Download Malthus provided a key idea to both Darwin and Wallace in the

Evolutionary Genetics Midterm 2006
Name
Student #
Signature
The Rules:
P1:
/15
P2:
/22
P3:
/16
P4:
/25
P5:
/22
TOT:
/100
(1)
Before you start, make sure you’ve got all five pages of
the exam, and you’ve written your name legibly on each one.
(2)
Show your work on calculation questions. Feel free to use
the back of the paper if required, but make sure to tell us when
to look on the back. Calculations are required for full credit
where asked.
(3)
If a calculation question requires you to start with the
result from a previous calculation, and you were unable to
answer the earlier question, you may make up a reasonable number
to start with and still potentially get full credit on the
second question.
1. (3 points) True or False: "Evolution requires that some types of individuals are more likely to
survive." False: evolutionary change can occur without differences in survival.
2. (3 points) True or False: "Because mutation rates are very low, mutation contributes little to the
process of evolution." False: mutation is the ultimate source of evolutionary novelty.
3. (3 points) True or False: "The main difference between Darwin’s and Wallace’s views of evolution
was that Darwin considered biogeographical patterns to be important." False: both were struck by
biogeographical patterns and compared species present in different locations.
4. (6 points) Provide three reasons why the frequency of allele A might increase from one generation
to the next within a haploid population. [Explain each answer in one sentence.]
Possibilities include:
a) Selection – Carriers of A are more fit. (If survival, fecundity, or mating success were separated
clearly, this was acceptable. If “fitness” and “selection” were given, only one was counted.)
b) Mutation – Carriers of a might mutate to A, increasing the frequency of A.
c) Drift – By chance, carriers of A might happen to survive and reproduce disproportionately
often.
d) Migration – Carriers of A might migrate into a population.
e) Genetic hitchhiking – Carries of A might increase in frequency if associated with a favorable
allele at another locus.
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Evolutionary Genetics Midterm 2006
5.
Name
(10 points) In the Cody and Overton (1996) study, the authors found that Vp/Va (= volume of the
pappus to the volume of the achene shown at left) (10 points) In the Cody and
Overton (1996) study, the authors found that
Vp/Va (= volume of the pappus to the volume of
the achene shown at left) was greatest for newly
colonized islands and decreased over time since
colonization, as shown in the graph to the right
(note that “mld” denotes the mainland
population). For island populations of the same age since
colonization:
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
a) How would you predict the trends in Vp/Va to differ on bigger islands versus smaller islands?
[Justify your answer in one or two sentences.]
Sample answers given full points (some alternative answers were accepted if logically sound):
“Vp/Va would decrease more slowly in bigger islands (VP/Va is larger in bigger islands). This is
because, in bigger islands, there is more space for the achenes to move around (get blown)
without leaving the island and getting lost at sea (die, cannot reproduce). Thus, large Vp/Va is
more fit on big islands than pn small islands, and [larger papuses] persist longer due to natural
selection.”
“Vp/Va would be larger on bigger islands vs. smaller islands because the seeds that dispersed
further on the big islands would still hit land whereas the seeds that dispersed further on the small
islands would be lost to the ocean.”
b) Would you predict Vp/Va to be larger or smaller on newly colonized islands that are closer to
the mainland? [Justify your answer in one sentence.]
Sample answers given full points (some alternative answers were accepted if logically sound):
“Vp/Va would be smaller in closer islands (to the mainland), because individuals with a smaller
Vp/Va are capable of getting to and colonizing closer islands but not further islands.”
“Vp/Va would be smaller on closer islands than on further islands because seeds that newly
colonize close islands from the mainland do not need to travel as far as those that newly
colonize further islands.”
6.
(12 points) You are considering the long-term effects of the ozone hole on the genetic health of a
diploid plant population. In the following, consider a gene subject to recurrent mutation, where the
deleterious mutations have selection coefficient, s, and dominance coefficient, h. [Support your
answers with calculations. Do not use your answer to part a) in the other parts.]
a) If more solar radiation enters through the atmosphere, the mutation rate  at the gene might
increase by a factor F to F . What is the new equilibrium frequency of heterozygous
individuals carrying the deleterious mutation?
F
, so that the frequency of
hs
 F F
F
heterozygotes becomes 2 pˆ qˆ  21
(or approximately 2 pˆ qˆ  2
), implying that the

 hs  hs
hs
ozone hole might increase the frequency of heterozygous
 mutants.
The equilibrium frequency of the deleterious mutation becomes qˆ 


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Evolutionary Genetics Midterm 2006
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b) Imagine that the deleterious mutation reduces the evaporative cooling ability of the plant. In
this case, the selection coefficient, s, might increase by a factor C to C s. What is the new
equilibrium frequency of heterozygous individuals?
The equilibrium frequency of the deleterious mutation becomes qˆ 

Chs
, so that the frequency of

  

heterozygotes becomes 2 pˆ qˆ  21
(or approximately 2 pˆ qˆ  2
), implying that the

