Question

1. create a punnett square for a cross of a female who is a hemophilia carrier with a normal male. hemophilia is a recessive trait. what are the odds of their daughter having hemophiliac? what about their son?

Explanation:

Step1: Determine Genotypes

Hemophilia is X - linked recessive. Let \(X^H\) be normal allele, \(X^h\) be hemophilia allele. Carrier female: \(X^H X^h\), normal male: \(X^H Y\).

Step2: Set Up Punnett Square

	\(X^H\)	\(Y\)
\(X^h\)	\(X^H X^h\)	\(X^h Y\)

Step3: Analyze Daughter Odds

Daughters get \(X\) from each parent. Genotypes: \(X^H X^H\) (normal), \(X^H X^h\) (carrier). No \(X^h X^h\) (hemophiliac) for daughters. Odds: \(0\) (or \(0\%\)).

Step4: Analyze Son Odds

Sons get \(X\) from mother, \(Y\) from father. Genotypes: \(X^H Y\) (normal), \(X^h Y\) (hemophiliac). 1 out of 2. Odds: \(\frac{1}{2}\) (or \(50\%\)).

Answer:

• Odds of daughter having hemophilia: \(0\) (or \(0\%\)).
• Odds of son having hemophilia: \(\frac{1}{2}\) (or \(50\%\)).

Punnett Square:

	\(X^H\)	\(Y\)
\(X^h\)	\(X^H X^h\)	\(X^h Y\)