Question

the pedigree on the right shows the inheritance pattern for an x - linked recessive disorder. if the female circled in pink married an affected male and had a female child, what is the chance that the child would also be affected? express your answer in a percentage.% what is the chance that the child would be a carrier?% true or false? the child could be standard.

Explanation:

Step1: Determine the genotype of the female

Since it is an X - linked recessive disorder (let the normal allele be \(X^A\) and the mutant allele be \(X^a\)), and the female is a carrier (half - filled circle in pedigree for X - linked recessive, which means \(X^AX^a\)). The affected male has the genotype \(X^aY\).

Step2: Analyze the cross for female child

The cross for a female child (\(XX\)):
The possible gametes from the female (\(X^AX^a\)) are \(X^A\) and \(X^a\), and from the male (\(X^aY\)) for the \(X\) - chromosome contribution (since for female child, male contributes \(X\)) is \(X^a\).
The Punnett square for the cross (considering only \(X\) - chromosome contributions for female child):

Female Gametes	Male Gamete (\(X^a\))
\(X^a\)	\(X^aX^a\) (affected)

The probability of getting \(X^aX^a\) (affected female child) is \(\frac{1}{2}=50\%\)
The probability of getting \(X^AX^a\) (carrier female child) is \(\frac{1}{2} = 50\%\)
For the statement "The child could be standard (\(X^AX^A\))", since the male parent only has \(X^a\) to contribute to a female child (and the female parent has \(X^A\) and \(X^a\)), there is no way to get \(X^AX^A\). So the statement is False.

Answer:

• The chance that the female child is affected: \(50\%\)
• The chance that the female child is a carrier: \(50\%\)
• True or false (child could be standard): False