Question

8.6. calculate the probability of inheriting a particular genetic disorder by using a punnett square.
a man possesses an allele for a recessive x - linked genetic condition, and his wife does not possess the allele. what are the odds that the couples son will express the condition?
0%
25%
50%
100%

Explanation:

Step1: Analyze the inheritance of X - linked genes

Men have the sex chromosomes \(XY\). If a man has a recessive X - linked allele (\(X^{a}Y\)), and women have \(XX\) chromosomes. The wife does not possess the allele (\(X^{A}X^{A}\)).

Step2: Determine the possible genotypes of offspring

When having children, the man can pass either \(X^{a}\) or \(Y\), and the woman can only pass \(X^{A}\). For a son (who gets \(Y\) from the father and \(X\) from the mother), the genotype of the son is \(X^{A}Y\) (because the father's \(Y\) is passed to the son, and the mother's \(X^{A}\) is passed to the son).

Answer:

A. \(0\%\)