Answers to Quiz 3
1. The key here is to identify the fact
that if we are looking at a diploid
heterozygote segregating, then we
would expect a 3:1 ratio of
dominant:recessive. If we look at
the results of Trisomic B, both the
D/d locus and the R/r locus show
a 3:1 ratio.
But for the L/l locus, we see close to a 10:1 ratio. What would be expected if we had
a trisomic undergoing self fertilization? Since the trisomic chromosome is dominant,
then the genotype would be LLl x LLl, and there would be a 2/3 probability of a
gamete inheriting a dominant allele and generating a diploid. This should produce a
9:1 ratio of dominant to recessive:
2/3 L
4/9 LL
1/3 l
2/9 Ll
2/3 L
2/9 Ll
1/3 l
1/9 ll
2/3 L
1/3 l
Ans: L (d)
2. Mammals are usually quite intolerant to aneuploidy- therefore, it would be
unlikely that they would produce viable offspring that were monosomic or
trisomic and definitely not triploid. It is also very unlikely that all horse
chromosomes might segregate to the same gamete (232), although there has
been speculation that there may be a phenomenon termed ‘affinity’, where
the donkey chromosomes all are partitioned to a polar body and only horse
chromosomes end up in the ovum. In the cases that have been examined
cytologically, the mule-like hybrids have contained both horse and donkey
chromosomes. The answer is probably that the donkey and horse
chromosomes are similar enough for animals to exist as chimeras, but
multiple chromosomal rearrangements (inversions, translocations, etc.) lead
to difficulties in generating normal eggs. It seems that normal
spermatogenesis in not possible since only fertile females are obtained. Ans:
I will accept any answer, since cytogenetic analysis of different offspring
show different karyotypes.
3. In this problem, the problem is due to chromosomal loss of an X chromosome
carrying the wild type alleles early during development. This will give rise to
an embryo that is XX for half the cells, and X0 for the rest. Remember that in
Drosophila, XX relative to a complete autosome complement is female; and
XO relative to a complete autosome complement is male. Therefore, this fly is
a gynandromorph- half female and half male, with the male portion
expressing the recessive alleles on the single X chromosome. Ans: (e), none of
the above.
4. The drone honey bees contain a single chromosome set- they are monoploid.
Since that set is complete, they are euploid, the same category as the diploid
workers. Ans: false.
5. Del 1 (a, b, c)
a
b
c
d
e
f
Del 2 (a, b)
Deletion 1
−
−
−
+
+
+
Del 3 (a, b, d)
Deletion 2
−
−
+
+
+
+
Deletion 3
−
−
+
−
+
+
Deletion 4
−
+
+
−
−
+
Deletion 5
+
+
+
−
−
−
Del 4 (a, d, e,)
Del 5 (d, e, f)
5. order: c b a d e f
Ans: none of the above (e).
6. The location of a is defined by the right
breakpoint in deletions 1 &2 and the left !
breakpoint in deletion 4
(c, b, a)
(b, a)
(b, a, d)
(a, d, e)
(d, e, f)
Located near band 3, Ans: (c)
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 1, 4 and 5 must be F- cells
Strain 2, 6, and 8 must be Hfrs
Strains 3, 7 must be F+ cells
Ans: none of the above (e)
8. Crosses 1 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
9. Time of entry:
b+ (5 min)  c+ (10 min) d+ (12.5) a+ (15 min)
Ans: bcda (e) none of the above.
10. The measure of distance in an interrupted mating experiment is time.
The distance between b and c is 5 minutes. The distance between c and d is
2.5 minutes. Therefore, b and c are about twice the distance from each other
as c and d.
Ans: false.