BCOR 102 Exam 2
3/15/06
Be sure to show your work!
Name:
1. A population has three alleles (A, B, and C) at a particular locus. The frequency of the A
allele is 0.1, the frequency of the B allele is 0.2 and the frequency of the C allele is 0.7.
Assume the population is in Hardy-Weinberg equilibrium.
a. What is the probability of finding a CC homozygote?
0.7^2=0.49
b. What is the probability of finding a BC heterozygote?
2*0.2*0.7 = 0.28
2. An ecologist studying the flower color of lupines examines 100 plants and finds the
following results. Use a chi-square
red
pink
white
(X2) test to determine if the
Observed
30
40
30
population is in Hardy-Weinberg
Expected under HWE 25
50
25
equilibrium (Note: the critical value
of X2 for this test is 3.84)
a. What is the value of X2?
 (O-E)2/E = 52/25 + 102/50 + 52/25 = 1+2+1 = 4.0
b. Are the genotype frequencies consistent with the expectations?
No. The chi-square value is greater than the critical value, so the deviations are
“significantly” large. The frequencies are not consistent with HWE.
3. Populations of the alpine sky pilot have different flower sizes in different locations.
Tundra populations have larger flowers than meadow populations.
a. Briefly describe an experiment you could do to determine if the difference is
genetic in origin, or if the difference is simply the result of growing in a different
habitat?
1) You could do a reciprocal transplant experiment and move plants between
populations. If the alpine plants have smaller flowers when they are moved to the
meadow , then flower size is probably determined by habitat.
2) Or you could grow plants from both populations in the same environment,
perhaps a greenhouse. If they retain their differences when grown in the same
environment, it is probably a genetic difference.
b. How could you measure the heritability of flower size?
Measure the flowers of parents and their offspring. The slope of the regression of
offspring flower size on parent flower size equals the heritability.
4. The following genotypes were observed at the enzyme locus PGM in the flowering plant
Impatiens capensis. Three different
N=
70
30
50
patterns were observed when 150 different
individuals were analyzed by gel
------electrophoresis. The results are
------summarized in the following figure. (The
numbers above each column are the
number of individuals that had that
particular banding pattern.)
a. What is the observed heterozygosity for this population?
30/150 = 0.2
b. What is the expected heterozygosity under HWE?
First find the allele frequencies:
p = 70*2 + 30 / (2*150) = 170/300 = 0.566
q=1-p = 0.433
then, Hexp = 2pq = 0.491
c. What is the inbreeding coefficient (F)?
F = (Hexp – Hobs)/Hexp = (0.491 – 0.2) / 0.491
= 0.592
d. What biological process could account for that pattern of heterozygosity?
The high inbreeding coefficient is most likely a result of selfing in this plant, or at
least mating between close relatives.
5. Fill out the following table showing the effects of each evolutionary process on allele
frequency: (increase or decrease or neither or both)
Effect on
Effect on
Affects all loci
Process
variation within a variation between equally? (Y/N)
population?
populations?
Migration
increase
decrease
Yes
Mutation
increase
increase
No
Drift
decrease
increase
Yes
Selection
Usually decreases
variation, but both
are possible.
Both (can
increase or
decrease)
No
6. A population contains three genotypes at the “A” locus. The three genotypes have the
following frequencies and relative fitnesses.
AA
Aa
aa
Frequency
0.1
0.4
0.5 (these are NOT Hardy Weinberg proportions)
Relative fitness
0.8
1
0.6
a. What is the starting frequency of allele “A”?
0.1 + 0.4/2 = 0.3
b. What will be the frequency of the A allele next generation?
First calculate mean fitness = 0.1*0.8 + 0.4*1 + 0.5*0.6 = 0.78
Then calculate p’ = (0.1*0.8 + 0.2*1)/0.78 = 0.359
c. Bonus: what will be the eventual frequency of allele A after many generations of
selection?
call the fitnesses 1-s, 1 and 1-t. Then, s = 0.2 and t=0.4. The equilibrium frequency
will be t/(s+t) = 0.4/0.6 = 0.666
7. Mosquitoes with a dominant R allele have increased resistance to the pesticide DDT.
Homozygous recessive rr mosquitoes have 60% lower survival. Initially the frequency
of the R allele in the population is p=0.20. What is the expected frequency of R after one
generation of selection by DDT?
First calculate the genotypes frequencies for RR, Rr and rr: 0.04, 0.32, 0.64
Then assign relative fitnesses: 1, 1, and (1-0.6) = 0.4
Mean fitness = 0.04*1 + 0.32*1 + 0.64*0.4 = 0.616
p’ = (0.04*1 + 0.16*1 )/ 0.616 = 0.325
8. A population geneticist studying populations of grizzly bears in the Canadian Rockies
analyzed the pattern of variation at several loci. She
Region
FST
estimated different values for FST for a set of northern
Northern
0.04
populations compared to a set of southern populations near
populations
the US border.
Southern
0.23
a. Which set of populations (northern or southern)
populations
experiences more migration among the populations?
How do you know?
Northern populations have more migration.
Fst is a measure of differentiation among populations so a low Fst means the
populations have similar allele frequencies. High migration will lead to similar allele
frequencies in each population and a low Fst value.
Also, Nm appears in the denominator of the formula Fst=1/(4Nm+1). If Nm is high,
then Fst will be low and vice versa.
9. In a classic study of natural selection, Bumpus (1917) found a population of sparrows
that had washed up on a beach after a particularly severe fall storm. Some of the birds
were still alive and some were dead. The overall average body length of the entire
population was 160 mm, which was presumably a good estimate of the entire population
before the storm. The average body length of the surviving birds was 159 mm and the
average body length of dead birds was 160.5 mm.
a. What is the selection differential (S) caused by that storm?
S = 159-160 = -1
b. If the heritability of body length is 0.5, what was the predicted response to
selection (R)?
R=h2S = 0.5*-1 = -0.5 mm
c. What was the predicted length for this population in the next generation?
160 – 0.5 = 159.5 mm
10. Buri studied genetic drift in some laboratory populations of fruit flies and produced this
figure of his results:
a. How would his final results have differed if the
starting allele frequencies had been 0.2 and 0.8
(instead of 0.5 and 0.5)?
The only real difference would be that more populations
would become fixed for the allele that was initially at
higher frequency. The figure would look about the same
but it would be skewed toward the allele that was in
higher frequency.
b. How would his final results have differed if he had
used populations of 50 flies, instead of populations of 16 flies?
If he had used larger populations, drift would be slower and there would be smaller
changes in allele frequency. There would be many more popuations with
intermediate allele frequencies at the end of the experiment and fewer populations
would have become fixed.
c. If all of the flies at the end of the experiment were combined into a single large
population, how would the allele frequency differ from the starting frequency in
generation 1?
There would be no change in allele frequency. On average, the same number of
populations become fixed for each allele. So the allele frequency in the combined
population would be 0.5, just as it was in the beginning.