetic counselling:

Thalassemia is a type of genetic disorder where haemoglobin Is produced In decreased amounts. The decreased amount of haemoglobin In the blood causes anaemia, which reduces
oxygen carrying capacity of blood. RBC becomes fragile and breaks down easily (haemolysis).

To compensate loss of RBC, frequent blood transfusion is done.

All the above conditions may result into Iron overload, which means excess Iron in the body. Excess iron in vital organs Increases the risk for liver disease (cirrhosis, cancer), heart
attack, diabetes mellltus, osteoarthritis, osteoporosis as Well as problem of endocrine system like hypothyroidism, defective gonads etc.

Iron overload may be inherited or it may be acquired by numerous blood transfusion, iron injection etc. When a person is receiving blood transfusion on a regular basis, iron may build
up to toxic level in the body.

Role of genetic counselling In prevention of thalassemia :

Those persons who contain gene for thalassemia in recessive condition are called carriers or minors. They have normal dominant gene. So, they will not suffer from any problem
of thalassemia.

If both father and mother are carrier, then in F1 generation, probability of thalassemia may be 25%.

Role of genetic counselling In prevention of thalassemia

Before marriage of any two man and woman, family history must be studied for genetic counselling to find out any thalassemia carrier in the proposed male and female or their
relatives, parents etc. Two thalassemia carrier should not get married.

There are method of blood test for detection “haemoglobin disorders” which can be done before marriage to find thalassemia carrier (thalassemia minor). These premarital precautions
will help to prevent the Inheritance of thalassemia major (homozygous).

Thus birth of thalassemia babies can be checked that would justify—’prevention Is better than cure’.

Class 10 Science Notes For Evolution And Adaptation

Evolution (How do life form change over time ?)

Concept of Evolution :

The word ‘evolution’ means change of life forms over of wide variety of plants and animals on earth along with the. remains of dead or extinct organisms that lived in remote past, arises questions—how and from where all these organisms came into existence.

Evolution is a process of gradual unfolding of the new organisms from the preexisting primitive organisms through slow and steady changes. Thus enormous variety of plants and animals have come to exist on earth as a result of constant evolutionary process.

Evolution is the centralmost important idea in.Biology that complex living organisms evolved on earth from relatively simple ones. Hence, evolution is the cause of biodiversity on earth.
In fact, biodiversity (diversity of living organisms on earth) is due to genetic mutation.

That’s why, it is said, “Mutation is the raw material of evolution”.

Definition: Evolution  is a slow but gradual progressive process by which a simple form of organism gradually becomes complex in course of time by the process of reproduction, variation and heredity resulting into origin of new species.

Origin of life

Evolution (latin,evolvere=unfold) means ‘unfolding of life’ like the opening of a floral bud into a flower. probable steps of chemical origin of life (theory of abiogenesis)
are as follows:

Scientists believe that the earth at its time of formation was like a ‘fireball’. There were innumberable atoms of different elements like hydrogen, oxygen, carbon, nitrogen, sulphur, phosphorus etc.

Read and Learn More Class 10 Science

Free atoms combined to form H2, N2, H2O (water vapour), CH4 (methane), NH3, CO2 etc. H2 combined with O2 to from H2O and there was no free O2. H2 was burning and O2 helped in burning.

Diagram showing budding in microsphere

Thus primitive earth was without free O2 (reducing atmosphere). [However, present day earth is with free O2 (oxidising atmosphere)]. H2 also combined with N2 to form IMH3. Probably H2O and NH3 were first compounds on primitive earth.

Early inorganic molecules combined together to form simple organic molecules like glucose, amino acids, glycerol, fatty acids etc. These organic molecules assembled together to form microsphere that could grow by budding.

As earth started slowly cooling down, water vapour condensed to form cloud from where rainfall started and thus initiated water cycle. Rainwater was flowing from higher to lower level—thus formed fountain, hilly river, big river, sea and finally ocean.

During this downward flow of water, various salts, minerals and other chemical compounds get mixed with water and carried to oceanic water.

During rainfall, there was definitely thundershower. Thunder or lightning caused high voltage electric current that could be one of the major sources of external energy for various chemical reactions.

In this way probably a variety of complex organic molecules like protein, fatty acids, nucleotides etc. were formed and accumulated in sea or oceanic water.

Coacervate growth and division

Theory of ahiogenesis or chemical origin of life (Oparin and Haldane, 1923)): Definition: The formation of complex organic molecules from simpler inorganic molecules, through chemical reactions in the oceanic water during early history of earth is known as chemical evolution.

Origin of life on earth (Oparin-Haldane Theory)

Experiment of Miller and Urey :

Stanley Miller and Urey (1953) provided an experimental proof of the theory of abiogenesis. (Miller was student of Urey). They created a laboratory condition similar to probable early earth.

Experimental procedure :

Water in a closed small flask was taken (that resembled primitive sea). Water was heated to produce water vapour that moved into a second flask placed at a higher level.

Second big flask contained a mixture of methane (CH4); ammonia (NH3) and hydrogen (H2) gases in the ratio of 2:2:1. This resembled the primitive atmosphere.

Experimental apparatus of miller and urey

The gas mixture in the upper flask was subjected to high voltage electric discharge/ current (60,000 volts) at 800°C temperature from tungsten filament. This resembled
vigorous lightning and violent electrical storms in early earth.

The mixture was then condensed in raining water and dissolved molecules were collected at the bottom of the condenser. Miller continued the experimental set up and procedure for one week.

Samples formed in the condenser were collected and analysed.

Observation: Miller identified a variety of organic molecules like some amino acids (alanine, glycine etc.), long chain carbohydrates, formaldehyde etc. in the formed samples.

Inference: F.rom the above observation, Miller concluded that organic molecules might have been synthesized abiotically in early earth probably by the energy of high
voltage electric current of thundershower and other external sources of energy like solar radiation etc. (There was no ozone layer in pre-historic earth).

Therefore, the “theory of abiogenesis of origin of life” has been proved experimentally.

Major evolutionary events: Evolution has involved gradual transformation from simple, unicellular, underdeveloped form to complex multicellular, well differentiated forms of plants
and animals.

Major evolutionary events can be explained with a schematic diagram as follows :

Major evolutionary events can be explained with a schematic diagram as follows

Major evolutionary events

Theories of Evolution :

A number of theories have been put forward in order to explain evolutionary process taking place in plants and animals. However, Linnaeus proposed a scientific explanation of this evolutionary process, stating that, each and every species originates or evolves from some preexisting forms.

The theories of Evolution were put forward by Jean Baptiste de Monet Lamarck and Charles Robert Darwin. These theories are known as Lamarckism and Darwinism respectively.

Lamarckism

Jean Baptiste de Monet Lamarck was born at Bajastin in France in 1744. His theory on evolution was published in his book “Philosophic Zoologique” in 1809. His theory is known as ‘Theory of Inheritance of Acquired Characters’, and popularly known as Lamarckism. Lamarck proposed two ideas in order to explain his theory.

Jean Baptiste Lamarck

The ideas are : (i) Law of use and disuse of parts and (ii)Inheritance of acquired characters.

Lamarck’s theory :

Continuous increment in size—All living organisms increase in size due to growth.

Conscious effort and Environmental effect—All organisms are influenced by environment. So change in environment brings about changes in organisms that develop new
demand to produce new structures in the organisms.

Thus the organism can adapt in the changing environment.

Use and disuse of organs—Lamarck proposed that by constant use; a particular organ in organisms gradually develops more and more whereas by constant disuse, the organ
becomes smaller in size and ultimately degenerates.

Inheritance of acquired characters—Due to continuous increment in size, environmental effect together with use and disuse of organs, new characters may develop in an organism
known as acquired characters.

These characters are inherited to successive generations. Gradually these acquired characters are accumulated over the generations to give rise to new species. . .

Examples in favour of Lamarckism: Lamarck explained his theory citing following examples :

Effect of use::

Long neck of giraffe—The ancestors of present day giraffe had very short neck and forelimb like horse. They used to live in the places having poor surface vegetation. So they tried to stretch their neck and forcllmb in an effort to reach to the foliages of tall trees.

Lamarck's Theory of Evolution

This resulted into little longer neck that was inherited to next generation. Over the generations by similar effort the neck and forelimb gradually stretched more and more thus present day giraffe with long neck and forelimb has been evolved.

Webbed feet in ducks –Oucks swim in water. All the fingers are covered by a flap of skin called webbed feet that help in swimming. This proves that because of constant use
of fingers for swimming, webbed feet was evolved.

Effect of disuse :

Loss of limbs in snake-The present day snakes are without any forelimb or hindiimb. But the fossil records proved that snakes in the prehistoric time had limbs. (Even present
day snakes like Boa, Python have vestigial pectoral girdle and pelvic girdle though they limbless).

Snakes used to crawl on the abdominal muscle (since they had very long body). So their limbs became useless (constant disuse), gradually degenerated, became vestigial and ultimately disappeared.

Criticism against Lamarckism :

Germplasm theory (Weismann)—August Weismann, a German biologist strongly criticised Lamarck’s theory. He experimentally cut off the tails of rats for nearly 22 generations
and allowed the tailless rats to breed, but no tailless offspring was born.

If tailless condition is an acquired character, it was not inherited to any generation. (So acquired character does not always inherit.)

Weismann proposed that in the body of organisms, there are two types of protoplasm germplasm (protoplasm of germ cell) and somatoplasm (protoplasm of somatic cell). Only
germplasm is inherited from one generation to next but not the somatoplasm.

Neo-Lamarckism :

According to modern concept of genetics, Lamarck’s theory has been newly modified called Neo-Lamarckism, that can be stated as follows :

  1. Genes are responsible for ail characteristics of an organism.
  2. If by constant use and disuse, a genetic mutation develops, then only an acquired character may rise newly.
  3. For inheritance of acquired character, the mutation must be present in germ cell (germplasm) because only germ cells are inherited from one generation to next.

Darwinism

Charles Robert Darwin (1809-1882) was born in England on 12th February, 1809. He published his theory of evolution in his famous book, “On the origin of species by means of natural selection”.

The theory of evolution is known as The Theory of Natural Selection! It is also called Darwinism.

Charles Robert Darwin

Darwin made arrangements for his assignment scientist in a voyage for exploration in the famous historical H, M. S. Beagle. The ship went on voyage on 27th December, 1831. It visited many islands of the Atlantic Ocean and South Pacific and some coasts of South America.

The Beagle returned safely on 2nd October, 1836 after 5 years of extensive survey. Darwin took extensive notes and collected lot of specimens of the Galapagos island and other places.

The name of the island was Galapagos (Spanish word for tortoise) as there were plenty of giant tortoise. Darwin observed Giant tortoise, Iguanas (Reptiles), Insects, Lizards, and thousands of birds (called Finches).

He stated his observations in his book entitled, “On the origin of species by means of Natural Selection” in 1859..

Darwin’s theory of Natural Selection: The theory is a mixture of some observations and inferences drawn from it.

Darwin's theory of Natural Selection

Principles of Darwin’s Theory of Natural Selection are discussed as follows :

Rapid multiplication and over production—Every living organisms multiply by the process of reproduction. Thus they increase in number in geometric progression.

Prodigality of production—Even though all species produce a large number of offsprings, population remains more or less constant. This means more youngs are produced than to survive.

For example, a female Ascaris (round worm) produces about 7,00,000 eggs in 24 hours; a salmon fish produces about 28,00,000 eggs in one breeding season; an oyster produces 114,000,000 eggs and so on.

All these eggs will not hatch or grow upto adult. So many eggs are eaten up by other predators; some of eggs may be decomposed; some offsprings may be eaten up by other predators of food chain etc.

Darwin's idea of evolution of modern long necked giraffe

Constancy of food and space—Amount of food and space in a particular area remains more or less constant although the organisms increase in number.

Struggle for existence—A struggle between members of same species and of different species for food and Space is called struggle for existence. Usually there are three types of struggle found in nature, such as—

  1. Intra-specific struggle—It is the struggle for food and space between the organisms of their own kind,
  2. Inter-specific struggle—It is the struggle for food and space between the individual of different species,
  3. Environmental struggle—It is the struggle against physical factors of the environment, like excess of moisture, heat, cold, rainfall and against geological conditions.

Variation—No two organisms are exactly same (except the identical twins). So, there are differences among the organisms and these differences are called as variation. Darwin
proposed that these variations are continuous (gradual).

Some variations are advantageous and can adjust better with the environment than that of others and these variations (adaptive features) are known as favourable variation that are inherited generation after generation.

Survival of the fittest—The individuals, which can environmental conditions are successful in struggle for existence. Struggle for existence eliminates the unfit individuals. The fit individuals possessing favourable variations survive and reproduce. So there will be survival of fittest and elimination of unfit.

Natural selection—The individuals possessing favourable variation enjoy a competitive advantage over the others. They are better adapted to their environment, survive more and
produce more offsprings.

The individuals with disadvantageous variation fail to adapt properly to their environment and therefore get eliminated by natural selection.

Inheritance of useful variation and origin ofmew species (Speciation)—The favourble variations of selected organisms gradually accumulate by reproduction generation after
generation and ultimately this may give rise to a new species known as “Origin of Species”.

Examples in favour of Darwinism :

According to Darwin’s theory of Evolution, the ancestors of giraffe showed necks and forelimbs of different lengths. As giraffes were forced to reach leaves on tall trees, the giraffes with longer neck and forelimbs had advantage over others.

Thus these giraffes had better chances of survival (fit) and were selected by nature. When they reproduced, the offsprings possessed the same advantageous variation. This explained how present day giraffes with long neck and forelimbs came into existence.

Giraffes with short neck and forelimbs (unfit) starve and die. This example justifies the idea of survival of fit and elimination of unfit.

Criticism against Darwinism :

Darwin had no idea about chromosome, DNA, gene, mutation. So, he could not explain the cause of variation, mechanism of inheritance of variation. On the contrary, he proposed
theory of pangenesis that was not accepted by the scientists.

Neo-Darwinism: In the light of modern genetics, Darwin’s theory has been newly interpreted known as Neo-Darwinism (also called as synthetic theory of evolution), which may be explained briefly as follows :

  1. Genes are responsible for all characteristics of an organism.
  2. Due to genetic mutation, new characteristic (variation) may develop which is transmitted through germ cell (reproduction) from one generation to next.
  3. Accumulation of mutation by reproduction over the generations may give rise to origin of new species.

Comparison between Darwinism and Lamarckism :

Comparison between Darwinism and Lamarckism

Evidences for the theory of Evolution: (What are the evidences of evolution ?)

Palaeontological (Gr. Palaeos = ancient) evidences of evolution :

Palaeontology is the branch of science (Geology) that deals with the fossils. Thus palaeontology links geology with biology.

Definition of Fossil: The fossils are the remnants of any hard part of the body of prehistoric organisms or its impression on any layer of earth strata, that are preserved by nature.

Fossil of Dinosaurus

Importance of fossils :

  1. Fossils are important means for identifying rock’s strata.
  2. Fossil gives us evidence of steps of evolution,
  3. Age of fossils reflects the age of evolution of a particular organism,
  4. Fossil reflects the idea of structure and function of prehistoric organisms,
  5. Fossil also provides some idea about possible cause of extinction of the prehistoric organisms.

