DNA, genes, mitosis & meiosis

Year 10 answers

Fluency

Adenine, thymine, guanine, cytosine. A pairs with T; G pairs with C.
Body cell: $46$ chromosomes. Gamete: $23$ chromosomes.
Gene — a length of DNA coding for a trait. Allele — one version of a gene. Genotype — the pair of alleles carried. Phenotype — the observable trait.
Heterozygous (two different alleles).
Growth and repair of tissues; replacing old cells (e.g. skin, blood).
Mitosis produces two identical diploid cells; meiosis produces four genetically different haploid gametes.

Fluency

$3 : 1$ brown to blue.
All offspring $Bb$ ; all brown.
$1 : 1$ tall to short.
$\dfrac{1}{4}$ , or $25\%$ .
All children will have attached earlobes. Both parents are $ee$ , so every child must inherit $e$ from each parent.

Reasoning

Her mother was $ww$ , so the woman inherited $w$ from mother and $W$ from her widow’s-peak parent; genotype $Ww$ . Cross $Ww \times ww$ : $\dfrac{1}{2}$ widow’s peak, $\dfrac{1}{2}$ straight.
Expected $\dfrac{1}{4} \times 4 = 1$ affected child. The actual number can differ because each child is an independent event; outcomes follow a binomial distribution, not a guaranteed quota.
Offspring genotypes $BB : Bb : bb = 1 : 2 : 1$ . Because $BB$ and $Bb$ share the dominant phenotype, the three “brown” outcomes combine, giving $3 : 1$ .
Father is $X^c Y$ , mother is $X^C X^C$ . All daughters are $X^C X^c$ (carriers, not colour-blind). All sons are $X^C Y$ (not colour-blind).
Ratio $\approx 3 : 1$ suggests both parents are heterozygous $Rr$ . Punnett: $RR, Rr, Rr, rr$ , giving $3$ red $: 1$ white.

Problem solving

Meiosis shuffles maternal and paternal chromosomes by independent assortment and crossing over, producing gametes with new combinations. Random fertilisation then pairs two such gametes, so each child represents one of millions of possible genotypes.
$4$ chromosomes per gamete; $4$ gametes from one starting cell.
Son has haemophilia with probability $\dfrac{1}{2}$ . Daughter is a carrier with probability $\dfrac{1}{2}$ (the other $\dfrac{1}{2}$ are $X^H X^H$ ).
$1 : 2 : 1$ red : pink : white.

Reasoning

Each conception is independent; meiosis in the father randomly produces an $X$ -bearing or $Y$ -bearing sperm with probability $\tfrac{1}{2}$ . Previous children do not change the odds for the next.
His father must have been $Hh$ (or he could not pass the allele), so the son’s probability of being $Hh$ is $\dfrac{1}{2}$ . If he is $Hh$ , each child has probability $\dfrac{1}{2}$ of inheriting $H$ .
Ratio $9 : 3 : 3 : 1$ (both dominant : yellow round; yellow wrinkled; green round; green wrinkled). The $9$ comes from $\dfrac{3}{4} \times \dfrac{3}{4} = \dfrac{9}{16}$ having at least one dominant allele for each gene.
(i) With $23$ chromosome pairs, independent assortment gives $2^{23} \approx 8.4 \times 10^6$ gamete combinations per parent. (ii) Random fertilisation squares this: $\approx 7 \times 10^{13}$ possible zygote genotypes, before crossing over. Variation fuels adaptation to changing environments.