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NCERT Solutions · Class 12 Biology Principles of Inheritance and Variation

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Exercises 4.1–4.10 (part 1 of 2)

  1. Exercise 4.1

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

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    The garden pea gave Mendel clear-cut pairs of contrasting traits in a single species, and full control over pollination — the two things his method needed.
    Seven pairs of contrasting characters were available: stem height (tall/dwarf), flower colour (violet/white), flower position (axial/terminal), pod shape (inflated/constricted), pod colour (green/yellow), seed shape (round/wrinkled) and seed colour (yellow/green).
    Each character was manifested as two opposing traits with no in-between form, so every plant could be scored without doubt.
    True-breeding lines were available: a line that has undergone continuous self-pollination shows stable trait inheritance for several generations. Mendel selected $\displaystyle 14$ such varieties, as pairs alike except for one character.
    The pea flower can self-pollinate, yet artificial cross-pollination is easy, so Mendel could decide the parentage of every cross himself.
    The plants allowed a large sampling size, which gave greater credibility to his data and let him apply statistical analysis and mathematical logic.
    Results could be confirmed on successive generations (F1, F2, F3, F4), proving the results were general rules and not chance.

  2. Exercise 4.2

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

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    (a) Dominance and Recessive
    Dominant — in a dissimilar pair of factors, the member that expresses itself and masks the other; e.g. T (tall) shows in the heterozygote Tt.
    Recessive — the member that stays hidden in the heterozygote and is expressed only in the homozygous condition (tt), reappearing in the F2 as $\displaystyle 1$/$\displaystyle 4$ of the plants.
    At the level of the gene product, the unmodified allele making the normal functional enzyme is the dominant one; the modified allele making a non-functional enzyme, or no enzyme, is generally the recessive one.
    (b) Homozygous and Heterozygous
    Homozygous — the two alleles at a locus are identical (TT or tt); such a plant is true-breeding and forms only one kind of gamete.
    Heterozygous — the two alleles are dissimilar (Tt); the plant shows only the dominant phenotype but forms two kinds of gametes, T and t, in equal proportion.
    Externally TT and Tt cannot be told apart — a test cross is needed.
    (c) Monohybrid and Dihybrid
    Monohybrid — a hybrid heterozygous for the gene controlling one character (Tt). The cross TT × tt is a monohybrid cross; F2 gives a phenotypic ratio \(\displaystyle 3:1\) and a genotypic ratio \(\displaystyle 1:2:1\).
    Dihybrid — a hybrid heterozygous for two characters (RrYy). The cross RRYY × rryy is a dihybrid cross; F2 gives a phenotypic ratio \(\displaystyle 9:3:3:1\).
    A monohybrid F1 makes $\displaystyle 2$ kinds of gametes; a dihybrid F1 makes $\displaystyle 4$ kinds (RY, Ry, rY, ry), each $\displaystyle 1$/$\displaystyle 4$ of the total.

  3. Exercise 4.3

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

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    $\displaystyle 16$ types of gametes.
    At each heterozygous locus the two alleles segregate, so a gamete receives one or the other — $\displaystyle 2$ possibilities per locus.
    The loci assort independently of one another, so the possibilities multiply: \(\displaystyle 2^{n}\) where \(\displaystyle n\) is the number of heterozygous loci.
    Here \(\displaystyle n = 4\), so \(\displaystyle 2^{4} = 16\).

  4. Exercise 4.4

    Explain the Law of Dominance using a monohybrid cross.

