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I have been given this question:

In a population of 2500 mice, light-colored fur is dominant to dark-colored fur. 2400 mice originally have light-colored fur. However, an owl species moves into their environment, preying mainly upon the light-colored mice as they are easier to spot when hunting at night. When the owls finally leave, 80% of the dark-colored mice are left, and only 10% of the light-colored mice. After many generations, Hardy-Weinberg equilibrium is re-established. Approximately what percentage of mice now have dark-colored fur?

Answer choices are:

A) 10%

B) 15%

C) 20%

D) 25%

E) 30%

F) 35%

My thought process goes as such:

When the owls leave, the groups are 240 light mice and 80 dark mice.

$q^2 + 2pq + p^2 = 1$

q is recessive allele, p is dominant allele.

$0.5^2 + 2*0.5*0.5 + 0.5^2 = 1$

Apply fitness:

$0.8*(0.5^2) + 0.1*(2*0.5*0.5) + 0.1*(0.5^2) = 0.275$

(0.8 and 0.1 correlate to the fitness of dark and light mice, respectively)

Now divide everything by 0.275 to figure out new allele frequencies:

$0.727 + 0.182 + 0.091 = 1$

This means that 0.727 = 73% is the new frequency for homozygous recessive (dark) mice.

However, the correct answer is B) 15%

Any idea what I did wrong/how the correct answer is achieved?

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  • $\begingroup$ After many generations, Hardy-Weinberg equilibrium is re-established. This is a pretty awful sentence (not your fault). HW equilibrium is reached after in a single generation. $\endgroup$ – Remi.b Aug 26 '17 at 23:11
  • $\begingroup$ I agree it's not written well. But as long as it doesn't alter the meaning of the question... :) $\endgroup$ – Little Dragon Aug 26 '17 at 23:21
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Revised to add complete solution

Before the owls arrive, 100 out of the 2500 mice are dark-colored. This ratio provides the frequency of the recessive allele

$$q^2 = \frac{100}{2500}$$

So $q$ is 10/50 or 0.2. This means that the frequency of the dominant allele $p$ is 1 - 0.2 or 0.8.

Knowing these frequencies, it is possible to calculate the number of each allele

$$AA = (0.8)^2 \times 2500 = 1600$$ $$Aa = 2 (0.8) (0.2) \times 2500 = 800$$ $$aa = (0.2)^2 \times 2500 = 100$$

Of the light-colored mice, 2/3 are AA and 1/3 are Aa. The same proportions hold after the owls leave, but the total humber has been reduced. The dark-colored mice are all aa.

$$AA = \frac{2}{3} \times 240 = 160$$ $$Aa = \frac{1}{3} \times 240 = 80$$ $$aa = 80$$

From these values, the total number of each allele may be calculated

$$A = 2 \times 160 + 80 = 400$$ $$a = 80 + 2 \times 80 = 240$$

These numbers provide the frequency of each allele post-selection

$$p = \frac{400}{640} \approx 0.63$$ $$q = \frac{240}{640} \approx 0.37$$

Applying Hardy Weinberg for the new equilibrium

$$aa \approx (0.37)^2 \approx 0.14 $$

which is, "approximately" 15%.

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  • $\begingroup$ 0.8*(0.2^2) + 0.1*(2*0.2*0.8) + 0.1*(0.8^2) = 0.128. Dividing homozygous recessive (0.8*0.2^2 = 0.32) by 0.128 leads to a frequency after selection of 0.25 = 25%. That's still not the correct answer... unless the answer key is incorrect. $\endgroup$ – Little Dragon Aug 26 '17 at 23:17
  • $\begingroup$ I'm sorry, but I don't understand what that calculation is supposed to be saying. I updated my answer with the complete solution. $\endgroup$ – Stephen Thomas Aug 27 '17 at 12:24

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