EEB 245: Evolutionary Biology
Spring 2008
Problem Set 3
1. Enzyme variation in a species of grasshopper suggests that, out of 120 individuals
sampled, genotype frequencies are 62 A1A1, 49 A1A2 and 9 A2A2. Show that p = 0.72 and
q = 0.28 (where p and q are the frequencies of alleles A1 and A2), and that the genotype
frequencies of A1A1, A1A2 and A2A2 are approximately 0.51, 0.41, 0.08, respectively.
Demonstrate that the genotype frequencies are in Hardy-Weinberg equilibrium. (adapted
from Futuyma 2005)
Calculating allele frequencies:
p = 0.72, q = 0.28
freq (p) = {2(62) + 49}/2*120 = 0.72
freq (q) = {2(9) + 49}/2*120 OR 1 - 0.72 = 0.28
Calculating genotype frequencies:
freq (A1A1) = 62 individuals/120 individuals = 0.52
freq (A1A2) = 49 individuals/120 individuals = 0.41
freq (A2A2) = 9 individuals/120 individuals = 0.07
Hardy-Weinberg equilibrium satisfied? YES
Expected (given Hardy-Weinberg)
p2
(0.72) 2 = 0.52
2pq (2) (0.72) (0.28) = 0.41
q2
(0.28) 2 = 0.07
Observed (above genotype frequencies)
0.52
0.41
0.07
2. In an experimental population of Drosophila, a sample of males and virgin females
includes 66 A1A1, 86 A1A2 and 28 A2A2 flies. Each genotype is represented equally in
both sexes and each can be distinguished by eye color. (adapted from Futuyma 2005)
a) Determine the allele and genotype frequencies and whether or not the locus is
in Hardy-Weinberg equilibrium.
Calculating allele frequencies
freq (p) = {2(66) + 86}/2*180 = 0.61
freq (q) = {2(28) + 86}/2*180 OR 1 - 0.61 = 0.39
Calculating genotype frequencies:
freq (A1A1) = 66 individuals/180 individuals = 0.37
freq (A1A2) = 86 individuals/180 individuals = 0.48
freq (A2A2) = 28 individuals/180 individuals = 0.15
Hardy-Weinberg equilibrium satisfied? YES
Expected (given Hardy-Weinberg)
p2
(0.61) 2 = 0.37
2pq (2) (0.61) (0.39) = 0.48
q2
(0.39) 2 = 0.15
Observed (above genotype frequencies)
0.37
0.48
0.15
b) Now suppose you discard half the A1A1 flies and breed from the remainder of
the sample. Assuming the flies mate at random, what will be the genotype
frequencies among their offspring? (Hint: the proportion of A2A2 should be
approximately 0.23)
with ½ A1A1 flies:
A1A1 = 33
A1A2 = 86
A1A2 = 28
Calculating new allele frequencies
freq (p) = {2(33) + 86}/2*147 = 0.52
freq (q) = {2(28) + 86}/2*147 OR 1 - 0.51 = 0.48
Calculating new genotype frequencies under H-W:
Expected (offspring generation)
p2
(0.52) 2 = 0.27
2pq (2) (0.52) (0.48) = 0.50
q2
(0.48) 2 = 0.23
Observed (parental generation)
0.37
0.48
0.15
c) Now suppose you discarded half the A1A2 flies instead of A1A1 flies. What will
be the allele and genotype frequencies in the next generation? Why is the
outcome so different?
with ½ A1A2 flies:
A1A1 = 66
A1A2 = 43
A1A2 = 28
Calculating new allele frequencies:
freq (p) = {2(66) + 43}/2*137 = 0.64
freq (q) = {2(28) + 43}/2*137 OR 1 - 0.64 = 0.36
Calculating new genotype frequencies under H-W:
Expected (offspring generation)
p2
(0.64) 2 = 0.41
2pq (2) (0.64) (0.36) = 0.46
q2
(0.36) 2 = 0.13
Observed (parental generation)
0.37
0.48
0.15
When discarding A1A1 flies, you are decreasing the number of A1 alleles and
increasing the relative number of A2 alleles contributing to the next generation
resulting in a decrease in the frequency of A1 alleles and a relative increase in A2
allele frequency in the genotypes of the next generation. When discarding A1A2 flies,
the effect is greatest on the genotypes that include A2 considering this allele is at a
lower frequency to begin with (p = 0.64 and q = 0.36). There is little effect on A1
considering this allele enters this selected gene pool with a frequency greater than in
the original parental population.
