More Population Genetics Practice Problems

1. A) The selective coefficient for the AA genotype is 0.1. The selective

coefficient for the aa genotype is 0. There is complete dominance of the A

allele over the a allele. What are the fitnesses of the three genotypes?

WAA=0.9 WAa= 0.9 Waa= 1.0

B) If 40 out of a hundred individuals with the AA and Aa genotypes survive

to reproductive age and 80 out of 100 individuals with the aa genotype

survive to reproductive age what are the fitnesses of the three genotypes?

WAA=0.4/0.8=0.5 WAa= 0.4/0.8=0.5 Waa=0.8/0.8=1.0

2. A) Phenylketonuria (PKU) is an example of a genetic disease whose frequency may be due to an equilibrium between selection and mutation. If the mutation rate to PKU is 1x10^-5 and the fitness of individuals with PKU is half the fitness of individuals without PKU what is the frequency of the PKU allele at equilibrium?

The frequency of rare deleterious alleles in the population is influenced by both selection and mutation. Selection continually removes alleles from the gene pool, while mutation continually adds alleles to the gene pool ( see your text page 622). The balance between these two forces leads to an equilibrium frequency which depends on both the strength of selection against the allele and the rate of mutation to the allele. This equilibrium frequency can be calculated using the following formula:

B) At equilibrium, what is the frequency of the individuals with PKU in the population?

Frequency of homozygous recessive individuals= q²= (0.0045)² =0.00002

C) PKU has become much less severe with modern medical treatment. If the fitness of individuals with PKU increases to 0.8 what will be the new equilibrium frequency of the allele?

If the fitness of PKU individuals is now 0.8, then the selection coefficient is s=0.2. The new equilibrium frequency is:

D) What will the new equilibrium frequency of individuals with PKU ?

Frequency = 0.0071² = .00005

3. Ectrodactyly, also known as “lobster claw” is an autosomal recessive condition in humans in which the fingers and toes are fused. Otherwise, individuals with this condition are healthy. Suppose the frequency of ectrodactyly among newborns is approximately 1/10,000 (when the parents are unrelated).

(a) What is the risk of a child with ectrodactyly if the parents are siblings?

If the frequency of ectrodactyly among newborns is 1/10,000 when the parents are unrelated, then the allele frequency (q) of the disease allele (a) is

q = (1/10,000) = 0.01, p=0.99

To estimate the risk of getting an ectrodactyly affected child from a brother-sister mating, we need to calculate the inbreeding coefficient (F) for such a mating:

F = P(homozygous by descent for full siblings) = 1/4

The frequency of homozygous recessives for full sib mating is: q² + pq(1/4)= (0.01)² +(0.01)(0.99)(1/4 ) = 0.0026 = approximately three in every thousand

(b) What is the risk of a child with PKU if the parents are second cousins?

F for second cousins is 1/64, so following the same calculations as above:

The frequency of homozygous recessives for second cousin mating is: q² + pq(1/64)= (0.01)² +(0.01)(0.99)(1/64 ) = 0.00025 = approximately two-three in every ten thousand

4. In a large, randomly mating population of giraffes with no movement of individuals in and out of the population and no mutation, the two alleles at one gene, A and a, do not affect survival or reproduction of the giraffes. The two alleles at the other gene, S and B, do affect survival to adulthood by affecting the degree to which giraffes are subject to parasitism by ticks and biting flies. BB individuals have bitter blood, which repels parasites, and survive best. SB individuals have neutral tasting blood and survive 80% as well as do BB individuals. SS individuals have sweet blood and attract parasites; they survive only 4% as well as do BB individuals.

a) In a population of ADULT giraffes, the frequency of individuals with the AA genotype is 0.04. Calculate the frequency of individuals with the Aa genotype.

At the A,a gene, the population is in Hardy-Weinberg Equilibrium. So:

Adult Frequency of AA = 0.04 = p²

Allele frequency of A=p=square root of 0.04 = 0.20

Allele frequency of a=q=1-p=1-0.20=0.80

Genotype frequency of Aa=2pq=2*(0.20)*(0.80)=0.32

b) In a population of gametes at the start of a generation, the frequency of the B allele is 0.14. Calculate the frequency of zygotes with the SB genotype

Allele frequency of B=0.14=p Allele frequency of S=1-p=1-0.14=0.86

Zygote genotype frequencies are in Hardy-Weinberg proportions, so:

Genotype frequency of SB in zygotes =2pq=2(0.14)(0.86)=0.24

c) Continuing from part (b), calculate the frequency of the adults that survive from the zygotes that have the SB genotype.

w_BB =1 w_SB = 0.80 w_SS = 0.04

wbar=w_BBp²+w_SB2pq+w_SSq² = (1)(0.14)² + (0.80)(2)(0.14)(0.86) + (0.04)(0.86)² =0.24

Adult genotype frequencies: Freq(BB)=w_BBp²/wbar = (1)(0.14)² /0.24 = 0.08

Freq(SB)=w_SB2pq/wbar = (0.80)(2)(0.14)(0.86)/0.24=0.80

Freq(SS)=w_SSq² /wbar = (0.04)(0.86)² /0.24=0.12

d) Continuing from part (c), calculate the frequency of the B allele in the gametes that will produce the next generation

Allele frequencies produced by adults to start the next generation:

Frequency of B=Freq(BB)+(1/2)Freq(SB) = 0.08+(1/2)(0.8)=0.48

5. Consider two genes in a large, randomly mating population of turtles with no movement of individuals in and out of the population and no mutation. The two alleles at one gene, L and M, do not affect fitness. The two alleles at the other gene, T and t, do affect fitness -- they affect the thickness of turtle shells, and the degree to which they are protected from predation. TT individuals have thick shells, which repel predators, and survive best. Tt individuals have medium shell thickness and survive 88% as well as do TT individuals. tt individuals have thin shells and are easy for predators to eat; they survive only 14% as well as do TT individuals.

a) In a population of ADULT turtles, the frequency of individuals with the MM genotype is 0.06. Calculate the frequency of individuals with the LM genotype.

