Principles Of Inheritance And Variation Class 12th Biology CBSE Solution

Class 12th Biology CBSE Solution

Exercises
Question 1.

Mention the advantages of selecting pea plant for experiment by Mendel.


Answer:

The major advantages of selecting pea plant for experiment by Mendel are as follows:

i. Pea plant has 7 contrasting characteristics which can be phenotipically distinguished.


ii. Every contrasting character has only 2 alleles and none of them are linked.


iii. The plant undergoes self pollination thus easily produces the offsprings with same traits generation after generation.


iv. Due to self pollination homozygous plants can be obtained easily.


v. Cross pollination for experimentation can be easily performed by emasculating the stamen of the flower without affecting the flower.


vi. They have a short life span and produce many seeds at a time.



Question 2.

Differentiate between the following –

Dominance and Recessive


Answer:

The differences between dominance and recessive are:



Question 3.

Differentiate between the following –

Homozygous and Heterozygous


Answer:

The differences between homozygous and heterozygous are:



Question 4.

Differentiate between the following –

Monohybrid and Dihybrid.


Answer:

The differences between monohybrid and dihybrid are:



Question 5.

A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?


Answer:

If a diploid organism is heterozygous for 4 loci the different types of gametes that can be produced can be calculated by the formula 2n, where “n” is the number of loci. Thus,

n = 4, therefore the number of gametes produced are 2n = 24 = 16 types.


Therefore a diploid organism is heterozygous for 4 loci produces 16 different types of gametes.



Question 6.

Explain the Law of Dominance using a monohybrid cross.


Answer:

In a monohybrid cross, for any character, the F1 generation individual derived from crosses between the two different varieties having alternative characters, showed only one of the character and never the other. This feature was expressed as dominance of one trait over the other. The trait which appeared in the F1 generation was called the dominant and the other which did not appear in the F1 population was called recessive. It is obvious that though, in F1 generation the dominant phenotype appears, the recessive is not lost but re appears in the F2 generation. This suggests that there is no blending of Mendelian factors in the F1 generation but they stay together and only one is expressed.

Therefore, two or more forms of a single character can exist in a single gene locus of a homologous chromosome within a species that may put different effects on the phenotype of an organism. In the hybrids between two individual displaying different phenotypes only one character is observable. This phenotype or the character (allele) which is expressed in the hybrid is said to be dominant and the other to be recessive, whose phenotype remains masked in heterozygous condition, but is only expressed in homozygous condition. This is known as the law of dominance.



Here the character tall is the dominant trait whereas dwarf being the recessive one.



Question 7.

Define and design a test-cross.


Answer:

A test cross can be defined as the cross between the offsprings of any generation with its recessive parent so as to determine the composition of the offspring. The progenies of such a cross can be easily analysed to predict the genotype of the test organism.



Question 8.

Using a Punnett Square, workout the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus.


Answer:

The phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus, say height, is as follows:



Question 9.

When a cross in made between tall plants with yellow seeds (TtYy) and tall plants with green seed (Ttyy), what proportions of phenotype in the offspring could be expected to be

(A) Tall and green.

(B) Dwarf and green.


Answer:

If we design the cross between the tall plants with yellow seeds (TtYy) and tall plants with green seed (Ttyy) the proportion can be determined thus,



Question 10.

Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in F1 generation for a dihybrid cross?


Answer:

If the two loci of characters are linked then the distance separating them will be very small and recombination will not be able to take place in between them. Thus no segregation can be observed. As the parent is heterozygous thus both dominant and recessive alleles are present on the two homozygous chromosomes and one chromosome is obtained by the gamete. Thus, in the F1 generation for a dihybrid cross, if this gamete contains the dominant traits and is fertilized by either dominant or recessive alleles of the same trait from the other parent the dominant phenotypic character are observed. The recessive linked phenotype of the traits are only observable only when the both the gametes from the parents that are fertilized contains the recessive trait. Also due to no segregation between the two linked traits, they are always found together in both the cases.



Question 11.

Briefly mention the contribution of T.H. Morgan in genetics.


Answer:

The contributions of T.H. Morgan are:

i. Morgan was the first to provide experimental verification for the chromosomal theory of inheritance by his work on Drosophila melanogaster.


ii. He defined linkage between two genes located close together which are always inherited together.


iii. The identified linkage to be of two types tightly linked genes, where both the genes are passed onto the next generation and loosely linked genes in which recombination may take place due to a large distance present within the two genes.


iv. He also defined the term recombination of non – parental gene recombination.


v. His findings of linked genes pavemented the path for genome mapping that is done today.


vi. He also contributed to the understanding of sex – linked inheritance


vii. He has also worked on mutations.



Question 12.

What is pedigree analysis? Suggest how such an analysis, can be useful.


Answer:

Pedigree analysis is the process by which heritability of certain characteristic features in families can be determined. Thus, pedigree analysis is the practice of analysing inheritance pattern of traits in human beings. It is represented in a family tree over generations.

