BIOL/PBIO 3333
Genetics
3/29/13
For answers to the quiz, click here:
1. True or false: Monoploid plants can be produced from pollen grains by stimulating haploid
embryoids to grow with colchicine.
2. The picture on the right represents a karyotype analysis of
chromosome 3 from four species: orangutan (O); gorilla (G);
chimpanzee (C); and human (H). The rearrangement shown
here indicates:
a) the human chromosome 3 contains a Robertsonian
translocation; b) the chromosomal pattern is strikingly
unstable, indicating little synteny for this chromosome; c) the
orangutan chromosome looks like it has undergone a paracentric inversion relative to the other
chromosomes; d) all of the above; e) none of the above.
In Drosophila, seven
partial deletions (1
to 7), shown as gaps
in the diagram on
the right, have been
mapped on a
chromosome.
Intact
dele ons
1
2
3
4
5
6
7
Heterozygotes are constructed where the homologous chromosome contains seven recessive
mutations, identified in the table below as a through g. These deletion heterozygotes "uncovered"
(see p. 496/3e; p. 435/4e in text) the following
Deletion a b c d e f g recessive mutations:
1
2
3
4
5
6
7
+ m +
m + +
m + +
m + +
+ m +
m m m
m + +
m + + +
+ + m +
+ + + m
m + m m
m m + +
m + m m
+ + + +
3. True or false. In heterozygotes constructed
with deletion 6, gene e shows pseudodominance.
4. The correct order of these genes on this
Drosophila chromosome is:
a) cafdgbe; b) dgbeacf; c) facdbe; d) ebdgafc; e)
none of the above.
5. The distance from gene b to gene g:
a) is smaller than the distance between d to gene g; b) is smaller than the distance between gene g
and gene f; c) is larger than the distance between gene g and gene c; d) all of the above; e) it is
impossible to determine gene distances from this analysis.
6. An allotetraploid amphidiploid is backcrossed back to one of its progenitor species and a sterile
progeny is produced. This sterile individual can best be represented by:
a) n1 +n2; b) n1 + n2 +n3; c) 2n1 +2n2; d) 2n1; e) 1n1 +2n2.
7. True or false. A gal− mutant cannot grow without the sugar galactose.
H
C
G
O
a+b−
Four E. coli strains of genotype
are labeled 1, 2, 3, 4. Four
strains of genotype a−b+ are labeled 5, 6, 7 and 8. The two
genotypes are mixed in all possible combinations and (after
incubation) are plated to determine the frequency of a+b+
recombinants. The results indicated in the table are obtained,
where M = many recombinants, L = low numbers of
recombinants, and 0 = no recombinants. The strains can be
classified as 3 sex types: either F−, F+ or Hfr with regard to a and
b gene transfer.
strains
1
2
3
4
5
L
L
0
M
6
0
0
M
0
7
L
L
0
M
8
L
L
0
M
8. The Hfr strains present in this data set are represented by:
a) strain 3 only; b) strain 4 only; c) strains 3 and 4; d) strains 5, 6, 7, 8; e) none of the above.
Questions 9-10 pertain to the following. An Hfr strain of the genotype a+b+c+d+strs is mated with a
female strain of the genotype a−b−c−d−strr. At various times the culture is disrupted in a blender to
separate the mating pairs. The cells are then plated on agar of the following four agar types (see
table below, right), where nutrient A allows the growth of a− auxotrophs, nutrient B allows for the
growth of b− autxotrophs, etc. As indicated, all types contain streptomycin (Str). A (+) indicates the
presence of the nutrient or drug, a (-) indicates its absence.
agar 
entry time
0
2.5
5
7.5
10
12.5
15
17.5
20
25
30
35
1
2
3
4
0
0
0
443
725
954
1166
1323
1400
1437
1450
1462
0
0
0
0
0
0
0
0
0
86
167
200
0
0
0
0
0
343
622
776
986
1090
1130
1130
0
0
0
0
0
0
0
0
220
345
440
430
Agar Type
Str
A
B
C
D
1
+
+
+
+
−
2
+
−
+
+
+
3
+
+
−
+
+
4
+
+
+
−
+
The table on the left shows the number of
colonies on each type of agar for samples
taken at various times after the samples
are mixed, then disrupted and plated:
9. From the gene closest to the origin of replication, the order of these genes is:
a) d a b c; b) b a c d; c) c a b d; d) d c b a; e) none of the above.
10. True or false. On the basis of this data, gene b is closer to d than it is to c.