Download Key for the midterm exam

EVOLUTIONARY GENETICS (Genome 453)
Midterm Exam – KEY
Name
PLEASE: Put your name on every page and SHOW YOUR WORK. Also, lots of space is provided, but you do not have to
fill it all!
(The details of these problems are fictional, for exam purposes only; they don’t represent real data.)
1. (20 pts) In a large, isolated population of feral cats, a new mutation in gene H arises. Call the wild type allele H+ and
the mutant H−. Heterozygous H + /H− cats are resistant to the FeP virus, which gives them a fitness advantage of
25%, as shown below:
Genotype H+/H+ H+/HFitness
0.8
1.0
(a) (5 pts) Predict the approximate probability that the H− mutation will survive and spread in this population.
The advantage g is 0.25, so the chance is roughly 2g = 0.5. I only took a point off for people who said the advantage
was 0.2; this wasn’t very clear in class.
(b) (5 pts) When the H− allele becomes more common, we find that many H − /H− kittens suffer from kidney failure
within a few weeks of birth, so their fitness is only 0.15. Using this additional information, predict the expected
long-term frequency of H− in the population, assuming that it is not lost early on due to drift.
This is an overdominant equilibrium. s=0.2, t=0.85, t/(s + t) = 0.809. This is p(H+) so p(H-) is 0.1901.
(c) (5 pts) A large number of these cats are caught and bred in captivity. Distressed by the loss of kittens due to
kidney failure, the breeder attempts to get rid of the “kidney failure allele” by eliminating all kittens with the
H − /H− genotype. Is this strategy likely to eliminate the H− allele? Why or why not? (Assume FeP is just as
severe in the captive population as in the feral population.)
This changes the equilibrium as the fitness of H − /H− is now 0.0. Surprisingly, the new equilibrium is p(H−) =
0.167 (though it was not necessary to calculate that to get full points). So the breeder is unlikely to succeed as there
will still be quite a few copies of H− maintained in the more fit heterozygotes. I did not give full points to answers
that suggested if he just kept trying he’d eventually succeed; this is a stable equilibrium. If he wants to succeed he
should either develop a test for heterozygotes, or eliminate the parents and/or siblings of H − /H− kittens as well
in order to get rid of some heterozygotes.
(d) (5 pts) A mutation in the FeP virus makes it more deadly, leading to the following new fitnesses:
Genotype H+/H+ H+/H- H-/HFitness
0.6
1.0
0.15
After a long period, what percentage of newborn kittens in the feral population would you expect to be H − /H−?
Using the same formula, p(H-)=0.32, so p(H-/H-)=0.1024. I gave 1 point for people who worked out just one
generation.
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EVOLUTIONARY GENETICS (Genome 453)
Midterm Exam – KEY
Name
2. (20 pts) In a population of aphids with sexual reproduction and recombination, the following two genes are on the same
chromosome at a distance of 20 centiMorgans (i.e., 20% chance of recombination between them each generation).
Gene F
Fitnesses
FF Ff
ff
1.0 0.8 0.6
Gene G
Fitnesses
GG Gg gg
1.0 1.0 0.6
Frequencies
F
f
0.6 0.4
Frequencies
G
g
0.6 0.4
Throughout this problem, assume that the fitness of an individual is their F-locus fitness multiplied by their G-locus
fitness.
(a) (5 pts) The disequilibrium coefficient D is -0.1 in the initial population. What proportion of gametes do we expect
to be of genotype F G?
p(FG) = p(F)*p(G)+D = 0.36-0.1 = 0.26. I gave some credit for people who tried to work out what it would be
next generation, though this was not what I meant to ask.
(b) (5 pts) Alleles f and g are equally bad in the homozygote. Will one decrease more rapidly than the other, and if
so, which one? Explain briefly.
Allele g is recessive and can hide in the heterozygote. Allele f is revealed in the heterozygote, so it will decrease
faster. The linkage does not matter as 20% recombination is plenty to produce all four gametes.
(c) (10 pts) We create a population of aphids with only F g and f G chromosomes (in equal numbers), and also knock
out the genes that allow genetic recombination. Assume that the lack of recombination itself does not harm the
aphids. After a long time, what allele frequencies do you predict for loci F and G? Show your work.