Fossil history of horse :

Many complete fossils of horse have been discovered from different parts of the earth, Modern horse have reached the highest grade of cursorial adaptation. Whole of structural
organisation of horses is primarily due to food-getting mechanism and to attain speed. It took nearly 60 million years to complete the phylogeny of horse. The first fossil horse was
discovered in USA.

Continuous change of character during evolution is referred to as evolutionary trend Major evolutionary trends of horses were-

  1. Increase in size.
  2. Lengthening of limbs and feet.
  3. Reduction of lateral digits.
  4. Increase in length and thickness of the third digit.

Evolutionary history of horse may be briefly described as follows :

Stage 1—Eohippus: First fossil of horse is named tohippus (‘dawn horse’) that was discovered from North America. It was evolved nearly 60 million years ago.The size of the
animal was like a fox. It was 11″ high at the shoulders, with short head and neck.

The forelimb was with four complete fingers (2, 3, 4 and 5) and one it of finger 1 whereas hindlimb was with three functional fingers (toes)-2, 3 and 4 with two splints of finger 1 and 5. (Splints are vestigial side fingers of horse).

Stage 2— Meso’nippus : Mesohippus evolved from Eohippus. This is intermediate horse, evolved nearly 40 million years ago. The size of the animal was like that of present day sheep, about 24″ high at the shoulders.

Forelimb with three fingers (2, 3 and 4) and 1 splint of finger 5 whereas hindlimb had three fingers (2, 3 and 4) where finger 3 was longest and supported most of body weight.

Stage 3— Merychippus: This was evolved from Mesohippus, nearly 25 million years ago, about 40″ high at shoulders, with longer neck. Both forelimb and hindlimb had three fingers where middle finger (3rd) was longest and supported entire body weight.

Evolution of horse.

Stage 4—Pliohippus: Pliohippus evolved from Merychippus about 10 million years ago. It was nearly 50″ high at the shoulders. Both forelimb and hindlimb had strong complete finger 3 with splints of 2 and 4. Thus Pliohippus might be considered as first onetoed horse.

Stage 5— Equus: This is modern horse that was evolved from Pliohippus nearly 1 million years ago in North America and later spread throughout the world (except Australia). It is
about 60″ high at the shoulders with a long head and neck. Each forelimb and hindlimb has one very strong complete finger (finger 3) with two splints.

Therefore the line of progression of evolution of horse is

Eohippus —> Mesohippus —> Merychippus —> Pliohippus —> Equus (Modern horse)

Evidences from comparative anatomy and morphology :

Homologous Structures: Definition: The organs that have similar origin and structure but differ in shape, size and functions are called homologous organ.

Examples :

Forelimb of man, whale, bat, bird. The forelimb of all these animals have same basic structure but different shape and functions. In man, forelimb is used for grasping, in whale for
swimming (paddler/flipper), in bat for flight (patagium), in bird for flight (wings).

Homologous organs in plants

Tendril of pea plant and spine in cactus. Both of them are modified leaves but tendril helps to provide support whereas spine helps in protection, reduces transpiration.

Analogous organs/structures : Definition:The organs having different origin and structure but perform similar function are called analcgous organ.”

Examples :

Wings of birds and wings of insects. The have different structure but perform same function of flight.

Analogous organs

Phylloclade in cactus (modified stem) and stipules in pea plant; Tendril in pea plant (modified leaf) and tendril in passiflora (modified axillary bud of stem)

Differences between Homologous and Analogous organs :

Differences between Homologous and Analogous

Convergent evolution: Definition:lt is the process where different groups of organisms independently evolve analogous structures having similar functions to adapt in similar
environment Analogous structures are the result of convergent evolution.

Example: In aquatic environment, so many animals are living, such as, Invertebrates (e.g. prawn, lobster), so many Fishes, Amphibia (e.g. toad and frog), Reptilia (e.g. crocodile, turtle), birds (e.g. ducks), Mammals (e.g. whale, dolphin) etc.

Aquatic environment

Any aquatic animal must have the common function of swimming, steering and balancing. All these animals have developed different structures to serve those common functions.

Divergent evolution: Definition:lt is the process where similar groups of organisms develop different functional structures to adapt in different environment. Homologous
structures are the result of divergent evolution.

Example: The forelimb of man, tiger, whale and bat have same basic structural plan—i.e. same type of bones, muscles etc. But the forelimbs of these animals have different shapes
and functions.

In man, hands are used for various functions as well as grasping; in tiger for running as well as catching the prey; in whale the forelimb is modified into flipper or paddler for swimming; in bat it is modified into patagium (wing) for flight.

Fore limb of mammal

Differences between Convergent and Divergent evolution :

Differences between Convergent and Divergent evolution

Vestigial organs: Definition: The organ which is useless and functionless but present in the body in much reduced form is known as vestigeal organ. (The organ is believed to be
fully formed and functional in the ancestor).

How does vestigial organ justify evolution?

These rudimentary functionless organs found both in animals and plants are evidences of organic evolution. Biologists believe that these vestigial structures can be explained only on the basis that, previously (in the ancestors) these structures were present in their full-form, but now they have turned completely functionless.

Vestigial organs in Human being

The vestigial organs that have lost their adaptive functions still continue to persist in a reduced condition. It seems as if ition is a filterin and by its automatic operation useful structure will develop more and more whereas useless structure will be gradually atrophied and ultimately disappear in the long run.

Examples of vestigial organs :

Some examples in animals—Weidersheim has listed nearly 100 such characters in human, few are mentioned here.

Vermiform appendix—The vermiform appendix of the caecum of man is perhaps the best example of vestigial organ. It serves no useful purpose in man. It often becomes infected and inflammated and has to be removed by surgical operation.

In other primates, however, this structure is much longer. In the rabbit, for example, the appendix of the caecum is a functional part of the digestive system. The appendix of man, hence, is understandable as a degenerating inherited structure from his ancestor.

Caecum and appendix In man.

Nictitating membrane- It is the rudimentary structure present in the inner corner of the eye. It Is another example of a vestigial structure in man. This is the remnant of the fully functional nictitating membrane of other vertebrates. These structures, however, has no function in man.

Other vestigial organs- In human being, there are a nurmber of vestigial organs such as —a vestigial tail or coccyx at The end of vertebral column, mammary gland in male/segmental abdominal muscle, third molar teeth, muscles at the back side of the pinna, etc.

Vestigial wing in Ostrich

It has been observed that cow, dog and horse for example, are able to move their pinna for the more efficient detection of sound. In man the muscles necessary to move the pinna are rudimentary and functionless.

Some examples of vestigial organ in plants :

Cutin-covered stomata present on the stems of the Cacti.

Sterile stamen (staminode) (only filament without anther) of Mango,Canna, Cashew nut flowers, etc.

Sterile carpel in Coconut flower.

Functionless synergids and* antipodals ‘n the embryo Sac of the mature megagametophyte of the angiosperms.

Vestigial organ in Plants

Useless flagella on the-cycad sperms (that are passively transported to the egg), etc.

Structure of vertebrate heart: Heart shows interesting homologies,

In fish— Two chambered heart Of fish (venous heart) consists of one auricle (atrium) and one vertricle only. It is transformed into four chambered heart in birds and mammals through the intermediate forms like amphibians and reptiles, the two chambered heart of fishes is not only simple, but also efficiently adapted to the aquatic environment.

Comparative study of heart of vertebrates (From Fish to Ma.mmal)

ln amphibia—With the change of habitat the amphibian heart became three chambered to prevent the admixture of the oxygenated (arterial) and deoxygenated (veneous) blood. But even then, it was, not sufficient to serve the purpose.

In reptiles—The heart of reptiles is basically three chambered, but the single ventricle is partially partitioned by a vertical septum. Thus it is also called 3} chambered heart. (But in crocodiles the ventricular partition is complete but there is an opening in the heart called Foramen of Panizza through which oxygenated and deoxygenated blood get mixed.)

In birds and mammals—The ventricular partition is complete, thus the heart is completely four chambered. Due to this arrangement, the arterial and venous blood remain completely separated.

Above examples of evolution of heart in vertebrate series (Fish, Amphibia, Reptiles, Birds and Mammals) signified that simple form of heart in fish (2-chambered) gradually becomes
complex (4-chambered) in Birds and Mammals for greater advantage—such that poikilothermal or cold-bloodedanimals (Fish, Amphibia and Reptilia) become homeothermal or warm-blooded animals (Birds and Mammalia) due to complete separation of oxygenated and deoxygenated blood.

Evidences from /Comparative embryology :

Embryology deals with the embryonic developmental stages since zygote till birth of the young one. (newborn). These evidences are based on comparative study of the embryos of various animals.

Similarities of embryonic form of different vertebrates

If a comparative study is made with embryos of different vertebrates (such as fish, amphibia, reptilia, aves and mammalia), striking similarity is observed among the embryos as follows :

  1. Similar external branchial grooves (visceral clefts) in the pharyngeal region.
  2. Presence of a series of internal paired gill pouches.
  3. Presence of segmental myotomes (muscle blocks) in the tail-like structure.

Survival strategies: Adaptation

Introduction: A wide variety of living organisms are present in this planet Earth. They survive in a vast range of habitats. By constant interaction with environment, they survive
through adversities.

In order to gain maximum benefit from surrounding environment to ensure their survival, their long interactions with the environment bring about.certain morphological, anatomical, physiological, and behavioural changes.

The favourable changes which help an organism to thrive in their habitat are called adaptive features and the process bringing about their adjustment to their surroundings is called adaptation.

Cumulative adaptive features over the generation may lead to origin of new species called evolution.

The living organisms react directly to their environments in which they grow and live. As the environment is constantly changing, in order to survive in that changed condition, living
organisms may undergo certain structural modifications, otherwise there is a chance of their extinction.

These external and internal structural changes enable the living organisms to survive in their struggle for existence on earth.

Relation between adaptation and evolution :

The adaptive features enable the organisms to make the best of the conditions under which it lives. Organisms which are not able to adapt themselves to particular environment gets
eliminated, as they become weaker and weaker with successive generations, and gradually become extinct.

Organisms adapting themselves to new environment, may undergo genetic changes (mutation) which may result in production of some new varieties of organisms.

The variety of living organisms we see today is a consequence of evolution, where each variety is being modified gradually by adaptation for living in its own way. Thus adaptation is the cause and evolution is the effect/outcome.

Behaviour and adaptation : (Why is behaviour important in the process of evolution ? How does adaptation evolve in the population of an organism ?)

An adaptation can bestructural which means it is a physical part of the organism but an aptation can also bebehavioural which means the way an organism acts.

A behaviour is an action carried out by an organism under the control of the nervous system in response to an environmental stimulus (cue) or to the actions performed by an organism.

A cue (stimulus) may be an odour (eg. pheromone), sound (eg. call of Cuckoo) or visual signal behaviour is what an animal does. Behaviour allows animals to survive and reproduce and is, thus, extremely important (critical) to evolutionary process.

Behavioural adaptations include activities that help an animal to survive. Examples-Special behaviour (some animals live in groups eg. honey bee, ants etc. called social insect );Protec
behaviour (that helps to protect the animal, eg. opossum plays dead after watching beer), animal migration (for better climate, food, safe place to live and reproduce).

Behavioural adaptations can be inherited or learnt eg. swarming behaviour of honey bee. Adaptation to extreme climate (eg. cold like snowy region, dry like desert etc.) needs special
behaviour and physiology.

Behavioural pattern describes an animal’s dominant way of life eg. arborea’ An’ma* like monkey lives in trees, noctural animals like owl are active at night and so on.

Adaptive features develop because of genetic mutation. Some mutations help an organism to survive better than others (nonmutants). Adaptive character or adaptation evolve in a
population of organisms by the process of natural selection.

Definition: Proper structural, functional, physiological and behavioural modifications of an organism over generations in response to environmental change in order to survive and
reproduce is called adaptation.

Examples of adaptation :

(How does adaptation occur in organisms living in different environments ?)

Morphological adaptive features :

CACTUS (Conversion of leaf into spine or reduction in the number of leaf) :

In cactus, leaves are small, much less in number, scaly and often modified into sharp pointedsP’nes- This modification helps in checking loss of water by transpiration (since in desert, there is acute scarcity of water). Spines also help indefensive mechanism of the plant.

Spines and Phylloclade in cactus

In some xerophytes, leaves are thick succulent and the epidermis is covered by a waxy coating.

Swim Bladder Of Fish :

Swim bladder of bony fish is a thin walled sac, elongated in shape and filled with gases. It is located in the dorsal side of the body cavity below the vertebral column.

The Swim bladder (or air bladder) consists of two chambers, of which the anterior one is smaller than the posterior. The bladder is supplied with blood capillaries called retia mirbbilia
or red gland. The swim bladder is connected with the gut by a small duct called pneumatic duct.

Swim bladder and other adaptive features of Rohu fish

The bladder is called hydrostatic organ. When the fish absorbs gases into the sac, body becomes lighter in weight, buoyancy increases and the fish floats up. On the contrary, if

gases go out of the air sac, body becomes heavier in weight, buoyancy decreases and the fish sinks down. Contraction of body muscle causes increase or decrease of volume of air in the sac allowing the fish to swim freely at different desired depth of water.

Air Sac Of Bird :

Lungs of flying birds (e.g. Pigeon) are supplemented by thin wall sacs called air sacs. There-are generally nine major air sacs and four minor air sacs. In the air sac, there is no alveoli. Hence there is no gaseous exchange in the air sac but it can only store warm air.

This air makes the body light in weight that helps to increase buoyancy for flight in air ocean.

Air Sacs in Pigeon

Extra energy is required for volant adaptation that comes- from double respiration. The lungs in bird are comparatively smaller in size and contain alveoli where gaseous exchange takes place. During inspiration, air comes in contact with alveoli causing first gaseous exchange and then air enters into air sacs.

During expiration, air goes out of air sac and flows over the alveoli when air comes in contact with the alveoli for the second time causing second gaseous exchange. Thus, in one complete breathing cycle (inspiration and expiration), gaseous exchange takes place twice.

So the process is called double respiration. This provides increased functional efficiency of lungs, greater oxygen supply, higher rate of respiration and energy production—that are all useful for flight.

Physiological adaptive features :

Mechanism Of Salt Adaptation In Sundri :

Sundri (Heritiera sp.) is halophyte since it grows in saline soil. It is also known as ‘looking glass tree’ (as ventral surface is pale green in colour and dorsal surface is dark green). The plant grows in Sundarban delta (but now a threatened species).

High salt content interferes with cellular metabolism and high soil salinity makes it difficult to extract water from soil. The plant deals with this strong salinity of soil and water in several ways as follows :

Breathing root (Pneumatophore) in Sundrinormal view and magnified view.

Salt exclusion—They have significantly impermeable roots that are highly suberised which prevent entry of sodium salts. It has been shown that approximately 90% of salt has been excluded at the roots by special enzymatic mechanism.

Salt secretion- Most of the salts that enter the roots is transported with water through the xylem to leaves. Sometimes structure like salt glands eliminate excess salts by active transportation with the help of special enzyme system.

A.Leaves of Sundari plants (with salt deposits); B. Salt glands in the leaves to excrete excess salt.