3. The common morning glory (Ipomoea purpurea) is highly polymorphic for flower
color, producing either purple or pink flowers. Flower color depends on a functional
gene that produces an anthocyanin (pigment) synthesizing enzyme. Purple coloration is
due to one or two copies of the functional allele (A) while two copies of the
nonfunctional allele for this gene (a) results in plants that cannot synthesize anthocyanin,
producing pink coloration instead. A population of I. purpurea was found to have the
following genotypes and their associated phenotypes for flower color:
genotype:
AA
Aa
aa
phenotype:
Purple
Purple
Pink
number of
individuals:
200
80
120
a) What are the genotypic frequencies for each flower color type?
For genotype AA: 200 individuals/400 total individuals = 0.5
For genotype Aa: 80 individuals/400 total individuals = 0.2
For genotype aa: 120 individuals/400 total individuals = 0.3
b) What are the phenotypic frequencies for each flower color type?
For phenotype of Purple: 280 individuals/400 total individuals = 0.7
For phenotype of Pink: 120 individuals/400 total individuals = 0.3
**Remember, frequencies are calculated by dividing similar types of values by one
another, so that the units cancel. Frequencies are unitless values (i.e. p = 0.45)
c) What are the allele frequencies of A and a?
Remember, when we’re dealing with allele frequencies, we are now thinking about
the gene pool, not individuals anymore.
To get the frequency of A, which we’ll call “p”:
freq (A) = {2(200) + 80}/2*400 = 0.6
The first term 2(200) is the component contributed by the homozygote AA (thus 2
copies of A), 80 is contributed by the heterozygote (only one copy), and we divide by
the total number of alleles, and since each individual contributes 2 alleles, it’s 2 *400
total alleles in the gene pool.
Similarly:
freq (a) = {80 + 2(120)}/2*400 = 0.4
You could have also calculated the freq (a) by simply taking 1-freq(A), since these
two frequencies must always equal out to 1.
4. Now suppose that you cannot tell the difference between the homozygote and
heterozygote purple individuals from problem #3. Assuming that the original morning
glory population is in Hardy-Weinberg equilibrium:
a) What is the genotype of plant that has pink flowers?
aa
b) Using the equation for genotypic frequencies at HW equilibrium, calculate q
(allele frequency of a):
We already calculated the genotypic frequencies above for aa (this was 0.3);
and fitting this into the HW equilibrium equation for genotypic frequencies,
p2 + 2pq + q2 , where q2 equals the genotypic frequency of the aa genotype:
q2 = 0.3
q = 0.548 (taking the square root of both sides of the equation
to solve it)
c) Now calculate p (allele frequency of A):
If we feel confident about the above answer, p is simply 1-q.
p = 1-0.548 = 0.452
d) Using the answers from b and c, what are the genotypic frequencies of the
three possible genotypes at equilibrium?
freq (AA) = p2 = (.452) 2 = 0.204
freq (Aa) = 2pq = 2(.452)(.548) = 0.496
freq (aa) = q2 = (.548) 2 = 0.300
e) How many individuals are there of each genotype in the population? How do
these differ from the original population?
Expected (population under H-W) Observed (population)
AA = 0.204 (400) = 82
AA = 0.5 (400) = 200
Aa = 0.496 (400)= 198
Aa = 0.2 (400) = 80
aa = 0.300 (400) = 120
aa = 0.3 (400) = 120
There are many more heterozygotes (Aa) in this population under H-W
equilbrium!
f) When the next generation of morning glory plants flower in this population, it
is observed that genotype frequencies have changed – the number of
heterozygotes (Aa) have decreased from 80 to 60! Is this new generation in
Hardy-Weinberg equilibrium?
