Zoo/Bot 3333
Genetics
Quiz 3
Fall 2009
For answers to the quiz, click here
1. In Drosophila, the white gene locus is located at the very tip of the X chromosome, far from the
centromere. In certain X chromosome inversions, an inversion breakpoint near the white gene locus will
move it from its normal position to a region close to the centromere, adjacent to a segment of centromeric
heterochromatin. What are the expected results of this inversion?
a) expression of the white gene should increase in all cells where it is normally expressed; b) expression of
the white gene should increase in some of the cells where it is normally expressed; c) expression of the
white gene should decrease in some cells where it is normally expressed; d) expression of the white gene
would spread to cells where it is not normally expressed; e) expression of the white gene should remain the
same, since the breakpoint is not located within coding sequence.
Questions 2-3 pertain to the following. A son with Klinefelter Syndrome is born to a mother who is
phenotypically normal and a father who has the X-linked skin condition anhidrotic ectodermal dysplasia.
The mother's skin is completely normal with no signs of the skin abnormality. In contrast, her son has
patches of normal skin and patches of abnormal skin.
2. In this case, the abnormal chromosome complement of the son arose as a result of:
a) nondisjunction during meiosis I in the father; b) nondisjunction during meiosis I in the mother; c)
nondisjunction during meiosis II in the father; d) nondisjunction during meiosis II in the mother; e)
nondisjunction could occur at either stage of meiosis in either parent.
3. The patchy pattern of anhidrotic ectodermal dysplasia expression is best explained by:
a) nondisjunction during embryogenesis; b) chromosome loss during embryogenesis; c) mitotic
recombination during embryogenesis; d) a new mutation; e) dosage compensation.
4. How many chromosomes would be found in somatic cells of an amphidiploid allotetraploid derived
from two plants, one with N = 8 and the other with N = 10?
a) 18; b) 36; c) 40; d) 72; e) 80.
5. Deletions can be used to determine the orientation of genes on a chromosome. A series of 5 deletions
"uncovered" the following recessive mutations in deletion heterozygotes, allowing them to show
pseudodominance (see text 496 3e):
deletion 1:
deletion 2:
deletion 3:
deletion 4:
deletion 5:
o, n
m, h, r
o, m, n
o, t
h, m
The order of the genes can be represented:
a) mhrotn; b) tonrhm; c) tonmhr; d) monrth; e) more than one gene order is possible from this data.
Questions 6 and 7 pertain to the table on the right. 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
strains 1 2 3 4
genotypes are mixed in all possible combinations and (after incubation) are
+ +
0 0 M M
5
plated to determine the frequency of a b recombinants. The results
0 0 L L
6
indicated in the table are obtained, where M = many recombinants, L = low
M L 0 0
7
numbers of recombinants, and 0 = no recombinants. The strains can be
0 0 M M
8
classified as three different conjugation types: either F−, F+ or Hfr with regard
to a and b gene transfer.
6. True or false: The only way that we can get zero recombinants is if an F– strain is crossed to an F– strain.
7. Which of the following can be classified as F– cells?
a) strains 1, 2, and 7; b) strain 6 only; c) strains 2 and 3; d) strains 1, 5 and 8; e) none of the above.
Questions 8-9 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 agar types (see table
below, right), where nutrient A allows the growth of a− auxotrophs, nutrient B allows for the growth of b−
autxotrophs, etc. As indicated, all types contain streptomycin (Str). A (+) indicates the presence of the
nutrient or drug, a (-) indicates its absence.
Agar Type Str A B C D
Timings of Samples Number of Colonies of Agar of Type
+
1
− + + +
+
+ + + −
2
1
2
3
4
+
+ − + +
3
0
0
0
0
0
+
+ + − +
4
2.5
0
4
0
0
5
5
60
0
0
7.5
60
132
0
0
10
98
220
0
0
12.5
127
315
0
6
15
153
370
0
67
17.5
170
398
4
104
20
190
414
35
125
25
205
416
40
138
30
210
420
42
140
35
215
425
48
142
The table above shows the number of colonies on each type of agar for samples taken at various times after
the samples are mixed:
8. From the gene closest to the origin of replication, the order of the genes is:
a) a b c d; b) b a c d; c) c a b d; d) d c b a; e) none of the above
9. True or false. On the basis of this data, gene b is closer to a than c is to d.
10. Reciprocal
Donor markers Recipient Markers thr allele in leu+ recombinants Percent thr+
transduction
1. thr+leu1
thr–leu2
350 thr+:349 thr–
50
crosses were used
2. thr+leu2
thr–leu1
60 thr+:300 thr–
17
to determine the
order of two mutations, leu1 and leu2 in the leuA gene relative to the thrA gene of E. coli. In each cross,
leu+ recombinants were selected on minimal medium containing threonine, but no leucine, and then tested
for a thr+ or thr- phenotype by replica plating onto plates containing no threonine. The results are given in
the above table:
What is the order of the leu1 and leu2 alleles relative to the thrA gene?
a) leu1 is closer to thr; b) leu 2 is closer to thr; c) the two leu alleles are equidistant from thr; d) the leu1 and
leu2 alleles can not be mapped in this assay.