Solutions For An Introduction to Genetic Analysis Chapter 4 Problem 63P

Giving the genotypes of the individuals:

The mother of the two women III-4 and III-5 would produce the following gametes:

• 0.45 – HB

• 0.45 – hb

• 0.05 – hB

• 0.05 – Hb

The woman III-4 can be either Hb/HB [0.45% chance] or Hb/hB [0.05% chance] since she received the B allele from her mother.

The chance of her being Hb/hB, would be:

If her genotype is Hb/hB, then she will produce the same gametes as her mother.

Her child has:

• A 45% chance of receiving hB

• A 5% chance of receiving hb

• A 50% chance of receiving Y chromosome from his dad

Hence, the probability that her child will be a hemophiliac son would be:

The woman III-5 can either be Hb/Hb [0.05% chance] or Hb/hb [0.45% chance] since she received the b allele from her mother.

The chance of her being Hb/hb, would be:

Her child has:

• A 50% chance of receiving the h allele

• A 50% chance of receiving Y chromosome from his dad

Hence, the probability that her child will be a hemophiliac son would be: