BIOL/PBIO 3333
Genetics
Quiz 3
10/24/14
For answers to the quiz, click here:
1. In an animal bearing the heterozygous inversion ABCDEFGHI/ABGFEDCHI, in one meiocyte a crossover occurred
between the D and E loci and another crossover occurred between the F and G loci. These crossovers involved the same two
non-sister chromatids. What percentage of the crossover products from this event will show duplications and deficiencies for
these linked genes?
a) 0%; b) 25%; c) 50%; d) 100%; e) it depends on the size of the D-E and F-G gene intervals.
2. In fragile X syndrome:
a) females carrying the chromosome have a greater chance for passing the disorder on to their children than their brothers
carrying the chromosome; b) individuals with over 50 trinucleotide repeats will show mental retardation; c) there is a linear
relationship between the number of trinucleotide repeats and the severity of the disorder; d) all of the above; e) none of the
above.
3. The cartoon on the right represents an ideogram of the GBanding patterns for two primate species. In comparing the
patterns, we see evidence of which events occurring in these
primates?
a) a Robertsonian translocation for chromosome two; b) a
pericentric inversion for chromosome 4;
c) a pericentric inversion for chromosome 5; d) all of the
above; e) none of the above.
4. Four strains of Drosophila melanogaster are isolated from
different locations. The banding patterns of a particular
region of salivary chromosome 2 have configurations shown
below (each letter denotes a band):
strain 1: acefgjkz; strain 2: acgjefkz; strain 3: akfejgcz; strain 4: acgfejkz
Assuming the chromosome in strain 1 is the ancestral sequence, the evolutionary ancestry leading to the other chromosomal
rearrangements is:
a) 1234; b) 1342; c) 1423; d) 1324; e) none of the above.
Seven partial deletions (1 to 7), shown as gaps in the diagram on the left below, have been mapped on an intact chromosome
(shown above deletions. Heterozygotes are constructed where the intact homologous chromosome contains seven recessive
mutations, identified in the table on the right below. The deletion heterozygotes "uncovered" (see p. 496/3e; p. 435/4e in text)
these recessive mutations and allowed them to show pseudodominance, indicated by a minus sign in the table:
1
2
3
4
5
6
7
chromosome
deletions
1
2
3
4
5
6
7
a
+
+
+
+
+
−
+
b
−
+
−
+
+
+
+
c
+
+
+
+
−
+
+
genes
g m n
− + −
+ + +
+ + +
+ − +
+ + −
+ − +
+ − +
r
+
+
−
+
+
+
+
s
+
−
+
−
+
+
+
5. The order of the genes on the intact chromosome is:
a) brcngxsvac; b) rbgncxsvma; c) rbngcmvsxa; d) grnbavmxcs; e) more than one gene orientation is possible from these
experiments and you would need additional overlapping mutants.
v
+
−
+
−
+
+
−
x
+
−
+
+
+
+
+
6. From the above data set, one can determine:
a) r is further from b than v is from a; b) c and s are the closest genes to one another; c) a heterozygote between deletions 1 and
2 would likely be inviable; d) all of the above; e) none of the above.
Questions 7 and 8 pertain to the following. Four E. coli strains of genotype a+b− are labeled 1, 2, 3, 4. Four strains of
genotype a−b+ are labeled 5, 6, 7 and 8. The two genotypes are mixed in all possible combinations and (after incubation) are
plated to determine the 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 sex types: either
F−, F+ or Hfr with regard to a and b gene transfer.
strains
5
6
7
8
1
L
0
0
L
2
M
0
0
M
3
0
M
L
0
4
M
0
0
M
7. The F− cells present in this data set are represented by:
a) 1 and 7; b) 2 and 4; c) 6 and 7; d) 7 only; e) none of the above.
8. True or false. Suppose after mixing strains 5 and 2 the culture was left to grow on medium containing the nutrients needed
by both the a- and b- mutants. The progenitor cells arising from this experiment would be expected to be able to form pili if
examined under the electron microscope.
Questions 9-10 pertain to the following. An Hfr strain of the genotype s+t+u+v+strs is mated with a female strain of the
genotype s−t−u−v−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 agar types (see table below, right), where nutrient S allows the growth of s− auxotrophs,
nutrient T allows for the growth of t− autxotrophs, etc. As indicated, all types contain streptomycin (Str). A (+) indicates the
presence of the nutrient or drug, a (-) indicates its absence.
Timings of
Samples
0
2.5
5
7.5
10
12.5
15
17.5
20
25
30
35
Number of Colonies of Agar of Type
1
2
3
4
0
0
0
0
0
0
0
0
0
32
44
48
0
0
0
3
60
113
139
164
180
183
186
189
0
0
0
0
2
56
87
114
122
135
143
146
0
4
68
136
222
318
371
396
414
416
420
425
Agar Type
1
2
3
4
Str
+
+
+
+
S
+
+
−
+
T
+
+
+
−
U
−
+
+
+
V
+
−
+
+
The table on the left shows the number of colonies on each type of agar for samples taken at various times after the samples are
mixed:
9. Relative to their proximity to the F factor origin of replication (the first gene listed below is closest) the gene order of these
four genes is:
a) s-t-u-v; b) t-v-s-u; c) t-u-s-v; d) v-s-t-u; e) none of the above.
10. The genes that map closest together in this experiment are:
a) v and u; b) t and v; c) v and s; d) t and s; e) there is a closer gene pair that is not listed.