All Solutions

Important

Tuesday, March 15, 2022

NCERT Class 10 Science Book Activities Solutions Chapter 9

 In this post, you will find NCERT Class 10 Science Book Activities Solutions Chapter 9 which is completely based on Chapter 9 Heredity and Evolution


NCERT Class 10 Science Book Activities Solutions Chapter 9

S.N.

Contents

1

Activity 9.1 Science Chapter 9

2

Activity 9.2 Science Chapter 9

 

 

Chapter 9 Heredity and Evolution Activity 9.1

Activity - Observe the ears of all the students in the class. Prepare a list of students having free or attached ear lobes and calculate the percentage of students having each




        Find out about the ear lobes of the parents of each student of the class. Correlate the ear lobe type of each student with that of their parents. Based on this evidence, suggest a possible rule for the inheritance of ear lobe type.

Chapter 9 Heredity and Evolution Activity 9.1Chapter 9 Heredity and Evolution Activity 9.1



Discussion – The presence of free and attached ear lobes in an example of an insignificant variation of human population.

  Free ear lobe – The lowest part of ear (ear lobe) is free from the side of the head.

Attached ear lobe – The lowest part of ear lobe is closely attached to the side of the head.

When you observe the ear lobes of your classmate you would find that majority  of the students have free ear lobes. Some have attached ear lobes as well.

When the ears of the parents of students having free ear lobes are observed its is revealed that either both or atleast one of their parent must have free ear lobe. For students having attached ear lobes their parents may have free or attached ear lobes.

  Like  any other trait inheritance of ear lobe is controlled by a gene. For the trait of ear lobe type, there are two alternate forms. Free ear lobe gene is dominant over the attached ear lobe gene. If we give a symbol of ‘D’ to the free lobe gene and ‘d ’ to the attached ear lobe gene, then the following three genotypes are possible in the human population.

     1.   Genotype – DD [for free ear lobe]

     2.   Genotype Dd [for free ear lobe and D is dominant over d]

     3.   Genotype dd [for attached ear lobe]

Now on the basis of the above genotypes, there may be the following conditions in its inheritance

(i)    If father is Dd and mother is Dd

 

D

d

D

DD

Dd

D

DD

Dd

 

In the above chart, all the children have free ear lobes because D is dominant.

(ii).  If both parents are Dd

 

D

d

D

DD

Dd

d

Dd

dd



Three children have free ear lobes while one will have an attached ear lobe.

(iii). If both parents are dd then their children will have attached ear lobe only.


Chapter 9 Heredity and Evolution Activity 9.2

 

 Activity -In the figure shown here, what experiment would we do to confirm that F2  generation did in fact have 1 : 2 : 1 ratio of TT, Tt and tt trait combination.

Chapter 9 Heredity and Evolution Activity 9.2

Discussion – In a monohybrid cross the F2 ratio is 3 : 1. It means 3 plants are tell and 1 is dwarf. This ration is called phenotypic ration or monohybrid ratio.

 To know whether a tall plant is true breeding or not or  to know that a plant is pure or hybrid we do Test cross.

 Test cross is a cross in which a plant of unknown genotype is crossed with a recessive parent.

 

If the unknown plant is pure (TT) and it is crossed with the recessive plant (tt) , we will get  all tall plants

 

 

t

t

T

Tt

Tt

T

Tt

Tt

 

If the unknown plant is a hybrid(Tt)  and it is crossed with a recessive plant(tt), we will get 50% tall plants and 50% dwarf plants.

 

t

t

T

Tt

Tt

t

tt

tt

 

 So with the help test cross, we can prove that phenotype 3:1 ratio is genotypic ratio 1:2:1.

so these activities are based on Chater 9 Heredity and Evolution.

 Related searches

1. Solution of Chapter 9 

2. Extra Questions  of Chapter 9