Heredity — CBSE Class 10 Science Important Questions

Answer: (b) 7. Mendel selected seven pairs of contrasting visible characters of Pisum sativum (e.g. tall/dwarf, round/wrinkled seeds, yellow/green seeds) for his breeding experiments.

Answer: (c) TtWW.

• All the progeny have violet flowers, so the tall parent must be WW. If it were Ww, about half the progeny would be white (Ww × ww).
• Almost half the progeny are short, so the tall parent must be Tt: Tt × tt gives 1 Tt (tall) : 1 tt (short).

Answer: (c) 9 : 3 : 3 : 1. A dihybrid cross between double heterozygotes gives an F2 phenotypic ratio of $9$9 (both dominant) $: 3 : 3 : 1$: 3 : 3 : 1 (both recessive), which supports the law of independent assortment.

Answer: (b) 1 : 2 : 1. The F2 genotypes are $1\,TT : 2\,Tt : 1\,tt$1 TT : 2 Tt : 1 tt. (Note the phenotypic ratio is $3:1$3:1 because TT and Tt both look tall.)

Answer: (a) Both A and R are true, and R is the correct explanation of A. A section of DNA that provides information for one protein is a gene. Plant height depends on the amount of a growth hormone, which depends on how efficiently an enzyme makes it, so the gene for that enzyme controls height.

A dominant trait is the one that expresses itself in the F1 hybrid even when only one copy of its factor is present — e.g. tallness (T) in pea plants. A recessive trait is the one that remains hidden in the F1 hybrid and reappears only when both factors are of the recessive type (e.g. dwarfness (t)), as in the genotype tt. Thus in a Tt plant, tall is dominant and dwarf is recessive.

No. The daughter gets one copy of the gene from each parent, and both possibilities fit the data:

• If A is dominant, the father could carry one A and one O copy and pass O to the daughter. She would then have two O copies and blood group O.
• If O is dominant, the daughter would show blood group O as soon as she received an O copy from her mother.

One child is also far too few to see a ratio. We would need the results of many such crosses to decide which trait is dominant.

Humans have 23 pairs of chromosomes — 22 pairs of autosomes and one pair of sex chromosomes. Females are XX and males are XY.

• The mother produces eggs that all carry an X chromosome.
• The father produces two kinds of sperm: half carry X and half carry Y.

At fertilization:

• If an X-bearing sperm fuses with the egg → XX → girl.
• If a Y-bearing sperm fuses with the egg → XY → boy.

Since the type of sperm decides the outcome, the father determines the sex of the child, and there is an equal ($1:1$1:1) chance of a boy or a girl.

Mendel selected garden pea because:

1. It has several easily observable pairs of contrasting characters (e.g. tall/dwarf, round/wrinkled seeds, yellow/green seeds).
2. It is naturally self-pollinating, giving pure-breeding lines, yet it can be artificially cross-pollinated by the experimenter, so crosses could be controlled.
3. It has a short life cycle and produces many seeds (offspring) in one generation, allowing large numbers to be analysed statistically in a short time. (Also, the flowers are bisexual and normally do not cross naturally, keeping lines pure.)

A $3:1$3:1 ratio in the offspring is the classic F2 monohybrid ratio, which is only obtained when both parents are heterozygous tall, i.e. Tt × Tt.

Cross:
Parents: Tt × Tt
Gametes: T, t (from each parent)

Offspring genotypes: $1\,TT : 2\,Tt : 1\,tt$1 TT : 2 Tt : 1 tt.
Phenotypes: 3 tall (TT, Tt, Tt) : 1 dwarf (tt).

This matches the observed $3:1$3:1 ratio, so both parents were Tt (heterozygous tall).

• Every body cell has two copies of each chromosome, one inherited from each parent. In humans that is 23 pairs, or 46 chromosomes.
• When germ cells (sperm and egg) are formed, each takes only one chromosome from each pair, so a human sperm or egg has 23 chromosomes. The chromosome taken from each pair may be of maternal or paternal origin.
• At fertilisation, a sperm and an egg fuse. The zygote gets 23 chromosomes from the father and 23 from the mother, so both parents contribute equally, and the normal number (46) is restored. This keeps the DNA of the species stable from one generation to the next.

Parents: Round-Yellow (RRYY) × Wrinkled-Green (rryy).

F1 generation: all seeds are Round-Yellow (RrYy), showing round (R) and yellow (Y) are dominant over wrinkled (r) and green (y).

F2 generation: F1 plants are self-pollinated. Each F1 produces four kinds of gametes — RY, Ry, rY, ry — which combine as shown in the 4×4 Punnett square, giving 16 combinations.

F2 phenotypic ratio:

• 9 Round-Yellow
• 3 Round-Green
• 3 Wrinkled-Yellow
• 1 Wrinkled-Green

So the ratio is $9:3:3:1$9:3:3:1.

Significance: the appearance of new combinations (round-green and wrinkled-yellow), which were not present in the parents, shows that the two pairs of traits are inherited independently of each other. This establishes Mendel's Law of Independent Assortment.

(a) Heredity is the transmission of characters (traits) from parents to their offspring through genes, generation after generation.

(b) The cross:

P generation: TT (tall) × tt (dwarf)
Gametes: T and t → all offspring Tt.
F1 generation: all Tt — 100% tall (T dominant over t).

F1 selfed: Tt × Tt

F2 generation:

• Genotypic ratio: $1\,TT : 2\,Tt : 1\,tt$1 TT : 2 Tt : 1 tt
• Phenotypic ratio: $3$3 Tall $: 1$: 1 Dwarf

(c) Laws illustrated:

1. Law of Dominance — in the F1, only the dominant trait (tall) appears, the recessive (dwarf) stays hidden.
2. Law of Segregation — the paired factors (T and t) separate during gamete formation, each gamete receiving only one, so the dwarf trait reappears in the F2.

(i) The father's sperm are of two kinds — 22 autosomes + X and 22 autosomes + Y (i.e. X-sperm and Y-sperm in equal numbers). The mother's egg is of one kind only — 22 autosomes + X.

(ii) The probability of a son is $\frac{1}{2}$1/2 (50%). The egg always carries X; if it is fertilized by a Y-sperm the child is a boy (XY) and if by an X-sperm a girl (XX). Since X- and Y-sperm are equally likely, each birth has a $1:1$1:1 chance — past daughters do not change this probability.

(iii) No. The mother's eggs all carry only X, so she cannot influence the sex of the child. The child's sex depends entirely on whether an X- or Y-bearing sperm from the father fertilizes the egg. Blaming the mother is scientifically wrong and unjust.

(iv) Because the father produces equal numbers of X and Y sperm, each fertilization has an equal chance of giving a boy or a girl. Over a large number of births, these equal probabilities average out, keeping the sex ratio close to $1:1$1:1.