1.
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
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The cross A x B indicates that the two genomes contain different (i.e.
nonhomologous chromosome sets- they do not pair (20 univalents).
The A x C and A x D crosses show that each species has a chromosome set (x =
5) that pair with A (bivalent), and one set that does not pair. C x D also shows that
the 5 chromosomes are not homologous.
Therefore, A must have a chromosome set derived from both C and D. Since A
has twice the number of chromosomes as C and D and these chromosomes have
pairing partners in A, then A must be a tetraploid amphidiploid, derived from the
C and D genomes.
Ans: (c)
2. The monoploid number, or the number of chromosomes in a single set for species C
and D is 5. Species A is a tetraploid and contains 4 sets of 5 chromosomes.
Ans: (a)
3. Hunter Syndrome is an X-linked recessive, for a female to inherit it, it must be present
in two copies. Therefore, the carrier mother must have undergone a non-disjunction
event during Meiosis II, and passed both copies of the mutant X chromosome to a
gamete. In addition, the daughter’s karyotype is not indicated. The daughter may be
triplo-X, where the paternal allele is X-inactivated; this is probably most likely. It is also
possible, but more unlikely, that nondisjunction occurred in both parents, and that the
daughter was fertilized by a sperm containing no sex chromosome.
Ans: (d), but I will accept (e).
4. If (a) is the original sequence: 12345678;
Then an inversion would give the sequence (c) 15432678;
A duplication of 2 would yield (f) 154322678;
A deletion of 5 would yield (d) 14322678;
An inversion would yield (e) 16223478.
And another inversion would yield (b) 12263478.
Ans: a,c,f,d,e,b. (c).
5. Amphidiploid allohexaploid: 6 chromsome sets that have been doubled.
Three sets: 8 + 4 + 10 (number donated in gametes).
22 x 2 = 44.
Ans: (c)
6.
z+
x y
z
Pseudodominance:
phenotype of deletion
heterozygote is mutant
(-) for x and y; wild
type (+) for z
Each deletion “uncovers” several mutations, indicating that they must be located in the
same region of the genome. By comparing overlapping deletions and the mutations they
uncover, a map can be constructed. It is easiest to start with the smallest deletions first
and build from them.
r,s
k,r,s
prs
o,z
n,z
prsk
p,r
zoprsk
nzoprsk
o,p,z
Ans: (e)
7. See your textbook, chapter 7 (193-94; ) All of the above are true.
Ans: (d).
8.
•
•
An F- cell would be one that could form
– 1) low numbers of colonies (when conjugated with F+)
– 2) many colonies (when conjugated with Hfr)
– 3) no colonies (when conjugated with F-)
Additionally, donor cells can only conjugate with F- cells...
strains
5 F+
6 F+
7 F8 Hfr
1 FL
L
0
M
2 F+
0
0
L
0
3 Hfr
0
0
M
0
4 FL
L
0
M
9.
•
•
Mixing strains 1-8 is a cross between an Hfr and a F- cell. An Hfr cell will
transfer certain genes at high frequency, but the conjugation bridge will usually
break before all of the F factor is transferred. Thus, ex-conjugants can’t be
donors.
Ans: false.
10.
Cross
Hfr A x F
Hfr B x F
Hfr C x F
Hfr D x F
Hfr E x F
Entry time (in minutes) for Hfr wild type genes
x+ gene
y+ gene
35
55
30
10
5
25
65
45
40
20
The x and y genes are located 30 minutes and 55 minutes from the intergration site for A.
This allows us to identify unambiguously the integration sites for the other Hfrs:
D
E
C
x
y
B
10. Doing this, we find B is integrated 10 minutes away from E.
Ans: (b)
11. A is being donated in a clockwise direction; B is being donated in a counter
clockwise direction.
Ans: false