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), her genotype is \(X^AX^a\). The affected male has the genotype \(X^aY\).

Step2: Analyze the cross for a female child

The cross is \(X^AX^a\times X^aY\). For a female child (\(XX\)), the possible genotypes are obtained from the following combinations:

• From the mother (\(X^A\) or \(X^a\)) and from the father (\(X^a\)).

The possible genotypes of the female child are \(X^AX^a\) (carrier) and \(X^aX^a\) (affected).
The probability of getting \(X^aX^a\) (affected) is \(\frac{1}{2}\) (because when considering the \(X\) - chromosome contribution to a female, the mother can give \(X^A\) or \(X^a\) (50% chance for each), and the father gives \(X^a\) always for a female).
The probability of getting \(X^AX^a\) (carrier) is \(\frac{1}{2}\)

Step3: Determine if the child can be standard

A standard (non - carrier and non - affected) female would have the genotype \(X^AX^A\). Since the father has \(X^a\) (and must give an \(X\) - chromosome to a female child) and the mother has \(X^A\) and \(X^a\), the child cannot have \(X^AX^A\)

Answer:

50%
50%
False