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18
Population Genetics
WORKING WITH THE FIGURES
1.
Which individual in Figure 18-3 has the most heterozygous loci, and which
individual has the fewest?
Answer: Individual 3 has three heterozygous loci (microsatellites 1, 3 and 4),
while individual 1 has none.
2.
Suppose that the seven haplotypes in Figures 18-1 and 18-4a represent a
random sample of haplotypes from a population.
a. Calculate gene diversity ( GD) separately for the indel, the microsatellite
locus, and the SNP at position 3.
b. If the sequence was shortened so that you had data only for positions 1
through 24, how many haplotypes would there be?
c. Calculate the linkage disequilibrium parameter (D) between the SNPs at
positions 29 and 33.
Answer:
a. GD (gene diversity) is calculated as:
1- (p 12 + p2 2 +... pn2) where p is allele frequency.
2/7 chromosomes are different at the indel site, so GD = 1-0.59 = 0.41
4/7 at the microsatellite locus, so GD = 0.49
1/7 differences based on SNP in position 3 and GD = 0.26
b. There would be five haplotypes, instead of six: A, B, C, D (lia and lie)
and E (both lib).
c. D stands for the difference between the observed frequency of a haplotype
and the expected frequency of a haplotype (with random association among
alleles at the two loci). D = P AB - PAPB· At position 29, G is present in 6
of 7 haplotypes for a frequency of 0.857. A is present at position 33 in
out
Chapter Eighteen 435
3 out
of 7 haplotypes for a frequency of 0.429. A random combination of G
and A would be expected to occur with a frequency of 0.857 x 0.429 = 0.368,
so PAPB would equal 0.368. The observed frequency of the GA haplotype,
is 3 out of 7, which equals 0.429. P AB - PAPs equals 0.429 - 0.368 =
P AB,
0.061 = D. D is different than 0, so population is in linkage disequilibrium for
the given haplotypes at the positions 29 and 33.
3.
Looking at Figure 18-6, can you count how many mitochondrial haplotypes
were carried from Asia into the Americas? How many of these made it all the
way to South America?
Answer: Five mtDNA haplotypes carried over from Asia to North and South
America: A x , A, B, C and D. Four of these made it to South America (A x did
not), with haplotype B having a somewhat different path than A, C, and D.
4.
In Figure 18-13, the "general" column for Japan is higher than the "general"
column for Europe. What does this tell you?
Answer: This column tells us that the frequency of first-cousin marriages is
higher in Japan (0.05) than in Europe (0.02) and that consequently, frequency of
genetic disorders caused by inbreeding depression is higher in Japan.
5.
In Figure 18-14, some individuals have unique SNP alleles-for example, the T
allele at SNP4 occurs only in individual 12. Can you identify two individuals
each of whom have unique alleles at two SNPs?
Answer: These individuals are: 21 (Native American) and 14 (South African).
Individual 21 has AT in SNP positions 6 and 7, while individual 14 has a T in
SNP 11 and A in SNP 13. Both with two unique alleles compared to the rest of
the sample.
6.
Looking at Figure 18-20, do people of the Middle East tend to have higher or
lower levels of heterozygosity compared to the people of East Asia? Why might
this be the case?
Answer: Yes, people in Middle East appear to have higher heterozygosity
(haplotype and microsatellite ), when compared to people in East Asia. This is
most likely due to the anthropological circumstances, considering that Middle
Eastern populations had strong migratory history and the present populations
have genetic signatures from a number of ancestral populations. East Asia had
been somewhat isolated, with less migration from Middle East, Europe or
Central Asia. It is also likely that genetic drift played a role. Human
populations migrated out of Africa and Middle Eastern populations were
436 Chapter Eighteen
established before East Asian populations. East Asian populations were
established from later migrations out of the Middle East. With each migration,
a reduction in population size would lead to bottlenecks and a greater influence
from drift, which would lower heterozygosity. Those populations established
after two successive migrations would be expected to have less heterozygosity
than those established after one migration.
BASIC PROBLEMS
7.
What are the forces that can change the frequency of an allele in a population?
Answer: The frequency of an allele in a population can be altered by natural
selection, mutation, migration, and genetic drift (sampling errors).
8.
In a population of mice, there are two alleles of the A locus (A 1 and A 2 ). Tests
showed that, in this population, there are 384 mice of genotype A 1/A 1, 210 of
A 1/A2, and 260 of A2/A 2. What are the frequencies of the two alleles in the
population?
Answer: There are a total of (2)(384) + (2)(21 0) + (2)(260) = 1708 alleles in the
population. Ofthose, (2)(384) + 210 = 978 areAl and 210 + (2)(260) = 730 are
A2. The frequency of A 1 is 978/1708 = 0.57, and the frequency of A2 is
730/1708 = 0.43.
9.
In a randomly mating laboratory population of Drosophila, 4 percent of the flies
have black bodies (encoded by the autosomal recessive b), and 96 percent have
brown bodies (the wild type, encoded by B). If this population is assumed to be
in Hardy-Weinberg equilibrium, what are the allele frequencies of B and b and
the genotypic frequencies of BIB and Bib?
Answer: The given data are l = 0.04 and p2 + 2pq = 0.96. Assuming HardyWeinberg equilibrium, if q2 = 0.04, q = 0.2, and p = 0.8. The frequency of BIB is
2
p = 0.64, and the frequency of Bib is 2pq = 0.32.
10.
In a population of a beetle species, you notice that there is a 3 : 1 ratio of shiny
to dull wing covers. Does this ratio prove that the shiny allele is dominant?
(Assume that the two states are caused by two alleles of one gene.) If not, what
does it prove? How would you elucidate the situation?
Answer: No, a 3: 1 ratio stands for F2 offspring from heterozygous F 1 genotypes
(with two alleles of one gene and dominant-recessive interaction). This
observation only proves that the shiny wing phenotype is more common among
the beetles observed.