Salt storage—Within the cells, salt is stored in the vacuole (vacuolization) whereas organic solutes are stored in cytoplasm. The vacuolar membrane (tonoplast) of cells of Sundri has a modified lipid composition to prevent leakage of Na+ back to cytoplasm.

In the leaves, there are large vacuoles where the salt is stored and later eliminated during shedding of leaves.

Water conservation—There are thick leaves with leaf hair, waxy cuticle to prevent water loss by transpiration so that spit concentration is maintained in dilute form.

Camel’s Ability To Withstand Extreme Water Loss And The Shape Of Rbc In Camel :

Osmoregulation and Thermoregulation:

Camel can travel great distance in desert without food and water for several days (almost a week). Hump is the source of energy and water. There are two types of camel-(l) With one hump (Dromedary) found in Middle East and Africa and (2) With two humps (Bactrian) found in Central Asia (Baby camels are born without any hump).

Camel 'The ship of desert'.

Hump stores fat (about 10-15 kg) and also protects other tissues from heating due to sunburn. Hump fat is metabolised to produce water (about 1111 gm of water per 1000 gm of humpfat).

There are several adaptive features in camel to withstand extreme water loss and thus to adjust osmoregulation and thermoregulation.

Thick skin—Camel’s thick skin insulate them from intense heat radiated from desert sand.

Tolerance of fluid loss-Camels can lose up to25% of their body fluid without showing any sign of dehydration. This provides; an extra tolerance against long term water shortage. However, they can tolerate water loss up to40% of their body weight.

Formation of dry faces—Camel reduces water loss by removal of almost dry faeces (used as fuel by Bedouins directly without further drying) by active absorption of water and salts in colon.

Body temperature and sweating—Camel can regulate body temperature in a very special way. Their body temperature ranges from 34°C at dawn upto a steady increase.to 40°C
by sunset and cool off at night again. Sweating will occur beyond 40°C temperature.

This type of unusual tolerance in extreme heat helps them to preserve approximately 5 litres of water per day.,

Role of kidney—Kidney decreases glomerular filtration rate (GFR) and can reabsorb water tremendously. If they don’t drink water, urine volume will be 500 gm/day. On the other hand, if water is abundant urine volume may go up to 7 litres/day.

After reabsorption,the urine becomes as thick as syrup and has twice the salt content of sea water.

Role of nostrils—During expiration water vapour is trapped in the nostrils and reabsorbed into the body to conserve water.

Food habit—Camel eats green herbs and thus can ingest sufficient moisture to maintain body’s hydration.

Shape of RBC and Osmoregufation :

RBC of camel is nucleated, oval in shape (found in no other mammal) which resist clumping in waterloss (dehydration). Plasma volume is maintained by absorbing tissue fluid, so that circulation is not impaired

Water is scarce in desert and needs to be stored in large amount when available. Amazingly, camels can drink upto 150-200 litres of water at once, (A 600 kg camel can drink 200 litres 3 minutes). They do this to compensate previous fluid loss.

Comparison between human RBC and camel RBC

Drinking so much of water in such a short time could be a problem that may induce water intoxication and severe osmotic probiem. The oval shape of camel’s RBC and their ability to swell upto doub adjust this situation.

Water is absorbed very slowly through their stomach and intestine, allowing time for equilibrium. Moreover, RBC can swell upto 240% of normal size without bursting (while other species can only go upto 150%).

Behavioural adaptive features :

Problem-Solving In Chimpanzees :

Termites are favourite food of chimpanzee. They take a twig of a tree, remove the leaves from the twig. Then they use the stick like a “fishing rod” to “fish” the termites. Chimpanzee inserts the leafless stick (twig) into one of the holes in termites mound, waits for a moment, then slowly pulls it out.

The termites sticking to the twig are eaten up by chimpanzee.

Chimpanzees—A. Playing with computer; B. Breaking of nut; C. Feeds on captured termites;

Chimpanzee also can crack open nuts using pieces of woods in a hammer and anvil” technique.

It has been also observed that chimpanzee eats leaves of medicinal plants when infected with certain parasites.

Communication In Honeybees :

Honeybees go out to locate food source.’ A honeybee returns to the hive after successfully locating a source of food. The foraging worker bee dances at a dance floor close to the entrance of bee hive.

Two different types of bee-dance.

The worker bee (collector). is a sterile female who performs a set of skilful movements (manoeuvres) on the honeycomb that resembles a figure 8 (8) while waggling her abdomen. Based on the way of her dance, other bees are able to leave the hive and quickly locate the food she is dancing and the duration
source.

Class 10 Science Notes For Environment, Its Resources And Their Conservation

Nitrogen Cycle:

Life cannot exist without nitrogen. The atmospheric air contains 77-17% nitrogen. It is an essential component of all proteins. It is required- immensely for the synthesis of amino acids, enzymes, chlorophylls,’ nucleic acid, etc.

Nitrogen gas forms approximately 4/5th of.the atmosphere and neither plants nor animals can fix N2. Plants can take nitrogen in the form of nitrates which they absorb from the soil, J and animals obtain nitrogen by eating plants.

The mechanisms which replace nitrates in the soil may be divided into 2 types—

  1. those which transform atmospheric nitrogen into soil nitrates (N2 fixer), and
  2. those which transform plant and animal protein into nitrates (decomposers).

Definition of Nitrogen cycle: The complete series of cyclical events which occur partly in the micro-organisms of the soil and partly in the tissues of higher plants and animals are collectively known as the Nitrogen cycle.

Nitrogen cycle-a diagrammatic sketch.

Stages of Nitrogen cycle: Nitrogen cycle consists of the various steps such as,

  1. Nitrogen fixation,
  2. Nitrogen assimilation,
  3. Ammonification,
  4. Nitrification,
  5. Denitrification and
  6. Sedimentation.

Nitrogen fixation; The process of conversion of nitrogen of atmosphere into the biologically acceptable form or nitrogenous compounds is referred to as nitrogen fixation.

Read and Learn More Class 10 Science

This process is of two types such as physico-chemical or non-biological nitrogen fixation and biological nitrogen fixation.

Physico-chemical process of nitrogen fixation—In this process the atmospheric nitrogen combines with oxygen during lightning of electrical discharges in the clouds and produces different nitrogen oxides.

Physico-chemical process of nitrogen fixation

These nitrogen oxides get dissolved’ in rain water and on reaching earth surface react with mineral.compounds to form nitrates and other nitrogenous compounds.

Physico-chemical process of nitrogen fixation 1

Biological nitrogen fixation—The process of conversion of molecular nitrogen of the atmosphere into nitrogenous compounds through the agency of some living organisms is called biological nitrogen fixation.

Certain bacteria, blue-green algae, leguminuous plants etc. can fix atmospheric nitrogen and are grouped as follows—

Autotrophic—

  1. Aerobic e.g, blue-green algae (Nostoc),
  2. Anaerobic e.g., Rhodospirillum, Chromatium etc.

Heterotrophic—

  1. Aerobic e.g., Azotobacter,
  2. Anaerobic e.g., Clostridium.

Symbiotic bacteria-Bacteria of this type use carbohydrate and atmospheric nitrogen to make compounds which are eventually released into the soil as nitrates.

They live inside the root cells of leguminous plants e.g., peas, beans, clover, etc. where they cause tiny swellings called root nodules.

They obtain carbohydrate from the plant cells and in return release nitrates into the plant tissues and the soil. An association of this kind in which two different organisms benefit from living together is called symbiosis, e.g., Rhizobium leguminosarum.

Many non-leguminous plants are able to fix atmospheric nitrogen. They are—Casuarina, Alnus, Podocarpus, Pinus mycorrhiza, Pavetta, Chomelia, etc.

The whole process of biological nitrogen fixation is controlled by the action of different enzymes like nitrogenase, nitrate reductase, hydroxylamine reductase and so on.

These enzymes are present in some bacteria, few blue green algae etc. but not in other plants and animals (including man). So these microbes can fix nitrogen whereas animals and plants not.

Moreover, a special Nif gene is present in N2-fixing organisms but not in other plants and animals.

The overall process of nitrogenÿ fixation may be represented as follows :

The overall process of nitrogenÿ fixation may be represented as follows

Industrial nitrogen fixation—Nitrogen fixation is essential fc>r agriculture and manufacture of fertilizer. Ammonia is a required precursor to fertilizers. The most common industrial method is the Haber process (Haber—Bosch process)

Industrial nitrogen fixation

Ammonification: The process of release of ammonia and its formation to ammonium ions is known as ammonification.

The proteins of the dead plants and animals pass into the soil, that are acted upon by soil micro-organism, which decompose protein with the liberation of NH3 (ammonia).

Soil water contains a large number of hydrogen ions (H+) which chemically unite with the free ammonia to form’ ammonium ions (NH4+). The micro-organisms influencing the process are mainly, Bacillus mycoides, Bacillus ramosus and Bacillus vulgaris.

Nitrification: The process of converting ammonia to nitrate via nitrite is known as nitrification.

The ammonium ions are oxidised by a group of bacteria in the soil known as the nitrifying bacteria. Nitrosomonas converts ammonium (NH4+) ions into nitrate ions (NO2+), and Nitrobacter converts nitrite ions to nitrate ions (NO3 ).

nitrification

Nitrogen assimilation: The process by which inorganic nitrogen in the form of nitrates, nitrites and ammonia are absorbed by green plants and are converted into nitrogenous organic compounds is called nitrogen assimilation.

Nitrates are converted into ammonia which combines with organic acids to form amino acids. Amino acids are used for the synthesis of protein, enzymes, nucleic acids, chlorophylls, etc.

During digestion, plant proteins are broken into amino acids which are transformed into animal proteins, nucleic acids, etc.

Denitrification: The process which involves conversion of nitrates and nitrites Into ammonia, nitrous oxide and nitrogen is called denitrification.

The process is accomplished by the denitrifying bacterias lik Bacillus denltrificans, species of Pseudomonas, Micrococcus, etc. Several autotrophs like Thiobacillus denltrificans, Thiobacillus thioparus, etc., also take part in this process.

Denitrification

Human activities and nitrogen cycle :

Human activities (such as use of fertilizers) greatly increase the amount of nitrogen in the environment. This excess quantity of nitrogen cycles between the living world and biosphere (soil, water and air).

This may result into some disastrous effects as follows :

Increased global concentration of nitrous oxide (N2O) may cause serious greenhouse effect (since N2O is a potent greenhouse gas).

Increased regional concentration of other nitrogen oxides like NO may result into bad air pollution that may cause bronchitis, pneumonia and other lung diseases.

Due to deposition of acids of nitrogen, substantial acidification of soil and water may cause soil pollution and water pollution respectively.

Nitrogen oxides (NOX) damages leaves of plant, growth of saplings, reduction in the rate of photosynthesis.

Excessive NOX is responsible for acid rain which is very harmful to flora and fauna.

Significance of Nitrogen cycle :

  1. To maintain nitrogen balance in nature.
  2. Nitrogen is an essential component of all protein. The cell which is the unit of living organism is made up of protein. Hence, life cannot exist without nitrogen.
  3. Nitrogen is the most important component of protoplasm.
  4. Nitrogen is th.e essential component of deoxyribonucleotide of DNA and ribonucleotide of RNA.

So, nitrogen cycle represents an excellent example—how nitrogen circulates around and through the physical and biological world, restoring (the balance of nature.

Environmental pollution

Concept of pollution. (What is pollution ?):

Environment denotes the sum total of physical and biological factors that directly influences the survival, growth, development and reproduction of organism. Environment means biosphere which includes atmosphere, hydrosphere and lithosphere.

The whole of the earth including all its living organisms and nonliving substances are together called as biosphere.

Any unfavourable change in physical, chemical and biological characteristics of surroundings (consisting of air, water and soil) due to several human activities which ’cause harmful effects on Our or other desirable species and cultural assets is known as environmental pollution.

Pollution is an undesirable change in the physicochemical and biological characteristics ofÿhe biosphere that have adverse effects over living organisms and the environment.

Types of pollution

(HGW does pullution affect our daily life ?)

Air Pollution :

Many pollutants (either natural or man made) may cause disruption of norrnal composition of atmosphere known as air pollution. There, are many causes and effects of air pollution.

cause of air pollution

cause of air pollution 1

Water Pollution :

WATER POLLUTION

Water pollution means the adverse changes in the composition and condition of water such that it becomes unsuitable for use.

various causes and effects of water pollution

Soil Pollution :

Soil pollution (land pollution) is the deterioration of the earth’s land surface, naturally or man made, which decreases the quality and productivity of plants and ground water.

Cause of soil pollution

Noise Pollution :

 

Unwanted, irregular, unpleasant and annoying sound which is caused by the vibration of matters is known as noise and the pollution caused by them is called noise pollution.

NOISE POLLUTION

common causes and effect of noise pollution

Environment and Human Population

Problems of ever-increasing population:

(What are the problems caused by ever-increasing population ?)

Environment in the sum total of living and nonliving things which exert influence on living organisms present in a particular area. The natural surroundings of an organism, which affect its life by directly influencing its activities is called art environment.

It consists of lithosphere (soil, rocks etc.), atmosphere (air), hydrosphere (water).

Overexploitation of forest resource.

Throughout the world, all the problems may be summarised as Bp’s—pollution, population and poverty. Ever increasing human population (over population) is creating lot of devastating pollution and overconsumption that can be summarised as follows :

Overexploitation and depletion of natural resources—Man is overexploiting natural resources like fossil fuel (coal and petroleum).

Deforestation and loss of ecosystem-forest trees contributes valuable quantity of oxygen to global atmosphere and thus helps to maintain O2-CO2 balance. By deforestation huge amount of forest area is decreasing causing lot of pollution hazards.

Deforestation—before and after

About eight million hactares of forest are lost each year.

Shrinking of agricultural land Because of extensive urbanisation, industrialisation, establishment of townships etc. agricultural land is decreasing 3/4th. of the earth is water and only 1/4th is land.

This area barren land, mountain, desert, forest, township, industry etc, So very small land area than 10%) is- available for agricultural purpose but this area is still squeezing by progressive urbanisation.

This is a big threat to global ecosystem.

Shortage of fresh water- Due to overpopulation, there is acute shortage of fresh water for drinking, regular household works and for agricuIturaL crops.

Garbage causing pollution.

Air and water pollution—Because of overpopulation, there is parallel pollution of air, water, soil and so oh. Overexploitation of underground water by shallow and deep tubewells results into arsenic poisoning.

A. Effect of Global Warming Drought. B. Global Warming

Changes in atmospheric condition and global warming—Due to excessive production of greenhouse gases(CO2, CH4, CFC, N2O -etc.) by industrialisation, modernisation, atmos pheric condition is gradually changingresulting into green house effect (global warming).

Average percentage of major greenhouse gases

Destruction of wetland and its consequences—Natural or manmade aquatic habitat where water may be running (lotic) or stagnant (lentic) is called wetland. Wetlands are very useful as water reservoir, ponds for cultivation of fish etc.

They are nature’s kidney but wetlands are also threatened.

Warming effect of greenhouse gases

Scarcity of food-Qverpopulation leads to competition, ‘struggle for existence’ causing scarcity of food ‘and increase of price.

Environment and human health :

Various lung diseases, cancer etc. may develop due to environmental pollution and increasing population.

