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Genetics
Quiz 3
November 2, 2012
For answers to the quiz, click here
1. In Burkitt’s Lymphoma:
a) a hybrid ABL protein is produced as a result of a translocation; b) a proto-oncogene is transferred into an
immunoglobulin gene cluster and expressed at high levels; c) the drug Gleevec has been approved for
treating this disorder; d) all of the above; e) none of the above.
2. 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 could therefore have occurred:
a) in the mother; b) in the father; c) you can not tell just on the basis of this data.
3. 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.e. they show pseudodominance see text p. 496 3e/ p. 435 4e):
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
4. 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.
5. The gene closest to band III in the figure would be gene:
a) a; b) b; c) c; d) d; f) none of the above.
Questions 6 and 7 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
L
0
2
M
0
M
0
3
0
M
0
L
4
L
0
L
0
f
+
+
+
−
+
6. True or false: The only way that we can get many recombinants is if an Hfr strain is crossed to an Hfr
strain.
7. Which of the following can be classified as Hfr cells?
a) strains 3, 5, and 7; b) strains 2, 4, and 8; c) strains 5 and 8; d) strains 2 and 6; 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
85
0
0
7.5
60
152
0
0
10
112
231
0
0
12.5
140
315
0
6
15
158
370
0
67
17.5
173
398
4
104
20
196
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 it is to d.
10. In a cross between an Hfr that has the genotype x+y+z+ and an F– that is x–y–z–, the x
x+y+z+ 220
gene is known to be transferred later than y and z. To determine the order of y and z with
x+y–z– 60
respect to x, x+ exconjugants are selected and screen for their y and z phenotypes. The
x+y–z+ 18
results of these tests are listed in the chart on the right:
x+y+z– 0
These results indicate:
a) that z is between x and y; b) that the distance between x and z is much smaller than the distance between
x and y; c) that z is closer to the F factor origin than y; d) all of the above; e) none of the above.