There are only three genotypes possible, with fitnesses as follows:
Genotype
Fg/Fg Fg/fG fG/fG
Raw fitness
0.6
0.8
0.6
Normalized fitness
0.75
1.00
0.75
This is an overdominant situation with equally bad homozygotes, so the frequency of both F and G will go to 0.5
and stay there. A population with p(F)=p(G)=1.0 would be much more fit but without recombination there is no
way to get there.
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EVOLUTIONARY GENETICS (Genome 453)
Midterm Exam – KEY
Name
3. (30 pts) Chris breeds purebred Greyhounds dogs (a breed that has been kept from mixing with other dogs for many
generations). Pat breeds dogs whose ancestors came from the animal shelter and were a random mix of many different
dog breeds. Both raise their animals in the exact same controlled environment.
(a) (5 pts) Chris and Pat both measure the heritability (h2 ) of height in their dogs. Who is likely to measure a larger
heritability, and why?
Pat. His/her dogs will have more genetic variation, and since the environment is the same, this will lead to higher
h2 .
(b) (5 pts) Chris breeds her tallest animals together, and after years of patient work produces giant Greyhounds who
are as tall as the largest known dogs, with a mean height of 40 inches. She selects her best dogs, with a mean
height of 42 inches, but the next generation has a mean height of 40 inches again. What is the heritability of
height in these giant Greyhounds?
Zero. Response to selection is intensity of selection times h2 . The intensity was 2 inches, but the response was
zero, so heritability must be zero.
(c) (10 pts) Chris sells some of her dogs, with a mean height of 40 inches, to pet breeders who do not select for height.
After many generations, the pet population still has a mean height of 40 inches. Based on this result plus all
previous data, what is a likely reason for Chris’s failure to get taller animals by continued selection?
Lack of useful genetic variation in her dogs. If there was still variation but natural selection was opposing artificial
selection, the dogs would probably get shorter when selection was relaxed. Chris’ dogs could either be genetically
identical at all loci controlling height, or have only useless variation such as tall-heterozygote loci or taller males
than females.
(d) (10 pts) Could Chris make her dogs taller by mixing in some dogs obtained from Pat? Pat’s dogs have a wide
variety of heights, but are all under 35 inches tall. If this could be helpful, explain how; if not, explain why not.
(Assume that it is possible to breed dogs together even if they are of very different heights.)
Pat’s dogs may be able to help. They could have tallness alleles at loci where the Greyhounds are fixed for short
alleles, and when these are combined with the tallness alleles already present in the Greyhounds, taller Greyhounds
could result. We can’t know for sure without trying.
Some students argued that the zero heritability means no further improvement is possible, but heritability is defined
for a particular gene pool (in this case, the giant Greyhounds) and if we add genetic variation from another source
we may be able to increase heritability.
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EVOLUTIONARY GENETICS (Genome 453)
Midterm Exam – KEY
Name
4. (20 pts) The following story contains at least two errors. Identify these errors and describe why they are wrong. (You
are not looking for arithmetic errors–all of the numbers are correctly computed–but for logic errors, bad assumptions,
or wrong use of equations.)
On his first assignment as a field biologist, Chuck conducts a large study of a salmon population. He examines 10,000
fish, finding that 9095 are dull-colored and 905 are bright-colored. All he knows about the underlying genetics is that
this color difference is a single-gene trait.
√
He calls the dull allele D and the bright allele d, and computes the frequency of d as 0.09 = 0.3 (so the frequency of
D is 0.7). From this he determines that his genotype frequencies are as follows:
Genotype
Frequency
DD
0.49
Dd
0.42
dd
0.09
Since he cannot distinguish DD from Dd, he lumps them together:
Genotype
Expected Frequency
Expected Number
Observed Number
D?
0.91
9100
9095
dd
0.09
900
905
He plugs these numbers into a χ2 calculator and gets a χ2 of 0.0305. He notes that this is not significant at the 5% level
for 1 degree of freedom (see table on last page of exam), and concludes that the fish are in Hardy-Weinberg proportions.