Lung diseases :

Asthma: This is a chronic lung disease where respiratory airways become narrow,produces extra mucus which results into difficulty in’ breathing, triggers coughing and wheezing (a whistling sound during breathing), develops chest congestion.

Effect of air pollution on children

Some environmental triggers for asthma may be—

  1. Animal dander—which I is composed of tiny flecks of skin of animals (like cat, dog, rabbit, guineapig, birds etc.) with hair or feather. This may cause allergic condition to develop asthma,
  2. Pollen grains of plant (like parthenium).
  3. Dust and mould.
  4. Manmade chemicals like smoke of fossil fuel combustion, SO2, NO2, O3, cigarette smoke etc.

Bronchitis: It is inflammation of the membrane in the bronchus of respiratory system, which causes pain (spasm) and coughing. Bronchitis may be either acute or chronic Common causes of bronchitis may be-

  1. Smoking cigarettes or other forms of tobacci.
  2. chronic inhalation of air pollutants or fumes or dusts (from occupational hazards ) such as coal minig,grain handling ,textile industry ,farming of livestock, metal moulding or welding etc. All these may result into COPD (Chronic Obstructive Pulmonary Disease).

Cancer: Various environmental toxins or poisons may cause cancer,

  1. Pesticides, herbicides and radioactive substances have the potential to cause cancer,
  2. Smokers are at increased risk for developing lung cancer. (In fact, smokers are more susceptible to cancer).
  3. Tobacco chewers are at increased risk for developing oral cancer.

Smoking and Lung Cancer

Smoking causes cancer.

Tobacco smoking or chewing are actually ‘wrong life style’ that can increase the risk of cancer.

Biodiversity and Conservation

Biodiversity and its importance :
(How the variety of fife forms in the environment help us ?)

The term “biological diversity” was first used by Dasmann (1968). Later the term is modified into “biodiversity” by Rosen in 1985. However, the term ‘biodiversity’ is popularised by Wilson to describe combined diversity of millions of organisms on earth.

Definition: The variety of iife on earth which includes piants,- animals and other living organisms’ along with their genotype and ecological interaction are together called as biodiversity.

Few examples of biodiversity on earth.

Total number of known plant and animal species on earth is approximately 15 million of which nearly 70% are animals and only 22% plants.

The species found to grow in a definite geographic area and not found elsewhere, is said to be endemic to that area. Example—one horned Rhino (Rhinoceros unicornis) is found in Jaldapara, Kaziranga forest zone but nowhere else in the world.

Beauty of biodiversity

Importance of biodiversity :

Production of food—Different plants and animals provide us food.

Production of ‘ drugs and medidnes-Most of the plants are medicinal plants that produce drugs and medicines. Some animal products, fungus and microbes also provide us medicines.

Medicinal plant—(a) vero (with flower), (B) Indian Pennywart (Brahmi)

Maintenance of ecological balance—Ecological balance is maintained by plants (producers), animals (consumers) and microbes (decomposers) through food chain, food web and energy flow.

Climate control-Climatic condition is maintained by the interaction of diverse plants,animals and microbes through biogeochemical cycles as well as conservation of biodiversity.

Economic importance—Various plants and their products
of house, household furniture, production of paper, gum, resin etc.

A. Paper industry B. Logs used in paper industry

Various animal products like honey, beewax, dairy products,leather, silk, wool, fishery pearl etc. are very useful to us.

Economic Importance of biodiversity A Sericulture B. peari Culture

Influence on art and literature— Biodiversity helps us to study natural history.

Birdwatching, gardening, fishkeeping are due to diverse plants/flowers, animals. Biodiversity inspires musicians, painters, writers, other artists with its natural beauty. Biodiversity has intrinsic aesthetic and spiritual value.

Biodiversity

Biodiversity hotspots

Biodiversity—Rich, spots in India

Definition: The geographic areas that contain high level of species diversity but are endangered to extinction are known as biodiversity hotspots.

34 biodiversity hotspots have been identified all over the world of which four important hotspots are in India, they are Eastern Himalayas, Indo-Burma, Western Ghats and Srilanka, Sundaland.

The Eastern Himalayas—This is the region that includes Bhutan, North-Eastern India, Nepal. This is wide mountain range from where major rivers arise. This hotspot has lot of globally threatened species viz. one horned Rhinoceros, Golden langur, Wild dogs, Sloth bears, Black bear etc.

A. National flower of India (Nelumbo nucifera); B. National bird of India (Povo cristatus);

Indo-Burma—This region includes North-Eastern India, Eastern Bangladesh, Myanmar, Malayasia, Vietnam, Thailand. This region is endemic to several animals like Monkeys, Langurs,Gibbons and many birds. Many plant species are also endemic to this region. Ginger is native to this region.

The Western Ghat—These are a chain of Hills that run along the Western edge of Peninsular India upto Srilanka.

They have high rainfall, close to Indian ocean. There is moist deciduous forest as well as rain forest having lot of species diversity. Varieties of amphibians and reptiles are endemic to this area.

Sundaland—This is a biogeographical region that comprises Indonesian islands of Borneo,Sumatra, Java, Bali and Malay peninsula. (However, these area do not belong to India). This ecoregion contains a diversity of freshwater habitats including hill streams.

So many fresh water fishes are endemic here.

Loss of biodiversity ;
(What are the threats to biodiversity ?)

Primarily, because of human interference (anthropogenic), large number of plant and animal species are extinct from earth or become endangered on the verge of extinction. Some common reasons behind loss of biodiversity can be explained as follows :

Loss of biodiversity

Destruction of habitat due to change in land use pattern— This is the most important cause for loss of biodiversity. Overpopulation, urbanisation,industrialisation require huge land area. So there is destruction of natural habitat by.

filling wetlands, ploughing grasslands, cutting down trees, burning forest. So there is loss of wild natural habitat. So many plantsand animals die out. Animals requiring large area (eg. mammals and birds) are badly affected. Migrating animals are also threatened.

Deforestation Loss of biodiversity

Parallelly there is also spread of agriculture—so grassland, wetland, forests are destroyed that results into extinction wi S0, there is a,so reduction in biodiversity.

Deforestation Cause of Global warming. Deforestation Cause of Pollution.

Hunting and poaching—Once upon a time hunting in the forest was an example of bravery. Over the years many wild animals like tiger, cheetah, leopard, panther, deer, elephant, dolphins, whales, rhinoceros have been brutally killed.

Even today, the poachers are still killing the wild animals. Thus many animals become extinct or endangered. Presently by Government rules and laws, hunting is strictly prohibited.

Global warming and climate change—There is change of temperature, humidity, rainfall etc.—so climatic change occurs. In the history of earth, there was iceage when many organisms became extinct.

Presently, due to greenhouse effect, there is global warming when there is gross change in climate—decrease of rainfall; expansion of desert, melting of iceberg. Thus
penguins, seals, whales are also threatened.

Effect of global warming melting of iceberg

Pollution— Excessive use of pesticide has polluted waterbodies, ground water, where many species died. Overuse of pesticide may cause biorriagnification which is more dangerous for biodiversity. There is sharp decrease of carnivorous birds like falcon, kingfisher etc.

“Run off” of fertiliser from cultivating field into nearest waterbody results into eutrophication which will cause excess demand of oxygen (BOD—Biological Oxygen Demand) which is harmful for the primary aquatic animals like

Overexploitation—Overexploitation of any species will lead to reduction in the size of population (eg. Hilsa fish and other marine fishes) so that they may become vulnerable to extinction. In last 500 years—Dodo and others have become extinct due to overexploitation by man.

Natural calamities—Cyclone, Tornado, Tsunami, volcanic eruption and other devastating natural calamities caused tremendous loss of biodiversity all over the world at different time.
Those are irrepairable, irreversible damage of biodiversity that can not be compensated.

Forest fire causing loss of biodiversity

Example—During Tsunami, coral bed at Andaman area has been badly fragmented
and swayed away at different parts of Indian Ocean.

Introduction to exotic species—Exotic species (Alien species) are those that have been imported from foreign countries of world which are growing and reproducing well in our climate.

Sometimes they become very invasive and drive away the local species to extinction.Exotic species may be harmful to both aquatic and terrestrial ecosystem.

Example—Water hyacinth (Eichhornia) was brought from Amazon Basin of South America and introduced in Indian water to resuce pollution but it has resulted into death of many indigenous aquatic plants and animals causing oss o hiodiversity.

Environmental problems of the Sundarbans
(What ails the environment of the Sundarbans ?}

Sundarban covers around Sundarbansquare kilometremeans “beautifulout of whichforest”.60% belongsThere areto BangladeshSundri trees.nd 40% in India. Sundarban is the largest mangrove wetland ecosystem in the world.

Sundarban is included under UNESCO world heritage site. It has the largest Royal Bengal Tiger reserve. It contains diverse fauna—about 35 species of reptiles, more than 270 species of birds, 42 mammals (including last population of tiger inhabiting mangroves in the world).

Sundri plants in Sundarban

Some common local species are – tiger (Panthera tigris), Saltwater Crocodile (Crocodylus porosus), Water Monitor (Varanus sp.), Rhesus Monkey (Macaca sp.), Jackal (Canis aureus).

However, this unique coastal tropical mangrove is most threatened in the world. The threats to the mangrove ecosystem are arising partly due to natural biotic pressure from the surrounding environment and partly due to human induced (anthropogenic).The causes of threats can be outlined as follows :

Destruction of mangroves due to urbanization—Lot of forest trees are cut off for building, shops and urban development.

Agriculture—Agricultural expansion results into clearance of mangrove trees.

Eresh water crisis— Reduced flow of sweet water into Sundarban mangrove system results into fresh water crisis.

Destruction of habitat –There is continuous deforestation and destruction of habitat by cutting trees to meet the demand of small timbers and fuel wood for local consumption.

Natural deforestation

Pollution — Chemical pollution by I marine points, hydrocarbons, sewage and industrial pollution results serious threat to flora and fauna at Sundarban.

Disbalance in prey-predator number-uncontrolled collection of prawn seedlings,uncontrolled fishing, trampling of river banks by fisherman, poaching of tiger, spotted deer, wild boar, marine turtle, horse shoe crab etc. resulted problem in food chain and food web of the ecosystem.

Submergence of islands due to rising sea level —This has made worst effect on human settlement in nearby blocks. Their main livelihood—agriculture and fishing are also affected.

Climatic factors-Various natural disaster like cyclone, flood, storm etc. seriously affect this mangrove ecosystem.

Conservation of biodiversity
(What measures can be taken to conserve biodiversity ?)

Biodiversity is threatened by reduction in space, smaller and fragmented habitats, over exploitation by man, climatic change, pollution etc.

Conservation of biodiversity is a method for protection and scientific management of biodiversity to its optimum level and derive sustainable benefits for the present as well as future generations.

A. Afforestation of Shegun tree, B. Teak plantation. C. Sal forest in red soil

Various methods could be applied for conservation of biodiversity. These methods are divided into two major categories depending on the place of conservation. In one type life forms are conserved in their original habitat (ip-sfcu conservation) and in. another type conservation is done outside the natural habitat of the organisms (Ex-situ conservafton)

Royal Bengal Tiger.

In-situ conservation: The method of conservation of biodiversity in the natural habitat of the organism is known as in-situ conservation, eg. conservation of rhinoceros in Gorumara, Jaldapara, Kaziranga forest; Royal Bengal tiger in Sundarban forest etc.

Ex-situ conservation: The method of conservation of biodiversity outside their natural habitat in the man-made system is known as e>esitu conservation, eg. conservation of different plants in Botanical garden, conservation of different animals in Zoological garden etc.

Different methods of Ex-situ conservation :

Zoological garden-

  1. Made by man (government) to keep endangered animals and other animals to exhibit biodiversity.
  2. Hunting, disturbance is strictly prohibited but visitors are allowed.
  3. Animals can breed and reproduce.
  4. The endangered animals as well as others are provided with adequate nutrition, medical care. Some research works are also done here with wild animals.

Ex-situ conservation.

Example: Alipur Zoological Garden in Kolkata, West Bengal.

A. Zoological Garden, Alipur, Kolkata (ex-situ conservation); B. Birds in Zoo (ex-situ conservation)

Botanical Garden—

  1. Endangered and rare species of plants are maintained with proper care for successful rearing.
  2. Also lot of different types of plants are kept for explaining biodiversity.
  3. Common people can visit the garden and will be aware of useful plants, eg. Acharya Jagadish Chandra Bose Indian Botanical Garden, Shibpur, Howrah, West Bengal.

A. Botanical Garden (ex-situ conservation), Shibpur, Howrah (W.B).; B. Giant Banyan tree

Cryopreservation— Some plant tissues, seeds, embryos or other parts of plant may be preserved at -196″C in liquid  called cryopreservation. The cryopreserved material is revived through special technique as and when required.

In order to prevent extinction, parts of endangered organisms may be cryopreserved to help in conservation.

A. Cryopreservation-A; B. Cryopreservation-B.

Role of JFM -aptd -P-BR in -conserving biodiversity :

(How can common people participate in conservation efforts ?)

Joint Forest Management (JFM): This was instituted by the 1980s. It was initiated to meet the demands of the people and save forests from destruction. It involves the participation of common people in the protection and management of forests.

A. Joint Forest Management (JFM); B. Involvement of villagers for JFM.

It was realised in 1980s by Government of India that participation of local communities around the forest area is urgently needed for conservation of forest and its natural resources. So the concept of JFM developed.

This is a joint committee of Government people and people of local communities who work together for protecting and managing forests. In return for their services to the forest, the communities get benefit of various forest”products like fruits, gum, rubber, medicine etc. and thus the forest can be conserved in a sustainable manner.

The pioneer project of JFM started in 1971 for restoration and management of degraded Sal forests (Shorea robusta) by the local inhabitants of the Arabari forest in Paschim Midnapore district in West Bengal.

JFM is the official and popular term in India for partnership in forest movement involving both the state forest departments and local JFM are communities.

The policies and objectives of in 1990. detailed Usually a in village National committee Forest Policy known of as India Forest (1988) Protection and the Committee guidelines were(FPC)prepared and the Government Forest Department enter into JFM agreement.

JFM originated accidentally at Arabari Forest Range in West Midnapore in 1971. The Major hardwood of Arabari is Sal which is commercially profitable forest crop. Ajit Kumar

Banerjee, a silviculturist, working in the Forest Department as Divisional Forest Officer (DFO), conducting trials which were constantly being disturbed by grazing animals and illegal harvesting by the local population.

The forest officials formed an adhoc Forest Protection Committee which include eleven local villagers and negotiated the terms of a contract. The contract agreement was—

  1. Poor villagers will get work in the forest;
  2. They will protect the forest resource;
  3. They will get 25% share of total profit;
  4. The vanishing Sal forest tree will be restored and preserved;
  5. An overall cooperation of the local people and forest official to preserve the forest resource and to prevent random deforestation.

The experiment on Arabari Sal Forest was very successful and was expanded to other parts of West Bengal and India as well. After initial success in West Bengal, followed by Haryana, JFM schemes received national inportance in the legislation in 1988.

Presently, nearly 63,000 FPCs involved in JFM of over 140,000 km2 of forested land in India.

