Biology 336
Midterm answer key
1
General exam notes
120 points were possible. High score was 114.5 out of 120. Mean was 60%
Revised score:
Take original marks, minus question 3 marks, plus (0.4 * question 3 score), then divide by 102.5
Note: if this adjustment led to a lower mark, the original scoring was used for that individual.
Mean is now 65%.
Exam queries?
Read this key completely, then feel free to come to office hours to ask.
Poor performance?
If your mark is higher on the final, the final will be weighted more heavily. If your mark is
better on the midterm, the weighting will remain 25% for midterm, 35% for the final.
The population genetic models are challenging: things may ‘click’ if you continue to work on
them.
If you were careful and thorough in your studying, and still did very poorly, come to office hours
to discuss how you studied.
Remarking of exams
Time and care were taken to award partial credit. Remarks should be requested only for major
errors. The entire exam will be re-marked if any question is re-marked: scores may drop if any
of the original marks are found to be generous. Re-mark requests must be made in writing and
submitted with the exam. Individuals requesting re-marking will not be eligible for a higher
weight for the final exam.
There were two versions of some questions: answers to both versions are given below.
Biology 336
Midterm answer key
2
Question 1: questions and answers are the same for both exam versions.
Mean score: 14.7 / 20. 18 individuals scored 20 / 20
1a (10 pts). HIV will quickly evolve resistance to AZT in patients that are treated with just AZT. Explain the
requirements for evolution by natural selection and describe how drug resistant AZT fulfills the requirements. (3
sentences)
Natural selection requires variation in a phenotype, selection favouring individuals with one
phenotype over those with another, and heritability, so that the offspring of favoured parents
will share their phenotype to some degree. HIV generates heritable variation, including
variation in polymerase effectiveness when AZT is present, due to its high mutation rates.
Virions with mutations that allow them to replicate in the presence of AZT will produce more
copies and will thus become more common.
2 marks each for variation, selection, and heritability. 1 mark each for a concrete link between
these concepts and the AZT example.
Common error: not mentioning heritability.
b. (5 pts) Explain the role played by stratigraphy in convincing Darwin and his contemporaries of the plausibility of
evolution. (2 sentences)
Stratigraphy demonstrated that the layers of rock were laid down over many millions of years.
This convinced Darwin and others that sufficient time had passed for natural selection to have
led to the existing diversity of organisms.
2 marks for demonstrating an understanding of stratigraphy, 1 mark for noting that stratigraphy
demonstrates that the world is older than previously thought, and 2 marks for saying that the old
age was necessary for the plausibility of evolution by the gradual process of natural selection.
Common errors:
- irrelevant discussion of homology and fossils
- missing the importance of the age of the earth to natural seelctoin
c. (5 pts) Identify the two major problems with this description of evolution of beak depth during the drought.
Explain why each of these is a misconception. (2 sentences)
During the drought in 1979 the finches faced a crisis that threatened the survival of the species. Unless the finches
mutated to have larger beaks, the species would go extinct. Fortunately, individuals were able to adapt to the new,
dryer conditions, and the species survived with larger beaks.
First, mutations do not occur when required, but occur randomly, so the variation for larger
beaks must have been present in the population. Second, individuals do not evolve,
populations do, so the change in beak size did not occur in individuals but over time in the
population.
2.5 marks for mutations occurring randomly not as needed. 2.5 marks for stating that individuals
do not evolve, populations evolve. Many students missed the second misconception.
Biology 336
Midterm answer key
3
2a. (20 points). Researchers have developed a mouse model for cystic fibrosis, a disease in humans that was
typically lethal until recently. Mice homozygous for the cf mutation have poorly developed lungs and are more
susceptible to infection, just as human cystic fibrosis patients are. Evolutionary biologists have collaborated with
the medical geneticists to test hypotheses. In a (hypothetical) study, the scientists bred populations with a low
frequency of cf, allowed them to randomly mate for a few generations, then exposed them to cholera (or the mouse
version of it).
There were two versions of this question.
A
Genotype
Starting population
Population post disease
cf / cf
5
2
CF / cf
130
84
CF / CF
865
514
Starting population
Population post disease
5
1
130
94
865
495
B
a. (10 pts) Calculate the genotype and allele frequencies for this population after the epidemic. Is it in HWE? Be
sure to show your work.
