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Zoo/Bot 3333
Genetics
Quiz 3
April 1, 2005
For the answers to the quiz, click here:
Questions 1 and 2 pertain to the following: Mules are the sterile progeny of a male donkey (2N = 62
chromosomes) with a female horse (2N = 64). Assume that mules are sterile because of a failure of
chromosome pairing and segregation during meiosis.
1. How many chromosomes are present in a somatic cell in a mule?
a) 62; b) 63; c) 64; d) 128; e) none of the above.
2. What would be the probability that two mules could produce a fertile ‘amphidiploid’, assuming no
pairing of horse and donkey chromosomes during meiosis and amphidiploids were viable?
a) (1/2)10; b) 1/16,389; c) (1/2)63; d) (1/2)126; e) (1/2)128.
3. A child was born with trisomy 18. When a gene on chromosome 18 was examined in both parents, the
mother was found to be Aa and the father was observed to be aa, while the child was aaa in genotype.
Nondisjunction must therefore have occurred:
a) in the mother; b) in the father; c) you can not tell just on the basis of this data.
4. Rearrangements in chromosomes may affect gene expression or gene transmission by altering the
________________________ of certain genes in the genome.
a) position; b) linkage group; c) ability to pair and segregate properly during meiosis; d) all of the above; e)
none of the above.
Six bands (I-VI) in a salivary gland chromosome of Drosophila are shown in the figure below, along with
the extent of five deletions (numbered below the figure).
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, where (+) indicates a
wild type phenotype, and a (-) indicates a mutant phenotype:
I II III IV V VI
Deletion 1
Deletion 2
Deletion 3
Deletion 4
Deletion 5
1
a
+
+
−
−
−
b
−
−
−
+
+
c
+
−
−
+
+
d
−
+
+
+
+
e
−
+
−
+
+
2
3
4
5
5. The correct order of genes relative to chromosome bands I-VI is:
a) acebdf; b) defcab; c) afcdeb; d) bdecaf; e) none of the above.
6. The gene closest to band IV in the figure would be gene:
a) a; b) b; c) c; d) d; f) 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
f
+
+
+
−
+
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
M
0
0
M
2
0
M
L
0
3
L
0
0
L
4
M
0
0
M
7. True or false: The only way that we can get zero recombinants is if an Hfr strain is crossed to an F +
strain.
8. Which of the following can be classified as Hfr cells?.
a) strains 5, 8, 2 and 4; b) strains 1, 2 and 4; c) strains 5 and 8; d) strains 1, 4 and 6; e) none of the above.
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 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
0
60
0
0
7.5
10
132
0
0
10
60
220
0
0
12.5
110
315
0
6
15
140
370
0
67
17.5
165
398
0
104
20
180
414
0
125
25
182
416
4
138
30
185
420
35
140
35
187
425
40
142
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 (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) a-d-c-b; d) b-d-a-c; e) none of the above.
10. True or false: Gene d is closer to c than c is to b.