No, the new generation has a different genotype frequency which means at
least one of the assumptions underlying the Hardy-Weinberg principle has
been violated or else the genotype frequency from the parental generation to
this next generation of morning glories should have remained the same.
g) Preliminary field experiments suggest that heterozygote morning glory plants
do not produce as many seeds as either of the homozygotes in the low
nutrient soil the population inhabits. Which of the four assumptions necessary to
produce the Hardy-Weinberg equilibrium has been violated?
The assumption that there is no difference among genotypes in survival or
fecundity has been violated. It appears that there is selection for
individuals with higher fecundity under low nutrient conditions in this
population and the heterozygotes are at a disadvantage, producing less
progeny then either of the two homozygotes.
5. Assume a population of Hawaiian picture-wing flies (Drosophila mulli) has a gene
controlling wing size for which there is two alleles: W (large wings) and w (reduced
wings), and that counts of each genotype in the population are as follows:
genotype:
WW
Ww
ww
number of
Individuals:
120
100
10
What is the probability that allele w will go to fixation by genetic drift? (Hint: see page
231 in textbook)
The probability that an allele goes to fixation is EQUAL to that allele’s initial
frequency in that population.
Freq (w): {2(10) + 100}/2(230) = 0.26
The probability of allele w going to fixation is 0.26 (26% chance of fixation)
6. Describe why the rate of change in heterozygosity (2pq) is often used as a measure of
genetic drift within a population. Completing the table below should make the trend
apparent.
p
q
2pq
0.74
0.88
0.95
0.99
It can be noted that as the frequency of one of the alleles approaches 1.0
(fixation), the frequency of the heterozygous genotype declines. The rate of
decline in heterozygosity can be used to measure the rate of genetic drift
within a population. This is because large fluctuations in allele frequencies
resulting in an increased frequency of one allele will concurrently result in
decreased frequency of the other allele and therefore decrease the potential
of these two alleles combining to result in a heterozygous genotype. (This
result is similar to extensive inbreeding)
p
0.74
q
0.26
2pq
0.3848
0.88
0.95
0.99
0.12
0.05
0.01
0.2112
0.0950
0.0198
7. A single, fertilized female Drosophila melanogaster is carried by a storm and
randomly lands on a small, newly formed oceanic island and deposits her eggs on the
vegetation present. Within several generations, there is a small, isolated population
present on the small island. What is this phenomenon called? Does it result in an
increase or decrease in genetic variation? How does the effect on genetic variation
generated by this phenomenon differ from the effect produced by large-scale migration of
a species to an island (say, populations of seabirds that can move between an island and
the mainland)?
When a new population is established by a small number of colonists (a mating pair,
an inseminated female, a handful of seeds, etc) the random genetic drift that follows
establishment is a type of bottleneck referred to as the founder effect. Considering
only a small, isolated population is present on the island after a few generations; it is
likely that the population will be affected by genetic drift resulting in alteration of
allele frequencies and a decrease in genetic variation. If the population persists and
increases in size, the appearance of new mutations should eventually restore
heterozygosity (i.e. increase genetic variation) over time. While bottlenecks involve
a small number of individuals contributing alleles to the gene pool, regular
migration of individuals to an island population results in the potential introduction
of new alleles into the gene pool (of course it is necessary that migrating individuals
reproduce with individuals in the population), increasing the genetic variation
within an island population and possibly reducing the genetic divergence of the
mainland and island populations (i.e. gene flow via migration). With enough
migration between the mainland and island, any initial increase in genetic variation
should level off as the separate gene pools come to resemble each other.
8. A team of population geneticists are interested in determining the effect of genetic
drift on the genetic divergence of several small, isolated populations of Japanese
macaque (Macaca fuscata). However, there are several underlying assumptions that they
must account for in order to make predictions about the magnitude of genetic drift in the
populations. What evidence in the life history account of this species (below) suggests
that the effective population size (the number of individuals producing offspring) is much
smaller than the population census size (the total number of individuals counted)?
Briefly explain your choice of evidence.
