Sample problems for final exam – population genetics, etc. (not to be turned in, won’t be
graded, answers are on last pages of this handout)
1. Huntington’s chorea causes neurodegeneration and ultimately death. Onset of symptoms is
usually between the ages of 30 and 50. Huntington’s is inherited as an autosomal dominant.
The frequency of Huntington’s is approximately 1/20,000. Assume the population meets the
criteria for Hardy-Weinberg, and that there are only two alleles of the Huntington’s gene, H
(the allele responsible for the disorder) and h.
A. Calculate the frequency of H (H = mutant allele) in the population.
B. Calculate the frequency of h (where h refers to the wild-type allele).
C. Calculate the frequency of homozygous wild type, heterozygotes and homozygous mutant
within the population.
2. Patterning of the anterior-posterior body axis in Drosophila involves a series of protein
interactions, where early acting proteins regulate the expression of subsequent proteins.
A. Briefly describe the series of events that divides the embryo into polarized segments and
the provides the identity for each segment. Include the names of each class of genes, and give
one example of a member of each of those classes of genes.
B. Early Drosophila development differs from many animals, including mammals, in at least
one significant way. Because of this difference, the earliest steps in establishing the anterior-
posterior axis must differ between Drosophila and mammals. How is early patterning in
Drosophila likely to differ from mammal?
3. Explain briefly how sex determination differs between C. elegans, Drosophila and mammals.
4. ß-thalassemia is an autosomal recessive disorder resulting from a mutation in the gene
encoding ß-globin. In some Mediterranean populations ß-thalassemia occurs at a frequency of
1/400 (i.e. 1/400 people are affected.) Assume the population meets the criteria for HardyWeinberg, and that there are only two alleles of the ß-globin gene, B and b.
A. Calculate the frequency of b (b = mutant allele) in the population.
B. Calculate the frequency of B (where B refers to the wild-type allele).
C. Calculate the frequency of homozygous wild type, heterozygotes and homozygous mutant
within the population.
5. The pedigree to the right shows a family in which one boy is
affected by Hurler syndrome, an autosomal recessive disorder that
affects about 1/100,000 births. It is typically diagnosed soon after
birth. What is the probability that the individual denoted by the question mark will be affected
by Hurler syndrome?
6. Hemophilia A is an X-linked recessive disorder that affects approximately 1/10,000 boys.
Using the Hardy-Weinberg principle what do you estimate the frequency of carriers to be?
7. The frequency of Tay Sachs disease (TS) in the Ashkenazi Jewish population is
approximately 1 in 3600. A couple from this population seeks advice from their genetic
counselor. With a previous husband, the woman had a child with TS. The two men are
unrelated to each other and her current spouse does not have any family history of TS. What is
the risk that their first child will have TS?
8. X-linked hypophosphatemia is transmitted as an X-linked dominant disorder. The disorder
affects 1/20,000 male births. Based on this, the frequency of affected females can be estimated
to be …
9. Assume for the purposes of this problem that the order of and distances between three
linked genes are the following:
unc-32 10mu vab-7 20mu dpy-17
What is the probability of obtaining a phenotypic unc-32 vab-7 + offspring from a cross
between a heterozygous fly(unc-32 vab-7 dpy-17/+ + +) and the triple recessive nematode?
10. While on an African safari, you find two true breeding strains of a unique giraffe. One
strain has bright purple spots and the other has turquoise spots. When the two strains are
crossed, all the F1 individuals have bright purple spots. When F1 individuals are crossed, you
find:
595 purple spots
153 purple and turquoise spots
52 turquoise spots
A. Please give the genotypes of the F2s.
B. If you cross a purple and turquoise spotted F2 with a turquoise spotted F2, what is the
probability that the offspring will have purple and turquoise spots?
Answers:
1A. The key to this problem is to realize that, since Huntington’s chorea is dominant, effected
individuals are either HH or Hh. Affected individuals occur with a frequence of 1/20,000 or
0.00005. Therefore, unaffected individuals, i.e. hh, are 1 – 1/20,000 or 0.99995, and this equals
p2. Therefore, p (the frequency of h allele in the population) = (0.99995)1/2 or 0.999975. q = 1 – p
or 1 – 0.999975 = 0.000025.
B. The answer is 0.999975 as described above.
C. The frequencies are:
hh = p2 = 0.99995 (see part A for an explanation of this)
Hh = 2pq = 2(0.000025)(0.999975) = 0.0000499
HH = q2 = (0.000025)2 = 6.25 X 10-10
2A.
morphogen (eg. bcd, bh, nos)
gap genes (eg. kruppel, knirps)
homeotic genes (eg. antp, btx)
pair rule genes (eg. eve, ftz)
segment polarity genes (eg. wg, en)
B. The key here is to remember that early Drosophila embryos form a syncitium prior to
cellularization. This means that mammals, which are cellularized, must use secreted molecules
and receptors for those molecules to establish a gradient of morphogens, whereas flies don’t
have this problem because all nuclei are in a common cytoplasm
3A.
worms
acts at level of cell
flies
acts at level of cell
negative regulatory cascade
regulate level/activity
X:A ratio
positive regulatory cascade
primarily regulated splicing
X:A ratio
4A.
q2 = freq. of bb = 1/400
q = freq. of b = (1/400)1/2 = 1/20 or 0.05
B.
p = 1 – q = 1 – 0.05 = 0.95
C.
BB = p2 = 0.9025
Bb = 2pq = 2(0.095)(0.05)
bb = 1/400 or 0.0025
mammals
acts at level of individual
animal
production of testosterone
presence of sry on Y chromo.
5. q2 = 1/100,000 so q = 0.0032 Therefore the probability that the woman who marries in is a
carrier is 2pq = 0.0064. The probability of the couple having an affected child is then
0.0064(0.67)(0.25) = 0.001 (The 0.67 is the probability that the father is a carrier, given that he is
unaffected. The 0.25 is the probability of a homozygous mutant child from two carrier
parents.)
6. q = 1/10,000 p~1 The frequency of heterozygous females is 2pq = 1/5,000
7. q2 = 1/3600 so q = 1/60, 2pq = 2(59/60)(1/60) ~ 1/30, the probability that her current
husband is a carrier. The probability that their child will be affected is ¼(1/30)(1) – 1/120
8. p (the frequency of the dominant, mutant allele) = 1/20,000, q ~1
Therefore 2pq = 2(1/20,000)(1) = 1/10,000, which is the frequency of heterozygous females
who will be affected by the disorder. Although this number ignores homozygous mutant
females, their frequency is negligible (1/20,000)2.
9. 0.20 = freq. SCO + freq. DCO; freq. DCO = 0.1(0.2) = 0.02
0.20 = freq. SCO + 0.02 so freq. SCO = 0.18
Half of the SCO’s between vab-7 and dpy-17 produce the requisite chromosome so answer is
0.09
10A. 595 purple spots P–T– and ppT–
153 purple and turquoise spots P–tt
52 turquoise spots pptt
B. P–tt X pptt
1/3 PPtt X pptt => Pptt
2/3 Pptt X pptt => ½ Pptt and ½ pptt
Probability of purple and turquoise = 1/3(1) + 2/3(1/2) = 2/3