People’s Biodiversity Register (PBR): Biodiversity Management Committee (BMC) will be constructed involving local people along with government officials who will prepare People’s Biodiversity (PBR) in consultation with local people.

The register contains comprehensive information on availability and knowledge of different
local biological resources—their medicinal use,commercial use and so On as per Biodiversity Act, 2003.

People's Biodiversity Registers (PBR).

This documented register is prepared to know the history of the biological resources of a particular place, changes in resource and people’s opinion about management of that resource.

A number of PBRs have been prepared in different parts of India since 1995 with the initiatives of many NGOs and educational institutions along with local communities and village councils (Panchayat).

PBR is actually a documentation of local biodiversity (flora and fauna). It includes—cultivating land, water bodies (pond, lake, canal, river etc), forest (any special variety of plant), plants (crops, vegetable, fruits medicinal plant), animals (wild animals, domestic animals, varieties of insects, birds), any other important notable feature of the place and so on.

Significance of PBR :

  1. PBR of a particular locality is very significant as follows :
  2. Identification of spots rich in local flora and fauna.
  3. Demarcation of areas that are biologically significant (e.g. a lake where lot of migrating birds come in every year).
  4. Identification of local endemic species.
  5. Identification of local endangered plants or animals (if at all).
  6. Information about local geographical history and its panoramic changes in different seasons of the year.
  7. Effect of natural calamity (like flood in rainy season, drought in summer) on local flora and fauna.
  8. Dependence of life style of local people on the local flora and fauna.
  9. Past history of the locality—how soil erosion has affected the area; so many flora and fauna are no more; how various types of pollution are affecting ecology of the locality.

These attempts have motivated common people to create awareness about management of natural resources. PBR will have an important role to piay in promoting conservation, sustainable use and equitable sharing of benefits of biodiversity resources in coming decade.

Some endangered species ofIndia and their conservation :
Conservation efforts of Tiger, Rhinoceros, Lion, Crocodile and Red Panda in India : (Name and place of the projects undertaken by Government of india).

For the proper conservation of all the above mentioned endangered animals, many projects have been adopted by the Government. In those projects, hunting and poaching are strictly banned.

Forest guards, pfficers, scientists are appointed to look after the wild animals.with proper care.

Wild life traditionally refers to undomesticated animal species but now it includes all plants, fungi and other organisms that grow or live wild in an area without being introduced by humans. Wildlife can be found in any ecosystem.

The four most general reasons that lead to destruction of wild life includes—

  1. Overkill
  2. Habitat destruction and fragmentation
  3. Impact of introduced species
  4. Chains of extinction (extinction of one species may cause extinction of other).

Conservation of tiger (Panthera tigris):

Tiger is the national animal of India, the most majestic cats in the world. Tiger conservation aimed to prevent the animal from becoming extinct and preserving in its natural habitat.

Tiger ( Panthera tigris)

Conservation efforts: In India, Project tiger was started in 1972 to conserve tiger in-situ.(in its natural habitat). At the beginning 0f 20th century, tiger population in India was approximately 40,000 whereas in 1972, it has become only 1827.

Project undertaken: Project tiger signifies national ban on tiger hunting and poaching. Project tiger was initiated by Mrs. Indira Gandhi in 1973.

Site of project/conservation: Today, there are 27 Project Tiger wildlife Reserves in India, covering an area of 37,761 km2.

Some important sites are—

  1. Sundarban Tiger Reserve in West Bengal
  2. Bandipur Tiger Reserve in Karnataka
  3. Kanna Tiger Reserve in Madhya Pradesh
  4. Palamou Tiger Reserve in Jharkhand

Methods of conservation:

  1. Hunting of tiger is strictly prohibited,
  2. Poachers are penalised according to Indian Forest Act.
  3. Preservation of forest for breeding and reproduction of tiger in natural habitat (in-situ).
  4. Checking of deforestation and increase of afforestation,
  5. To deploy forest officer, forest guards etc.

Present situation/status: As a result of this program of project tiger, the population of Bengal tiger had increased from about 1200 in 1973 to an impressive 3500+ in 2007. Presently (2015) the tiger population is approximately 2226 in India.

According to report of West Bengal Forest Department, 2013, the minimum identified tiger count in Sundarbans stands at 103.

Conservation of Rhinoceros, (One horned Rhino, Rhinoceros unicornis) :

Among terrestrial land mammals In Asia, Indian Rhinoceros is second in size only to the Asian elephant They have thick grey—brown skin and a single black horn, present in both male and female.

The horn is made of keratin protein (which is present in our hair and fingernails). Rhinoceros is listed as Vulnerable on the IUCN Red list. In 2015, a total of 3555 Indian Rhinoceros are estimated to live in the wild.

They are found in Duars, Assam, North East India. Indian Rhinos were hunted relentlessly; poaching of Rhinoceros horn was the most important reason.

Conservation efforts: Indian Government has taken major steps towards conservation of Rhinoceros especially with the help of WWF (World Wild Life Fund for Nature) and other NGOs.

Project undertaken: Rhinoceros project.

Site of project/conservation :

  • Jaldapara National Park in West Bengal
  • Kaziranga National Park in Assam

Methods of conservation :

  1. Hunting and poaching strictly prohibited,
  2. In-situ conservation.
  3. Preservation of forest for natural breeding and reproduction,
  4. Poachers are punished according to pff Indian Forest Act.
  5. To deploy forest officer, forest guard etc.

One horned Rhinoceros

Present situation/status: In Kaziranga National Park 2048 Rhinos were estimated in 2009; in Jaldapara National Park, 108 in 2002. They must be properly preserved in order to increase their population, to protect them against extinction.

Conservation of Asiatic Lion (Panthera leo persica) :
(Also known as Indian lion or Persian lion.)

The Asiatic lion is a lion subspecies that exists as a single population in Gujrat state of India. It is listed as endangered by IUCN. Since 2010,the lion population in the Gir Forest National Park has steadily increased.

Lion (Panthera leo persica)

Conservation efforts: The Asiatic Lion Reintroduction Project is an initiative by j Government of India to provide safeguards to Asiatic lion from extinction in the wild by means of reintroduction.

The only single population in Gujrat faces threats of epidemics, natural disasters etc.

Project undertaken: Leo project to preserve Asiatic lion.

Site of project/conservation :

  • Kuno Wildlife Sanctuary in Madhya Pradesh
  • Gir Forest National Park in Gujarat

Methods of conservation :

  1. Hunting of lion is strictly prohibited,
  2. Poachers are penalised according to Indian Forest Act.
  3. Preservation of forest for breeding and reproduction of lion in natural habitat (in-situ).
  4. Checking of deforestation and increase of afforestation,
  5. To deploy forest officer, forest guards etc.
  6. The project aims to establish a second independent population of Asiatic lions at the Kuno Wildlife Sanctuary in Madhya Pradesh.

Present situation/status: As per census in May, 2015, the lion population was estimated as 523 individuals of which 109 were adult males, 201 adult females and 213 cubs.

Conservation of Crocodile (Crocodylus porosus) :

Crocodile population decreased and threatened by loss of riverine habitat, depletion of fish resources and entanglement in fishing nets. They are enlisted as critically endangered on IUCN Red list.

Crocodile (Crocodylus porosus)

In 1974, a survey was made by Government of India on the three species of Crocodiles present in India. They are—

  1. Gharia! (Gravialis gangeticus)—Found in river of North India.
  2. Estuarine Crocodile (Crocodylus porosus)— Found in delta region of Orissa, Sunderban of West Bengal.
  3. Mugger(Crocodylus palustris)—Marsh crocodile, depleted in number and very rare.

Conservation efforts: In response to declining crocodilian populations, Government of India launched a Crocodile Conservation Program in 1975.

Project undertaken: Crocodile project

Site of project/conservation :

  • Bhagabatpur Crocodile Project in Sundarban, West Bengal;
  • Bhitarkanika in Orissa etc.

Methods of conservation :

Collection of eggs with subsequent incubation and rearing of young until they become suitable in size to release in water—thus to boost reproductive output,

To locate, establish and manage a series of crocodile rehabilitation centres and sanctuaries in suitable habitats.

Present situation/status: 16 crocodile rehabilitation centres and 11 crocodile sanctuaries have been established. A total of 879 Gharials, 190 Estuarine Crocodiles and 493 Mugger are reared and released..

The greatest achivement is the reestablishment of viable Gharial breeding population.

Conservation of Red Panda (Ailurus fulgens) :

The red panda (also called, the lesser panda, the red bear-cat, red cat-bear) is a mammal native to eastern Himalayas. It has reddish brown fur, a long shaggy tail. It is slightly larger than a domestic cat.

A. Red Panda; B. Enlarged view of Red Panda.

It is arboreal, feeds mainly on bamboo leaves and shoots. It is active at night from dusk to dawn. It is classified “endangered’ by IUCN in 2008. They are found in Assam, Sikkim. The primary threat to red panda is uncontrolled killing or poaching, habitat destruction (due to deforestation).

Conservation efforts: India has 20 protected areas with known red panda population in Sikkim, Arunachal Pradesh. In West Bengal, there are Kanchanzonga National Park, Singalila
National Park, Namdapha National Park etc. for conservation of Red Panda.

Project undertaken: Red Panda Project (as mentioned above).

Site of project/conservation: In Sikkim, Arunachal Pradesh, West Bengal.

Methods of conservation :

  1. Hunting of Red Panda is strictly prohibited,
  2. Poachers are penalised according to Indian Forest Act.
  3. Preservation of forest for breeding and reproduction of red panda in natural habitat (in-situ).
  4. Checking of deforestation and increase of afforestation.
  5. To deploy forest officer, forest guards etc.

Present situation/status: The population of red panda has increased considerably due to development of protected areas. They breed in captivity and then can be introduced to natural habitats.

Class 10 Science Notes For Continuity Of Life

Cell division Cell and Cell Cycle:

All living organisms are made up of cells, which are the structural and functional unit of life. It is independent and self reproducing under favourable condition. Shapes and sizes ofcells vary.

A division of labour exists among the cells forming a multicellular plant or an animal body. “Omnis cellula e cellula”, means every cell is derived from a pre-existing cell.

Starting with a single fertilized egg cell (Zygote), the first division produces two cells; when these daughter cells divide, four cells are produced—thus number of cells increases called hyperplasia.

Interrelationship among chromosome, DMA and gene (What are hromosomes? How are they formed ?) :

Inside the nucleus of eukaryotic cell, there is thread-like intertwisted network called chromatin network. These are actually nucieoprotein threads consisting of DNA and protein.
DNA is a large biomolecule.

Structure of Interphase nucleus.

Inside the nucleus, DNA remains in partly open and partly folded condition to form the network, also known as chromatin reticulum. In the folded condition, DNA is looped tightly around proteins.

This folded structure of DNA is termed as chromosome.

Read and Learn More Class 10 Science

So the difference between chromatin and chromosome is that chromatin is unfolded, uncondensed, extended DNA while chromosomes are condensed DNA.

Chromosomes are condensed chromatin and chromatin is unfolded chromosome. Chromatin is present in a cell normally in Interphase, while the chromatin threads get condensed into chromosome during cell division.

Thus chromatin or chromatin reticulum and chromosomes are actually different folding condition of DNA molecule.

Chromosome (Gr. Chroma-colour; soma—body) :

The intranuclear gene bearing, self dividing, fixed numbered, rod shaped structures which become dearly visible during metaphase and anaphase of cell division are known as
chromosomes that control all cellular activities.

Chromosome was first named by W. Waldeyer In 1888. During metaphase of the cell division the chromosomes are clearly visible and countable under compound microscope.

DNA double helix :

  1. Full form of DNA is Deoxyribo Nucleic Acid. It is usually double-stranded (double helical).
  2. Double helical molecular structure of DNA was first discovered by Watson and Crick (1953), for which they were awarded Nobel Prize.
  3. The DNA is formed of two helices or strands or chains named as a and (). Two chains are interlinked by Hydrogen bonds (H-bonds). Two chains are antiparallel. (Much details of DNA have been discovered by many scientists.)
  4. Each chain of DNA consists of number of structural units called deoxyribonucleotides. Hence DNA is the polymer of deoxyribonucleotides. Two consecutive deoxyribonucleotides are interconnected by phosphodiester bond.

Gene (Where are genes located ?) :

Gene is the hereditary unit consisting of a particular sequence of bases in DNA and specifying the production of distinct enzymatic protein in the cell. Gene Is the functional unit of DNA.

Pathway of 'Central Dogma'.

Genes are responsible for transmitting the hereditary traits from parents to offspring. Series of genes are present on each DNA.

How does gene act ?

Gene is a particular part of DNA that carries specific code. From DNA (gene) m RNA is synthesized by the process of Transcription. From m RNA specific protein Is synthesized by a process known as Translation.

This specific protein may be a structural protein or enzyme protein. The enzyme alters metabolic rate (profile) in the cell or organ of the body leading to the expression of different characters Is an organism.

Thus gene acts In the cell indirectly through the formation of enzymes.

Types of chromosomes: Functionally, chromosomes are of two types :

Autosome— The chromosomes that are responsible for controlling all the somatic characters of the body except sex determination are called autosomes.

Autosomes and Sex chromosomes in man

Example— in each human somatic cell,number of autosome is 22 pairs. Somatic character means height of the body, skin complexion, texture of hair etc.

Allosome (Sex chromosome)— The chromosomes that are primarily responsible for sex determination (maieness or femaleness) of an organism are called allosomes or sex chromosomes.

Example—In each somatic cell of man, number of allosome is 1 pair. In human male, two allosomes are named as X and Y whereas in human female, two allosomes are named as X and X.

So, in human female, two sex chromosomes are identical (one pair of X) but in human male, they are nonidentical—one X and one Y. X and Y differ in size and morphology.

Karyotype (type of chromosome) of human c? = 44A + XY
” ” ” ” ?                                                               = 44A + XX

Chromosome number :

The characteristic number of chromosomes present in each somatic cell of an organism which is constant for any particular species of plant or animal is known as chromosome number of that species.

e.g. chromosome number of man (2n = 46), Fruit fly (Drosophila nelanogaster) (2n = 8), Pea plant (Pisum sativum) (2n = 14), Honey Bee (Apis indica) (2n = 32).

Somatic cell (Body cell): Any cell of a multicellular organism other than the reproductive cell is called somatic cell. e.g. In human body all cells of liver, kidney, blood, brain etc. are called somatic cells.

Generally, somatic cell contains diploid (2n) number of chromosome.

Germ cell and germ mother cell: A cell that contains half the number of chromosomes (n) of a somatic cell and participates in fertilization of sexual reproduction is known as germ cell. e.g. sperm in male, ovum in female.

The mother cell from where the germ cell is formed is known as germ mother cell. e.g.Spermatogonia in testis, oogonia in ovary.

Diploid— When the nucleus of a cell contains two complete sets of chromosomes paternal set (from father) and one maternal set (from mother), the cell is called diploid.

It is represented by 2n. In man, diploid number of chromosome, 2n = 46, is found in the somatic cells like hepatocyte (in liver), WBC (in blood), neuron (in brain) etc.

Haploid — When a single set of unpaired chromosome is present in the nucleus of a cell,the cell is called haploid. It is represented by n. In man, haploid number of chromosome, n = 23, is found in the germ cell like sperm in testis, ovum in ovary.

