Question

a woman is a carrier for a recessive x - linked genetic condition, and her husband does not possess the allele. what are the odds that the couples son will express the condition? 0% 25% 50% 100%

Explanation:

Step1: Determine the genotypes

Let \(X^A\) represent the normal allele and \(X^a\) represent the recessive allele for the X - linked disorder. The woman (carrier) has the genotype \(X^AX^a\), and the man (not possessing the allele) has the genotype \(X^AY\).

Step2: Create the Punnett square

For a son, the man will contribute a \(Y\) chromosome. The woman can contribute either \(X^A\) or \(X^a\).
The possible combinations for a son are:

• If the woman contributes \(X^A\): \(X^AY\) (normal son)
• If the woman contributes \(X^a\): \(X^aY\) (son with the disorder)

Step3: Calculate the probability

There are 2 possible outcomes for the son's genotype (\(X^AY\) and \(X^aY\)). The number of favorable outcomes (son with the disorder, \(X^aY\)) is 1.
Using the probability formula \(P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\), we have \(P = \frac{1}{2}=50\%\)

Answer:

50%