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BIOL/PBIO 3333
Genetics
Quiz 3
11/1/13
For the answers to the quiz, click here
Questions 1-2 pertain to the following. Fertile varieties of the Golana melon are known that contain 14, 28,
42, 56, and 70 chromosomes, respectively. A variety that contains 21 chromosomes exists, but can only be
propagated through cuttings.
1. The monoploid number for these Golana melon species is:
a) 2; b) 7; c) 14; d) 21; e) it is different for every species listed.
2. The 21 chromosome variety:
a) has an aneuploid chromosome complement; b) would form synaptonemal complexes between nonhomolog chromosomes during first meiotic prophase; c) would segregate homologs randomly, producing
unbalanced gametes; d) all of the above; e) none of the above.
Questions 3-4 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 anhydrotic 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.
3. 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) we can
not specifically determine where non-disjunction occurred from the information given.
4. The expression of the dysplastic (i.e. mutant) tissue phenotype in the mutant sectors is best explained
by:
a) XIST mRNA production from the paternally inherited X early in embryonic development; b)
modification of chromatin leading to Barr body formation of the paternally inherited X during
embryogenesis; c) overexpression of the normal allele on the maternal X chromosome in the mutant tissue
sectors; d) all of the above; e) none of the above.
Seven partial deletions (1 to 7), shown as gaps in the diagram on the left, 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
genes
deletions
a b c g m n r s v x
1
2
3
4
5
6
7
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
5. The order of the genes on the intact chromosome is:
a) nrgbmvacsx; b) ambgnrcvxs; c) rgnbvamcxs; d) grnbavmxcs; e) more than one gene orientation is
possible form these experiments and you would need additional overlapping mutants.
6. From the above data set, one can determine:
a) r is further from b than g; 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
0
L
L
M
2
M
0
0
0
3
L
0
0
0
4
0
L
L
M
7. The F− cells present in this data set are represented by:
a) 1, 4, and 5; b) 2 and 3; c) 6 and 7; d) 7 and 3; e) none of the above.
8. True or false. Suppose after mixing strains 1 and 7 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.
Agar Type Str S T U V
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 (the first gene listed below is closest) the
gene order of these four genes is:
a) s-t-u-v; b) u-t-v-s; c) t-u-s-v; d) v-s-t-u; e) none of the above.
10. True or false: Gene s is closer to t than s is to v.