A
f(cf) = 0.073 f(CF) = 0.927
cf/cf
CF/cf
Observed genotype frequencies
0.003
0.14
Expected genotype numbers
3.2
81.5
deviates
0.47
0.07
X^2 = 0.54 << 3.84; not significantly out of equilibrium
CF/CF
0.857
515.2
0.003
f(cf) = 0.081 f(CF) = 0.919
cf/cf
CF/cf
CF/CF
Observed genotype frequencies
0.002
0.16
0.84
Expected genotype numbers
3.9
88.2
497.9
deviates
2.2
0.4
0.02
X^2 = 2.6 << 3.84; not significantly out of equilibrium
Common errors: using expected frequencies rather than expected numbers to perform the chisquare test.
b. (10 pts) Calculate the changes in allele frequency due to the epidemic and the relative fitnesses for the three
genotypes. Show your work.
A
p(0) = 0.07; ∆p = 0.003;
Absolute fitnesses
Rel fit
cf/cf
2/5 = 0.4
.62
CF/cf
0.65
1
CF/CF
0.59
.92
B
p(0) = 0.07; ∆p = 0.011
Absolute fitness
Relative fitness
cf/cf
0.2
0.3
CF/cf
0.72
1
CF/CF
0.57
0.8
Biology 336
Midterm answer key
4
Common errors:
- using the “after” frequencies in the formulae to calculate the after frequencies. (No
formulae were needed: the “after” frequencies were calculated in part a, and the
initial frequencies could be easily calculated.
- using the wrong formulae, particularly the haploid model
- calculating absolute fitness by dividing by the total number of individuals in the
population, thereby confusing frequency with fitness.
Mean score: 14.9 / 20. 35 individuals scored 20 / 20. Easy question!
Question 2b: 20 points
2c. (10 pts). Do these data support the idea that selection due to cholera might maintain the cf allele in human
populations? Explain how this might be true, and whether the data support it, being careful to refer to the data above.
(2 sentences)
Yes, the heterozygote has a higher fitness than either homozygote; heterozygote advantage will
maintain both alleles in the population even if one is harmful. If humans were similar to mice,
this could explain how cholera maintained the cf allele in the population.
If calculations in part 2b resulted in different estimates of relative fitness, and these data were
coherently discussed, partial credit was given.
Common errors:
- if the population is in HWE, selection is not acting. This is not a valid inference: a
population may be affected by selection but the genotype frequencies do not change sufficiently
to detect the process with a chi-square test
- confusing heterozygote advantage (heterozygote has highest fitness) with a dominant
favourable allele (deleterious allele in a heterozygote has high fitness, but not higher than one of
the two homozygotes). Heterozygote advantage will maintain two alleles in a population
indefinitely, while directional selection with a recessive deleterious allele will slow the loss of
the deleterious allele from the population.
- failing to mention the assumption that the mouse model applies to humans (1 point)
2d. (10 pts). An alternate hypothesis would propose that mutation maintains the cf allele in human populations.
Estimate the mutation rate that would be required, and evaluate whether this hypothesis is reasonable given your
knowledge of mutation. (Note: cf patients nearly always die before reproducing). Show your work. ~1 sentence
for the evaluation.
We will assume that cf is recessive (that is, without cholera, heterozygotes have same fitness as
CF/CF: this assumption was given in part a)). Then, the frequency of cf = square root of (-u /
s). We assume that s = -1 (zero fitness). So, f(cf) = sqrt(u) = 0.04. This means that u =
(0.04)^2 = 0.0016. This is high: we estimate mutation rates per locus at 10-5 to 10-6. This
makes it an unlikely explanation.
Biology 336
Midterm answer key
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Common errors:
- many students used the f(cf) from part a. Note that this question asked about humans.
An experimental population would not be a good one to evaluate for mutation /
selection balance.
- many students used s from part b, ignoring the information that cf patients die prior to
reproducing, giving a very low fitness. Also, s in part b was a case of heterozygote
advantage: mutation –selection balance is an alternate hypothesis, and assumes
directional selection.
- many students used the dominant equilibrium equation, rather than the recessive
equation.
10.4 marks out of 20. No perfect scores: 11 scored 19 out of 20.
3. (20 pts) Two graduate students are performing mutation accumulation experiments in bacteria. Each follows a
slightly different experimental protocol. One student starts with a bacterium in a large culture broth. Broth
continuously flows into the system and out of the system, so that a large population is maintained for 500 days. The
student maintains 50 different cultures during the experiment. The second student starts with a single bacterium in a
small broth culture. Each day, the student transfers a single bacterium to a new culture, and maintains 50 replicate
lines during the experiment. In both cases, the students freeze a sample every 50 days.
