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1. This is an example of position effect variegation. Moving the w+ locus near
heterochromatin can lead to a remodeling of the chromatin associated with the w+ locus;
this inactivation seems to occur early during eye development, leading to some cells
where the gene is not correctly expressed. Ans: (c) See page 480 in Hartwell text.
2. The disease is X-linked and being passed through the dad. The son must therefore
receive both the X and Y chromosome from the dad. They would normally segregate
from one another during 1st meiotic prophase; so nondisjunction occurs in the father
during the first division cycle. Ans: (a)
3. Anhydrotic displasia shows mosaicism because of random X chromosome inactivation
during early embryogenesis- a result of Barr body formation and dosage compensation in
mammals. Ans: (e).
4. This is essentially testing several definitions. An allotetraploid contains four
genomes, derived from different individuals; an amphidiploid consists of an organism
that has duplicated the genome sets from two parents of different species. Thus the
monoploid number for both parental species would be 8 + 10 = 18, and a duplication
would lead to 36 chromosomes (where each genome now has a pairing partner. Ans: (b).
z+
x y
z
Pseudodominance:
phenotype of deletion
heterozygote is mutant
(-) for x and y; wild type
(+) for z
5. Deletion mapping:
Look for deletions that uncover the lowest number of loci first…
o,t
o,n
h,m
→ not
omn
mon
mno
→
→ rhmnot or tonmhr
rhm
hmr
5. Ans: (c) tonmhr
To answer this question, the easiest way is to remember that F– cells can give rise to
either large numbers of recombinants, or small numbers of recombinants, depending on
whether they are undergoing conjugation with Hfr or F+ cells, respectively.
Since strains 3, 4, and 7 can give rise to BOTH large and small numbers of a+b+ cells,
they are the F–strains. Depending on the resulting number of recombinant cells produced
in a mating, the other strains can be deduced:
strains
5 (Hfr)
6 (F+)
7 (F–)
8 (Hfr)
1 (Hfr)
0
0
M
0
2 (F+)
0
0
L
0
3 (F–)
M
L
0
M
4 (F–)
M
L
0
M
6. False. Only donor crosses to F– strains yield recombinants, i.e. F+ x F+ , F+ x Hfr or
Hfr x Hfr will not yield recombinants.
7. The F– cells would be 3, 4 and 7.
The key to this problem is to determine which agar is
selecting for which gene. The compound missing in
the medium is selecting for the gene that will confer
prototrophy.
Agar 1 selects for a+, 2 for d+, 3 for b+ and 4 for c+:
Agar Type Str A
+ −
1
+ +
2
+ +
3
+ +
B
+
+
−
+
C
+
+
+
−
D
+
−
+
+
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 (a+) 2 (d+) 3 (b+) 4 (c+)
0
0
0
0
0
4
0
0
5
60
0
0
60
132
0
0
98
220
0
0
127
315
0
6
153
370
0
67
170
398
4
104
190
414
35
125
205
416
40
138
210
420
42
140
215
425
48
142
450
400
d+
350
300
250
a+
200
150
100
c+
50
b+
0
2.5
5
7.5
10
12.5
15
17.5
20
25
30
35
8. The order of the genes is d+ a+ c+ b+ Ans: (e) none of the above.
9. False. d-c interval is approximately 10 minutes; a-b interval is approximately 12.5
minutes; therefore c is closer to d.
If leu1 were closer to thr, than leu2, the following crossovers would need to occur to
generate colonies that could survive on minimal plates:
thr+
X
X
thr-
leu1
+
X + X leu X
2
thr+
+
thr-
leu1
leu2
X
+
Here we would expect cross #1 to give fewer colonies than cross #2.
If leu2 were closer to thr than leu1, however, we would expect the reciprocal result; the
first cross would require fewer cross overs and should generate the greater number of
colonies.
X
thr+
+
thr-
leu2
thr+
X
thr-
leu1
X+
leu2
X
+
+
X
leu 1
X
Therefore, leu2 is closer to thr. Ans (b).