LS4 Problem Set 2, 2010
Population Genetics
Corresponding Lectures: November 5, 8 and 10
Corresponding Reading: Hartwell et al., 757-773
Corresponding Quiz: November 15-19 (Along with positional cloning)
If you want to try additional problems, there are many at the end of chapter 21 that
pertain to this section of the course.
Problem 1.
a) What are the allele frequencies of M and N in this population?
Freq M = [(550*2)+150]/(2*1000) = 0.625
Freq N = [(300*2)+150]/(2*1000) = 0.375
b) If this population is at Hardy-Weinberg equilibrium, how many individuals with a
genotype of MN would you expect to see?
2pq = 2*0.625*0.375 = 0.469
0.469*1000 = 469 people
Problem 2.
a) What are the allele frequencies of R and r in population 1?
p = Freq R = [(2*10)+74]/(2*100) = 0.47
q = Freq r = [(2*16)+74]/(2*100) = 0.53
b) What are the allele frequencies of R and r in population 2?
p = Freq R = [(2*81)+18]/200 = 0.9
q = Freq r = [(2*1)+18)]/200 = 0.1
c) Which population is at Hardy-Weinberg equilibrium?
Population 2 is at H.W. eq
Population 1 expected genotype frequency
p^2 = 0.47^2 = 0.2209 è 0.2209 *100 = 22 RR individuals
2pq = 2*0.47*0.53 = 0.4982 à 0.4982 * 100 = 50 Rr individuals
q^2 = 0.53^2 = 0.2809 à 0.2809 * 100 = 28 rr individuals
Doesn’t match given observed values, not in H.W. eq
Population 2 expected genotype frequency
p^2 = 0.9^2 = 0.81 à 0.81*100 = 81 RR individuals
2pq = 2*0.9*0.1 = 0.18 à 0.18*100 = 18 Rr individuals
q^2 = 0.1^2 = .01 à .01*100 = 1 rr individual
This matches the observed values, so population 2 is at H.W. eq
Problem 3.
1. List the expected values for each genotypic class under your null hypothesis.
A1A1 = 306
A1A2 = 170
A2A2 = 24
2. Perform the chi-square calculation.
χ2 = .241
3. State your conclusion based on the data.
population is in HWE
(3-1 = 2 dof; 0.90 > p > 0.50)
Problem 4.
A) Is the population at H-W equilibrium for the A gene? How about the B gene? (Show
your work)
Determine allelic frequencies:
f(A) = p = ( 202 +202 + 101 + 372 + 372 + 186 + 166 + 166 + 83 ) / 2960 = .625
f (a) = q = 1 – p = 1 - .625 = .375
Now find the genotypic frequencies and the number of expected individuals out of
1480 to have this genotype and compare it to the observed number:
f (AA) = p2 = .6252 = .39 X 1480 = 577 expected and 740 observed
f (Aa) = 2pq = 2 X .625 X .375 = .47 X 1480 = 696 expected and 370 observed
f (aa) = q2 = .3752 = .14 X 1480 = 207 expected and 370 observed
Expected and observed are not similar so the A gene is not at H-W equilibrium.
Do the same thing for the B gene and you will find that the number of expected
and the number of observed are nearly identical. Thus, the population is at H-W
equilibrium for the B gene.
B) After one generation of random mating, what fraction of the next generation will be AA
(independent of the B gene)?
Since we are at H-W equilibrium allele frequencies do not change.
The f (AA) is simply p2 = .6252 = .39
C) After one generation of random mating, what fraction of the next generation will be
BB (independent of the A gene)?
This is the same as above and f (BB) is simply p2 = .522 = .27
D) What is the chance that the first child of a Aa Bb female and a AA Bb will be a Aa bb
male? (Hint: Do you need to consider allele frequencies to solve this problem? Think
back to earlier in the quarter.)
There is no need to consider allele frequencies. Allele frequencies apply to
populations and can assign the probability of a certain genotype. But we know
the genotypes of the parents in this problem. The probability of the child being Aa
is 1/2 (do the punnett square if you need convinced), the probability of being bb is
1/4 (again do the punnett square), and the probability of being male is 1/2.
1/2 X 1/4 X 1/2 = 1/16
Problem 5.
a) Find the allele frequencies in this generation of surviving flies:
+ = 0.7
m = 0.3
b) If you assumed HW equilibrium, what are the expected numbers of different
genotypes in the next generation if 200 viable offspring of the population in part (a) are
counted? Make sure to take into account the fact that one genotype dies! (write out all
possible genotypes and write numbers of flies next to them)
+/+ = 108
m/+ = 92
m/m = 0
Problem 6.
a.) What is the allele frequency of the M allele in the sailor population?
(450+450+45)/1000= 0.945
b.) 10 years later 1000 children have been born on the coastline. (All unions were
consensual!) If the population of 1000 young people on the coast mate randomly, how
many of the 1000 children would you expect to have MN bloodtype?
469.9 ~ 470. (Be careful with rounding too early!)
Tunisian Population:
45 MM
210 MN
245NN
Tunisian population+ Sailor Population New Genotype Frequency:
495 MM
255 MN
250NN
New Allele Frequency
M=0.6225
N= 0.3775
2pq= 2 (0.6225)(0.3775)= 469.9
c.) In fact, 50 children have MM bloodtype, 850 have MN bloodtype and 100 have NN
blood type. What is the observed allele frequency of the N allele among the children?
Frequency of N
(850+100+100)/ 2000= 0.525 = 52.5%
Problem 7.
a. What are the genotype frequencies in the population?
CR CR = 100/1000=0.10
CR CW = 740/1000= 0.74
CW CW = 160/1000= 0.16
b. What are the allele frequencies of CR and CW in this population?
