Human population genetics question

Question

4) (10 pts). Consider a population that is made up of all AA individuals. If all of our Hardy Weinberg assumptions hold, except for one-way mutation, in which A mutates to a at a rate of 0.0005, a) how many generations will it take until the population is only 25% AA individuals? (Hint: What allele frequency is the final population at?) a) __________ generations

For this question I ended up using natural logs to get the answer but I'm not sure if this was the right way to do it.

After finding that A=0.5 and a=0.5 I used the formula

p'=p-µp

Then:

p_t = p_i (1- µ)^t (Where p_i= 1)

=(1-0.0005)^t = 0.5

and then isolating the t and taking the natural log of both sides

t= ln (0.5)/ ln (1-0.0005) = 1385.5 ???

Is this right?? Any help is appreciated. I'm forever bitter that I can't just snap a picture of my work for easier clarification. Thank you in advance!

Answer 1

Your approach and reasoning are indeed correct! Let’s go through the steps to confirm:

Problem Recap:

1. You start with a population of only AA individuals, meaning the initial frequency of allele AAA (denoted as ppp) is 1.
2. There is a one-way mutation from AAA to aaa with a mutation rate μ=0.0005.
3. You want to know how many generations it will take for the population to reach only 25% AA individuals.

Steps to Solution:

1. Identify Target Frequency:
2. For the population to be 25% AA individuals, we need the frequency of genotype AA to be 0.25.
3. In Hardy-Weinberg equilibrium, the frequency of AA is p² so p² = 0.25

Solving for p, we get: p=0.√25 ⇒ p ​= 0.5

1. Therefore, we want the frequency of allele A, p to drop from 1 to 0.5.
2. Use the Mutation Formula:
3. With one-way mutation, the allele frequency after ttt generations can be described by:
4. p_t = p_i ( 1- μ)^t
5. where pi=1 is the initial frequency of A, μ=0.0005 and pt=0.5
6. Set Up the Equation and Solve for ttt:
7. 0.5=(1)(1−0.0005)t0.5 = (1)(1 - 0.0005)^t0.5=(1)(1−0.0005)t
8. Taking the natural log of both sides:
9. ln⁡(0.5)=tln⁡(1−0.0005)\ln(0.5) = t \ln(1 - 0.0005)ln(0.5)=tln(1−0.0005)
10. Solving for ttt:
11. t=ln⁡(0.5)ln⁡(1−0.0005)t = \frac{\ln(0.5)}{\ln(1 - 0.0005)}t=ln(1−0.0005)ln(0.5)​
12. Calculate t:
13. ln⁡(0.5)≈ −0.6931
14. ln⁡(1−0.0005)≈−0.0005 (for small values, ln⁡(1−x) ≈ −x
15. Substituting these values:
16. t=−0.6931 / -0.0005 ≈ 1386

Conclusion

Your answer of approximately 1385.5 generations (rounded to 1386) is correct. Great job using natural logs to solve this problem!