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Principles of Inheritance and Variation Test - 4

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Principles of Inheritance and Variation Test - 4
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  • Question 1
    1 / -0

    Johannson in 1903 formulated the concept of pure lines while working with Phaseolus vulgaris. Which of the following statements is true with reference to pure lines?

    Solution

    Pure line varieties are homozygous or homogeneous as they are genetically similar and true breeding, they produce only one type of gametes. Therefore, they, by selfing or inbreeding, give rise to progeny like themselves in successive generations.

  • Question 2
    1 / -0

    When a cross is made between the pure red-flowered and pure white flowered Mirabilis jalapa plants that yield hybrid plants with pink flowers, the phenotypic F2 ratio is found to be 1 : 2 : 1. This is a characteristic principle of

    Solution

    A cross between a plant pure for red flowers and a plant pure for white flowers yields hybrid plants with pink flowers in F1 generation. Neither red nor white is completely dominant.
    When two of the hybrid plants with pink flowers are crossed, the F2 generation includes plants with red, pink and white flowers.
    In the F2 generation of such cross, the phenotypic ratio differs from the Mendelian ratio, also being 1 : 2 : 1 instead of 3 : 1 (red : white).
    Hence, the phenotypic F2 ratio of 1 : 2 : 1 is a characteristic of incomplete dominance.

  • Question 3
    1 / -0

    Comb types (single, rose, pea and walnut) of chickens follow supplementary gene interaction, where F2 generation shows segregation in the ratio of 9 walnut-combed : 3 pea-combed : 3 rose combed : 1 single-combed, when the 2 walnut-combed F1 individuals are mated. Such a cross illustrates the kind of gene interaction when

    Solution

    When the pea-combed and the rose-combed chickens are crossed, all the offspring in F1 generation have an entirely new type of comb that resembles half a walnut kernel (walnut comb).
    When the 2 walnut-combed F1 individuals are mated, offspring in F2 generation show a ratio of 9 walnut-combed : 3 pea-combed : 3 rose-combed : 1 single-combed.
    The walnut-combed is thus not a unit character, but merely a phenotype of pea and rose when they act together.
    Hence, the combination of the alleles P and R respectively give pea and rose phenotypes due to their own simple dominance.

  • Question 4
    1 / -0

    Mendel studied the breeding experiment dealing with two characters at the same time (form of the seed and colour of the cotyledons), which led him to formulate the

    Solution

    Mendel's law of independent assortment states that the alleles of two (or more) different genes get sorted into gametes independently of one another. In other words, the allele a gamete receives for one gene does not influence the allele received for another gene.

  • Question 5
    1 / -0

    Hymenopterous insects show a unique phenomenon called haplodiploidy, where the sex of the offspring is determined by the

    Solution

    Hymenopterous insects, such as bees, wasps, saw flies, and ants, show a unique phenomenon in which an unfertilised egg develops into a male and is haploid (N) and a fertilised egg develops into a female and is diploid (2N). Eggs are formed by meiosis and sperms by mitosis.
    Fertilisation restores the diploid number of chromosomes in the zygote which gives rise to the female.
    Hence, the sex is determined by the number of sets of chromosomes an individual receives.

  • Question 6
    1 / -0

    The principle of linkage was put forward as a regular concept by Morgan in 1910 from his work on Drosophila melanogaster and noted that in a di-hybrid test cross involving complete linkage of genes, the frequency of

    Solution

    When a dominant female Drosophila homozygous for red eyes and normal wings is crossed with a double recessive male Drosophila homozygous for purple eyes and vestigial wings, the F1 flies are all heterozygous red-eyed and normal-winged.
    F1 female fly is then test crossed with a double recessive male having purple eyes and vestigial wings.
    F2 comprises only two types of flies: red-eyed, normal-winged and purple-eyed, vestigial-winged in the ratio of 1 : 1.
    No recombinant types are formed because linkage is complete and no crossing over occurs.

  • Question 7
    1 / -0

    Consider a monohybrid cross in pea plants, where a pure tall pea plant is crossed with a pure dwarf pea plant. All of the F1 offspring from the cross are tall. What would be the percentage of genotypes and phenotypes if one of the tall F1 plants is crossed with a pure dwarf pea plant?

    Solution

    A back cross between F1 heterozygote and a homozygous recessive parent produces offspring with both the genotypes (50% heterozygous tall and 50% homozygous recessive dwarf).

      T t
    t Tt tt
    t Tt tt


    Back cross progeny Tt, Tt (50% heterozygous tall) and tt, tt (50% homozygous recessive dwarf)

  • Question 8
    1 / -0

    Identify the characteristic feature pertaining to X-linked inheritance which is not true.

    Solution

    Only homozygous females express an X-linked recessive trait. Heterozygous females are carriers of X-linked recessive trait and themselves do not express the trait.

  • Question 9
    1 / -0

    Protanopia and deuteranopia have been described by Horner (1876) and are diseases that are more often found in man. These are the examples of

    Solution

    The red colour blindness is called protanopia and the green blindness is deuteranopia. X chromosome possesses a normal gene which controls the functioning and formation of colour sensitive cells in the retina.
    Its recessive allele fails to control the functioning of colour sensitive cells that results in colour blindness.
    Hence, colour blindness is an X-linked disease and a single gene in man results in disease colour blindness, whereas a woman needs two such genes for the same.

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