Pea plants heterozygous for seed shape and seed color are crossed. Determine the phenotypic ratio of the offspring.

Haemophilia is a sex-linked recessive disorder. Let $X^H$ be the allele for normal blood clotting and $X^h$ be the allele for haemophilia.
Woman (carrier): $X^H X^h$
Man (normal blood clotting): $X^H Y$

1. Step 3: Calculate the probability of a son with haemophilia. From the Punnett square, the possible male genotypes are $X^H Y$ (normal) and $X^h Y$ (haemophiliac). The probability of having a son with haemophilia ($X^h Y$) is 1 out of 4 total offspring. The probability is $\boxed{\frac{1}{4} \text{ or } 25\%}$.

2. Step 4: Calculate the probability of a daughter who is a carrier. From the Punnett square, the possible female genotypes are $X^H X^H$ (normal) and $X^H X^h$ (carrier). The probability of having a daughter who is a carrier ($X^H X^h$) is 1 out of 4 total offspring. The probability is $\boxed{\frac{1}{4} \text{ or } 25\%}$.

3. Step 5: Determine if they can produce a daughter with the disease. A daughter with haemophilia would have the genotype $X^h X^h$. For this to occur, she must inherit an $X^h$ allele from both her mother and her father. The mother ($X^H X^h$) can provide an $X^h$ allele. However, the father ($X^H Y$) only has an $X^H$ allele to pass on to his daughters. Therefore, he cannot contribute the $X^h$ allele necessary for a daughter to have the disease. No, this couple cannot produce a daughter with the disease because the father does not carry the recessive haemophilia allele ($X^h$) on his X chromosome to pass to a daughter.