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 %
Gm(a+) / Gm(a-)
Norbotten County / 139 / 55.40 / 44.60
Stockholm city and rural district / 509 / 57.76 / 42.24
Malmohus and Kristianstad counties / 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: q2 = 0.446

q = 0.668

p = 0.332

2pq = 0.444

Stockholm 2pq = 0.455

Malmohus 2pq = 0.442

2.  A sheep rancher in Iceland finds that the recessive allele y for yellow fat has become established in his flock of 1,024 and that about 1 out of every 256 sheep expresses the trait.

a)  The rancher wishes to know how many of the normal sheep carry the recessive allele. Assuming the population is randomly mating for this gene and all genotypes have the same reproductive fitness, what is this proportion?

Again, the proportion of individuals displaying the recessive phenotype is q2 if the population is in equilibrium (randomly mating for this gene and same fitness is good enough). So, using 1/256 = q2, you can calculate 2pq = 0.115

b)  How many of the 1,020 white animals can be expected to be homozygous?

p2 = 0.882

3.  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

=

= 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.

4.  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 Hardy-Weinberg 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

5.  The MN blood-group frequencies (in percent) in a certain population are MM = 28.38, MN = 49.57, NN = 22.05.

a)  Calculate the frequencies of the M and N alleles and determine whether or not this population is in Hardy-Weinberg equilibrium.

p = f(M) = .2838 + ½ (.4957) = 0.532

q = f(N) = .468

Expected MM = p2 = 0.283

Expected MN = 2pq = 0.498

Expected NN = q2 = 0.219

It’s pretty darn close. Without whole numbers you can’t really tell for sure.

6.  You wish by artificial selection to reduce the frequency of a recessive trait in a large randomly mating population in which the frequency of the recessive allele is 0.5.

a)  Show the initial types and proportions of the different phenotypes in this population, assuming Hardy-Weinberg equilibrium.

q = 0.5, p = 0.5

Dominant phenotypes = p2 + 2pq = 0.25 + 0.5 = 0.75

Recessive phenotypes = q2 = 0.25

b)  Determine the frequency of the alleles in the population after one and two generations of complete selection against the recessive allele (s=1).

Generation 1:

= 0.75

= 0.67; q= 0.33

Generation 2:

= 0.89

= 0.75; q = 0.25

7.  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

= 0.75

= 0.67; q’ = 0.33

b)  sa/a = 0.70

= 0.825

= 0.61; q’ = 0.39

c)  sa/a = 0.10

= 0.975

= 0.51; q’= 0.49

8.  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 31 cncn 171 Cncn 60 CnCn Total 262

Population B 182 BB 391 Bb 152 bb Total 725

Population C 100 ByBy 372 Byby 40 byby. Total 512

Compare these distributions with those expected for a population at Hardy-Weinberg 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 Observed 60

Expected Cncn = 2pq = 0.49 x 262 = 129 Observed 171

Expected cncn = q2 = 0.19 x 262 = 51 Observed 31

Population B: p = 0.52; q = 0.48

Expected BB = 196 Observed BB = 182

Expected Bb = 362 Observed = 391

Expected bb = 167 Observed = 152

Population C: p = 0.56 ; q = 0.44

Expected ByBy = 161 Observed = 100

Expected Byby = 252 Observed = 372

Expected byby = 99 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.

9.  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