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1. How many chromosomes in a somatic cell of the hybrid?
1n of horse = 32
1n of mule = 31
Fertilization would result in 63 chromosomes in zygote
Ans: 63 (b)
2. The likelihood that all chromosomes would migrate in any given gamete cell would be
(1/2)63
This would have to occur twice for a gamete cell to find a partner with chromosomes
from both species
Since this would require two independent events, the likelihood is the product of their
independent probabilities:
(1/2)63 * (1/2)63 = (1/2)126; Ans: (d)
3. Nondisjunction in the homozygous parent (father) can give rise to an aa gamete.
What about heterozygote?
Suppose cell undergoes nondisjunction in 2nd meiotic metaphase:
a
a
Both heterozygote (mom) and dad can undergo nondisjunction to yield a trisomic
child…Ans: (c)
4. Position can influence expression of genes or transmission (e.g. position effect
variegation in Drosophila: bringing a euchromatic gene in the vicinity of heterochromatin
can influence its expression; myc gene in Burkitt lymphoma)
Linkage group can influence gene expession or transmission; (e.g. abl gene; effects of
adjacent segregation in reciprocal translocation heterozygotes).
Ability to pair and segregate can also influence expression and transmission; (e.g.
inversion heterozygotes).
Ans: all of the above; (d)
z+
x y
Deletion 1
Deletion 2
Deletion 3
Deletion 4
Deletion 5
z
Pseudodominance:
phenotype of deletion
heterozygote is mutant
(-) for x and y; wild
type (+) for z
b
−
−
−
+
+
c
+
−
−
+
+
d
−
+
+
+
+
e
−
+
−
+
+
f
+
+
+
−
+
For example, Deletion 1 “uncovers”
genes b, d, and e.
d e b c a f
I
a
+
+
−
−
−
5.
Comparing the overlapping deletions, the
order or the genes can be deduced:
II III IV V VI
debcaf
1
(b,d,e)
2
(b,c)
3
(a,b,c,e)
4
(a,f)
a
5
5. The order of genes is debcaf or facbed; Ans: (e) none of the above.
6. The gene closest to band IV would be gene c. Ans: (c)
7.
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
Expected characteristics:
F+ will transfer chromosomal gene markers at LOW frequency (L = low recombinants)
only to F− cells
Hfr will transfer chromosomal gene markers at HIGH frequency (M = many
recombinants) only to F− cells
F− cells will yield no recombinants if crossed to themselves, but could show either L or
M colonies depending on whether they are crossed to F+ or Hfr strains…
Using these criteria:
Strain 2 is F−
Strain 5 is F−
Strain 8 is F−
Strain 1 is Hfr
Strain 4 is Hfr
Strain 6 is Hfr
Strain 3 is F+
Strain 7 is F+
7. false. We can get zero recombinants by crossing F− x F−.
8. Strains 1,4, and 6 are Hfr strains. Ans: (d).
Agar Type
Str
A
B
C
D
1
+
+
+
+
−
2
+
−
+
+
+
3
+
+
−
+
+
4
+
+
+
−
+
Timings of Samples
0
2.5
5
7.5
10
12.5
15
17.5
20
25
30
35
Agar type 1: will select for d+ exconjugants
Agar type 2: will select for a+ exconjugants
Agar type 3: will select for b+ exconjugants
Agar type 4: will select for c+ exconjugants
Number of Colonies of Agar of Type
1
2
3
4
0
0
0
0
0
4
0
0
0
60
0
0
10
132
0
0
60
220
0
0
110
315
0
6
140
370
0
67
165
398
0
104
180
414
0
125
182
416
4
138
185
420
35
140
187
425
40
142
If the data set in the above table is graphed the following time of entry mapping data is
obtained:
450
400
350
300
250
Series1
Series2
200
Series3
Series4
150
100
50
0
0
5
10
15
20
25
-50
9. The gene order is ori- a – d – c – b
Ans: (c)
10. True. d is closer to c than c is to b. Ans: (a)
30
35
40