Practice problems for exam 2
1.
In a population of 800 people, 32 individuals suffer from a recessive inherited disorder.
Calculate the frequencies of the “disease” and “normal” alleles. Using the allele frequencies you
calculated above, calculate how many individuals you would expect to be homozygous for the
“normal” allele and how many individuals you would expect to be heterozygous for the
“normal” and “disease” alleles.
Let d = frequency of disease allele and D be the frequency of non-disease allele.
32/800 = 0.04. This is the frequency of the homozygous recessive genotype dd.
The HW term equal to this is q2. So q2 = 0.04. To calculate q we take the square root of 0.04, which is =
0.2.
Therefore, frequency of D is 1-0.2 = 0.8.
Expected frequency of homozygous individuals is p2, which is 0.8X0.8 = 0.64 which we multiply by the
population size 800 to get an expected number of homozygous “normal” individuals of 512.
Number of carriers is 2pq times 800, which is 2*0.8*0.2*800 = 256 carriers
To check your math, see if genotypes numbers add to 800. Add 256+512+32 which is indeed 800.
2. In humans the COL1A1 locus codes for a certain collagen protein found in bone. The normal
allele at this locus is denoted with S. A recessive allele s is associated with reduced mineral
density and increased risk of fractures in both Ss and ss women. A recent study of 1,778 women
showed that 1,194 were SS, 526 were Ss and 58 were ss. Showing your calculations, determine
whether these two alleles are in Hardy-Weinberg equilibrium in this population.
The counts of genotypes in the question are observed data. We want to see if they match what we
would have expected to find in the population based on the observed allele frequencies. Hence, we
need to figure out the allele frequencies and calculate expected genotype frequencies.
Given that we know the how many individuals there are of all three genotypes the way to calculate this
is to calculate the total number of S alleles and the number of s alleles in the population and estimate
allele frequencies that way and then calculate predicted genotype frequencies from that.
Hence: Total number of S alleles is 1194 + 1194 [remember each SS individuals has two alleles] + 526
[each Ss individual has one S allele]. This gives a total of 2914 S alleles. The total number of alleles (both
S and s) in the population is 1778*2 or 3556. Therefore the frequency of S is 2914/3556 = 0.8194.
The frequency of s is 58+58+526 = 642/3556 = 0.1806.
Check. p+q should add to 1. Do they? 0.8194 + 0.1806 = 1.
Use our p and q values to calculate the number of heterozygotes we would expect. This 2pq* population
size which is 2*0.8194*0.1806 * 1778 = 526.2 (close enough to the observed 526, the difference is due
to rounding decimals).
When you calculate expected SS and ss frequencies (p2 and q2, respectively) you get
p2 = (0.8194)*(0.8194) * 1778 = 1193.8, same as observed value of 1194 [slight difference is due to
rounding of decimals]
q2 = (0.1806)(0.1806)* 1778 = 57.99, same as observed value of 58
Population is in HW equilibrium.
3. On a remote archipelago in the Pacific Ocean lives a population of large flightless pigeons that
occurs on several islands in the archipelago. You visit the first island in the chain and are able to
take blood samples from the pigeons. From your analysis of these samples you determine the
genotypes of all the birds for a particular gene. In the population there are 200 birds with the
following genotypes:
88 AA
64 Aa
48 aa
Showing your calculations demonstrate that this population is not in Hardy Weinberg equilibrium. You
will need to figure out the allele frequencies from the data above and then use the Hardy-Weinberg
formula.
Total number of A alleles is 88+88+64 = 240. Frequency of A allele is 240/400= 0.6 (p).
Total number of a alleles is 48+48+64 = 160. Frequency of a allele is 160/400 = 0.4 (q).
[Total number of A and a alleles is 400]
Expected genotype frequencies are
p2 = 0.6*0.6 = 0.36. Expected number of AA individuals is 0.36*200 =72
2pq = 2*0.6*0.4= 0.48. Expected number of Aa individuals is 0.48*200= 96
q2 = 0.4*0.4 = 0.16. Expected number of aa individuals is 0.16*200 = 32
Compare observed and expected genotypes frequencies
Observed
Expected
AA
88
>
72
Aa
64
<
96
aa
48
>
32
We observe fewer heterozygotes than expected and more homozygotes. This would suggest that either
heterozygotes are being selected against or that the population is inbred.
4. On a remote island in the South Atlantic Ocean lives a population of red-eared squirrels. You
visit the island and find the population has 400 adults with the following genotypes:
144 AA
192 Aa
64 aa
Showing your calculations, tell me whether or not this population is in Hardy Weinberg equilibrium. You
will need to figure out the allele frequencies from the data above and then use the Hardy-Weinberg
formula. Show your calculations.
The next year you return to the island to examine the offspring of the previous year’s red-eared squirrels
(young red-eared squirrels don’t have red ears so you can easily identify them). You discover the
following genotypes among the 200 offspring you examine.
68 AA
105 Aa
27 aa
Using your allele frequency data from the previous generation show that this generation (the offspring
generation) is not in Hardy-Weinberg equilibrium Show your calculations.