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    The Law of Dominance states that in a dissimilar pair of factors one member dominates the other, so only the dominant trait appears in the F1 while both traits reappear in the F2 in the ratio \(\displaystyle 3:1\).
    The law has three parts: (i) characters are controlled by discrete units called factors; (ii) factors occur in pairs; (iii) in a dissimilar pair of factors one member dominates (dominant) the other (recessive).
    The monohybrid cross that shows it:
    Parents: true-breeding tall TT × true-breeding dwarf tt.
    Gametes: T from one parent, t from the other; all zygotes are Tt.
    F1: all plants are tall, none dwarf and none of in-between height — the dwarf factor is hidden, not lost or blended.
    F1 selfed (Tt × Tt) gives F2 genotypes $\displaystyle 1$ TT : $\displaystyle 2$ Tt : $\displaystyle 1$ tt.
    F2 phenotypes: $\displaystyle 3$/$\displaystyle 4$ tall (TT and Tt together) : $\displaystyle 1$/$\displaystyle 4$ dwarf — the \(\displaystyle 3:1\) ratio, with a genotypic ratio of \(\displaystyle 1:2:1\).
    The law thus explains two things at once: why only one parental character shows in F1, and why the hidden character comes back unchanged as one-quarter of the F2.

  5. Exercise 4.5

    Define and design a test-cross.

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    ncert_XII_Biology_fig_4-5
    Definition: a test cross is a cross in which an organism showing the dominant phenotype, and whose genotype is to be determined, is crossed with the homozygous recessive parent instead of being self-crossed; the progeny then reveal the unknown genotype.
    Why it is needed: a tall plant may be TT or Tt, and the two cannot be distinguished by looking at them.
    Designing it (draw this as a Punnett square):
    Take the tall plant of unknown genotype (T_) and cross it with a dwarf plant, tt.
    Draw a $\displaystyle 2$ × $\displaystyle 2$ square. Write the gametes of the tested plant along the top row and the gametes of the dwarf parent (t and t) down the left column. Fill each box with the pair of alleles from its row and column, and write the phenotype under each box.
    If the tested plant is TT: its gametes are T and T, every box is Tt, and all the offspring are tall — no dwarf appears.
    If the tested plant is Tt: its gametes are T and t, the boxes read Tt, tt, Tt, tt, giving $\displaystyle 1$ tall : $\displaystyle 1$ dwarf, a \(\displaystyle 1:1\) ratio.
    Reading the result: the appearance of even one recessive (dwarf) offspring shows the tested plant is heterozygous; an all-dominant progeny shows it is homozygous dominant.
    The chapter's Figure $\displaystyle 4.5$ designs the same cross with violet flower colour (V) dominant over white (v).

  6. Exercise 4.6

    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 a single locus.

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    All the offspring show the dominant phenotype — $\displaystyle 100$ per cent tall, no dwarf — when the homozygous female is homozygous dominant.
    The cross and the square to draw:
    Female TT (homozygous) × male Tt (heterozygous).
    Draw a $\displaystyle 2$ × $\displaystyle 2$ Punnett square. Write the male's two gamete types T and t along the top row; write the female's gametes T and T down the left column. Label the top row with the male symbol and the left column with the female symbol.
    Fill the four boxes: row $\displaystyle 1$ gives TT and Tt; row $\displaystyle 2$ gives TT and Tt.
    Distribution in the F1:
    Phenotypic: $\displaystyle 4$ out of $\displaystyle 4$ tall — a single phenotype, since every offspring carries at least one T.
    Genotypic: $\displaystyle 2$ TT : $\displaystyle 2$ Tt, that is \(\displaystyle 1:1\).
    Note: the question does not say which homozygote the female is. If she is homozygous recessive (tt), the same square gives $\displaystyle 1$ Tt : $\displaystyle 1$ tt, i.e. $\displaystyle 1$ tall : $\displaystyle 1$ dwarf (\(\displaystyle 1:1\)) — that cross is the test cross of Section 4.2.

  7. Exercise 4.7

    When a cross in 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.

