1. 1. Ans: a) 0%: a double crossover within the inversion will be balanced- no
duplications or deficiencies.
E
D X
C
e
d
c
f
g
A B C
F
X
A
B
G
H I
a
b
h
a b
D
g
e
f
G H I
F
E
d c h
i
i
2. a) Is true; daughters of transmitter males have a greater chance of passing on the
disorder due to premutations in their father’s germline. Although there IS a
correlation between an increase in the trinucleotide repeat involved in the disease
and phenotype, it is not strictly linear- individuals with between 50-200 repeats can
still be normal with regard to mental disability.
3: (d) All of the above.
Chromosome 2 has evidence
that the metacentric
chromosome in one species
shares chromosome arms
from two acrocentric
chromosomes from the other
species. Chromosomes 4 and
5 show banding patterns
consistent with pericentric
inversions.
Robertsonian
translocation:
4. A single inversion can change the gene
orientation from one species to the other: If
species 1 were the ancestral species, those
changes would occur in the following order:
Pericentric inversions
1: acefgjkz 4: ac(gfe)jkz
3: a(kfejgc)z
4. Ans: c) 1
4
2
3
2:acg(jef)kz
5. In a heterozygote paired with a chromosome that
contains a recessive mutation within the region of the
mutation, each deletion chromosome ‘uncovers’ the
recessive phenotypes of mutations contained by its
partner- but we can not order those mutations within
an single mutation; to do this we need overlapping
mutations.
chromosome
deletions
1
2
3
4
5
6
7
a
+
+
+
+
+
−
+
b
−
+
−
+
+
+
+
c
+
+
+
+
−
+
+
g
−
+
+
+
+
+
+
genes
m n
+ −
+ +
+ +
− +
+ −
− +
− +
r
+
+
−
+
+
+
+
s
+
−
+
−
+
+
+
v
+
−
+
−
+
+
−
x y ?
y x ?
x
+
−
+
+
+
+
+
Matching regions where overlaps occur allows genes to be placed in a consistent
array:
r
1"
2"
3"
4"
5"
6"
7"
g
b
n c
x
s
v
m
a
(b,g,n)
(s,v,x)
(b,r)
(c,n)
(m,s,v)
(a,m)
(m,v,)
5. Ans: b) rbgncxsvma
6. From the above data set, one can determine that only (e) is true. c) is false, since
deletions 1 and 2 do not overlap and a heterozygote should have a wild type copy of
each gene.
7. Ans: e) none of the above.
Strains that give zero, low, or high numbers of colonies when crossed with other
strains must be F− strains.
F−: Strains #3, 5, 8:
F+: Strains #1, 7, must be F+. They give low numbers when mated with F− cells.
Hfr: Stains #2,6,4 must be Hfr strains. They give high numbers when mated with F−
cells.
8. Ans: false. This is an Hfr x F− cross. Hfr cells can transfer genes close to their
integrated F factor origin of replication in high frequency, but DO NOT transfer
donor status, since the conjugation bridge breaks before all plasmid genes needed
for conjugation can be transferred.
9. To first answer this
question, one needs to
identify which agar types are
selecting for which genes.
Then the time of entry maps
will identify the order in
which the genes are donated.
Those genes that are nearest
the origin of replication will
be transferred in greatest
frequency and will be most
likely to convert the recipient
cell to prototrophy.
Timings of
Samples
Number of Colonies of Agar of Type
0
2.5
5
7.5
10
12.5
15
17.5
20
25
30
35
1
2
3
4
0
0
0
0
0
0
0
0
0
32
44
48
0
0
0
3
60
113
139
164
180
183
186
189
0
0
0
0
2
56
87
114
122
135
143
146
0
4
68
136
222
318
371
396
414
416
420
425
Strains 1-4 select for genes u,
v, s and t respectively, where t is donated first, followed by v, s and finally u:
u+
s+v+ t+
9. origin- t….v..s……u
Ans: (b)
Graphing the data from the above
chart, we can generate the
following time of entry map for
each gene. The time of entry on
the X axis is a measure of how
close the genes are to one
another- genes that are closest
together will be donated closest
in time.
450
400
350
u
300
s
250
200
v
150
t
100
50
0
10: v and s are closest together on
the time of entry map…
Ans: c)
0
2.5
5
7.5
10 12.5 15 17.5 20
25
30
35