List two things that Chuck did wrong, and explain why they are wrong. (You can list three if you want and I will grade
the best two, but no more than three, please.)
(a) (10 pts) First error
Chuck assumed that dull was dominant, but this may not be true. Bright color could be a rare dominant, or even
a heterozygote.
(b) (10 pts) Second error
Chuck assumed that the frequency of bright (d) was the square root of the proportion of homozygotes. This is only
true if the population is in Hardy-Weinberg equilibrium, and he doesn’t know that (in fact he is trying to test it).
(c) (Optional) third error
Chuck’s χ2 test has zero degrees of freedom, not one. He lost one for counting the fish and one for estimating the
allele frequencies, and he only had two to start with. So he can’t do a valid test. In a mathematical sense this may
be the same answer as #2 but I did accept both of them.
If you answered that Chuck should not have accepted the H-W hypothesis but only failed to reject it, that’s true but
minor compared to his other errors. If you answered that his χ2 is not significant you are interpreting the results
wrong: high χ2 here means rejection of H-W. Think about what χ2 would equal if the data fit H-W perfectly, if you
need help remembering this.
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EVOLUTIONARY GENETICS (Genome 453)
Midterm Exam – KEY
Name
5. (15 pts) Human females go through menopause, which is a (fairly) orderly shutting down of the reproductive system
and requires the action of multiple genes. On average this occurs between ages 45-55. Pregnancy is unlikely during
menopause and practically impossible after it.
It has been argued that menopause is an “evolutionary accident” which could not be selected for, as deliberately
becoming infertile must reduce an individual’s reproductive fitness.
Describe a way in which genetically encoded menopause might not be an accident, but might be due to
natural selection. Be sure to specify how the selection acts.
(Note: It may help to know that while the popular press says that pre-historic humans only lived to be 40-45 years
old, this number is an average and includes high infant mortality. In a modern group of hunter-gatherers, 25% of the
population survived past 60.)
Two plausible arguments:
(1) A female may maximize her lifetime surviving offspring by not getting pregnant late in life. Such pregnancies are
risky and if she dies, her orphaned previous children may die too. Late pregnancies also produce offspring with a lower
survival chance due to birth defects so the loss of fertility is less than it appears.
(2) If a female stops reproducing she may have more resources to put into insuring the survival of her children, grandchildren, and other relatives. This may give her genes a bigger boost than having another (possibly sickly) child of her
own, if the benefit to her kin times their degree of relatedness is larger than the cost to her.
Both of these are kin selection: the first one focuses on her own offspring (which are certainly kin) whereas the second
can also include her other relatives.
Two not so plausible arguments:
(3) Women are optimizing having fewer but healthier babies by only breeding when young. But couldn’t they have
healthy babies when young and then a few sickly ones later on–wouldn’t that still be better than stopping? Why would
reproducing late in life retroactively make your early babies worse? (It could make them worse if you still needed to care
for the earlier babies, which is argument #1; but I did want to see the whole argument.)
(4) A tribe with sickly babies is less fit than one with only healthy babies. How would group selection overcome the
individual selection for higher fertility?
A really implausible argument:
(5) It’s just an accident, because people didn’t live long enough for selection on this trait; menopause arose by drift.
This contradicts the problem (prehistoric people did often live this long, if they survived infancy–at least, that’s what
the modern hunter-gatherers tell us) and it would be hard for a complicated, coordinated shutdown to happen by chance.
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EVOLUTIONARY GENETICS (Genome 453)
Midterm Exam – KEY
Name
Equation sheet
You may tear this sheet off to use as scratch paper if you wish. Nothing on it will be
graded.
χ2 table
df p=0.05 p=0.01
1
3.84
6.64
2
5.99
9.21
3
7.81
11.35
p2 + 2pq + q 2 p(A) =
p(A) =
ω=
1
Ne
DN
DS
=
h2 =
ν
µ+ν
P
VA
VT
R = h2 S
t
s+t
C < Br
ŵ = p(AA)f (AA) + p(Aa)f (Aa) + p(aa)f (aa)
1
N /k
VT = VG + VE = VA + VD + VGE + VE
2g
D = (pAB ∗ pab) − (pAb ∗ paB)
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