In human genetics, pedigree study provides a strong tool, which is utilized to trace the inheritance of a specific, abnormality or disease. Also, alterations, known as mutations, that are inherited can be identified. It is also used to know the possibility of passing of a heritable disease from parents to offsprings and also used for genetic counselling to avoid diseases in children.



Question 13.

How is sex determined in human beings?


Answer:

Humans have 46 chromosomes thus they have 23 pairs of chromosomes. Of these 23 pairs 22 pairs are autosomes and the remaining 1 pair is the sex chromosome that determines the sex of the child. In humans sex chromosome is of two types X and Y. The genetic composition of females is 44+XX whereas males have a genetic composition of 44+XY. As humans are diploid organisms and during fertilization the child receives half of its chromosomes from one parent and the other half from another the diploid state is restored. Now, the gametes formed by the female is all 22+X where as males produce two types of gametes, 22+X and 22+Y. During fertilization if the female gamete with 22+X chromosome is fertilized by the sperm bearing 22+X chromosome the resulting zygote has the genetic composition as 44+XX thus giving rise to a female child. But, if the female gamete with 22+X chromosome is fertilized by the sperm bearing 22+Y chromosome the resulting zygote has the genetic composition as 44+XY thus giving rise to a male child. Thus the sex of the child is determined by the type of sperm that fertilizes the ovum and not the other way round.



Question 14.

A child has blood group O. If the father has blood group A and mother blood group B, work out the genotypes of the parents and the possible genotypes of the other offsprings.


Answer:

If a child has blood group O it has the genotype of “ii” which is recessive to both blood group A (IA) and blood group B (IB). As we know recessive traits only occur in homozygous condition thus both the parents must have the recessive allele making them heterozygous for their blood group. Thus the following punnett square can determine the possible genotypes and phenotypes of the other offsprings.



Question 15.

Explain the following terms with example

Co-dominance


Answer:

Sometimes both the alleles of genes in a heterozygote lack the dominant and recessive relationship, that is, each allele is capable of some degree of phenotypic expression. In a sense, co-dominance is no dominance at all, the heterozygotes showing the phenotypes of both homozygotes. Hence heterozygote genotype gives rise to a phenotype, distinctly different from either of the homozygous genotypes of parent.

Example: The cote colour of the short horn breed of cattle represents a classical example of co-dominance. When cattle of red coat (CRCR) are crossed with cattle of white coat (CwCw), the F1 heterozygote is found to posses roan coat (CRCw). In roan coat the red and white hair occurs in definite patches but no hair a intermediate colour of red and white. In such a case the F2 phenotypic ratio and genotypic ratio are the same as follows:




Question 16.

Explain the following terms with example

Incomplete dominance


Answer:

Sometimes in the heterozygote, dominant allele does not completely mask the phenotypic expression of recessive allele and there occurs an intermediate phenotype in the heterozygote. This is called incomplete dominance.

Example: When a red – flowered 4 o’ clock plant (RR) is crossed with a white – flowered (rr) 4 o’ clock plant the F1 hybrid 4 o’ clock plants were observed to have pink flowers (Rr). It shows that the genes for red colour could not completely dominate the genes for white colour. In such a case the F2 phenotypic ratio and genotypic ratio are the same as follows:




Question 17.

What is point mutation? Give one example.


Answer:

Point mutation is often caused by chemicals or malfunction of DNA replication where exchange of a single nucleotide occurs. For example, sickle cell anaemia where defect is caused by the substitution of a single amino acid Glutamic acid by Valine at the sixth position of the β globin chain of the haemoglobin molecule.



Question 18.

Who had proposed the chromosomal theory of the inheritance?


Answer:

The chromosomal theory of inheritance was independently proposed by Walter Sutton and Theodore Boveri in 1903 and 1902 respectively. The experimental verification of this theory was put forth by Thomas Hunt Morgan in 1915 by his work on Drosophila melanogaster.



Question 19.

Mention any two autosomal genetic disorders with their symptoms.


Answer:

Sickle cell anaemia: This is an autosomal linked recessive trait that can be transmitted from parents to offsprings only when both the parents are the carrier for the gene governing the trait. Thus only homozygous recessive individuals show the diseased phenotype. This defect is caused by the substitution of the amino acid Glutamic acid by Valine at the sixth position of the β globin chain of the haemoglobin molecule.

Symptoms: The different symptoms of sickle cell anaemia are- rapid heart rate, breathlessness, weakness, excessive thirst, chest pain, prone to diseases such as jaundice, delayed growth and puberty and decreased fertility.


Down’s Syndrome: It was first discovered and published in 1866 by a British physician John Langdon Brown. This is a trisomic condition of chromosome 21 where an extra copy of chromosome 21 is found. It is the result of asymmetrical division in meiosis of the parents.


Symptoms: They are characteristically short; have protruding, furrowed tongue; short and broad hands with fingers showing characteristic palm and fingerprint patterns. Physical psychomotor and mental development is retarded and poor muscle tone is characteristic. Children affected by Down’s syndrome are prone to respiratory diseases and heart malfunctions and they show incidence of leukemia approximately 20 times higher than the normal population. Its estimated occurrence is approximately 1 in every 800 live births.


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