Download Zoo/Bot 3333 Genetics Quiz #3 10/28/11 For the answers to the quiz

Zoo/Bot 3333
Genetics
Quiz #3
10/28/11
For the answers to the quiz, click here:
In tomatoes, J.W. Lesley determined the location of
several gene loci by self-fertilizing three different
plants, each trisomic for a different chromosome (A,
B, or I) carrying a dominant allele, and heterozygous
for the following allele pairs on the other homologs:
1. On the basis of this data, the gene(s) that are
located on chromosome B are:
a) D; b) D, L, R and A; c) Y and R; d) L; e) none of
the above.
2. Mules result from a mating between a horse (2n = 64) and a donkey (2n = 62). Although very rare, there
are cases of female mules giving birth to healthy offspring. In these situations, when a female mule mates
with a male horse, the offspring is horse-like in appearance; when she mates with a male donkey, they are
mule-like in appearance. What is the best explanation for the origin of the horse-like offspring?
a) the offspring are monosomic for only horse chromosomes; b) the offspring are trisomic for an extra
horse chromosome; c) the offspring have lost an extra horse chromosome; d) during meiosis in the female,
all of the horse chromosomes and none of the donkey chromosomes segregated to the ovum that was
fertilized; e) the father donated two sets of horse chromosomes to an ovum that had one set of donkey
chromosomes.
3. The picture on the right depicts a fruitfly that has the following phenotype. The
body pigmentation, as well as body size, is asymmetrical. The left side of the animal
also has a white eye and a miniature wing (these mutations were discussed during
our lectures on recombination). What would explain this phenomenon?
a) only half of the fly’s cells contain a Y chromosome; b) the fertilized egg that gave
rise to this fly must have been homozygous for the wild type alleles for white and
miniature, since these are recessive mutations; c) mutations for white and miniature
occurred early during embryogenesis, generating a mosaic; d) all of the above; e)
none of the above.
4. True or false. Since drone honey bees contain only one set of chromosomes and worker honey bees
contain two sets of chromosomes, only workers can be considered euploid.
Questions 5-6 pertain to the following. Six bands in a salivary gland chromosome of Drosophila are shown
in the figure below (left), along with the extent of five deletions:
Deletion 1
Deletion 2
Deletion 3
Deletion 4
Deletion 5
a
−
−
−
−
+
b
−
−
−
+
+
c
−
+
+
+
+
d
+
+
−
−
−
e
+
+
+
−
−
f
+
+
+
+
−
Recessive alleles a, b, c, d, e, and f are known to be in the region, but their order is unknown. When the
deletions are heterozygous with each mutant allele, the following results are obtained (above right), where
(+) indicates a wild type phenotype, and a (-) indicates a mutant phenotype:
5. The correct order of genes relative to chromosome bands 1-6 is:
a) cedabf; b) decfab; c) cabfde; d) bacedf; e) none of the above.
6. The a gene locus would lie closest to band:
a) 1; b) 2; c) 3; d) 4; e) none of the above.
Questions 7 and 8 pertain to the following. Four E. coli
strains
1
2
3
4
strains of genotype a+b− are labeled 1, 2, 3, 4. Four
0
M
L
0
5
− +
strains of genotype a b are labeled 5, 6, 7 and 8. The
M
0
0
M
6
two genotypes are mixed in all possible combinations
L
0
0
L
7
and (after incubation) are plated to determine the
+ +
M
0
0
M
8
frequency of a b recombinants. The results indicated
in the table are obtained, where M = many
recombinants, L = low numbers of recombinants, and 0 = no recombinants. The strains can be classified as
3 conjugation types: either F−, F+ or Hfr with regard to a and b gene transfer.
7. Which of the following can be classified as Hfr cells?
a) strains 2, 3, 7, 6 and 8; b) strain 2 only; c) strains 2 and 3; d) strains 2 and 7; e) none of the above.
8. True or False. We would expect the majority of bacteria arising from the cross of stain 1 and strain 7 to
produce cells that were capable of being donors.
Questions 9-10 pertain to the following. An Hfr strain of the genotype a+b+c+d+strs is mated with a female
strain of the genotype a−b−c−d−strr. At various times the culture is disrupted in a blender to separate the
mating pairs. The cells are then plated on agar of the following four types (see table below), where nutrient
A allows the growth of a− auxotrophs, nutrient B allows for the growth of b− autxotrophs, etc. A (+)
indicates the presence of the nutrient or
Timings of Samples Number of Colonies of Agar of Type
drug, a (-) indicates its absence.
1
2
3
4
0
0
0
0
0
2.5
0
0
0
0
5
15
0
0
0
7.5
104
0
0
0
Agar Type Str A B C D
+
+ − + +
10
215
8
0
0
1
+
+ + − +
2
12.5
301
75
6
0
+
+ + + −
3
15
347
150
67
5
+
− + + +
4
17.5
400
202
104
50
The table (above right) shows the number
20
402
205
150
90
of colonies on each type of agar for
25
398
203
153
90
samples taken at various times after the
samples are mixed:
9. Relative to their proximity to the F factor origin of replication (first gene listed below is closest) the
gene order of these four genes is:
a) a-b-c-d; b) b-a-c-d; c) d-a-c-b; d) b-d-a-c; e) none of the above.
10. True or false: Gene b is closer to c than c is to d.