Inversion loops will form in inversion heterozygotes. A cross over within the inversion
loop will produce duplications and deficiencies in the resultant recombinants. These
would give rise to cells that might not complete gametogenesis or would produce
aneuploid fertilization products.
1. Numbers 2 and 3 are inversion heterozygotes. Ans (b).
2. If the two homologs are both inversions, then the loci should pair up along their
length and no loops would form during 1st meiotic prophase. Crossovers that occur
between homologs 1 and 4 would not generate duplication and deletion gametes because
no inversion loop would form. The other two nonsister chromatids would segregate
without crossovers. Ans (d)
3. Both 3 and 4 are inversion heterozygotes, but only 3 contains a paracentric inversion.
Therefore, only 3 would generate acentric and dicentric fragments if a crossover occurred
in the inversion loop. Ans: none of the above (e)
4. 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)
5.
b+
a-
Deletion ‘uncovers’
recessive mutation: will see
an a- mutant phenotype
in deletion heterozygote
(pseudodominance; p.
496)
b-
If area of deletion does not
contain b gene, then a wild
type copy of b will be on
deletion
chromosome…phenotype of
heterozygote will be wild type
Deletion heterozygotes can
therefore be used for mapping….
deletion 1:
deletion 2:
deletion 3:
deletion 4:
a,d,e
c,d,f
b,c
d,e
Start with deletions
covering smallest area
b must be next to c
d must be next to e
(bcdf) or
(bcfd)
(eda) or
(dea) or
(ade)
bc
and
de
bcfdea or
eadfcd the
consistent
possibility
Ans: (c)
6. As discussed in class, females have a greater chance of passing on the disorder than
males- they are more likely to amplify a premutation to levels that lead to symptoms.
The other statements are not true, because premutation alleles are subject to variable
expressivity. Please see pp. 216-17 in text.
Ans: (a)
7. False. F+ x F+ or Hfr x Hfr or F+ x Hfr will not yield recombinants. Only donor x
recipient crosses will.
8. The key to this question is to remember that F– cells can produce either zero (0), low
numbers (L) or many recombinants (M), depending on what strain they are crossed to:
So strains 3, 6, and 8 must be F– strains.
Given this information, strains 2 and 7 produce many recombinants. They are the Hfr
strains.
Strains 1, 5 and 4 are F+ strains.
Ans: (d)
9.
Agar
Type
1
2
3
4
Str A B C D
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
-
Timings of Samples
0
2.5
5
7.5
10
12.5
15
17.5
20
25
Number of Colonies on Agar of Type:
1 B
2 A
3 C
4 D
0
0
0
0
0
0
0
0
15
0
0
0
104
0
0
0
215
0
8
0
301
6
75
0
347
67
150
5
400
104
202
50
402
150
205
90
398
153
203
90
Note: Plate 1 selects for the
transfer of the b+ gene, since it does not contain nutrient B, Plate 2 selects for the transfer
of the a+ gene, since it does not contain nutrient A, etc.
9. The time of entry map indicates the order is bcad. The b+ gene is transferred first,
since b+ exconjugants appear earliest(~ 5 min), and in the greatest overall frequency.
This is followed by c+ (~10 min), then a+ (~12.5 min) then d+ (~15 min). Ans: none of
the above (e)
10. False, the distance between b and c is about 5 min. The distance between c and a is
about 2.5 min.