# 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 think it is a case of the wording in the question: 'will have a daughter' still means we must consider all the offspring in ou probability. This means we must look at all the children they could have an including boys and girls and the probability that the daughter has haemophilia (with all of your working). If they had said: What is the probability that a random man and random woman's daughter has haemophilia? your answer would be correct, as we would have excluded the males from the offspring (as a conditional probability) which you had done correctly.
– Hen
Jun 13 at 4:44

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! 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%