X Linked Hardy Weinberg Equilibrium Problem

In a given population under Hardy Weinberg equilibrium, 40.0% of men have hemophilia. What is the probability that a random man and random woman will have a daughter with haemophilia?

I think the answer is 16%, but the answer given is 9.6%. According to Hardy-Weinberg principle, p2 + 2pq + q2 = 1. In order to inherit the disease, the mother must either be a carrier of have the disease, which occurs with probability 1-q2 = 0.72

Therefore, the odds of having a child with the disease is (0.84)(0.4). Since it asks for the probability of a girl, the total must be divided by two, so the answer is 0.168

Where am I wrong?

I make it 8%. Here is my reasoning.

40% mutant males, so freq(mutant allele) = p = 0.4, and freq(wt allele) = q = 0.6

To get a mutant female we have to have a mutant male parent, probability = 0.4 Of these matings one half will produce a female offspring so 0.4*0.5 = 0.2

i.e. 20% of matings derive from a mutant male and produce a female offspring.

Now look at the female mate:

probability(mutant) = p2 = 0.16

probability(carrier) = 2pq = 0.48

probability(wt) = q2 = 0.36

so our 20% of matings that have the potential to produce a mutant female offspring partition as:

mating with a mutant female: 0.2 x 0.16 = 0.032 mutant female offspring

mating with a carrier female: 0.2 x 0.48 = 0.096 of which:

0.048 mutant female offspring

0.048 carrier female offspring

mating with wt female: 0.2 x 0.36 = 0.072 carrier female offspring

(sanity check - 0.032 + 0.096 + 0.072 = 0.2)

Thus the probability of random mating producing a mutant female is 0.032 + 0.048 = 0.08 (8%)

Incidentally there is another way of thinking about this. Note that the H-W frequency of mutant females in the population is 16%. One of the assumptions of H-W is random mating. So the probability of a random mating producing a mutant female = p(female) * p(mutant if female) = 0.5 x 0.16 = 0.08

So - where have I gone wrong?

• I get the same result. These type of questions can throw people off very easily, apparently including the people writing the answer keys! – A. Kennard Mar 8 '14 at 1:01

In a given population, 40% of men have hemophilia – an X-linked recessive disorder. What are the odds that a random woman and a random man from that population will have a daughter with hemophilia? Hemophilia is X-linked and recessive, so the frequency of males having the disease = q. So, q = 0.40. To determine the frequency of the dominant allele in the population, use… p + q = 1 p + 0.4 = 1, p = 0.6 Use these allele frequencies to calculate the genotype frequencies in the females using the Hardy-Weinberg equation: P2 + 2pq + q2 = 1 0.36 + 0.48 + 0.16 = 1 Now use these frequencies in two separate Punnett squares 1) All of the offspring from a homozygous recessive woman and a hemophiliac man will have hemophilia. Thus, (0.16)(0.4) = 0.064. Half of these offspring will be daughters, so 0.064/2 = 0.032

2) Half of the offspring of a cross between a heterozygous woman and a hemophiliac man will have hemophilia. Thus, (0.48)(0.4)(.5) = 0.096. Half of these offspring will be daughters, so 0.096/2 = 0.048.

3) Add the two possibilities together, so 0.032 + 0.048 = 0.08, or 8%

Frequency of allele for Haemophilia (q) = 0.4
Frequency of normal allele (p) = 0.6

Cross between Heterozygote female and hemophiliac male by punnett square:
Probability of hemophiliac daughter = 0.5
P(hemophiliac male) = 0.4
P(heterozygote female) = 2pq = 2 × 0.4 × 0.6 = 0.48
P(hetero female+hemophiliac male+hemophiliac daughter) = 0.5 × 0.4 × 0.48 = 0.096

Hence 9.6%

In a given population, 40% of men have hemophilia – an X-linked recessive disorder. What are the odds that a random woman and a random man from that population will have a daughter with hemophilia? Hemophilia is X-linked and recessive, so the frequency of males having the disease = q. So, q = 0.40. To determine the frequency of the dominant allele in the population, use… p + q = 1 p + 0.4 = 1, p = 0.6 Use these allele frequencies to calculate the genotype frequencies in the females using the Hardy-Weinberg equation: P2 + 2pq + q2 = 1 0.36 + 0.48 + 0.16 = 1 Now use these frequencies in two separate Punnett squares 1) All of the offspring from a homozygous recessive woman and a hemophiliac man will have hemophilia. Thus, (0.16)(0.4) = 0.064. Half of these offspring will be daughters, so 0.064/2 = 0.032

2) Half of the offspring of a cross between a heterozygous woman and a hemophiliac man will have hemophilia. Thus, (0.48)(0.4)(.5) = 0.096. Half of these offspring will be daughters, so 0.096/2 = 0.048.

3) Add the two possibilities together, so 0.032 + 0.048 = 0.08, or 8%