Genetics Class 10 ICSE Important Questions with Answers

Answer: (b) Phenotype.

Phenotype is the set of visible or expressed traits; genotype refers to the underlying genetic constitution.

Answer: (b) Garden pea (Pisum sativum).

Mendel chose the garden pea for its contrasting characters, ease of cultivation and self-pollinating nature.

Answer: (c) $9:3:3:1$9:3:3:1.

A dihybrid cross between two heterozygotes ($RrYy \times RrYy$RrYy × RrYy) gives an F2 phenotypic ratio of $9:3:3:1$9:3:3:1.

Answer: (b) $\dfrac{1}{4}$1/4.

The F1 are all $Tt$Tt. Selfing $Tt \times Tt$Tt × Tt gives $1\,TT : 2\,Tt : 1\,tt$1 TT : 2 Tt : 1 tt, so dwarf ($tt$tt) plants are $\dfrac{1}{4}$1/4 of the F2.

Answer: (c) A is true but R is false. The father is heterogametic ($XY$XY) and produces X- and Y-bearing sperms, so he determines the sex; the mother ($XX$XX) produces only X-bearing eggs, so R is incorrect.

(i) Allele: One of the two or more alternative forms of a gene that occupy the same position (locus) on homologous chromosomes and control contrasting forms of a character, e.g. $T$T for tallness and $t$t for dwarfness.

(ii) Heterozygous: An organism having two different (unlike) alleles of a gene for a character, e.g. $Tt$Tt.

(i) Mutation: a sudden, inheritable change in a gene (or chromosome). Example: the change in a gene on the X chromosome that causes haemophilia.

(ii) Variation: the differences between individuals of the same species, including between parents and offspring and among brothers and sisters. Example: differences in height or eye colour between children of the same parents. Variations arise from the reshuffling of genes in sexual reproduction and from mutations.

P generation: $TT$TT (tall) $\times$× $tt$tt (dwarf).

F1: All plants are $Tt$Tt — tall (tallness is dominant).

F1 selfed: $Tt \times Tt$Tt × Tt. Gametes: $T$T and $t$t from each parent.

F2 genotypes: $1\,TT : 2\,Tt : 1\,tt$1 TT : 2 Tt : 1 tt.

F2 phenotypes: $3$3 tall $: 1$: 1 dwarf, i.e. the phenotypic ratio is $\mathbf{3:1}$3:1.

In humans, of the 23 pairs of chromosomes, one pair is the sex chromosomes. Females are $XX$XX (homogametic) and males are $XY$XY (heterogametic).

The mother produces only X-bearing eggs, while the father produces two kinds of sperm — X-bearing and Y-bearing.

• If an X sperm fertilises the egg $\rightarrow XX \rightarrow$→ XX → girl.
• If a Y sperm fertilises the egg $\rightarrow XY \rightarrow$→ XY → boy.

$XX\ (\text{mother}) \times XY\ (\text{father}) \rightarrow XX : XY = 1:1$XX (mother) × XY (father) → XX : XY = 1:1

Thus there is a $50\%$50% chance of a boy and $50\%$50% of a girl, and the father determines the sex of the child.

1. Law of dominance: when two individuals differing in a pair of contrasting characters are crossed, only one character, the dominant one, appears in the F1 generation; the other, the recessive one, stays hidden.

2. Law of segregation (law of purity of gametes): the two alleles of a character separate during gamete formation, so each gamete receives only one allele of the pair; they come together again at fertilisation.

3. Law of independent assortment: in a cross involving two or more pairs of characters, the alleles of each pair are inherited independently of the others. This is why a dihybrid cross gives new combinations and the $9:3:3:1$9:3:3:1 ratio.

P: $RRYY$RRYY (round, yellow) $\times$× $rryy$rryy (wrinkled, green).

F1: All $RrYy$RrYy — round and yellow seeds (both dominants expressed).

Gametes of F1 ($RrYy$RrYy): $RY,\ Ry,\ rY,\ ry$RY, Ry, rY, ry.

Punnett square (F1 $\times$× F1):

F2 phenotypes:

• $9$9 round, yellow
• $3$3 round, green
• $3$3 wrinkled, yellow
• $1$1 wrinkled, green

F2 phenotypic ratio $= 9:3:3:1$= 9:3:3:1. This illustrates Mendel's law of independent assortment — the two characters (seed shape and colour) are inherited independently of each other.

(a) Carrier woman: $X^HX^h$X^HX^h. Normal man: $X^HY$X^HY.

(b) Half of the sons ($X^hY$X^hY) are expected to be haemophilic. None of the daughters is haemophilic, but half of them are carriers ($X^HX^h$X^HX^h).

(c) A man has only one X chromosome, so a single recessive allele $X^h$X^h shows its effect. A woman has two X chromosomes and is haemophilic only if both carry $h$h, which is rare; with one normal allele she is only a carrier.

(d) No. A son receives his father's Y chromosome, not his X; the colour-blindness allele is on the X chromosome. The son gets his X chromosome from his mother.

(i) Man (heterozygous roller) $= Rr$= Rr; woman (non-roller) $= rr$= rr.

(ii) Cross $Rr \times rr$Rr × rr. Gametes: man $R, r$R, r; woman $r, r$r, r.

Children: $2\,Rr$2 Rr (rollers) $: 2\,rr$: 2 rr (non-rollers), i.e. genotypes $Rr$Rr and $rr$rr in a $1:1$1:1 ratio.

(iii) Tongue-rollers ($Rr$Rr) are $\dfrac{2}{4} = \mathbf{50\%}$2/4 = 50% of the children.

(iv) Tongue-rolling is a dominant character (it is expressed in the heterozygous $Rr$Rr children).