Difference between haploid and diploid :

Difference between haploid and diploid

Structure of chromosome (what are the parts of a typical chromosome ?) :

Morphology of Eukaryotic rchromosome: Chromosome is microscopic threadlike structure.

Diagramatic structure of Chromosome

Function of chromosome :

  1. Chromosomes are responsible for cell division.
  2. Chromosome contains DNA which contains genes that are responsible for inheritance of characters from parents to offspring.
  3. Chromosome controls cellular metabolism by the synthesis of necessary enzyme indirectly through mRNA (Central Dogma).
  4. Chromosome undergoes crossing over causing mutation and thus results into formation of new variation.
  5. Sex chromosome plays role in sex determination.
  6. Autosome controls all somatic characteristics of the body.

Internal structure of Chromosome.

Chemical components of Chromosome :

The major chemical components of eukaryotic chromosomes are nucleic acids and proteins (basic proteins).

The nucleic acids are of two types i.e. about 45% Deoxyribonucleic acid (DIMA) and about 5% Ribonucleic acid (RNA).

Chemical composition of chromosome

DNA combines with histone, a simple basic protein, to form nucleoprotein which is conjugated protein They remain with the ratio 1:1. The RNA and acid protein vary in amount quite widely from one kind of cell to another.

Structure of DNA double helix

Primarily chromosome contains 90% DNA + basic protein and 10% RNA + acid.protein

Sugar phosphate backbone and base pairs in DNA doble helix

DNA:

  1. The full term of DNA is,deoxyribonucleic acid.
  2. It is usually double stranded (helical), consisting of a number of unit called deoxyribonucleotides,
  3. A nucleotide consists of a deoxyribose pentose sugar, phosphoric acid and a nitrogen base (purine or pyrimidine).
  4. The pyrimidine bases are cytosine (C), thymine (T) and the purine bases are adenine (A) and guanine (G).
  5. DNA molecule forms the structures called genes

RNA :

  1. The full form of RNA is ribonucleic acid.
  2. It is usually single stranded (helical).
  3. The structural unit of RNA is ribonucleotide.
  4. The pentose sugar is ribose.
  5. The nitrogenous bases are Adenine (A),Guanine (G),Cytosineand (C) Uracil (U).
  6. RNA helps in protein synthesis.

Single stranded RNA

Ribonucleotide

Types of RNA: Mainly RNA are of two types—Genetic RNA and Non-Genetic RNA.

Genetic RNA—Here RNA bears hereditary characters (genes) only e.g. Most plant virus and some animal virus like, influenza virus, HIV.

Non-Genetic RNA—Here DNA forms the genetic material. This type of RNA originates from the DNA of the cell and helps in protein synthesis.

They are of three types—

  1. Messenger RNA (mRNA)—Carries information from DNA for protein synthesis,
  2. Transfer RNA (tRNA)—Collects actual amino acids for protein synthesis,
  3. Ribosomal RNA (rRNA)—Helps in protein synthesis.

Protein: In addition to DNA and RNA, the chromatin (nucleoprotein) contains histone protein (basic protein) and non-histone protein (acid protein). The alkaline protein,histone, mainly contains amino acids like arginine, lysine, histidine, whereas, the acidic protein generally possess amino acids like tryptophan and tyrosine.

Nucleoside and Nucleotide :

Nucleoside—It is made up of pentose sugar molecule and a N2 base (without H3P04).

Nucleotide—It is made up of a pentose sugar molecule, N2 base and phosphate group.

Difference between DNA and RNA :

Difference between DNA and RNA

Euchromatin and Heterochromatin :

Euchromatin—The uncoiled part of chromosome, which is genetically active in interphase is called euchromatin.

Distribution of euchromatin and heterochromatic in a chromosome

Heterochromatin—The densely coiled part of chromosome, which is genetically inactive in interphase is known as heterochromatin.

Distribution of euchromatin and heterochromatin Heterochromatin and Euchromatin in chromosome

Difference between Euchromatin and Heterochromatin:

Difference between Euchromatin and Heterochromatin

Cell organelles and structures involved in cel! division :

Cell is the smallest organised unit of living body which is self-reproducing under favourable condition.

Prokaryotic arid Eukaryotic cells:

A. Bactarial (Prokaryotic) Cell; B. Plant (Eukaryotic) cell

Prokaryotic cell: The cell which lacks a nuclear envelope, nucleolus and well defined cytoplasmic organelles, such as endoplasmic reticulum, Golgi body, mitochondria, centriole, etc. is known as prokaryotic cell.

Examples—Bacteria (Azotobacter, Clostridium), Blue green algae [Nostoc, Anabdena), Mycoplasma.

Eukaryotic cell: The cell in which the nucleus has a definite nuclear membrane, nucleolus-ana well defined cytoplasmic organelles like endoplasmic reticulum, Golgi bodies, mitochondria, lysosomes, etc. is known as eukaryotic cell.

Examples— Cells of higher plants and animals.

Difference between Prokaryotic and Eukaryotic cell :

Difference between Prokaryotic and Eukaryotic cell

Role of cell organelles in cell division :

Some cell organelles play significant role in cell division as follows :

Nucleus—It is largest cell organelle where number of chromosomes are present in form of a chromatin network in interphase. During cell division, these chromosomes will be first divided into two chromatids and then separate chromosomes will be formed through different stages of karyokinesis. So, nucleus is the site of division of chromosomes.

Nucleus

Centrosome and microtubule—Centrosome is present generally in animal cells and few exceptional lower plant cells. One centrosome consists of two centrioles,perpendicular to each other.

Centrioles are formed of nine triplet sets of microtubule (9 + 0 cart wheel model). Each microtubule is formed of tubulin protein.

During cell division, a diamond shaped structure is formed in the middle of the cell called Spindle. The spindle is made of many spindle fibres. In animal cell, these fibres are formed from centriole.

During cell division, two centrioles are separated and pushed apart to opposite poles of the cell by the elongation of spindle fibres in between them.

In animal cells, the spindle having centriole with aster at two poles is called or astral spindle or amphiastral spindle but in most plant cells, spindle withoutcentricasterspindle(astral rays) Is called as acentric spindle or anastral spindle.

So, in animal cell, spindle is formed from centriole whereas In plant cell the spindle is formed from cytoplasm and nucleoplasm.

Ribosome-This is the cell organelle responsible for protein synthesis. Huge amount Of protein is required for cell division-like tubulin for spindle fibre, enzyme protein etc. All these are synthesized by ribosome.

Mitochondria—There are Lot of metabolic functions during cell division, for which sufficient quantity of energy is required which is supplied by mitochondria.

Cell division and its significance :

The active process by which a matured mother call duplicates itself to give rise to two daughter ceils is called cell division.

Starting with a single fertilized egg cell (zygote), the first division produces two cells. These two daughter cells divide to produce four cells and so on. Thus starting from a single cell, four consecutive divisions will produce 24 = 16 cells, 10 division will produce 210 = 1024 cells.

It would require only 43 divisioin cycles to produce 5 trillion cells in the body, starting from a single cell. This increase in number of cells by cell division is called hyperplasia.

Significance of cell division :

Growth—By repeated cell division, number of cells in multicellular organism increases (hyperplasia) which results into overall growth and development of organs and body of the organism.

During embryonic growth, the first formed cell is zygote which undergoes repeated cell division to form an embryo and then into newborn.

Reproduction—Unicellular organism (like protozoa, yeast, bacteria etc.) reproduces by the process of cell divison. Mitosis helps in vegetative reproduction in plants and asexual reproduction in plants and animals.

On the other hand, meiosis helps in sexual reproduction both in plants and animals.

Repair—By cell division, old decaying dead cells are replaced by new cells. At every moment, millions of cells are dying in our body and millions of cells are forming by cell’ division to compensate the loss.

Healing of wound of an injured area is done by repeated cell division of adjoining living cells. Thus the injured area is repaired.

Types of cell division :

In animals and plants three types of cell divisions have been distinguished Mitosis and Meiosis.

Amitosis or Direct cell division :

The process of direct cell division in which the nucleus first constricts in the middle to divide into two nuclei and then the cell body undergoes changes without spindle formation and ultimately divides into two daughter cells is known as amitosis.

Stages of Amitosis

Occurrence :

  1. In lower plants—Chara, yeast, bacteria, etc.
  2. In animalsAmoeba.

Mitosis or Indirect cell division :

Mitosis is a process of equational and indirect cell division that occurs in the somatic cells of the body where a mother cell divides itself into two daughter ceils that are qualitatively and quantitatively similar to the original mother ceil.

‘Qualitatively means type of genes (genotype) and ‘quantitatively’ means number of chromosomes of daughter cells.

Occurrence: Mitosis occurs in the somatic cells of the multicellular eukaryotes.

In plants—It occurs in the growing parts like root, stem, leaves, flower buds, apical buds, embryonic buds and in cambium,

In animals—It takes place in all the somatic cells except nerve cells, sex cells and muscle cells.

Mitosis is called an equational division because, the two daughter cells (diploid or 2n) which are produced, remain identical to the diploid (2n) parent cell.

DNA content during mitosis and meiosis.

Meiosis or reduction division :

Meiosis is a process of reduction division that generally occurs in the germ mother cell where a diploid (2n) mother cell undergoes two consecutive divisions (Meiosis I & II) to produce four haploid (n) daughter cells.

Occurrence: In higher plants- Meiosis occurs in spore mother cells(2) Of anther and ovule forming four haploid spores(n). In Lower animals and plants – Meiosis takes place in zygote. In higher animals – Meiosis takes place during gemetogenesis i.e in gamete (sperm and ovum ) formation.

Cell Cycle or Mitotic Cycle :

Definition: The sequence of cyclical events and changes through which grows (by G1; S, G2, stages), becomes mother cell and finally divides to fo cells is known as cell cycle.

Phases of cell cycle: In continuously dividing cells, an individual cell pass cyclically through two main phases of cell division. Every cell cycle has a long resting phase called ‘Interphase’, followed by a short divisionary phase called ‘M’ phase.

Interphase is divided into three subphases—G1phase, S phase and G2 phase.

G1 phase—It is a phase when a cell prepares itself for division. Cell grows in size, synthesizes RNA and proteins. DNA content remains unchanged.

S phase or Synthetic phase—In this phase new DNA is synthesized on the mother DNA template. So, the amount of DNA is doubled. Histone protein is also synthesized in S phase. Each chromosome now consists of two chromatids joined at the centromere.

Formation of two daughter cells

G2 phase—In this phase, cell growth, synthesis of RNA and protein continues.

Cell cycle time: The duration of cell cycle varies from one organism to another and from cell to cell. It ranges from few minutes to years. However, for fast dividing mammalian cells, the length of cell cycle is approximately 24 hours.

Significance of cell cycle :

Preparatory phase- During interphase of cell cycle (G1 S, G2) the cell gets itself prepared for cell division. There are many synthetic metabolic reactions inside the cell at this phase. This is wrongly (misnomer) called as ‘resting phase’. In fact, it is the most ‘active phase’ of cell cycle.

Interphase and Mitotic phase of cell cycle.

So, interphase is the ‘preparatory phase’ of cell cycle. In a pie chart of 360°, the interphase is nearly 320°-340n whereas the M-phase is approximately 20°-40°. Thus ‘preparation’ is much prolonged than that of ‘actual function’ (cell division).

Checkpoints in cell cycle—Many proteins (like cyclin etc.) and enzymes CDK (Cyclin Dependent Kinase) control cell cycle at different points called check points. First check point is in between G1 and S whereas second check point is in between G2 and M.

Thus the process of cell cycle and cell division is regulated, controlled and balanced.

Loss of control of cell cycle and formation of tumor—In our body at every moment large number of cells are dying called apoptosis. To compensate this loss, large number of cells are forming by cell cycle and cell division.

Thus a balance is maintained between ‘birth rate’ and ‘death rate’ of cell.

If the ‘check points’ in cell cycle do not work properly or if the control of cell cycle is lost,then the process of cell division and cell cycle becomes erroneous which may lead to formation of tumor. Tumor develops due to “uncontrolled cell division” (without differentiation) when a cell starts dividing at a vigorous rate.

Tumor is a type of unnatural swelling in any part of the body. This is richly supplied with blood vessels. Generally, tumor is of two major types—(a) Benign tumor, (b) Malignant tumor.

Benign tumor is noncancerous tumor. These cells do not spread over the body and remain restricted at a particular part of the body.

Malignant tumor is cancerous tumor that may result into cancer. Cancer cells (malignant cells) can spread from origin to different parts of the body called metastasis.

Mitosis :

Different phases of mitosis in plant cells.

Definition: Mitosis is a process of equational and indirect cell division that occurs in the somatic ceils of the body where a mother cel! divides itself into two daughter cells that qualitatively and quantitatively similar to the. mother cell.

Characteristics of Mitosis :

  1. Mitosis is called equational division because the chromosome number of mother cell and daughter cell remains same in mitosis.
  2. Mitosis is known as indirect cell division because during this process, chromatids are pulled towards opposite side by spindle fibre. So the chromatids do not get separate themselves. Hence it is indirect cell division.
  3. Mitosis is called somatic cell division because this process occurs only in the somatic cells of the body.
  4. By this process one mother cell is divided into two daughter cells.
  5. The daughter cells have same chromosome number (quantity) to that of mother cell as well as same genotype (quality) like the mother cell. Hence the two daughter cells produced by mitosis are qualitatively and quantitatively similar between themselves and to the mother cell.

The process of mitosis and its different stages :

The whole process of mitosis is completed in two phases— Nuclear division (karyokinesis) and cytoplasmic division (cytokinesis).

Karyokinesis—Definition: The process by which the nucleus of mother cell is divided into two daughter nuclei through Prophase, Metaphase, Anaphase and Telophase is known as Karyokinesis.

Prophase (Gr. Pro, First) :

The first stage of mitosis during which the chromosomes condense and become visible within the nucleus followed by the dissolution of the nuclear envelope is known as prophase.

In prophase, most dramatic changes take place both in the nucleus and in the cytoplasm.

It is the longest stage of mitosis.

Important features are as follows :

  1. The chromosomes condense, shorten, thicken and become stainable.
  2. Condensation occurs due to dehydration of water from chromatin fibre.
  3. Chromosome is coiled by spiralisation forming series of bead like structures called chromomere.
  4. Each chrbmosome consists of two chromatids that are in very close association with each other all along their lenght.
  5. Two chromatids coil around each other and are held together by centromere.
  6. During early prophase,the chrbmosomes are evenly distributed in nucleoplasm but as prophase progresses, the chromosomes migrateare towards the nuclear in membrane.
  7. Nucleolus and Nuclear membrane gradually disappear.
  8. In animal cell-The centrioles duplicate and move towards the opposite poles of the cell.
  9. The movement of centrioles occurs due to pushing apart by growth of the spindle fibres between them. In animal cells or in the cells of lower plants, fibrils appear like spokes of wheel around each centriole to form asters. The asters are formed from cytoplasm, whereas spindle fibres are formed mainly from nuclear material.

Different phases of mitosis in animal cell

The aster, the centrioles and the spindle together make up the structure called mitotic apparatus or achromatic figure.