 Chs Chs
Chs
ozone hole might decrease the frequency of heterozygous
 mutants.
c) As in b), imagine that the deleterious mutation reduces the evaporative cooling ability of the
plant, so that
is the new equilibrium mean fitness of the
 s increases by a factor C to C s. What
diploid population?
The equilibrium mean fitness in a diploid population is 1 2 , regardless of the selection
coefficient.
For those who tried to rederive this:
W  W AA pˆ 2  W Aa 2 pˆ qˆ  W aaqˆ 2

2


  2
 
   
 11
  1 Chs21
  1 Cs 
 Chs 
 Chs Chs 
Chs 
  2
  2
  2   2
  2


 1 2
    2
 2Chs
 2   2Chs      Cs   of
Chs  Chs 
Chs 
Chs Chs 
Chs
Chs Chs 

 1 2

Chs
2

Chs
 2Chs

Chs
 1 2
where terms of order  2 are dropped in the second to last step.

 7. (16 points) A group of researchers has developed a new breed of highly valuable black tulips.
They produce a large number of bulbs from parents of known genotype for unveiling at the annual
garden show. They expect all the flowers to be black, but when the flowers open, they discover
that 6 of their 15,000 plants have even more valuable pale blue flowers. Upon returning to the
nursery, they determine that their pale blue plants are heterozygotes for a white flower mutation
(that is, BB individuals are black, Bb individuals are pale blue, and bb individuals are white). [Show
calculations for parts a, b and d.]
a) Based on these data, what is the mutation rate, µ from the black allele to the white allele?
There are 15,000 diploid plants (carrying 30,000 alleles) and 6 Bb mutant plants arose, indicating
that 6/30,000 alleles mutated. This gives a mutation rate of   0.0002 .
b) The relative fitnesses of the three flower types are given by WBB = 1, WBb = 0.88, Wbb =0.76
Calculate the dominance coefficient h and the selection coefficient s for the white allele, b.
 and h = (1–0.88)/0.24 = 0.5.
WBb = 1–hs = 0.88, Wbb = 1–s = 0.76. Thus, s = 0.24
c) Using the above fitnesses, is the white allele, b, dominant, partially dominant, additive,
partially recessive or recessive to the black allele, B?
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Evolutionary Genetics Midterm 2006
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With h = 0.5, the mutant allele, b, is additive. (You can also see that the fitness of the
heterozygote is exactly mid-way between the two homozygotes.)
d) If artificial selection practiced by the farmers increases the fitness of pale blue flowers by 50%
(to WBb = 1.22), what would the equilibrium frequency of the black allele, B, be? [Show your
calculations and ignore mutation for this part.]
Using pˆ 
8.
W Bb  W bb
1.22  0.76

 0.676 .
2W Bb  W BB  W bb 21.22 1 0.76
(10 points) Barrett et al. (2006) maintained 200 lines of the haploid bacteria, Escherichia coli, for

10 days (110 generations). Each line started from a genetically identical stock and was maintained
at a population size of 800,000. The authors then searched the evolved lines for signs that a
beneficial sweep had occurred. Their method (using neutral markers) allowed them to detect
beneficial mutations that reached a frequency of 45% within this time frame. Assuming that a
beneficial mutation started out at a frequency of 1/800,000, what is the minimum fitness advantage
of a beneficial mutation that the authors could have detected? [Show your calculations.]
t
pt  p0  W mut 



Using
 , where t = 110 generations and p(t) is the frequency of the mutant
qt  q0 
W
 wildtype 
allele at time t. This can be solved for the relative fitness of the mutant as follows:

110
0.45
1/800000  W mut 




0.55 799999 /800000 
W
 wildtype 
0.45 799999 /800000 1/110  W mut 


 


W

0.55
1/800000 
 wildtype 
 W

mut 
1.129  
W

 wildtype 

Thus, the mutant must have been at least 12.9% more fit than the wildtype to be detected.
Note: Many students defined p(t) to be the frequency of the wildtype allele, but then their final
answer was 0.885, which they interpreted incorrectly to be the fitness advantage of the mutant. (1 –
0.885 = 0.115 would then represent the disadvantage of the wildtype allele.)
9. (15 points) A population of white-tailed deer that has existed on the small island of Vance is being
relocated to the neighbouring island, Juniper, by wildlife managers. 45 deer are moved from Vance
and added to the 78 deer already on Juniper. Genetic data show that these populations are
genetically differentiated at two loci. The Vance population is fixed for alleles Spota and Tailb. The
original Juniper population is fixed for the alternate alleles, Spotb and Taila .
a) Estimate the amount of linkage disequilibrium in the new population immediately after the
movement of adults from Vance to Juniper. [Support your answer with calculations.]
Letting
x1 = freq(Spota and Taila chromosomes) =
0
0
2  45  2  78
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
Evolutionary Genetics Midterm 2006
Name
2  45
 0.366
2  45  2  78
2  78
x3 = freq(Spotb and Taila chromosomes) =
 0.634
2  45  2  78
0

x4 = freq(Spotb and Tailb chromosomes)
=
0
2  45  2  78
the disequilibrium is D  x1x4  x
2 x3  0.366  0.634  0.232. If you defined the frequencies
differently, you might have gotten 0.232, which was acceptable if you showed your work.
Note: The 2  in the above calculations
is there because the individuals are all diploids and carry

two alleles. Full points were given if this wasn’t included.

