RowQ
The Vault
RowQ
The Vault
CBSE Class 12 Biology · 11 questions · 26 marks
Long before anyone had seen a gene, Gregor Mendel worked out the rules of heredity by counting peas — and those rules still hold. This chapter builds from his monohybrid and dihybrid crosses to the deviations that came later: incomplete dominance, codominance, multiple alleles, linkage, and sex-linked disorders. Expect to draw Punnett squares and read pedigree charts, so practise the arithmetic of ratios until it is automatic.
In snapdragon (Antirrhinum majus), a red-flowered plant crossed with a white-flowered plant gives all pink F₁ plants. What phenotypic ratio is expected in the F₂ generation?
Answer
1 red : 2 pink : 1 white is correct — this is a case of incomplete dominance, where the heterozygote has an intermediate phenotype. Because the heterozygote is visibly different from either homozygote, the phenotypic ratio matches the genotypic ratio of 1 : 2 : 1 instead of collapsing into the usual 3 : 1.
A man with blood group AB marries a woman with blood group O. Which blood groups can their children possibly have?
Answer
A or B only is correct — the father is I^A I^B and the mother is ii, so each child receives I^A or I^B from the father and i from the mother, giving genotypes I^A i (group A) or I^B i (group B) in equal proportion. Neither AB nor O is possible, because no child can inherit two dominant alleles or two recessive i alleles from this pairing.
Down's syndrome in humans is caused by:
Answer
An extra copy of chromosome 21 is correct — non-disjunction during gamete formation gives a trisomic individual with the karyotype 45 + XX or XY, that is, three copies of chromosome 21. Absence of an X causes Turner's syndrome (45, X0), a single base substitution in the beta-globin gene causes sickle-cell anaemia, and an extra X in a male causes Klinefelter's syndrome (47, XXY).
Why do two genes located very close together on the same chromosome fail to show the expected 9 : 3 : 3 : 1 ratio in the F₂?
Answer
They are tightly linked and tend to be inherited together is correct — the law of independent assortment applies only to genes on different chromosomes or far apart on the same chromosome. Closely linked genes rarely separate by crossing over, so parental combinations greatly outnumber recombinants and the dihybrid ratio departs sharply from 9 : 3 : 3 : 1. Physical proximity, not mutation rate or dominance, is the reason.
Assertion (A): Red-green colour blindness appears far more frequently in men than in women. Reason (R): The gene lies on the X chromosome, and a man has only one X, so a single recessive allele is enough to express the condition in him.
Answer
Both A and R are true and R is the correct explanation of A — a male is hemizygous for X-linked genes, so there is no second X carrying a normal allele to mask the defective one. A woman must inherit the recessive allele on both her X chromosomes to be colour blind, which is a far less likely event, so she is usually only a carrier.
A breeder has a tall pea plant but does not know whether it is homozygous or heterozygous. Describe the cross he should perform and how he should interpret the result.
Answer
He should perform a test cross, crossing the unknown tall plant with a homozygous dwarf plant (tt), since the dwarf parent can contribute only recessive alleles and therefore cannot mask anything. If the unknown plant is homozygous TT, every gamete carries T, so all the offspring are Tt and 100 percent are tall. If the unknown plant is heterozygous Tt, half its gametes carry T and half carry t, so the offspring appear in a 1 : 1 ratio of tall (Tt) to dwarf (tt). The appearance of even a single dwarf offspring proves the parent was heterozygous.
In a garden pea, seed shape is controlled by R (round, dominant) and r (wrinkled), and pod colour by G (green, dominant) and g (yellow). A true-breeding round green plant is crossed with a true-breeding wrinkled yellow plant. Work out the F₁ and the F₂ phenotypic ratio, showing the gamete types involved.
Answer
The parents are RRGG (round green) and rrgg (wrinkled yellow). The first parent produces only RG gametes and the second only rg gametes, so every F₁ plant is RrGg and is round with green pods — both dominant characters appear. When the F₁ is self-pollinated, each RrGg plant produces four kinds of gametes in equal proportion: RG, Rg, rG, and rg. A 4 × 4 Punnett square of these gametes gives 16 combinations. Counting the phenotypes gives 9 round green : 3 round yellow : 3 wrinkled green : 1 wrinkled yellow, the classic dihybrid 9 : 3 : 3 : 1 ratio. It arises because the two gene pairs assort independently of each other during gamete formation.
Distinguish between a Mendelian disorder and a chromosomal disorder, giving one example of each.
Answer
A Mendelian disorder is caused by an alteration or mutation in a single gene, and it is inherited according to Mendel's principles, so its transmission can be traced through a pedigree as dominant, recessive, autosomal, or sex-linked. Sickle-cell anaemia, caused by the substitution of valine for glutamic acid at the sixth position of the beta-globin chain, is an example. A chromosomal disorder is caused by the absence, excess, or abnormal arrangement of one or more whole chromosomes, usually resulting from non-disjunction during gamete formation rather than from a faulty single gene. Down's syndrome, caused by an extra copy of chromosome 21, is an example.
