Download Sample problems for final exam – population genetics, etc. (not to be

Sample problems for final exam – population genetics, etc. (not to be turned in, won’t be
graded, answers are on last page 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 a recessive disorder resulting from a mutation in the gene encoding ßglobin. 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 Hardy-Weinberg, 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.
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
mammals
acts at level of individual
animal
production of testosterone
presence of sry on Y chromo.
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