NCERT Solutions for Class 12 Biology Chapter 5: Principles of Inheritance and Variation

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

    Ans.Mendel selected pea plants for his experiment due to following reasons:
    (i) The plants have short life span of 3 months. Thus, gives results within 3 months. It is easy to grow and maintain.
    (ii) The flowers of this plants are bisexual. Therefore it undergoes self pollination and produces true breeding varieties.
    (iii) It is very easy to perform cross pollination.
    (iv) The different physical characteristics were easy to recognize and study

    2. Differentiate between the following:
    (a) Dominance and Recessive
    (b) Homozygous and Heterozygous
    (c) Monohybrid and Dihybrid.

    Ans. (a) The difference between Dominance and Recessive me:

    Dominance Recessive
    1. When a factor (allele) expresses itself in the presence or absence of its recessive factor called dominance.
    It can express itself only in the absence of its dominant factor or allele.
    1. It forms a complete functional enzyme that perfectly express it.
    It forms a incomplete defective enzyme which fails to express itself when present with its dominant allele, i.e., in heterozygous condition.

    (b) Differences between Homozygous and Heterozygous individuals:

    Homozygous Heterozygous
    1. It is pure of trait and breeds true i.e., give rise to similar homozygous individuals.
    It is seldom pure and produces offspring of different genotype.
    1. Both the alleles of traits are similar i.e., it is either dominant (TT) or recessive (tt)
    It carries dissimilar alleles i.e., (Tt).
    1. It produce one type of gametes either ‘T’ or ‘t’ type only not both.
    It produces two types of gametes containing ‘T’ and ‘t’.


    Monohybrid Dihybrid
    1. It is across between two pure organisms in order to study the inheritance of a single pair of allels.
    It is across between two pure organisms of species in order to study the inheritance of two pair of alleles belonging to two different characters.
    1. It produces phenotypic ratio 3 : 1 and genotypic ratio 1 : 2 : 1 in F2 generation.
    It produces phenotypic ratio 9 : 3 : 3 : 1 and genotypic ratio 1 : 2 : 2 : 2 : 4 : 2 : 1 :

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

    Ans. Type of gametes produced by an organism = 2n
    Where n = number of loci for which organism is heterozygous
    Given n = 4
    Then, types of gametes produces = 24 = 16

    4. Explain the Law of Dominance using a monohybrid cross.

    Ans. Mendel’s law of dominance: It states that a dominant alleles expresses itself in a monohybrid cross and suppresses the expression of recessive allele. However, this recessive allele for a character is not lost and remains hidden or masked in the progenies of F1 generation and reappears in the next generation.
    When pea plants with round seeds (RR) are crossed with plants with wrinkled seeds (rr), all seeds in F1 generation were found to be round (Rr). When these seeds (Rr) are self fertilised, both the round and wrinkled seeds appeared in F2 generation in 3 : 1 phenotypic ratio. Hence, in F1 generation, the dominant character (wrinkled seeds) got suppressed, which reappeared in F2 generation.

    Dominance using

    5. Define and design a test-cross.

    Ans. Test cross: It is a cross between an organism with unknown genotype and a recessive parent. It is used to determine whether the individuals homozygous or heterozygous for tract.

    Test cross helps in establishing hetero/ homozygosity of dominant trait.

    6. 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 single locus.

    All tall
    Genotype ratio: TT : Tt 2 : 2 or 1 : 1

    7. When a cross is made between tall plant with yellow seeds (TtYy) and tall plant with green seed (Ttyy), what proportions of phenotype in the offspring could be expected to be:
    (a) Tall and green.
    (b) Dwarf and green.

    Ans. Cross between tall plant with yellow seeds (TtYy) and tall plant with green seed (Ttyy) is given as:


    $$\\{Phenotypes: Tall and green = 3}\\{Tall and yellow = 3}\\{Dwarf and green = 1}\\{Dwarf and yellow = 1}\\(a)\space Tall and green proportion\\ =\frac{Tall and green phenotype}{Total phenotype}\\ =\frac{3}{3+3+1+1} =\frac{3}{8}\\ (b)\space Dwarf and green proportion\\ =\frac{Dwarf and green phenotype}{Total phenotype}\\ =\frac{3}{3+3+1+1} =\frac{3}{8}$$

    8. 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?