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    (a) Tall and green = $\displaystyle 3$/$\displaystyle 8$ of the offspring ($\displaystyle 37.5$ per cent).
    (b) Dwarf and green = $\displaystyle 1$/$\displaystyle 8$ of the offspring ($\displaystyle 12.5$ per cent).
    Working:
    The cross is TtYy × Ttyy. Because the two pairs of genes assort independently, treat each character as a separate monohybrid cross and then multiply.
    Height: Tt × Tt gives $\displaystyle 3$/$\displaystyle 4$ tall : $\displaystyle 1$/$\displaystyle 4$ dwarf.
    Seed colour: Yy × yy gives $\displaystyle 1$/$\displaystyle 2$ yellow : $\displaystyle 1$/$\displaystyle 2$ green.
    Tall and green \(\displaystyle = 3/4 \times 1/2 = 3/8\).
    Dwarf and green \(\displaystyle = 1/4 \times 1/2 = 1/8\).
    The whole progeny works out as $\displaystyle 3$ tall yellow : $\displaystyle 3$ tall green : $\displaystyle 1$ dwarf yellow : $\displaystyle 1$ dwarf green, a \(\displaystyle 3:3:1:1\) ratio.

  8. Exercise 4.8

    Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in \(\displaystyle F_{1}\) generation for a dibybrid cross?

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    The four phenotypes still appear, but not in the \(\displaystyle 9:3:3:1\) ratio — the ratio deviates very significantly from it.
    The two parental combinations of characters are present in a much higher proportion than the two non-parental (recombinant) combinations.
    Reason: linked genes lie on the same chromosome, so they do not assort independently; they tend to be transmitted together into the same gamete, and the parental combinations are therefore over-represented.
    The recombinant types arise only when the linked genes are separated, which is what recombination means.
    How far the ratio departs depends on the strength of the linkage: tightly linked genes give very few recombinants (in Morgan's flies, white and yellow showed only $\displaystyle 1.3$ per cent recombination), while loosely linked genes give many more (white and miniature wing showed $\displaystyle 37.2$ per cent).
    This is exactly what Morgan found: his dihybrid crosses in Drosophila with genes on the same chromosome gave a ratio that deviated very significantly from $\displaystyle 9$:$\displaystyle 3$:$\displaystyle 3$:$\displaystyle 1$, which is how linkage was discovered.

  9. Exercise 4.9

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

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    Morgan provided the experimental verification of the chromosomal theory of inheritance, and in doing so discovered the basis of the variation that sexual reproduction produces.
    He worked with the fruit fly Drosophila melanogaster, which suited such studies well: it grows on a simple synthetic medium, completes its life cycle in about two weeks, a single mating gives a large number of progeny, males and females are easily distinguished, and it shows many hereditary variations visible under a low-power microscope.
    He carried out dihybrid crosses in Drosophila to study sex-linked genes — for example, yellow-bodied, white-eyed females crossed with brown-bodied, red-eyed males.
    He found that genes on the same chromosome do not segregate independently: the F2 ratio deviated very significantly from \(\displaystyle 9:3:3:1\), and parental combinations far outnumbered non-parental ones.
    He coined the terms linkage, for the physical association of genes on a chromosome, and recombination, for the generation of non-parental gene combinations.
    He showed that linkage varies in strength — white and yellow were tightly linked ($\displaystyle 1.3$ per cent recombination) while white and miniature wing were loosely linked ($\displaystyle 37.2$ per cent).
    His student Alfred Sturtevant used these recombination frequencies as a measure of the distance between genes and mapped their positions on the chromosome; such genetic maps are still used as a starting point for sequencing whole genomes.

  10. Exercise 4.10

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

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    Pedigree analysis is the analysis of a trait across several generations of a family, in which the inheritance of that particular trait is represented as a family tree drawn with standard symbols.
    It is needed because the controlled crosses that can be made in pea plants or other organisms are not possible in human beings; the study of family history is the alternative.
    How such an analysis is useful:
    It traces the inheritance of a specific trait, abnormality or disease through the generations of a family.
    It shows whether the trait in question is dominant or recessive — for example an autosomal dominant trait such as myotonic dystrophy, or an autosomal recessive one such as sickle-cell anaemia.
    It shows whether the gene is on an autosome or on a sex chromosome: haemophilia gives the tell-tale X-linked recessive pattern, passing from an unaffected carrier female to some of her sons.
    Since Mendelian disorders are determined by a single gene and are transmitted along the lines of the principles of inheritance, the pedigree lets that pattern of transmission be worked out in a real family.