Question

1. sex - linked traits (short)

• traits on the x chromosome show different inheritance patterns in males and females.
• males have one x chromosome, so a single recessive allele on x causes expression in males.

quick question: if a mother is a carrier for an x - linked recessive trait and father is unaffected, what is the chance a son will express the trait?

Explanation:

Step1: Determine Mother's Genotype

A carrier for an X - linked recessive trait has the genotype $X^A X^a$ (where $X^A$ is the dominant allele and $X^a$ is the recessive allele). The father, being unaffected, has the genotype $X^A Y$ (since males have an X and a Y chromosome, and the dominant allele on the X prevents expression of the recessive trait).

Step2: Analyze Gamete Production

The mother can produce two types of gametes: $X^A$ and $X^a$, each with a probability of $\frac{1}{2}$. The father can produce two types of gametes: $X^A$ and $Y$, each with a probability of $\frac{1}{2}$.

Step3: Determine Son's Genotype Probability

For a son, he inherits the Y chromosome from the father and an X chromosome from the mother. The probability that he inherits the $X^a$ chromosome from the mother (which will cause the expression of the recessive trait) is the probability that the mother passes on $X^a$ (which is $\frac{1}{2}$) and the father passes on $Y$ (which is $\frac{1}{2}$). But since the father must pass on $Y$ for a son, the relevant probability is just the probability the mother passes on $X^a$, which is $\frac{1}{2}$.

Answer:

$\frac{1}{2}$ (or 50%)