Type
Insertion
Deletion
Substitution
Table 1A
Type
Insertion
Deletion
Substitution
Table 1B
Line 1
.10
.30
.60
Line 1
.10
.20
.70
Line 2
.40
.20
.40
Line 2
.30
.30
.40
Mutations observed per line
At the end of the experiment, both students sequence three genes from each of the 50 lines at every 50 days. The
data are shown below. The table at the left breaks down the frequency of each type of mutation found.
0.45
0.4
0.35
0.3
0.25
line 1
0.2
0.15
0.1
0.05
0
line 2
0
200
400
600
Days
Figure 1:
a. (10 pts). Explain why the differences in the methods led to the differences in the number of mutations observed
per line (Figure 1). (1-2 sentences)
Method 1 has high levels of competition (2 marks), which means that unfavourable mutations
will become less common through time (3 marks), while method 2, with limited mutation, will
maintain more unfavourable mutations (3 marks). This will result in lower levels of mutation
detected for line 1 compared to line 2. The Ne (effective population sizes are different with
Ne=v.small (1) for Method 2 and Ne=large (500) in Method 1. (2marks)
Biology 336
Midterm answer key
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2 marks only if you poined out that line 2 had more mutations
3 marks only if you recognized a difference in the competitive environment and made the
correct association with the appropriate method.
b.
(10 pts). Based on the comparison of the frequency of the different types of mutations observed in this
experiment (Table 1), explain the possible differences in the fitness effects of each type of mutation. Be
sure to connect your hypotheses about fitness effects to the data presented! (1-2 sentences)
A: The mutations seen in line 1 that are less common in line 2 are those that are more
harmful. This means that insertions are more harmful than deletions or substitutions.
B: The mutations seen in line 1 that are less common in line 2 are those that are more
harmful. This means that deletions and insertions are more harmful than substitutions.
2 marks only for just pointing out the differences ie A. insertions > others for Line 2 or B.
insertions and deletions>other.
The other marks come from making the connection between fitness and their frequencies,
given the method.
COMMON MISTAKES:
1. The assumption that a larger population size would have a larger number of mutations.
On its own this is true, but given the context of a selective environment in Method 1, this
statement is problematic. This hypothesis does not fit with the data presented.
2. Mixing up the lines. Line 1=Method 1 and Line 2=Method 2.
3. Common misconceptions about a haploid bacterial organism (which means no
heterozygotes!)
a. Line 1=homozygous, Line 2=heterozygotes
b. Bacteria have sex. They do not. They make exchange DNA via conjugation,
transduction or transformation.
c. Mating – implies sexual reproduction
d. Inbreeding.
4. Many students assumed the experiments were run for different periods of time. The
graph clearly shows this is not true.
5. Migration is occurring in Method 2. It is not.
6. Assumption that Method 2 exposed bacteria to different environments because they were
transferred to a new culture or that nutrients were somehow different in Method 2. This
is not in the question. Culture is the same in both methods one is simply using test tubes
and Petri dishes while the other uses a large broth of culture. The nutrient media is the
same in both.
7. Bacterial transfers in Method 2 DO NOT result in starting with a new line of bacteria.
The line or genetic background comes from the starting bacteria but you pick a single
bacterium and then allow bacteria to expand in population size each transfer. But always
bottlenecking each time step to a population size of 1.
Biology 336
Midterm answer key
7
Mean score: 7.7 out of 20. 21 individuals scored 20 out of 20. (This question was reweighted to 5 points due to its apparent difficulty.)
4. (20 pts). A friend is studying sunflowers growing in saline (= salty) areas next to the Salt Lake in Utah in the US
and has surveyed the population at one locus. He surveyed seedlings when they had 4 leaves, then followed the
plants to adulthood. Here are the number of seedlings found with each genotype, and the numbers that survived to
adulthood.
A.
Genotype
Seedlings
Adults
Genotype
Seedlings
Adults
B.
AA
491
245
AA
91
18
Aa
418
207
Aa
418
207
aa
91
18
aa
491
245
1
1
0.9
0.9
0.8
0.8
0.7
0.7
0.6
0.6
f(a)
f(a)
a. (10 pts) Assume that the relative fitness for each genotype remains constant over the next 20
years. Graph f(a) vs. time for that time period. (Ignore drift). (Do not calculate frequencies for
all 20 generations, just sketch the general shape of the graph). Don't forget useful items such as
labelled axes.
0.5
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
0
0
0
5
10
generation
15
20
0
5
10
15
20
generation
b. (10 pts) Explain the pattern you graphed, noting how each factor affects the rate of evolution, and how they did
so in this case. Show calculations where appropriate. (~3 sentences)
A. A is favourable, with s = -0.6, which is large, so the decrease is fast. The initial
frequency is 0.258, which gives many copies of the favourable allele, so it is fairly steep
at first. Allele a is recessive (or nearly so: h = 0.012) so the rate of increase slows as a
becomes more common as most of the copies of the unfavourable a alleles are in highfitness heterozygotes.
