Solutions to Practice Problems in Population Genetics

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since 5 March 1996

1. Suppose 25 out of 750 students are redheads. What is frequency of redheads? If a random student is choosen, what is the probability they are a redhead?
   • Freq (Redheads) = 25/750 = 0.033 or 3.3 percent

   • Probability of a Redhead = 25/750, or 3.3 percent

2. Consider a locus with two alleles, A and a. If the frequency of AA is 0.25, what is the frequency of A under Hardy-Weinberg?
   • Under H-W, if p = freq(A) , then p² = freq(AA), hence p² =0.25 or p = 0.5.

3. If the genotypes AA, Aa, and aa have frequencies 0.5, 0.25, and 0.25 (respectively), what are p = freq(A)? q = freq (a)? After a single generation of random mating, what is the expected frequency of AA? of Aa? of aa?
   • p = freq(AA) + (1/2) freq(Aa) = 0.5 + (1/2)(0.25) = 0.625.

   • q = 1-p = 0.375

   • freq(AA) = p² = 0.625² = 0.391

   • freq(Aa) = 2pq = 2*0.625*0.375 = 0.469

   • freq(aa) = q² = 0.375² = 0.140

4. Consider a locus with 12 alleles, A₁, A₂, ... , A₁₂. What is the frequency of allele A₁ if we know that freq(A₁A₁) = 0.10, and that the frequency of all heterozygote genotypes containing A₁ is 0.40. Under the assumption of Hardy-Weinberg, what is the expected frequency of A₁A₁? Of any heterozygote involving A₁?
   • freq(A₁) = freq(A₁A₁) + (1/2)(all A₁ heterozygotes) = 0.10 + (1/2)(0.4) = 0.3

   • freq(A₁A₁) = freq(A₁)² = 0.3² = 0.09

   • freq(A₁A_x) = 2*freq(A₁)(1- freq(A₁)) = 2*0.3*0.7 = 0.42

5. Assume Hardy-Weinberg and a locus with two alleles. If p = freq(A), what fraction of all copies of allele a. are found in heterozygotes?
   • freq(a) = p = (1/2)freq(Aa) +freq(aa), so that the fraction of a found in heterozygotes is (1/2)freq(Aa) / freq(a) = (1/2) 2p(1-p) / (1-p) = p

6. Assume that A is completely dominant over a, with aa individuals displaying a disease while all others are normal. An individual of genotype is Aa is called a carrier as they appear normal, but can have offspring with the disease.

   (i) Under Hardy-Weinberg, if freq(a) = 0.001 what fraction of all copies of a are in carriers?

   • From Problem 5, fraction in carriers= freq(A) = 1-.001 =0.999

   (ii) What fraction of all aa individuals had both parents as carriers (i.e., neither parent displays the trait).
   • The frequency of Aa x Aa matings is (because of random mating) (2pq)*(2pq), with (1/4) of the offspring being aa, hence since the frequency of aa is q², the fraction of these offpsring having only carriers as parents is (1/4)*(2pq)*(2pq) / q² = p² = 0.999² = .998

7. If the genotypes AA, Aa, and aa have fitnesses 1: 1.5: 1.6, what allele is fixed?
   • Allele a, as genotype aa has the highest fitness.

8. The genotype aa is lethal and yet the population has an equilibrium frequency for a of .40. If the fitness of Aa is 1, what is the fitness of the AA genotype?
   • Recall if the genotypes AA: Aa: aa have fitnesses 1-s : 1 : 1-t, then the equilibrium frequency of A is t/(s+t).
     • Here, t = 1 (as aa is lethal), so that freq(A) = 1- freq(a) =0.6 = 1/(1+s)
     • Solving gives 1+s = 1/0.6, or s = 1/0.6 -1 = 2/3
     • Hence fitness of AA = 1-s = 1-2/3 = 1/3

9. Suppose a population starts out with 10,000 AA individuals, 20,000 Aa, and 10,000 aa.

   (i) What is the frequency of A? Is this population in Hardy-Weinberg?

   • freq(AA)=freq(aa) = 1/4, freq(Aa) =1/2.
   • freq(A) =freq(AA)+(1/2)freq(Aa) = 1/4 + (1/2)*(1/2) = 1/2

   • Under HW expect freq(AA) = p² =1/4, freq(Aa) = 2pq =1/2, and freq(aa) = q² =1/4 which is indeed what we see.

   (ii) Suppose all aa individuals die before reproducing, while (on average) AA and Aa individuals leave the same number of offspring. What are the fitnesses of these three genotypes?

   • fitness of aa = 0, while fitness of AA and Aa are the same, and we set this equal to one.

   (iii) Following selection, what fraction of the surviving adults are AA? Aa? aa? What is the frequency of a?

   • After selection, freq(aa) = 0, while the proportion of of AA and Aa is 1:2 after selection, so that freq(AA) = 1/(1+2) = 1/3 and freq(Aa) = 2/(1+2) = 2/3. Hence, freq(a) after selection = freq(aa) + (1/2)freq(Aa) = 0 + (1/2)(2/3) = 1/3.

   (iv) If these surviving adults mate at random, what is the frequency of a in the next generation (before selection acts).

   • Here freq(a) = 1/3, freq(A) = 2/3, giving
     • freq(AA) = (2/3)² = 4/9,
     • freq(aa) = (1/3)² = 1/9,
     • freq(Aa) = 2*(1/3)(2/3)= 4/9.