Question

q6.if a female carries a gene for haemophilia on one of her x chromosome, marries a normal man, what is the probability of producing sons or daughters who have haemophilia? please draw chart for explanation comm /5

Explanation:

Step1: Determine the genotypes

Let \(X^h\) represent the X - chromosome with the haemophilia gene and \(X^H\) represent the normal X - chromosome. The female carrier has the genotype \(X^H X^h\), and the normal male has the genotype \(X^H Y\).

Step2: Create a Punnett square

Gametes of female (\(X^H X^h\))	Gametes of male (\(X^H Y\))
\(X^h\)	\(X^H X^h\) (carrier female)	\(X^h Y\) (haemophilic male)

Step3: Calculate probabilities

• For sons: There are 2 possible genotypes for sons (\(X^H Y\) and \(X^h Y\)). The probability of a son having haemophilia (\(X^h Y\)) is \(\frac{1}{2}\) or \(50\%\).
• For daughters: There are 2 possible genotypes for daughters (\(X^H X^H\) and \(X^H X^h\)). The probability of a daughter having haemophilia is \(0\) (since a daughter would need \(X^h X^h\) to have haemophilia, and there is no \(X^h\) from the father to pair with the \(X^h\) from the mother in a way to get \(X^h X^h\)).

Answer:

The probability of producing a son with haemophilia is \(50\%\) (or \(\frac{1}{2}\)), and the probability of producing a daughter with haemophilia is \(0\).