Problems and Discussion Topics with Sample Answers
by Joseph Lachance and Paul Bourdeau
to accompany
Evolution,
Second Edition
Chapter 9: Variation
1. In an electrophoretic study of enzyme variation in a species of grasshopper, you
find 62
A
1
A
1
, 49
A
1
A
2
, and 9
A
2
A
2
individuals in a sample of 120. Show that
p
= 0.72
and
q
= 0.28 (where
p
and
q
are the frequencies of alleles
A
1
and
A
2
), and that the
genotype frequencies of
A
1
A
1
,
A
1
A
2
, and
A
2
A
2
are approximately 0.51, 0.41, and 0.08,
respectively. Demonstrate that the genotype frequencies are in Hardy-Weinberg
equilibrium.
Answer:
Both alleles present in
A
1
A
1
homozygotes are
A
1
, while half the alleles present in
A
1
A
2
heterozygotes are
A
1
. There are 240 genes copies present in the population, 173
(2*62 + 49) of which are
A
1
. This gives
p
=
0.7208 (173/240) and
q
= 27.92 (67/240).
51.67% (62/120) of grasshoppers are
A
1
A
1
, 40.83% (49/120) are
A
1
A
2
, and 7.50% (9/120)
are
A
2
A
2
. If genotype frequencies are in Hardy-Weinberg equilibrium we would expect
51.96% (
p
2
) to be
A
1
A
1
homozygotes, 40.25% (2
pq
) to be
A
1
A
2
heterozygotes, and 7.79%
(
q
2
) to be
A
2
A
2
homozygotes. Note the close match between expected and observed
genotype frequencies.
2. In a sample from a different population of this grasshopper species, you find four
alleles at this locus. The frequencies of
A
1
,
A
2
,
A
3
, and
A
4
are
p
1
= 0.50,
p
2
= 0.30,
p
3
=
0.15,
p
4
= 0.05. Assuming Hardy-Weinberg equilibrium, calculate the expected
proportion of each of the ten possible genotypes (e.g., that of
A
2
A
3
should be 0.09).
Show that heterozygotes, of all kinds, should constitute 63.9 percent of the
population. In a sample of 100 specimens, how many would you expect to be
heterozygous for allele
A
4
? How many would you expect to be homozygous
A
4
A
4
?
Answer:
Homozygote frequencies are equal to the square of allele frequency, and
heterozygote frequencies are equal to two times the product of the allele frequencies of
each allele present in the heterozygote of interest.
Proportion
A
1
A
1
= 0.50*0.50 = 0.25
Proportion
A
2
A
2
= 0.30*0.30 = 0.09
Proportion
A
3
A
3
= 0.15*0.15 = 0.0225
Proportion
A
4
A
4
= 0.05*0.05 = 0.0025
Proportion
A
1
A
2
= 2*0.50*0.30 = 0.3
Proportion
A
1
A
3
= 2*0.50*0.15 = 0.15
Proportion
A
1
A
4
= 2*0.50*0.05 = 0.05
Proportion
A
2
A
3
= 2*0.30*0.15 = 0.09
Proportion
A
2
A
4
= 2*0.30*0.05 = 0.03
Proportion
A
3
A
4
= 2*0.15*0.05 = 0.015
Summing up the last six genotype frequencies above gives us the total proportion that is
© 2009 Sinauer Associates, Inc.
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