Frequency of allele A is 144+ 144 + 192/800 = 480/800= 0.6 (p)
Frequency of allele a is 64+64+192 /800 = 320/800= 0.4 (q)
Expected AA genotype frequency = p2 = 0.6*0.6 = 0.36. Expected number of AA homozygotes is
0.36*400= 144
Expected aa genotype frequency = q2 = 0.4*0.4 = 0.16. Expected number of aa homozygotes is
0.16*400= 64.
Expected Aa genotype frequency = 2pq = 2*0.6*0.4 =0.48. Expected number of heterozygotes is
0.48*400= 192.
Observed frequencies = expected frequencies. Therefore, population in generation one is in HW
equilibrium.
Part 2.
Observed genotype frequencies among offspring
68 AA
105 Aa
27 aa
The expected genotype frequencies using the allele frequencies of the parental generation as the
question specifies (p=0.6 and q = 0.4) are
p2 = 0.6*0.6 =0.36. 0.36 x 200 [population size] gives the expected number of AA individuals, which is 72.
q2 = 0.4*0.4 = 0.16. 0.16 X 200 = expected number of aa individuals, which is 32
2pq = 2*0.6*0.4 = 0.48. 0.48 X 200 = expected number of Aa individuals which is 96.
Observed
Expected
68 AA
<
72AA
105 Aa
>
96Aa
27 aa
<
32aa
The observed and expected frequencies differ so the population is not in HW equilibrium. If we assume
that the results are signifuicantly different from random i.e. not just due to chance events (which we
could test for using a chi-squared test) then one of the five assumptions of HW must have been
violated. There is an excess of heterozygotes, so inbreeding is not the cause. Because the squirrels live
on a remote island in the Pacific migration is very unlikely. Mutation is also almost certainly not
responsible as mutation rates are usually too small to have a strong influence on allele frequencies and
you would also expect an increase in the frequency of one of the homozygotes too, not just in
heterozygotes. It could be selection favoring heterozygotes or possibly non-random mating if
homozygotes preferentially avoided mating with homozygotes with the same genotype as them.
Selection is probably the most plausible answer.
5. Two alleles A1 and A2 are the only alleles present in a gene pool. The frequency of A1 is 0.9 and
the frequency of A2 is 0.1. The fitnesses of the different genotypes are w11 = 0.8, w12 = 0.8, w22 =
2.0, where w11 is the fitness of the A1A1 genotype, etc.]. Calculate the frequency of the A1 allele
after one generation of selection. What percentage change in the allele frequency of A1 is this?
Using the same fitness values, if the starting frequency of A1 was 0.6 and A2 was 0.4 what would the
frequency of the A1 allele be after one generation of selection? What percentage change in the allele
frequency of A1 is this? Is there a difference from what you found in the first situation? If so, explain
why that might be?
Part 1.
First we need to calculate the average fitness of the population (symbolized by w with a line drawn over
the top. I will refer to this as wbar as I can’t easily draw a line over a letter).
Freq A1 = p, freq A2 = q.
P= 0.9, q =0.1
Wbar
= (p2Xw11 ) + (2pq X w12 )+ (q2Xw22)
= (0.81X0.8) + (0.18X0.8) + (0.01 X 2.0)
= 0.648 + 0.144 + 0.02
= 0.812
Freq A1 (after one generation of selection) p(t+1) = [(p2 X w11) + (pq X w12)]/ wbar
= [0.81 X 0.8 + 0.09 X 0.08]/wbar
= [0.648 + 0.072]/0.812
= 0.72/0.812
p(t+1) = 0.886
Percentage change in A1 allele = (0.9-0.886)/0.9 = 0.015 (multiply by 100 to convert proportion to a
percentage) = 1.5%.
So A1 has declined slightly in frequency about 1.5% from a frequency of 0.9 to 0.886. A2 has obviously
also increased as A1 has decreased.
Part 2. Repeat same process as for part one, but with different starting frequencies of A1 and A2. P = 0.6,
q = 0.4.
Calculate average fitness of population (wbar).
Wbar
= (p2Xw11 ) + (2pq X w12 )+ (q2Xw22)
= (0.36 X 0.8) + (0.48 X 0.8) + (0.16 X 2.0)
= 0.288 + 0.384 + 0.32
= 0.992
Using wbar calculate frequency of A1 in next generation
Freq A1 (after one generation of selection) p(t+1) = [(p2 X w11) + (pq X w12)]/ wbar
= [ (0.36 X 0.8) + (0.24 X 0.8)]/ 0.992
= [0.288 + 0.192]/ 0.992
= 0.48/0.992
= 0.484
Percentage decline in A1 calculated as (0.6-0.484)/0.6 = 0.193.
So A1 has declined significantly in frequency about 19.3% from a frequency of 0.6 to 0.484. A2 has
obviously also increased from 0.4 to 0.516 as A1 has decreased.
The reason A1 decreased so much more in part 2 than in part 1 is that in part 2 A2 was much more
common to begin with (q= 0.4). With the higher starting frequency of the A2 allele, far more individuals
were homozygous for that allele and these individuals have the highest fitness (2.0). In part 1, A2
homozygotes were rare (only 1% of all genotypes) so even though they were successful they contributed
relatively few extra A2 alleles to the next generation. As a result the increase in A2 alleles (and decline in
A1 alleles) was small.