B. a is favourable, with s = -0.6, which is large, so it increases fairly quickly. The initial
frequency is 0.741, which gives many copies of the favourable homozygote so the rise is
fairly rapid at first. Since a is almost dominant (h = 0.012: Aa is like AA) the rate of
incline slows as A becomes rarer since it is mostly found in the high-fitness
heterozygotes.
Biology 336
Midterm answer key
8
For full marks the calculations for s & h must have been given. Using s, h and the allele
frequencies why the curve was shaped as it was (ie why did it rates start relatively fast and
why rates slowed).
Mean score: 13.5 / 20. 10 individuals scored 20 / 20.
5a. (10) Imagine that there are two populations of plants, one of size 100 and one of size 2000. (For the moment,
we will assume that N = Ne). How would the difference in population size would affect a) the fate of a slightly
beneficial rare mutation, and b) the fate of a slightly deleterious mutation? (2-3 sentences: please briefly explain
the process as well as the outcome).
In smaller populations the relative effects of drift are stronger since the change in allele
frequency due to sampling error of one individual is larger. This means that in a smaller
population, a favourable allele as a higher probability of be lost and an unfavourable allele
has a higher probability of fixing, although it is still more likely that favourable alleles will fix
and unfavourable will be lost.
5b. (5) A new mutation has occurred in a population of 400 haploid algae. How long will it take this mutation to
reach frequency of 0.8 if the mutation has a selective advantage of 30%? Show your work. (Version B: 600 algae,
selective advantage of 25%).
A: p(t) = 0.8; q(t) = 0.2; p(0) = 1/400; q(0) = 399/400; WA = 1, Wa = 0.7).
t
p (t ) p (0)W A
. Since WA = 1, (0.8 / 0.2) = 0.0025 / (0.9975 * .7t); t ln .7 = ln 0.0063.
=
t
q (t ) q (0)Wa
t = 20.7 – 21 generations.
B: p(t) = 0.8; q(t) = 0.2; p(0) = 1/600; q(0) = 599/600; WA = 1, Wa = 0.75).
t
p (t ) p (0)W A
=
. Since WA = 1, (0.8 / 0.2) = 0.002 / (0.998 * .75t); t ln .75 = ln 0.0004
q (t ) q (0)Wa t
t = 27.05 – 27 generations.
Note: the more precise fitness calculations of WA = 1, Wa = 0.77 or WA = 1, Wa = 0.83 were
fine.
Common errors:
- Choosing arbitrary initial frequency anywhere from 0.01 to 0.0000001. The question indicates
the starting frequency by saying “A new mutation has occurred in a population of 400 haploid
algae.” This means f(mutant)=1/400 and f(nonmutant)=399/400. Sinc this is a HAPLOID algae
there is no need to multiply the denominator by 2.
2 marks were taken off for choosing an arbitrary initial frequency
2 marks were taken off for making it a diploid.
1 mark for calculation errors that were horrendous.
Midterm answer key
Linkage disequilibrium (D) vs.
time
Linkage disequilibrium (D) vs.
time
0.2
0.2
0.15
0.15
D
5c. (5) You observe the following pattern of linkage
disequilibrium in a population (see graph). Suggest one
explanation for the increase in D (1 sentence), and estimate r.
Show your work.
9
D
Biology 336
0.1
0.1
0.05
0.05
0
0
0
5
10
Time (generations)
0
5
Time (generations)
D could increase due to migration from another population, selection on a new linked
mutation, or selection on a multi-locus genotype. D will
decrease due to r: D’ = (1-r)D.
For A: 0.12 = (1-r)(0.15); r = 0.2.
For B: 0.14 = (1-r)(0.15); r = 1/15 = 0.07. (more accurately, 0.1 = (1-r)4(0.15);
ln (0.67) = 4 ln(1-r); -0.1 = ln(1-r); 1-r = exp(-0.1) = 0.90; r = 0.1.
2 marks for a correct explanation for an increase in D, 3 marks for a complete solution for
estimating r.
Common errors:
- stating that D increases due a decrease in recombination without giving a potential
explanation / cause of it.
- not realizing that r will decrease d and so calculating r from time 0. (Recombination
cannot explain an increase in D!)
Average marks: 10.7 / 20. Nine individuals scored 20 / 20.
10