    Ans. Parents AaBb × AaBb

    Gametes AB, ab AB, ab

    Ab, aB Ab, aB

    Gametes Ab, aB are not possible as 2 loci are linked i.e., a dominant and recessive character does not come together.
    F1 generation

    dihybrid cross

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

    Ans. Thomas Hunt Morgan (1866-1945), an American geneticist and Nobel Prize winner of 1933, is considered as “Father of experimental genetics” for his work on and discovery of linkage, crossing over, sex linkage, criss cross inheritance, linkage maps, mutability of genes, etc. He is called fly man of genetics because of selecting fruit fly (Drosophila melanogaster) as research, material in experimental genetics. It was largely due to his book, “The Theory of Gene”, that genetics was accepted as a distinct branch of biology. In 1910, he discovered linkage and distinguished linked and unlinked genes. Morgan and Castle (1911) proposed “Chromosome Theory of Linkage” showing that genes are located on the chromosomes and arranged in linear order. Morgan and Sturtevant (1911) found that frequency of crossing over (recombination) between two linked genes is directly proportional to the distance between the two. 1% recombination is considered to be equal to 1 centi Morgan (cM) or 1 map unit. He worked on sex linked inheritance and reported a white eyed male Drosophila in a population of red eyed and proved that gene of eye colour is located on X-chromosome. The male passed its genes on X-chromosomes to the daughter while the son gets genes on X-chromosome from the female (mother): It is called criss-cross inheritance.

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

    Ans.Pedigree analysis: An analysis of inheritance of a trait over a several generations of a family is called, pedigres analysis.
    (i) It helps in analysis of transmission of character in family over generation.
    (ii) It helps in genetic counselling to avoid disorders children.
    (iii) To identify whether a particular genetic disease is due to recessive gene or a dominant gene.
    (iv) To identify the possible origin of the defective gene in the family or in a population.

    11. How is sex determined in human beings?

    Ans. In human sex is determined by a pair of sex chromosome XX in female and XY in male.

    Thus, the sex of body is determined by the father and possibilities of male or female child is 50% depending on which gamete is going to fuse.

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

    Ans. Father blood group = A, possible sets of gametes = IAIA or IAi
    Mother blood group = B, possible sets to gametes = IBIB or IBi
    Child blood group = O, possible set of gamete = ii
    Case I: When we cross homozygous father of blood group A (IAIA) with homozygous mother of blood group B (IBIB), the results will be:

    In this, all offspring having blood group AB. So genotype of parents is not possible because in question it is mentioned that one of child is having blood group O.
    Case II: When we cross heterozygous father of blood group A(IA i) with heterozygous mother of blood group B(IB i) the results will be:

    Offspring having all blood groups i.e., A, B, AB and O.
    Thus, the Possible genotypes of the parents will be IAi (Heterozygous) and IBi (Heterozygous mother) and the possible genotypes of the other offsprings will be IAi (A-blood group), IBi (B-blood group).

    13. Explain the following terms with example:
    (a) Co-dominance
    (b) Incomplete dominance

    Ans. (a) Codominance: It is a type of inheritance in which two alleles of the same gene expresses equally as in parents, called codominance.
    Example: ABO blood group in human.

    Genotype Blood group
    I A I A A
    I A i A
    I B I B B
    I B i B
    I A I B AB
    i i O

    (b) Incomplete dominance: In this, the genes of allelomorphic pairs are expressed as dominant and recessive by express them selves partially when present together in hybrid.
    Example: Flower colour in Mirabilis jalapa

    All pink colour flower

    14. What is point mutation? Give one example.

    Ans. Point Mutation: Point mutation is a change in single base pair of DNA substitution, delation or insertion of a single nitrogenous base.
    Example: Sickle cell anaemia. In this, the glutamic acid in short arm of chromosome get replace with valine at the sixth position.

    15. Who had proposed the chromosomal theory of the inheritance?

    Ans. Theodor Boveri and Walder Sutton are the scientists who proposed the chromosomal theory of the inheritance in 1903. They liked the inheritance of traits to the chromosomes.

    16. Mention any two autosomal genetic disorders with their symptoms.

    Ans. Sickle cell anaemia: Haemoglobin has less 02 transport, sickle shaped RBCs etc.
    Phenylketonuria: Mental retardation (due to accumulation of phenylalanine in brain), hypopigmentation of skin and hair, eczema etc.

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