Hardy Mouse

Hardy Mouse
Can someone help me on 3 Hardy Weinburg Problems?

I need help on these 3 Hardy Weinberg Problem..can someone go through the steps of each one to show how you got the answer? I’m totally confused!

Problem 4: Hardy-Weinberg says that p^2 + 2pq + q^2 = 1 when p represents the dominant allele and q is the recessive allele. The white mice are the q^2 because they must have two of the recessive alleles each.

q^2 = 0.49 Changed 49% to a decimal.
q = 0.7 Because 0.7 squared is 0.49.

Start substituting in the equation to find out what p equals.

p^2 + 2pq + q^2 = 1
p^2 + 2p(0.7) + 0.49 = 1
p^2 + 1.4 p – 0.51 = 0 Simplified and put into standard form.

Solve using the quadratic formula or factor and solve.
(p + 0.17)(p – 0.3) = 0
p = -0.17 or p = 0.3 (answer must be positive)

Either way you get p = 0.3 which is the frequency of the dominant allele.
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Problem 5: If 75% show free earlobes, then 25% show attached earlobes. p^2 + 2pq = 0.75 q^2 = 0.25 (so q = 0.5)
p + q = 1 This is an underlying principle of H-W
p + 0.5 = 1
p = 0.5

2 pq = 2(0.5)(0.5) = 0.5 So half of the population is heterozygous.

The population is 9375. Half of 9375 = 4687 or 4688 people are heterozygous.
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Problem 7: q^2 is the frequency of homozgous recessive (1/25000 in this problem, which is 0.00004).

q^2 = 0.00004, so q = 0.0063
p + q = 1, so p = 1 – 0.0063 = 0.9937

According to equation p^2 + 2pq + q^2 = 1
Proportion of the carriers = 2 pq
Substitute figures from above for p and q
2 pq = 2(0.0063)(0.9937) = 0.0125 which is just about 1 out of 800 since I had a bit of rounding along the way.

Answer 1/800 or 0.0125 of the population is heterozygous.

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