Metaphase (Gr. msta, between):

The stage of mitosis during which the chromosomes attach to the spindle fibres at the centromeric region and arrange in the equatorial plane of the cell is known as metaphase.

Important features are as follows :

  1. A diamond shaped structure is formed during metaphase called spindle. The spindle formation starts in prophase and is completed in metaphase.
  2. Two pointed ends of the spindle are called as poles and central broad part is known as equator/equatorial plate/metaphasic plate.
  3. The spindle is formed of spindle fibres that are made of tubulin protein. There are two types of spindle fibres—
    1. short fibre/discontinuous fibre/chromosomal fibre—which is extended from pole to equator,
    2. long fibre/continuous fibre/achromosomai fibre—which is extended from one pole to another pole.
  4. The chromosomes begin to proceed towards the equator. They reach the equator of the spindle and line up in one plane. Maximum contraction of chromosomes or coiling or condensation takes place at this stage and the chromosomes of a particular cell are easily countable.
  5. The centromere is attached with the short chromosomal fibre and the arms are directed towards the poles (the arms are repelling each other).
  6. In plant cell, the orientation of chromatids on equatorial plate is random but in animal cell, a pattern orientation occurs when larger chromosomes are attached at the periphery of the plate and smaller chromosomes are placed at the central part.

Anaphase (Gr. ana, back):

The stage of mitosis during which the centromeres and daughter chromosomes separate and begin to move towards opposjte poles of the cell is known as anaphase.

Important features are as follows :

  1. It is the shortest stage of mitosis.
  2. The centromeres of the chromosomes divide and the two chromatids of each pair separate. After separation, they are called daughter chromosomes.
  3. The daughter chromosomes migrate towards the poles (anaphase movement or metakinesis) due to shortening of spindle fibres attached to the centromeres.
  4. Pulling causes the chromosomes to assume their characteristic V shape or L shape, etc.
  5. Several forces are involved for this anaphase movement of the chromosomes like jerking force, pulling force, pushing force, repulsive force etc.

Telophase (Gr. teio, end):

The stage of mitosis during which the chromosomes uncoil and become surrounded by new nuclear envelope is known as telophase.

This stage begins at the end of the polar migration of the daughter chromosomes.

Important features are as follows :

  1. This stage is reverse of prophase.
  2. The chromosomes uncoil and despiralise in a process (despiralisation) comparable to a reversal of prophase and gather into masses of chromatin to form chromatin network.
  3. A new nuclear membrane is formed around each mass of chromatin from the endoplasmic reticulum.
  4. Nucleoplasm and cytoplasm are rehydrated.
  5. Nucleolus reappears in each cell.
  6. Spindle fibres are gradually disorganised.
  7. In animal cell, astral rays disappear.

Cytokinesis :

Definition: It is the process where cytoplasm of mother cell is divided into two daughter cells after Karyokinesis. Various cell organelles (mitochondria, golgi bodies) are also distributed
between two daughter cells.

If cytokinesis does not occur after karyokinesis, then the cell will be binucleate.

In plant cell: Commonest method of cytokinesis in plant cell is known as cell plate formation. Golgi bodies (dictyosome) forms small vesicles called phragmosome containing liquid cellulose, pectin etc.

Cytokinesis in plant cell.

These vesicles line up at the middle of the cell and fusetogether to form a thin film of (phragmoplast) cell plate from the middle towards periphery of cell (centrifugal).

Gradually more cellulose, hemicellulose, pectin etc. are deposited on both sides of cell plate and a complete cell wall is formed separating the cytoplasm of two daughter cells.

In animal cell: Cytokinesis in animal cell occurs by the process of furrowing or cleavage or invagination. During this process, a cleavage furrow develops at the middle of the cell between two daughter nuclei.

Cytokinesis in animal cell. 1

The furrow or invagination gradually deepens towards the centre of the cell (centripetal) and fuse with each other. Thus the cytoplasm is divided into two equal halves.

Difference between Plant and Animal cytokinesis :

Difference between Plant and Animal cytokinesis

Difference between Mitosis of Plant cell and Animal cell :

Difference between Mitosis of Plant cell and Animal cell

Dltiorence between Karyokinesis and Cytokinesis :

Difference between Karyokinesis and Cytokinesis

Difference between Amitosis and Mitosis :

Difference between Amitosis and Mitosis

Significance of Mitosis :

Growth: By repeated mitosis, number of genetically identical cells in multicellular organism increases (known as hyperplasia ), which results into overall growth, development of organs and body of the organism.

In sexual reproduction, the first formed cell is zygote (2n) which undergoes constant mitosis to form an embryo and gradually a newborn.

Cell replacement: By the process of mitosis, old decaying dead cells new are replaced by cells. For example, the skin constantly renews, itself by shedding dead cells from the surface and generating new cells from the deeper layen In most areas of the body, the epidermis has four layers and their transition from inner layer to surface layer takes place as follows :

epidermis has four layers and their transition from inner layer to surface layer

epidermis has four layers and their transition from inner layer to surface layer 1 Repair (Healing of wound): In any injured part of the body,repajr or heaiing of wound is done by repeated mitosis of the adjoining living cells.

Regeneration: Definition— It is the Process bY which some oreanisms replace or restore lost or amputated body parts by mitosis. Regeneration, of limbs occurs widely in the animals of phylum Arthropods (eg. legs in Crustacea).

This may happen naturally during moulting or removal of leg by any injury or autotomy (Spontaneous loss of body parts). Some Star fish (Phylum—Echinodermata) can reproduce asexually by breaking an arm.

Regeneration of leg in Crab (Crustacea).

Starfish is well known for its power of regeneration of lost part of the body. Moreover, a complete new animal can grow from a small fragment (e.g. Planaria). The loss of fragment from the body may be due to injury or autotomy (as in house lizard).

Regeneration of arm in star fish.

Reproduction: So many unicellular organisms reproduce by mitosis. Mitosis helps in vegetative reproduction in plants. Asexual reproduction in plants and animals also takes place by mitosis.

Meiosis:

Definition:The process 0f rec|U(;tion division by which a diploid (2n) germ mother cell divides twice to produce four haploid (n) daughter ceils is known as meiosis.

Characteristics of meiosis :

  1. Meiosis is called reduction division because chromosome number of mother cell is reduced to half in daughter cells. Thus diploid (2n) mother cell produces haploid (n) daughter cells.
  2. The whole process of meiosis is completed in two steps of division Meiosis I (Ml) and Meiosis II (Mil).
  3. One diploid mother cell produces four haploid daughter cells.
  4. The process of meiosis primarily helps in sexual reproduction of higher plants and animals.
  5. During meiosis, by crossing over there is mixture of paternal and maternal genes that results into formation of new genotype and variation.
  6. In higher plants and animals, meiosis occurs in gonads. So meiosis is also known as germ cell division.

Occurrence: In higher animals, meiosis occurs in testis in male and ovary in female. In higher plants, meiosis occurs in pollen mother cell and egg mother cell.

Homologous chromosome :

The corresponding paternal and maternal chromosomes that are identical in structure,size (morphology), bearing same genes or their alleles and pair with each other (synapsis) during meiosis are known as homologous chromosome.

Paternal chromosome (that is inherited from father through male gamete).

Maternal chromosome (that is inherited from mother through female gamete).

The genes present in the homologous chromosome are similar—but they may be same or may not be same.

Example—Let us consider gene for human skin complexion is white and black.

In this person, both the homologous chromosomes bear white gene for skin complexion. Both the genes indicate skin complexion (similar) and whiteness (same). So, here the genes are similar as well as same.

In this person, one of the homologous chromosome bears gene for blackness and other for whiteness. However, both the genes indicate skin complexion (similar) but black and white.

Thus, here both the genes are similar but not same. These contrasting genes in homologous chromosomes are called as allele.

Sister and Non-sister chromatid :

Sister chromatids are two identical copies of a single chromosome, that are interconnected by a common centromere.

A sister chromatid referes to either of the two identical copies (chromatids) formed by the division of a single chromosome. Sister chromatids are formed during ‘S’ phase of cell cycle and remain attached by centromere.

Two sister chromatids are separated by division of centromere during anaphase stage of mitosis and during anaphase II of meiosis.

Sister and Non-sister chromatid

Nonsister chromatid refers to the corresponding paternal and maternal chromatid of the homologous pair.

So in a homologous pair of chromosome two paternal chromatids (produced from one paternal chromosome) are paternal sister chromatids whereas two maternal chromatids (produced from one maternal chromosome) are maternal sister chromatids.

However, any one paternal and any one maternal chromatid of the homologous pair called as nonsister chromatid.

Important features of Meiosis :

Separation of homologous chromosome : Homologous chromosomes (each having one pair of sister chromatids) are separated in Anaphase I of Meiosis I. Anaphase II of Meiosis II is, however, like mitotic anaphase.

Separation of homologous chromosome

During Mitosis

During Mitosis

In Metaphase of mitosis arrangement of chromatids in linear pattern on the’equatorial plate.

In Anaphase of mitosis, centomere of each pair of sister chromatid divides and the seprated chromatids move towards opposite poles.

During Meiosis I

During Meiosis I

In Metaphase I of meiosis,two pairs of chromatids of two homologous chromosomes (known as tetrad) are arranged on a particular point of equatorial plate of spindle.

In Anaphase ! of meiosis, centromere of sister chromatids does not divide and two chromauds move together towards one pole. Actual reducdon of chromosome number in meiosis occurs at this stage.

Stages of Meiosis l and Meiosis II in an Animal cell

During Meiosis II

In Metaphase II of meiosis, only one pair of chromatid is arranged on a particular point on equatorial plate of spindle.

In Anaphase II of meiosis, centromere of sister chromatids divides and two chromatids are separated. This is exactly like mitosis

Meiosis II is similar to Mitosis.

Reduction in chromosome number: During meiosis, chromosome number of mother cell is reduced to half in daughter cells. So, one diploid (2n) mother cell produces four haploid (n) daughter cells.

Reduction in chromosome number

Crossing Over: Definition- During pachytene stage of prophase I Of meiosis , nonsister chromatids of the homologous chromosome may exchange their segments (genetic material) reciprocally known as crossing over.

So, there may be mixture of paternal and maternal genes that will result into a new genotype which is partly paternal and partly maternal. However, crossing over is a matter of chance that may occur or may not.

Crossing over of nonsister chromatids results into an ‘X’ like structure called chiasma. Hence crossing over is the cause and chiasma is the effect.

crossing over

Significance of Meiosis :

Maintenance of constant chromosome number of a species—By the process of meiosis, chromosome number of higher plants and animals (who reproduce sexually) is maintained constant.

Significance of Meiosis

From the above chart, it is clear that during meiosis, chromosome number of mother cell (2n) is reduced to half in gametes (n). So, haploid sperm and ovum are formed.

During fertilization (syngamy), two haploid sperm and ovum unite together to form diploid zygote (2n). Hence in meiosis, chromosome number is reduced where as in fertilization, chromosome number is doubled.

That’s why, meiosis is regarded as a compensatory mechanism opposite to syngamy or fertilization and thus constant chrbmosome number of a species is maintained.

Production of variation in organisms—During meiosis, there may be crossing over between nonsister chromatids which results into “shuffling and reshuffling” of paternal and maternal genes. Thus new genotypic variation may occur in the organisms.

Formation of gametes—In higher plants and animals, meiosis takes place in diploid germ mother cells (2n) to produce haploid gametes (n). The process is known as gametogenesis.
Haploid male and female gametes fertilize to form diploid zygote.

Alternation of generation—Meiosis helps in alternation of generation as follows :

Alternation of generation

  1. In haplontic alternation of generation, meiosis occurs in diploid zygote to form haploid adult e.g. Spirogyra. This is called post zygotic meiosis.
  2. In diplontic alternation of generation, meiosis occurs during gametogenesis to’ form haploid gametes e.g. all higher plants and animals. This is called gametic meiosis or prezygotic meiosis.
  3. In haplodiplontic alternation’ of generation, meiosis occurs in diploid spore mother cell to produce haploid spores e.g. Fern, Moss. This is known as sporic meiosis.

Difference between mitosis and meiosis :

Difference between mitosis and meiosis

Reproduction:

Concept of reproduction (Why is reproduction necessary for an organism?)

Reproduction is the production of a new generation of individuals of the same species. It is one of the most important fundamental characteristic of living organisms. By this process they can reproduce new individuals like ownself in order to maintain the continuity of the species and also to increase in number (multiplication).

Definition: Reproduction is the process of multiplication by which living parental organisms produce offsprings of its own kind by the transfer of genes.

Importance of reproduction :

Living organisms increase in number (multiplication) by reproduction.

By reproduction living organisms maintain the continuity of generation and race.

After a certain span of life, each organism will die. To compensate this loss, there must be new birth by reproduction. Thus death rate (Mortality) and birth rate (Natality) should be balanced.

The existence of living organisms on earth will be endangered if they are incapable of reproducing new ones.

For example, Dinosaurs are extinct from earth now. Why ? There are many theories about this. But what was the exact cause—cannot be concluded.

However, one thing is obvious—that due to some reason or other Dinosaurs failed to reproduce and the eggs did not hatch. Thus due to stoppage of reproduction, Dinosaurs were perished from earth.

Modes of reproduction :

General mode of reproduction in plants and animals is asexual and sexual.

Asexual reproduction: Definition—The process by which reproduction takes place without fusion of gametes is called asexual reproduction. Asexual reproduction occurs in both plants arid animals.

For Example—In animals, asexual reproduction in hydra by budding; in plants, spore formation in moss, fern, fungi etc.

Characteristics and Importance :

  1. It is the process by which offsprings arise from a single parental organism and inherit the genes of that parent only.
  2. It does not involve the fusion of gametes.
  3. Many offsprings can be produced from an individual parent.
  4. This is a simple and easier method of reproduction.
  5. This method may help an individual to regenerate.
  6. It also helps an organism to overcome unfavourable conditions.

Sexual reproduction: Definition—The process of reproduction which involves the fusion of unicellular units called gametes to develop genetically distinct offsprings is called sexual reproduction. Sexual reproduction occurs in both plants and animals.

For Example—Pollination and fertilization in higher plants; in animals, fertilization of sperm with ovum to form zygote.

Characteristics and Importance :

  1. During sexual reproduction, there is union between haploid (n) male and female gametes resulting production of diploid zygote (2n).
  2. By this process, genetic variations or diversity may be observed among the offspring.
  3. Alternation of haploid (n) and diploid (2n) phase may occur in the life cycle.
  4. This is more efficient method of reproduction.

Vegetative Propagation (How do plants reffroduce by vegetative means?):

The reproduction by ‘which daughter plants are produced simply from the vegetative parts of a plant is called vegetative reproduction. Higher plant body consists of vegetative parts (root, stem and leaf) and reproductive parts (flowers, fruits and seeds).

Characteristics and Importance :

  1. Vegetative reproduction is simple and rapid method of reproduction. New daughter organisms are produced from the already existing forms.
  2. Daughter individuals resemble their parents.
  3. Chances of survival is more.
  4. Good quality individuals can be obtained easily from good parental varieties.
  5. Artificial vegetative reproduction helps in improvement of plants in terms of quality and quantity.

Types of vegetative reproduction :

Vegetative reproduction (vegetative propagation) or multiplication is basically a special type of asexual reproduction in plants. This may be natural or artificial.