The Japanese macaque (Macaca fuscata) or snow monkey is a terrestrial old
world monkey species native to Japan. It is a diurnal species with an average lifespan of 30 years and overlapping generations. The prime habitat of the Japanese
macaque includes forests within the mountainous areas of Honshu. Diet includes
seeds, roots, buds, fruit, invertebrates, leaves and bark. Troops of 20-100
individuals are usually subdivided into matrilineal groups consisting of many
females to several males. On average, females outnumber males by 3 to 1.
There are several lines of evidence to suggest that the effective population size may
be smaller (and thus more prone to drift/random sampling error) than what studies
based only on the census population size would assume. For instance, the Japanese
macaque has a breeding system in which overlapping generations may breed
simultaneously. This may result in parent-offspring matings (inbreeding) that
would result in identical copies of the same alleles entering the gene pool, decreasing
heterozygosity and effectively decreasing the diversity of alleles in the gene pool.
There is also a skew in the sex ratio resulting in many females mating with the same
males which will also reduce the diversity of alleles entering the gene pool.
9. Slade et al. (1998) studied the population genetics of the southern elephant seal,
Mirounga leonine, and estimated that the breeding sex ratio was 5 females per male. A
researcher is studying a small secluded population of this animal, consisting of 50 males
and 50 females.
a) The observed population size is 100 individuals total. Based on Slade et al.’s work
however, what should the researcher expect the effective population size (NE ) to be?
Equation : NE = 4Nm*Nf/(Nm + Nf); where Nm and Nf are only the # of breeding
males and females, respectively.
Expected # of breeding males: With 5 females per male, only 10 males are expected to
breed
NE = 4(10)(50)/(10+50) = 33
b) What is the significance of a small effective population size? (consider the probability
of a deleterious allele going to fixation)
Drift is more likely to result in random sampling error of alleles such that some may
become fixed and others lost in a small population. If large fluctuations in allele
frequencies result in fixation of a deleterious allele, this may decrease
survival/reproductive success of individuals and possibly lead to extinction of the
population, especially if the deleterious alleles are lethal.
Genetic Drift Simulation Exercise:
We are going to explore factors that affect genetic drift, an evolutionary force discussed
in lecture on Thurs 3/20, using the simulation found at
http://darwin.eeb.uconn.edu/simulations/drift.html.
Initial allele frequency, population size, and time (the number of generations) can all
effect genetic drift.
Let’s explore the effect of population size:
1. Use the following settings:
p = 0.5
N(population size): 10
Generations: 1000.
Run the simulation 20 times.
For each question below make a prediction of what you’ll observe and then record what
you find.
A. How many times did an allele go to fixation? Which allele went to fixation?
B. How many generations did it take for the allele to reach fixation? (Give a
range, the fewest to the greatest number of generations.)
2. Now let’s look at a larger population.
Use the following settings:
p = 0.5
N: 1000
Generations: 1000.
For each question below make a prediction of what you’ll observe and then record what
you find.
A. How many times did an allele go to fixation? Which allele went to fixation?
B. How many generations did it take for the allele to reach fixation? (Give a
range, the fewest to the greatest number of generations.)
Think about the relationship between population size and the effect of genetic drift.
Compare your findings from questions 1 and 2. Is an allele more likely to go to fixation
in a small or large population? Which population (large or small) loses genetic diversity
more rapidly?
Let’s explore the effect of initial allele frequency.
3. Use the following settings:
p = 0.1
N(population size): 250
Generations: 1000.
Run the simulation 20 times.
For each question below make a prediction of what you’ll observe and then record what
you find.
A. How many times did an allele go to fixation?
B. Which allele went to fixation?
4. Use the following settings:
p = 0.9
N(population size): 250
Run the simulation 20 times.
Generations: 1000.
For each question below make a prediction of what you’ll observe and then record what
you find.
A. How many times did an allele go to fixation?
B. Which allele went to fixation?
Does the likelihood that an allele will go to fixation depend on the initial allele
frequency?
Take time to explore what happens when the number of generations ranges from 50 to
2500. Can you set the parameters so that no alleles are fixed (genetic variation is
maintained)?