Question

1. An isolated population of chimpanzees was discovered on an island. Biochemical studies show that 1 out of 9 chimpanzees in this population are homozygous for a mutation in the gene encoding the enzyme, hexokinase. This mutation completely inactivates the enzyme so that no activity is detected in the blood. A female who exhibits hexokinase activity mates with a male who is a carrier of the hexokinase mutation. What is the probability that their first baby chimpanzee will be homozygous for the hexokinase mutation? Assume that the population is in Hardy-Weinberg equilibrium.

2. Cystic fibrosis is an autosomal recessive, fully penetrant trait in humans.    A man who is unaffected by cystic fibrosis has two unaffected parents and a sister with cystic fibrosis.  The man plans to marry a healthy woman, who was adopted and has no knowledge about her biological parents except that they are both Caucasian. The frequency of cystic fibrosis in the Caucasian population is 1/2500. Assuming that this population is in Hardy-Weinberg equilibrium for the CF gene, what is the probability that this couple's first child will have cystic fibrosis?

Answer 1

1. It has been mentioned :

• the male is a carrier of the hexokinase mutation.
• the female exhibits hexokinase activity

This means that male also has hexokinase activity and only one of the alleles coding for the enzyme is active and the other is not. This means we should consider the female has hexokinase activity on both the alleles.

If we consider 'H' as normal allele and 'h' as the mutated one and cross them, we get:

HH × Hh

From this cross, we can predict that since only one of the parents has one mutated gene, there is Zero probability that any of their baby chimpanzees will be homozygous for the hexokinase mutation.

2. If the sister has CF but the parents do not, this means the parents are heterozygous for this gene. this means there is a probability of 0.5 that the man is heterozygous and a carrier of the CF gene.

If we consider that this population is in Hardy-Weinberg equilibrium for the CF gene. We knoe 1/2500 caucasians have cystic fibrosis. Since the woman in question has both caucasian parents, she is herself caucasian and there is 1/2500 chance that she was affected. But, since we know she is not, let us consider:

p²: Frequency of the population that is healthy and homozygous and have both alleles normal

pq: Frequency of the population to be heterozygous carriers of the CF gene

q² : Frequency of the population to be suffering from Cystic Fibrosis.

We know,

q² = 1/2500

implies, q = 1/50

in Hardy Wienberg population,

p +q = 1

Implies, p= 1- 1/50 = 0.98

Then, p² = 0.9604

Therefore,

p² + 2pq + q² = 1

implies, 0.9604 + 2pq + 0.0004 = 1

implies, 2pq = 0.0392

implies, pq= 0.0196

Hence the probability that the woman in question is heterozygous is 0.0196 and homozygous healthy is 0.9604

So, the total probability that the woman has CF gene = 0.0196

Then the probability that their first child had cystic fibrosis is :

0.0196 × 0.5 = 0.0098

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