Question

hemophilia is an x - linked recessive trait. the normal allele is xn, and the affected allele is xn.

1. what is the genotype of individual i - 1?
2. what is the genotype of individual ii - 4?
3. how is it possible that one child produced in generation iii is affected and the other is not affected?

Explanation:

Step1: Analyze genotype of I - 1

Since hemophilia is X - linked recessive, females have two X chromosomes. Individual I - 1 is a female (circle in pedigree). She has a shaded (affected) part. The genotype of a female with X - linked recessive trait (when she is a carrier or affected) is determined by alleles. Since the normal allele is \(X^N\) and affected is \(X^n\), and she is a carrier (half - shaded, assuming in this simple representation for X - linked in females where half - shaded can represent carrier in some basic pedigree setups for simplicity here), her genotype is \(X^NX^n\).

Step2: Analyze genotype of II - 4

Individual II - 4 is a male (square in pedigree). Males have one X and one Y chromosome. Since he is affected (shaded), and the affected allele is \(X^n\), his genotype is \(X^nY\).

Step3: Analyze offspring of II - 3 and II - 4

II - 3 is a female (circle, half - shaded, so genotype \(X^NX^n\)) and II - 4 is \(X^nY\).

• For females:
• The possible crosses are \(X^NX^n\times X^nY\). The possible genotypes of offspring:
• \(X^N\times X^n = X^NX^n\) (normal - carrier female)
• \(X^N\times Y=X^NY\) (normal male)
• \(X^n\times X^n = X^nX^n\) (affected female)
• \(X^n\times Y = X^nY\) (affected male)
• Using Punnett square:

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• The probability of getting an affected child (\(X^nX^n\) or \(X^nY\)) is \(\frac{1}{2}\) and probability of getting a non - affected child (\(X^NX^n\) or \(X^NY\)) is \(\frac{1}{2}\). When they have children, due to independent assortment of alleles during gamete formation (meiosis), 50% chance of passing the affected \(X^n\) (from mother II - 3) and \(X^n\) or \(Y\) (from father II - 4) to get an affected child (\(X^nX^n\) or \(X^nY\)) and 50% chance of passing \(X^N\) (from mother II - 3) and \(X^n\) or \(Y\) (from father II - 4) to get a non - affected child (\(X^NX^n\) or \(X^NY\)).

Answer:

1. \(X^NX^n\)
2. \(X^nY\)
3. II - 3 (\(X^NX^n\)) and II - 4 (\(X^nY\)) have a 50% chance of passing on the affected \(X^n\) allele (for an affected child) and 50% chance of passing on the normal \(X^N\) allele (for a non - affected child) due to independent assortment of alleles during meiosis and fertilization.