Exam 2 practice problems #2
1.
A gene has two alleles b and B. The allele B has a frequency of 0.8 and b has a frequency of 0.2.
Fitness values for the genotypes are as follow:
Wbb = 5
wbB = 4
wBB = 3.
Calculate the average fitness of the population (wbar). [wbar means w with a line over the top. I can’t write
that so I’m using wbar instead].
Answer:
Freq of B (p) = 0.8
Freq of b (q) = 0.2
To calculate the average fitness of the population there are three steps: (i) figure out the proportions of
genotypes in the population, (ii) multiply each genotype frequency by its associated fitness and (iii) add
all three results together.
Wbar = (P2 * wBB) + (2pq * wbB) +(q2 * wbb)
Wbar = [(0.8)2 * 3] + [(2*0.8*0.2) * 4] + [(0.2)2 * 5]
Wbar = (1.92 + 1.28 + 0.2) = 3.4
2.
A gene has two alleles d and D. The frequency of d is 0.1 and the frequency of D is 0.9. The
genotype fitness values are as follow:
wdd = 9
wdD = 6
wDD = 3
Given that the average fitness of the population (wbar) is 3.6 calculate (1) the average excess of fitness
(ad) for the allele d
(2) calculate delta q the change in the frequency of d from one generation to the next
and (3) calculate the frequency of the allele d in the next generation
Answer
Part 1.
Freq allele D (p) =0.9
Freq allele d (q) = 0.1
The average excess of fitness ad = q (wdd-wbar) + p (wDd – wbar)
ad = 0.1 (9 - 3.6) + 0.9 (6 – 3.6)
ad = 0.54 + 2.16
ad = 2.7
Part 2.
delta q = q (ad /wbar)
delta q = 0.1 (2.7/3.6)
delta q = 0.075
part 3.
q t+1 = q + delta q
q t+1 = 0.1 + 0.075 = 0.175
3.
A gene has two alleles d and D. The genotype fitness are as follow:
wdd = 4
wdD = 1
wDD = 1
The frequency of allele D is 0.9 and the frequency of allele d is 0.1.
(a) Calculate the average fitness of the population (wbar )
(b) Calculate the expected frequency of the allele d in the next generation (t+1)
Answer
(a) Let frequency of D = p and frequency of d = q.
wbar = (p2 *wDD) + (2pq * wDd) + (q2 * wdd)
wbar = (0.81 * 1) + (0.18 * 1) + (0.01 * 4)
wbar = 0.81 + 0.18 + 0.04 = 1.03
(b)
Expected frequency of allele d(t+1) = [(q2 *wdd)/wbar] + [(pq *wdD)/wbar]
= [(0.01 * 4)/1.03] + [(0.09 * 1)/1.03]
= (0.04/1.03) + (0.09/1.03)
= 0.039 + 0.087
= 0.126
4.
Using the same values as in question 3 for genotype fitnesses repeat the same calculations as
for question 3, but this time for starting allele frequencies of D = 0.5 and d =0.5.
wdd = 4
wdD = 1
wDD = 1
(a)Calculate the average fitness of the population (wbar )
(b) Calculate the expected frequency of the allele d in the next generation (t+1)
(c) Explain why the frequency of allele d increases much more in generation t+1 than it did in
question 3.
Answers
(a) Let frequency of D = p and frequency of d = q. p=0.5, q=0.5
wbar = (p2 *wDD) + (2pq * wDd) + (q2 * wdd)
wbar = (0.25 * 1) + (0.5 * 1) + (0.25 * 4)
wbar = 0.25 + 0.5 + 1 = 1.75
(b) Expected frequency of allele d(t+1)
= [(q2 *wdd)/wbar] + [(pq *wdD)/wbar]
= [(0.25 * 4)/1.75] + [(0.25 * 1)/1.75]
= (1/1.75) + (0.25/1.75)
= 0.0.571 + 0.143
= 0.714
(c) In question 3 the frequency of allele d increased from 0.1 to 0.126 an increase of 0.026. In
question 4 the frequency of allele d increased from 0.5 to 0.714, an increase of 0.214, which is a
much larger increase.
Looking at the fitness values of the genotypes it is clear that the fitness of the homozygote for the d
allele has by far the highest fitness (wdd = 4) [both the homozygote DD and heterozygote Dd have fitness
values of 1] so we would expect d to increase in frequency as it did in both cases.
The reason d increases more rapidly in question 4 is that with a starting allele frequency of 0.5 there are
far more dd homozygotes in the population (q2 = 0.25—i.e. 25% of the whole population) than when the
starting allele frequency is 0.1 (q2 = 0.01 – only 1% of the population are homozygous dd). The larger
number of dd homozygotes means that more copies of the d allele are visible to selection and so the
allele increases faster.