A woman whose father was haemophilic marries a man who does not have haemophilia. Using a Punnett square, work out the probability that their children will be haemophilic, and explain why the pattern of inheritance differs between their sons and daughters.
Answer
Haemophilia is an X-linked recessive condition. Write the normal allele as X^H and the haemophilia allele as X^h. The woman's father was haemophilic, so his genotype was X^h Y. A father passes his single X chromosome to every daughter, so this woman must have received X^h from him, and since she is not herself haemophilic she must have received a normal X from her mother. Her genotype is therefore X^H X^h — a carrier. Her husband is unaffected, so his genotype is X^H Y. Cross: X^H X^h × X^H Y. The mother's gametes are X^H and X^h; the father's are X^H and Y. The Punnett square gives four equally likely offspring: X^H X^H (normal daughter), X^H X^h (carrier daughter), X^H Y (normal son), and X^h Y (haemophilic son). So one quarter of all children, that is 25 percent, are expected to be haemophilic — and all of them are boys. Considering the sexes separately, 50 percent of the sons are haemophilic and 50 percent are normal, while none of the daughters is haemophilic although 50 percent of them are carriers. The difference arises because a son has only one X chromosome, inherited from his mother, so a single X^h allele is fully expressed with no second X to mask it. A daughter receives one X from each parent; here the father always contributes X^H, so even a daughter who inherits X^h has a normal allele that masks it, making her a carrier rather than a patient.
State Mendel's three laws of inheritance and explain, with an example for each, two situations in which the observed inheritance departs from what these laws alone predict.
Answer
Mendel's laws are as follows. The law of dominance states that characters are controlled by discrete units called factors occurring in pairs, and in a dissimilar pair one member dominates and is expressed while the other remains hidden. The law of segregation states that the two alleles of a pair do not blend but separate during gamete formation, so each gamete receives only one of them; this law has no exceptions. The law of independent assortment states that the alleles of one gene pair segregate independently of the alleles of another pair when gametes are formed. First departure — codominance. In the human ABO blood group system, a person of genotype I^A I^B does not show one allele dominating the other, nor an intermediate blend. Both alleles are fully and simultaneously expressed, so the red cells carry both A and B antigens and the person is blood group AB. This contradicts the expectation of the law of dominance that one allele should mask the other. The same system also illustrates multiple alleles, since three alleles — I^A, I^B, and i — exist in the population for a single gene, although any one individual carries only two. Second departure — linkage. Morgan's experiments with Drosophila melanogaster showed that when two genes lie close together on the same chromosome, they are transmitted together far more often than chance allows. A dihybrid cross involving such genes does not give the expected 9 : 3 : 3 : 1 F₂ ratio; parental combinations are strongly over-represented and recombinants are rare, because only an occasional crossing over separates the two genes. Mendel's law of independent assortment therefore holds only for genes on different chromosomes, or on the same chromosome but far enough apart for crossing over to be frequent.
A hospital investigates a disputed identity case. The mother, Kavitha, is blood group A; the child in question, an infant boy, is blood group O. Two men are put forward as the possible father: Ravi, who is blood group AB, and Sameer, who is blood group B. Records also show that Kavitha's own father was colour blind, while Kavitha herself has normal colour vision. (a) State the genotype of Kavitha and of the child with respect to blood group. (b) Which of the two men can be excluded as the biological father, and why? (c) Can the remaining man be declared the father with certainty on this evidence alone? Justify. (d) What is the chance that Kavitha's son is colour blind, and why does this not depend on the father?
Answer
(a) The child is blood group O, so his genotype must be ii — he has received one i allele from each parent. Kavitha is group A but has passed on an i allele, so she cannot be I^A I^A; her genotype must be I^A i. (b) Ravi, who is blood group AB, can be excluded. His genotype is I^A I^B, so every sperm he produces carries either I^A or I^B and none can carry i. He is therefore incapable of fathering a group O child with any woman. (c) No. Sameer is group B, and to have contributed the second i allele his genotype must be I^B i, which is entirely possible. But this evidence only shows that he cannot be excluded — a very large number of men in the population are also I^B i or I^A i or ii and would fit equally well. Blood grouping can exclude a man from paternity with certainty, but it can never establish paternity; DNA fingerprinting would be needed for that. (d) Kavitha's father was colour blind (X^c Y), and a father passes his X to every daughter, so Kavitha must have received X^c from him. Since she sees colour normally, her genotype is X^C X^c, a carrier. A son receives his single X from his mother, and her two X chromosomes are equally likely to be passed on, so the chance that her son is colour blind is 1 in 2, or 50 percent. It does not depend on the father because a father contributes the Y chromosome to a son, and the Y carries no allele for colour vision.
RowQ generates fresh questions on Principles of Inheritance and Variation, marks your answers, and explains every step.
Start free