Natural methods of Vegetative Reproduction :

It is the process of reproduction which occurs naturally by any vegetative (somatic) part of the plant body. Generally, vegetative part means root, stem and leaves in higher plants.
Some common examples are as follows—

Root: In sweet potato (Ipomea batatus), adventitious buds are
M produced from the root of the plant, that develop into new plant.

Growth of adventitious buds from the root of sweet potato

Stem: Water hyacinth (Eichhornia) is floating aquatic plant. It develops a short, thick, spongy, horizontal branch from the axil of leaves,known as offset. It is, in (Ipomea). fact, a small internode in between two consecutive nodes.

Vegetative reproduction by offset in water hyacinth

Each node bears a rosette of leaves above and a tuft of roots below. The offset often breaks away from the mother plant and develops into a separate daughter plant.

Subaerial Modified stem (A) Runner of Oxalis; (B) Stolon of Mentha; c sucker of chrysanthemum

Reproduction by underground modified stem—A. Ginger, B. Onion, C. Potato, D. 01 (Amorphophallus).

Leaf: In Bryophyllum, a series of adventitious buds’are formed on leaf margins, which usually develop into new plants.

Growth of adventitious buds

‘Artificial method of Vegetative Reproduction’:

It is the pruuÿss of reproducuon (Vegetative Propagation) where a plant part is removed from the mother plant and placed in a suitable environment so that it can grow into a whole new plant, which is genetically identical to the parent.

This process is usually practised for the reproduction (propagation) of those plants which produce either very few seeds or do not produce any viable seeds. There are various artificial methods like Cutting, Grafting, Layering, Gootee (Air layering) etc.

Cutting: It is commonest artificial method of vegetative reproduction practised by man. mature stem of a plant bearing nodes and lateral buds can be cut and planted in moist soil.

Cutting and grafting in plants.

After few days new roots develop from underground cut end and the cutting becomes a new plant. Root formation at underground cut end is stimulated by treatment Of plant hormone like IAA,IBA,NAA etc.

Now-a-days, many plants are raised by stem cutting like sugarcane, banana/pineapple, orange, grapes, rose,chinarose etc.

Gafting : Definition-It is the process of artificial method vegetative reproduction where the cutting of desired variety of plant (called scion or graft) is transplated on another rooted plant (called stock).

After few days the cut ends of two different plants of the same species) fuse by their tissues to become a single, unit. The fused unit starts forming shoots and grows into a new plant, e.g. mango, litchi, guava etc. In these plants, grafting is done for preservation of original characters of the scion parent.

Micropropagation: The production of large number of individual plantlets (genetically identical to parent) from a small piece of parental plant tissue, cultured in proper nutrient medium is called micropropagation.

Basically, micropropagation is the propagation of plants by tissue culture. This process is used for propagating plants like Orchids, Dahlia, Chrysanthemum etc.

Micropropagation of banana plant from bud culture

Outline of the process :

  1. In this process, a small piece of tissue of a desired plant is cut known as explant.
  2. The explant is placed in suitable nutrient medium (culture medium) which contains added nutrients, plant hormones particularly Auxin and Cytokinin.
  3. The tissue (explant) grows into an unorganised mass known as callus. Since tissues are cultured, it is also known as tissue culture.
  4. Small part of this tissue is put in another medium which induces formation of very small plant, called plantlets.
  5. The plantlets can be transplanted in soil or pots for further development.

Significance (Importance/Application) :

  1. This process ensures rapid production of desired varieties of plant.
  2. This method allows us to grow whole plant from cells or tissues collected from different parts of the plant body.

Alternation of generation : In the life cycle of some plants and animals, the phenomenon of alternation of two generations asexual or sporophytic or diploid and sexual or gametophytic or haploid,in a cyclic manner is called alternation of generation e.g. Fern, Moss etc.Diagrammatic life cycle of Dryopteris.

 

Example of alternation of generation from Fern :

  1. There is distinct alternation of two generations in the life cycle of Fern (Dryopteris)—asexual / sporophytic / diploid and sexual / gametophytic / haploid.
  2. The adult plant body represents sporophytic generation whereas the prothallus represents gametophytic generation.
  3. The adult sporophyte has sporangium which contains diploid spore mother cell (2n) that undergoes meiosis to produce haploid spores (n).
  4. Haploid spores (n) germinate to give rise to haploid gametophytic plant (n). Gametophyte contains male reproductive organ—Antheridium (n) and female reproductive organ—Archaegonium (n).
  5. Both Antheridium and Archaegonium produce haploid male gamete (sperm or antherozoid) and haploid female gamete (egg) respectively by mitosis.
  6. The sperm (n) and egg (n) unite together (oogamy) to form oospore (2n)—which develops into adult sporophytic plant (2n).
  7. The sporophyte and gametophyte possess different chromosome numbers. In sporophyte, it is diploid (2n) and in gametophyte, it is haploid (n). The reduction in chromosome number during spore formation marks the beginning of the gametophyte (n), whereas, the fertilization between male and female gamete restores diploid (2n) number of chromosome and initiates sporophytic generation (2n).
  8. Hence, alternation of generation in Fern is primarily operated by diploid and haploid chromosome number.

Word diagram showing alternation of generation in the life cycle of Dryopteris (Fern).

Sexual Reproduction in Flowering plants: (How does reproduction occur in flowering plants ?)

Flower Is sexual reproductive structure 6f plants, especially in angiosperms. It is modified reproductive shoot and helps in sexual reproduction of plants. It consists of thalamus on which four sets of floral whorls are arranged in orderly manner.

The arrangement of flower on the floral axis is called Inflorescence or Anthotaxy.

Different parts of a typical flower :

A typical flower consists of a thalamus with four sets of floral leaves—Calyx, Corolla, Androecium and Gynoecium that are successively arranged.

Different parts of a typical flower (Hibiscus rosasinensis).

Calyx: It is the outermost part of the flower and consists of some green leaf like structures known as sepals.

Function—Sepals protect the inner parts of flower in bud condition from heat, cold, rain etc.

Corolla: It is the second inner floral set, consisting of several leafy structures called petals. Generally they have bright colour or beautiful smell or both. Function-Petals protect the inner Androecium and Gynoecium.

Androecium: It is the third inner floral whorl consisting of a number of male reproductive organs known as stamens. Each stamen consists of a slender stalk called filament and a saclike structure at the top of the filament, known as anther (pollen sac).

The pollen sac produces pollen grains that are commonly yellowish in colour.

Function— Anther of stamen produces male reproductive units of flower called pollen grains.

Gynoecium or Pistil: It is the innermost whorl of the flower. It consists of one or more female reproductive part known as carpel. Each carpel consists of a receptive terminal portion called stigma, a long slender stalk. called style and a swollen basal portion called ovary.

Ovary contains one or more ovule and each ovule contains an egg cell.

Function- The egg cell of ovule is the female gemete that helps in fertilization. After fertilization, overy is modifies into fruit and ovule is modifies into seed.

Sexual reproduction and life cycle of mango tree

Pollination:

The flowers mature in due course of time when matured pollen grains are transferred from the anther to stigma. Fertilization does not occur until pollination takes place.

Definition
The transfer Of Pollen grains from the anther of the stamen to the stigma of carpel of the same flower or of a different flower of same species is called pollination.

A Types of Pollination :

Pollination is of two types- Self pollination or Autogamy and Cross-pollination or Allogamy.

Self-pollination of Autogamy: When the pollen grain is transferred from anther of one flower to the stigma of the same flower, it is called self-pollination. Self-pollination generally occurs in bisexual flower, e.g. Pea, Soyabean, Orchid, Sunflower, Wheat, Rice, Barley, Oats,Tomato, Potato etc.

Cross-pollination or Allogamy: When the pollen grain is transferred from anther of one flower to the stigma of another flower of a different plant (but of same species), it is called cross-pollination. Unisexual flowers are always cross-pollinated. Some common plants performing cross-pollination are apple, grapes, maize, orchid, guava etc.

Merits and demerits of self and cross-pollination:

Merits and demerits of self and cross pollination

Various agents of pollination with common example :

Agents of- pollination. The process of pollination requires agents (pollinators) that carry or move the pollen grains from the anther to stigma. Some common agents of pollination in flowers are—

  1. Biotic pollination (by organisms)—Pollination by insects, birds, bats, etc.
  2. Abiotic pollination (by nonliving substances)—Pollination by water, air etc.

Fertilization and Development of a Mew plant : (How is a new plant produced ?)

The process of union between two dissimilar haploid gametes (anisogametes) is known as fertilization or syngamy,

During fertilization male gamete (n) and female gamete (n) unite together to form diploid zygote (2n). The zygote undergoes repeated mitosis to form an embryo (2n) which gives birth to the adult plant after germination.

The sequence of events can be explained as follows :

Polination-ln this process pollen grains are transferred from anther to stigma. There is a method of selection by which the stigma selects a particular pollen grain of matchingtype.

Formation of male germ cell—Pollen grain produces haploid male gamete which passes down the style through the pollen tube. Male gamete enters the ovule in the ovary.

Formation of Female gamete—Haploid female gamete or egg is formed in the embryo sac of the ovule.

Fertilization –The haploid male gamete (n) unites with the haploid egg (n) in the embryo sac of ovule to form a diploid zygote (2n).

(A-G) Different stages and prpcess of fertilization in angiosperm (flowering plant)

Division of zygote to form embryo—The diploid zygote undergoes repeated mitotic division to form diploid embryo within the ovule.

Formation of fruit and seed—After fertilization inner ovule is gradually modified into seed and outer ovary is modified into fruit.

Formation and extension

Seed contains the futyre plant—Seed contains the embryo which is the future plant. Under appropriate conditions (in presence of proper quantity of water, air and sunlight), the embryo germinates into a seedling which grows into complete plant in course of time.

Growth And Development

Concept of growth (How does an organism grow ?) *

Growth Is a fundamental characteristic of all living organisms. This process Is complex In nature and regulated by protoplasm. Growth Is expressed by a permanent irreversible increase in length and volume.

A dry cork when soaked In water Increase In volume due to absorption of v/ater but it returns to original state on dehydration. Hence, the volume of cork v/as Increased temporarily which should not be referred to as growth.

Moreover, the dry weight of cork has never been changed. Here the cork Is formed of dead cells and the increase in volume is not controlled by protoplasm. In facts, most conspicuous sign of growth is an increase in dry weight accompanied by an increase in size and volume of the living o ganism.

Growth is an outcome of some metabolic processes, some of which are anabolic and some are catabolic In nature. When anabolism exceeds over catabolism, then growth takes place. Once growth occurs in a living body, it can not be reversed.

Development:Definition: Development can be defined as the increase in series of progressive, nonrepetitlve, organized changes complexity) through cellular differentiation and growth by which multicellular living organism is formed from the zygote.

Phases of growth :

The life of a multicellular organism starts from zygote (2n) which is formed bythe union of haploid male and female gamete.

The zygote grows through three distinct phases as follows:

Cell division—It is the process of increase in cell number (hyperplasia) as a result of repeated mitotic cell division. In plant body, meristematic cel! divide mitotically to form innumberable daughter cells.

In animals, zygote divides to produce bunch of cells (moruia) which form biastula, gastruia and other progressive embryonic stages.

Location of different meristematic tissue in plant.

Cell enlargement—It is the process by which daughter cells produced by mitosis enlarge both in length and volume (hypertrophy). This is mainly due to increased rate of anabolism than catabolism. The cells take in more water resulting into higher turgor pressure. At the same time,’various biomolecules are synthesized in the cells (anabolism) resulting into increase in dry weight of protoplasm

Cell differentiation—In this phase, full I grown cells are genetically differentiated to form various tissues that carry out diverse functions. So, during cell differentiation, a common type of cell becomes more specialized for a specific function.

A. Hypertrophy animal cell b.steps of plant cell growth

In plants and animals, growth of the somatic organs is called somatic growth and that of reproductive organ is called reproductive growth. Somatic growth in plants is the growth of root,stem, leaf etc. whereas in animals, somatic growth is the growth of heart, lung, liver, muscle m and all other organs.

Cell differentiation after origin from common mother cell

Reproductive growth in plants is the growth of floral bud, flower, fruit, seed animals reproductive growth signifies growth of testis in male and ovary in female

Several factors influence growth in both plants and animals—c.g. Internal factors {nutrition, enzyme, hormone, genes) and External factors (water, food, etc.)

Phase of human development :

There are five phases of human development—

  1. Infancy
  2. Childhood
  3. Adolescence (puberty)
  4. Adulthood
  5. Senescence.

Infancy—

  1. An infant in the very young offspring of man, usually since birth upto 2 years, unable to walk properly, early stage of growth and develop-ment.
  2. A great deal of initial learning occurs at this stage through parental behaviour and surrounding environment.
  3. They depend on adult to care for them.
  4. They use all sense organs for seeing, hearing, tasting, smelling and touching.
  5. They begin to sit up, crawl and walk.
  6. All the organs of the body grow at a very fast rate at this phase.

Childhood —

  1. Generally a child is of age 3-12 years of which 3-6 years is known as early childhood and 7-12 years is called middle age childhood.
  2. Tremendous growth of neural fibres occurs’in brain,
  3. At 5 years of age, the children speak properly with correct coordination of nervous system.
  4. Large muscle for walking,running and other physical activities develop well.
  5. Many psychological development start and the child repeatedly asks “why” ?

Phases of human development.

Adolescence—

  1. The term adolescence is derived from Latin word ‘Adeloscere’, which which means to and grow maturity .It is critical stage of development which lies between later childhood and maturity.(emerging from critical childhood stage and of merging development into adulthood).
  2. This stage is usually at the age of 12-18 years. However, there are differences due to sex, climate, genes and individual constitution.
  3. There is very rapid growth at this stage but rata of growth differ in boys and girls.
  4. Significant increase in growth of reproductive organs is observed.
  5. Many biological, psychological and emotional characteristics develop like sex-consciousness,self-consciousness, imaginative activity etc. They often focus on friendship and romantic relationship. A|! these chanSes are called secondary sex characters.
  6. Spermatogenesis in male and oogenesis in female start at this stage—so the reproductive system becomes functionally active (Sexual maturity).
  7. Masculine characters in male and feminine characters in female become prominent.

Often, puberty and adolescence are used synonymously. However, puberty refers to physical changes adolescence ‘s Period of psychological and social transition between childhood and adulthood. puberty is the process of bodily changes by which adolescent reaches sexual maturity.

Adulthood—

  1. Biologically, an adult is a human being who has achieved sexual maturity ana becomes independent and self sufficient.
  2. On an average, the age is 20 years + 1
  3. All primary and secondary sex characters are very well developed in both male and female with high level of secretion of GTH.
  4. Allsomatic organs of the body are also fully developed.

Senescence (Late adulthood)—

  1. The word senescence has been derived from the Word ‘senescere’ meaning ‘to grow old’. can reÿer t0 cellular senescence or senescence body.
  2. It is the natural condition or process of gradual deteripra«on of body’s structure and function due aging/ when the cell looses itsÿ power of division and growth. Hence, senescence is the outcome of aging i.e. aging is the cause and senescence is the effect.
  3. This process occurs generally sfter 60 years of age.
  4. Senescence ultimately leads to death.