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1. The specified karyotype has two X chromosomes and no Y chromosome, so the person is a
female. The female carries three copies of chromosome 21, i.e. trisomy 21 or Down Syndrome.
Ans: (c) female with Down Syndrome
2, Albinism is an autosomal recessive disorder- one copy of the wild type allele is sufficient for a
wild type phenotype
It also manifests as a morphological mutation, in that the mutant phenotype is detectable due to an
absence of melanocytes in the skin.
Ans (a) morphological
3. As discussed in class, chromosome 2 in humans looks like it represents a Robertsonian
translocation relative to other primate chromosomes
A close examination of the banding patterns on chromsomes 4 and 5 indicate that they look as if
breaks occurred to either side of the centromere for both chromosomes- i.e. pericentric inversions.
Ans: (d) all of the above.
4. Variegated position effects are caused when a gene in a euchromatic region of the
chromosome is transferred into or near a heterochromatic block…see pp. 430-31
Ans: (b) inversion that moves a gene from euchromatin into or near heterochromatin
5.
1: 143
2: 152
3: 160
4: 155
5: 148
6: 98
Contain 3/2 (150%) of
what is observed in diploid
cells, i.e. are carrying an
extra gene copy
Contains only a diploid amount of enzyme…
z+
Ans: (d) Band 3
x y
The previous quiz had a similar problem using deletions
to map gene location…pseudodominance.
Make sure you review this topic as well (pp. 448-50).
6. A double crossover must occur
between the same non-sister chromatids for an
exchange to generate balanced gametes.
Ans: (d)
z
Pseudodominance:
phenotype of deletion
heterozygote is mutant
(-) for x and y; wild type
(+) for z
D
C
c
d
b
B
a
e
E
A
a
d
C
b
e
A
B
c
D
E
7. Expected characteristics:
a) F+ will transfer chromosomal gene markers at LOW frequency (L = low recombinants) only
to F− cells
b) Hfr will transfer chromosomal gene markers at HIGH frequency (M = many recombinants)
only to F− cells
c) 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:
Strains 2, 5 and 6 must be F- cells
Strain 1 and 4 must be Hfrs
Strains 3, 7 and 8 must be F+ cells
Ans: (c)
8. Crosses 2 and 7 would represent a F- x F+ cross. In this cross, the F factor would be expected
to be transferred to virtually every cell, making them all F+ and consequently expressing the
genes required to produce pili.
Ans: (a) true
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
15
75
45
5
65
5
10
50
65
25
9. The integration site for B is 30 minutes away from the
integration site for A
Ans: (d)
10. Both the A and B F factors have integrated into the
bacterial chromosome in the same relative orientation and
their respective Hfr’s donate in the clockwise direction.
Ans: (a) true
x
C
y
B