Since there are no fitness differences, this gene is in Hardy-Weinberg Equilibrium.

Freq. MM = 0.06 = p²

Freq. M = p = the square root of p2 = the square root of 0.06 = 0.24

Freq. L = q = 1-p = 1-0.24 = 0.76

Freq. LM = 2pq = 2(0.24)(0.76) = 0.36

b) In a population of gametes at the start of a generation, the frequency of the t allele is 0.23. Calculate the frequency of zygotes with the Tt genotype.

Freq. t = 0.23 = q. Freq. T = p = 1-0.23 = 0.77

Zygote genotype frequencies are in Hardy-Weinberg proportions, so:

Freq. Tt = 2pq = 2(0.77)(0.23) = 0.35

c) Continuing from part (b), calculate the frequency of the adults that survive from the zygotes that have the Tt genotype.

To calculate adult genotype frequencies, we need to know the relative fitnesses of the genotypes. These are:

TT survives the best, and the highest relative fitness is always 1, so w_TT=1

Tt survives 88% as well as TT, so w_Tt = 0.88

tt survives 14% as well as TT, so w_tt = 0.14

Now, to calculate adult genotype frequencies, the first step is to calculate the average population fitness, wbar:

wbar = w_TTp² + w_Tt2pq + w_ttq²

wbar = (1)(0.77)² + (0.88)(2)(0.77)(0.23) + (0.14)(0.23)² = 0.91

Freq. Tt = w_Tt2pq / wbar = (0.88)(2)(0.77)(0.23) / 0.91 = 0.34

d) Continuing from part (c), calculate the frequency of the t allele in the gametes that will produce the next generation

To calculate the frequency of t, we need to use the formula Freq. t = Freq. tt + (1/2) Freq. Tt

To use this, we first calculate Freq. Tt = w_ttq² / wbar = (0.14)(0.23)² / 0.91 = 0.0081

Freq. t = Freq. tt + (1/2) Freq. Tt = (0.0081) + (1/2)(0.34) = 0.18

6. A population of 80 squirrels resides on campus, and the frequency of the Est¹ allele among these squirrels is 0.70. Another population of squirrels is found in a nearby woods, and there the frequency of the Est¹ allele is 0.50. During a severe winter, 20 of the squirrels from the woods population migrate to campus in search of food and join the campus population. What will be the frequency of the Est¹ allele in the campus population in the spring?

This case is analogous to the “one-island” migration model, with the campus population representing the island population (where migrants are moving to) and the woods population being the continent population. So p_I = 0.7 and p_C = 0.5. The new joint population is made up of 20 migrants and the original 80 campus squirrels. Therefore, in this new population the frequency of migrants (and migrant genes) is

m =20/(20 + 80) = 0.2 and the frequency of nonmigrants is 1-m = 0.8

The frequency of the Est1 allele after migration will be given by the equation:

= (0.8)(0.7) + (0.2)(0.5) = 0.56 + 0.1 = 0.66

7. Calculate the effective population size for a breeding population of 60 adult males and 40 adult females.

Effective population size is given by the equation:

So in this problem Ne = (4)(60)(40)/ (60+40) = 96

8. In a population, the Dàd mutation rate is 4x10^-6. If p=0.8 today, what will p be after 50,000 generations?

0.65

9. As discussed in class, the gene for sickle-cell anemia exhibits heterozygote advantage in environments with a high prevalence of malaria. An individual who is an H^AH^S heterozygote has increased resistance to malaria and has higher fitness than the H^AH^A homozygote who is susceptible to malaria, and the H^SH^S homozygote who has sickle-cell anemia. Suppose the fitness values for the genotypes in a population are as follows:

H^AH^A = 0.91 H^AH^S = 1.0 H^SH^S = 0.04

Give the expected equilibrium frequencies of the two alleles in this population.

For a gene that shows heterozygote advantage will reach an equilibrium frequency of:

where s is the selection coefficient against the homozygous dominant genotype (which in this case is 1-0.91 =0.09) and t is the selection coefficient against homozygous recessive genotype (1-0.04=0.96), so

q= (0.09)/ (0.09+0.96) = 0.086

10. Determine the ultimate fate of the A allele with random mating when the relative fitnesses of AA, Aa, and aa are: (2 pts.)

a) 1.00, 0.98, 0.96

A fixed

b) 1.00, 1.02, 1.02

A lost

c) 1.00, 1.02, 0.98

stable equilibrium

d) 1.00, 1.01, 1.02

A lost

11. The fitnesses of 3 genotypes are: w_AA = 0.9, w_Aa = 1.0, w_aa = 0.7.

a) If the population starts at the allelic frequency of p = 0.5, what is the value of p in the next generation? (1 pt.)

0.528

b) What is the predicted equilibrium frequency? (1 pt.)

= (0.1)/(0.1+0.3) = 0.25