b) If the rate of recombination between the two loci is 0.3, what formula predicts the level of
linkage disequilibrium after one generation of recombination and what is the predicted value
of D (assuming random mating and no selection)?
x2 = freq(Spota and Tailb chromosomes) =
Dt  1  1 rDt   1 0.30.232  0.162
Note: This formula kicks in after random mating has occurred. In the very first generation, all
individuals are double homozygotes and recombination will be ineffective.

c) In the absence of selection on either locus, do you expect the allele frequencies to change after
one generation of recombination? Explain in one or two sentences why or why not.
No. In the absence of selection, recombination will cause the chromosome frequencies to
change, but not the allele frequencies.
Note: At first, the population is not at Hardy-Weinberg, but it would approach HardyWeinberg immediately after the population randomly mates. But this will not change the
allele frequencies. It is the case that drift would act in the population, but the question was
focused on the role of recombination.
10. (12 points) A strain of mice was selected for decreased blood cholesterol levels. Over five
generations, the mean blood cholesterol level of the mouse population decreased from 2.16 mg/ 100
ml to 2.01 mg/ 100 ml. The selection differential over this time period averaged 0.07 mg/ 100 ml
per generation.[Support your answers with calculations.]
a) What is the narrow sense heritability, h2, for blood cholesterol levels in this strain?
The response is 2.01 – 2.16 = –0.15 mg/100ml in five generations or –0.03 mg/100ml per
generation. The selection differential (against blood cholesterol) was –0.07 mg/100ml. Using
R  h 2 S , we get 0.03  h 2 0.07, so that h 2  0.429 .

Note: h2 represents the fraction of phenotypic variance ascribed to additive genetic variance
within a population; it must lie between 0 and 1.


b) If the phenotypic
variance in the population
at the start of the experiment is 8.35, what is the
estimated additive genetic variance for blood cholesterol levels?
If VP  8.35 and h 2 


VA
 0.429 , then VP  3.58 .
VP

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Evolutionary Genetics Midterm 2006
Name
c) After another 100 generations of selection using the same selection differential, the strain
shows no further response to selection, and blood cholesterol levels plateau at around 1.92 mg/
100 ml. In two sentences, explain what could account for this phenomenon and what might
allow the population to respond to selection once again.
The population might well have lost additive genetic variance/heritability. Allowing time for
mutations to occur (or increasing the population size to allow for more mutations) might allow
a response to selection. Alternatively, crossing multiple lines with similar blood cholesterol
levels would increase the genetic variability.
Alternatively, the population might have reached a threshold level of cholesterol, where natural
selection opposes any further change. A creative suggestion by a student was to try placing
the species in a new environment, where cholesterol levels might not be so critical.
11. (10 points) Describe the principle introduced by Thomas Malthus and how Darwin used this
principle to argue for evolution by natural selection. [Do not use more than the space below.]
Note that for this question, your mark was based partly on knowing the right information and
partly on how you organized your answer. If you see “-2” written on the text, it means you
wrote something that wasn’t correct or complete. You still might have a total of 6 or 7,
depending on how well you organized and expressed your ideas..
Here are two answers from classmates who got full marks:
“Thomas Malthus introduced the principle of overproduction. His essay on this topic included
the following points: 1. That organisms tend to produce more offspring than are capable of
surviving, 2. When resources are plentiful, populations will still only grow geometrically until
resources are consumed, 3. Famine, poverty and disease will lead to a “struggle for existence”.
Darwin took these ideas from Malthus to explain how some individuals are more fit than others.
In other words, Malthus’ theory explained why not everyone survives and Darwin took this
idea and expanded further on it to say that those who are more fit will tend to survive and
reproduce more and pass these traits on to their offspring. Therefore changing the traits of
individuals seen within a population.”
“Malthus’ studies on population growth demonstrated and brought to light that organisms such as
humans produce more offspring than can potentially survive to reproductive age. As such
there is naturally a struggle for existence among individuals for limited resources. Indeed it is
because of these limited resources that populations do not show unlimited increase. Darwin
deduced that this struggle for existence can lead to evolution within a population as there is
differential success among individuals. That is, some have traits which enable them to better
survive and reproduce. Those that do not are eliminated from the population. This is survival
of the fittest. This constant change in composition of individuals making up a population is in
essence evolution…”
Key elements:
Malthus introduced the idea of overproduction – that human populations should grow
exponentially according to observed reproductive capacity. Given limited resources, many
more offspring are produced than can survive and thus ensues a “struggle for existence”,
leading to poverty, disease and famine. Malthus did not talk about differential fitness, nor did
he focus on organisms other than humans.
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Evolutionary Genetics Midterm 2006
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Darwin saw in Malthus’s principle the action of checks on population size, and noted that many
organisms overproduce. In the struggle for existence, deaths were not likely to be random.
Individuals possessing traits that gave them an advantage in acquiring resources, and
producing young would pass these traits on to their young. As a result, these traits would
increase their representation in the next generation.
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