Download Practice Problems: Population Genetics

Practice Problems: Population Genetics
1. The gamma globulin of human blood serum exists in two forms, Gm(a+) and
Gm(a-), spedified respectively by an autosomal dominant gene Gm(a+) and
its recessive allele Gm(a-). Broman et al. (1963) recorded the tabulated
phenotypic frequencies in three Swedish populations. Assuming the
populations were at Hardy-Weinberg equilibrium, calculate the frequency of
heterozygotes in each population.
Region
No. Tested
Phenotype %
Norbotten County
Stockholm city and
rural district
Malmohus and
Kristianstad counties
139
509
Gm(a+)
55.40
57.76
Gm(a-)
44.60
42.24
293
54.95
45.05
Since the populations ARE in equilibrium, this problem can be solved by defining the
proportion of individuals displaying the recessive phenotype (GM (A-) as q2. You can
then calculate q, and using 1-q = p, you can calculate p. Finally, the frequency of
heterozygotes in a population in equilibrium is 2pq.
Norbotten County:
Stockholm
Malmohus
q2 = 0.446
q = 0.668
p = 0.332
2pq = 0.444
2pq = 0.455
2pq = 0.442
2. Among 2,820 Shorthorn cattle, 260 are white, 1,430 are red, and 1,130 are
roan. Is this consistent with the assumption that the traits are controlled by
a single pair of autosomal alleles and that mating has been at random for this
allele pair?
To determine if a population is in equilibrium, you must calculate the allele
frequencies, use those allele frequencies to calculate the expected number of
white, red and roan cattle, and then use chi-square analysis to compare the
expected numbers with the observed numbers.
First: R1 = red allele
R2 = white allele
Allele frequencies:
p = f(R1) = (1430 + ½(1130)) / 2820 = 0.71
q = 0.29
Expected # Red = p2 x total cattle =( 0.712 )( 2820) = 1421.562
Expected # Roan = 2pq x total cattle = 2(0.71)(0.29)(2820) = 1161.276
Expected # White = q2 x total cattle = (0.292)(2820) = 237.162
c =å
2
(O - E )
=
2
E
(1430 -1421.562)
1421.562
2
+
(1130 -1161.276)
1161.276
2
+
(260 - 237.162)
2
237.162
= 0.05 + 0.84 + 2.199 = 3.089
df = 3-1 = 2
critical value = 5.991
Accept the hypothesis; The traits appear to be controlled by a single pair of
autosomal genes under random mating.
3. On the basis of allele-frequency analysis of data from a randomly mating
population Snyder (1934) concluded that the ability vs. inability to taste
phenylthiocarbamide (PtC) is determined by a single pair of autosomal
alleles, of which T for taster is dominant to T for nontaster. Of the 3,643
individuals tested in this population, 70% were tasters and 30% were
nontasters. Assume the population satisfies the conditions of HardyWeinberg equilibrium.
a) Calculate the frequencies of the alleles T and T and the frequencies of the
genotypes TT, TT and TT.
q2 = 30% = 0.30
q = f(T) = 0.55
p = f(T) = 0.45
p2 = f (TT) = 0.20
2pq = f (TT) = 0.50
b) Determine the probability of a nontaster child from a taster x taster
mating.
Probability (tt) child = P(Tt parent) x P(Tt parent) x P (tt child)
However, we know the parents are not tt, so we must calculate the frequency of Tt
parents within the taster population (not the population at large) as follows:
= frequency Tt/ total frequency of tasters (Tt + TT)
= 0.50 / 0.50 + 0.20
= 0.71
So P(tt child from taster parents) = 0.71 x 0.71 x 0.25 = 0.126
4.
In Drosophila melanogaster, Cncn (red vs. cinnabar eyes), Bb( gray vs. black
body, (and Byby (normal vs. blistery wing) are autosomal pairs of alleles.
Samples of three large natural adult populations, each classified for a
different pair of traits, are found to have the following genotypes:
Population A
Population B
Population C
31 cncn
182 BB
100 ByBy
171 Cncn
391 Bb
372 Byby
60 CnCn
152 bb
40 byby.
Total 262
Total 725
Total 512
Compare these distributions with those expected for a population at HardyWeinberg equilibrium. Propose a reasonable explanation to account for any
differences.
Population A: p = (60 + 85.5)/262 = 0.56; q = 0.44
Expected CnCn = p2 = 0.31 x 262 = 82
Expected Cncn = 2pq = 0.49 x 262 = 129
Expected cncn = q2 = 0.19 x 262 = 51
Population B: p = 0.52; q = 0.48
Expected BB = 196
Expected Bb = 362
Expected bb = 167
Population C: p = 0.56 ; q = 0.44
Expected ByBy = 161
Expected Byby = 252
Expected byby = 99
Observed 60
Observed 171
Observed 31
Observed BB = 182
Observed = 391
Observed = 152
Observed = 100
Observed = 372
Observed = 40
Without doing Chi-square analysis, all three populations show increased
heterozygosity compared to the expected, as well as decreased homozygosity of
both genotypes. There may be selection for the heterozygote, or outbreeding
(you’d need to see the same pattern of increased heterozygotes at all genes) or
negative assortative mating (if this pattern only existed for one gene and closely
linked genes). All three of these situations would give an increase in
heterozygotes. As a side note, whatever is going on is much more pronounced in
population C.
5. A yak population is in Hardy-Weinberg equilibrium with allele frequencies
p(A) = 0.5 and q(a) = 0.5 for a gene governing color differences. If a new type
of predator appears in the area, calculate the new values of q if:
a) sa/a = 1.0
b) sa/a = 0.70
c) sa/a = 0.10
To solve this problem, you must first calculate the fitness values for each
recessive genotype (1-S = W). Since AA and Aa are not mentioned, they have
a fitness of 1.0.
Then, calculate the mean population fitness using:
W = p2WAA + 2 pqWAa + q 2Waa
Then calculate the new values for q using the formula:
qa = q2Waa + pqWAa
W
W
a)Waa = 1-1=0
W  .52 (1)  2(.5)(.5)(1)  .52 (o)  0.75
qa = .52(0) + (.5)(.5)(1)
0.75
0.75
= 0.33
b) Waa = 1-.7=0.3
W  .52 (1)  2(.5)(.5)(1)  .52 (.3)  0.825
qa = .52(0.3) + (.5)(.5)(1)
0.825 0.825
= 0.39
c) Waa = 1-.1 = 0.9
W  .52 (1)  2(.5)(.5)(1)  .52 (.9)  0.975
qa = .52(0.9) + (.5)(.5)(1)
0.975 0.975
= 0.49
6. If the frequency of an allele d is 0.25 in a migrant population and 0.5 in The
recipient population, and if the migration rate is 0.1, what is the frequency of
d in the recipient population after one generation of migration?
= -0.025
new p = 0.5-0.025 = 0.475