CR = 200+740 / 2000= 0.47
CW = 1- 0.47 = 0.53
c. what are the expected frequencies of the genotypes if the population is at HardyWeinberg equilibrium?
P2 = (0.47)2 = 0.22 CR CR
Q2= (0.53)2 = 0.28 CW CW
2PQ= 2 (0.47)(0.53) = 0.50 CR CW
Problem 8.
Assuming that the population is at Hardy-Weinberg equilibrium, what proportion of the
males in this population will also have the condition?
Xs = dominant allele for condition. X= recessive, doesn’t have the condition.
Females with Xs_ have the condition and 36% are Xs Xs and Xs X genotypes. Thus,
64% is XX.
Since 64% is XX (p2), the X (p) allele frequency is 0.8.
Using p + q =1, Xs (q) allele frequency is 0.2
For male population:
Since male has only one X, the genotype frequencies are equal to the allele
frequencies..
Thus males with XsY have the condition and there are 20% since the Xs allele
frequency is 0.2.
Problem 9.
Assuming the population is at Hardy-Weinberg equilibrium, how many heterozygous
tongue rollers are there in the population?
Let p = allele frequency for rolling
Let q = allele frequency for non-rolling
q2 = 2600/10,000 = 0.26
q = 0.51
p = 1 - q = 0.49
2pq = 2(0.49)0.51 = 0.4998
Number of heterozygous rollers = 0.4998(10,000) = 4998 (5000 would be close
enough)
Problem 10.
1. q = square root of 1/10,000 = 0.0100
2. 2pq = 0.0198 = 1.98% of U.S. population
Problem 11.
Is the baby population at Hardy-Weinberg equilibrium?
There are two ways to do this problem. You can calculate expected allele
frequencies from genotype frequencies and compare them to actual allele
frequencies, as below:
q2 = 20,000/100,000 = 0.2
q = 0.45
p = 1 - q = 0.55
p2(calculated) = 0.30 p 2(actual) = 20,000/100,000 = 0.2
2pq(calculated) = 2(0.45)0.55 = 0.50 2pq(actual) = 60,000/100,000 = 0.6
Answer: No
Or you could use observed allele frequencies to calculate expected genotype
frequencies and compare these to actual genotype frequencies, as below:
q = 100,000/200,000 = 0.5; p = 0.5
expected genotype frequencies:
q2= 0.52 = 0.25; expected homozygous sickle cell is 0.25x100,000, or 25,000
2pq = 2 x 0.5 x 0.5x = 0.5; expected heterozygotes: 0.5 x 100,000, or 50,000
p2= 0.52 = 0.25; expected homozygous normal is 0.25x100,000, or 25,000
compare actual to calculated genotypes: 25,000 vs 20,000 sickle cell
sufferers; 50,000 vs 60,000 hets, 25,000 vs 20,000 normal
Answer: No
Problem 12.
A) Assuming Hardy-Weinberg equilibrium, what is the frequency of the resistance allele?
B) What is the frequency of the non-resistance allele?
C) Of the 100,000 mosquitoes, how many are expected to be carriers of the resistance
allele (heterozygous)?
D) What is the frequency of the resistance allele now?
The strategy here is to figure out how many resistance alleles exist in each population
and divide the sum by the total number of alleles in both populations.
where,
was from part a,
was from part b,
was 1,
, was 0.
E) Over time, the mosquito population grew back to its natural size of 100,000. Assume
there was no difference in fitness between the two alleles of the resistance gene. How
many mosquitoes are expected to be resistant to malaria?
F) If it turned out that being homozygous for the resistance allele caused mosquitoes to
become partially sterile, do you predict a change in the frequency of alleles in the
population? If so, which allele would increase in frequency?
Yes, the nonresistance allele would increase in frequency as partial sterility represents a
change in fitness.
Problem 13.
A) Determine the frequency of each allele.
CL = [(450x2)+500]/2000
CS = [(50x2)+500]/2000
CL = 0.7
CS = 0.3
B) Are the allele frequencies maintained at Hardy-Weinberg equilibrium? How do you
know? Show your work (No need for a statistical test, just estimate).
CLCL = (0.7)2x1000 = 490 ≠ 450
CLCS = 2x(0.7)(0.3)x1000 = 420 ≠ 500
CSCS = (0.3)2x1000 = 90 ≠ 50
NO, observed values do not equal expected Hardy-Weinberg values
Problem 14.
A) Under Hardy-Weinberg equilibrium assumptions, determine the frequencies of both
the wild-type and mutant alleles.
20 = Round-eared
40 = rr
(40/1000) = q2
0.2 = q
0.8 = p
B) The Vulcans send the 20 round-eared individuals to Earth to infiltrate the American
government. If the 980 remaining Vulcans mate randomly, how many round-eared
individuals will there be in the next generation of 1,000 Vulcan offspring?
r = 0.2 x 2000 = 400
new r = 400 - 40
new r = 360
new q = (360/1960)
new q = 0.184
new q2 = 0.034
rr = 0.034 x 1000
rr = 34
round-eared = 34 x 0.5 = 17
C) A deadly ear-ache pandemic sweeps through the Vulcan’s planet. However,
heterozygosity at the pointy-eared locus confers resistance to the disease, allowing only
those individuals to survive and reproduce (homozygotes for either allele die).
Determine the proportion of pointy-eared individuals in the first generation of offspring
from the survivors.
q = 0.5
q2 = 0.25
rr = 0.25
round-eared = 0.25 x 0.5 = 0.125
pointy-eared = 1 – 0.125 = 0.875