Brown eye allele $B$ is dominant over blue allele $b$. Both parents are heterozygous ($Bb$). What fraction of their children are expected to have blue eyes?

1. Each parent can pass on either $B$ or $b$ with equal probability.
2. Draw a 2 by 2 Punnett square: possible combinations are $BB$, $Bb$, $Bb$, $bb$.
3. Genotype ratio: $1 : 2 : 1$. Phenotype ratio (brown : blue) $= 3 : 1$.
4. Expected fraction with blue eyes: $\dfrac{1}{4}$, or $25\%$.

Key idea: a cross between two heterozygotes gives the classic $3 : 1$ phenotypic ratio whenever one allele is fully dominant.

A human skin cell has $46$ chromosomes. After mitosis, how many chromosomes does each daughter cell have? After meiosis?

1. Mitosis: DNA is copied, then divided equally. Each daughter cell has $46$ chromosomes.
2. Meiosis: chromosomes are copied once but the cell divides twice, so each gamete has $\dfrac{46}{2} = 23$ chromosomes.

At fertilisation, a sperm ($23$) joins an egg ($23$) to give a zygote with $46$ chromosomes — the full diploid set.

In pea plants, tall ($T$) is dominant over short ($t$). Cross a heterozygous tall plant ($Tt$) with a short plant ($tt$). What are the genotype and phenotype ratios of the offspring?

1. $Tt$ parent produces gametes $T$ and $t$ (each $\tfrac{1}{2}$).
2. $tt$ parent produces only $t$ gametes.
3. Punnett square gives offspring: $Tt, Tt, tt, tt$.
4. Genotype ratio $1\,Tt : 1\,tt$. Phenotype ratio (tall : short) $= 1 : 1$.

So about half the offspring are tall, half are short.

Cystic fibrosis is caused by a recessive allele $f$. Two heterozygous carriers ($Ff$) have a child. Find the probability the child (a) has the disease and (b) is an unaffected carrier.

1. Punnett square: $FF, Ff, Ff, ff$.
2. (a) Has disease $=$ genotype $ff$: probability $\dfrac{1}{4}$, or $25\%$.
3. (b) Carrier $=$ genotype $Ff$: probability $\dfrac{2}{4} = \dfrac{1}{2}$, or $50\%$.

Key idea: two unaffected carriers can produce an affected child because each has a one-in-two chance of passing on the faulty allele.

The colour-blind allele $X^c$ is recessive. A carrier mother ($X^C X^c$) has a child with a normal-vision father ($X^C Y$). Find the probability that a son is colour-blind.

1. Mother’s gametes: $X^C$ or $X^c$ (each $\tfrac{1}{2}$).
2. Father’s gametes: $X^C$ or $Y$ (each $\tfrac{1}{2}$).
3. Sons receive $Y$ from father. So son genotypes are $X^C Y$ or $X^c Y$, each with probability $\tfrac{1}{2}$.
4. Probability a son is colour-blind $= \dfrac{1}{2}$. Daughters all receive $X^C$ from the father and are unaffected, though half are carriers.

Key idea: X-linked recessive conditions are far more common in males because they only need one copy of the allele to express it.