BIOL/PBIO 3333
Genetics
Quiz 3
3/27/15
For Answers to the Quiz, click here:
1. How many chromosomes would be found in somatic cells of an amphidiploid allohexaploid derived
from three plants, one with N = 6, N= 8; N = 12?
a) 13; b) 26; c) 44; d) 52; e) 156.
2. In humans, Hunter Syndrome is a sex-linked recessive trait that shows complete penetrance. Two
phenotypically normal parents produce a phenotypically normal daughter, a son with Hunter Syndrome,
and a daughter with Hunter Syndrome. No other symptoms are evident. The origin of the affected daughter
most likely resulted from:
a) nondisjunction during meiosis I in the father; b) nondisjunction during meiosis I in the mother; c)
nondisjunction during meiosis II in the father; d) nondisjunction during meiosis II in the mother; e)
nondisjunction during meiosis in BOTH parents.
3. Each of six regionally overlapping populations of Drosophila had a specific arrangement of bands in
one of the large autosomes:
a: 12345678
d: 16543278
b: 143277568
e: 165772348
c: 16572348
f: 147237568
Assume that arrangement ‘a’ is the original one. In what order would the other arrangements most likely
arise?
a) a,d,e,c,f,b; b) a,d,f,c,e,b,; c) a,d,c,e,b,f; d) a,b,e,d,f,c; e) a,c,e,f,d,b.
Questions 4-5 pertain to the following. The table below presents chromosome data on five species of
plants and their F1 hybrids:
Meiosis I Metaphase
Species or F1 hybrid Root tip chromosome number Number of Bivalents Number of Univalents
A
20
10
0
B
20
10
0
C
10
5
0
D
10
5
0
E
10
5
0
AxB
20
5
10
BxC
15
5
5
AxD
15
5
5
CxD
10
0
10
BxE
15
5
5
4. The chromosomal origin of species B is:
a) a chromosomal doubling of species A; b) a chromosomal doubling of species D; c) an amphidiploid
derived from species C and D; d) an amphidiploid derived from species D and E; e) none of the above.
5. The monoploid number (x) for species B is:
a) 5; b) 10; c) 15; d) 20; e) 40.
6. Deletions can be used to determine the orientation of genes on a chromosome. A series of 6
overlapping deletions "uncovered" the following recessive mutations in deletion heterozygotes, allowing
them to show a recessive phenotype in the heterozygotes (i.e. ‘pseudodominance’):
deletion 1:
r, s
deletion 4:
deletion 2:
k, r, s
deletion 5:
deletion 3:
o, p, z
deletion 6:
The order of the genes can be represented:
a) krsopzn; b) nozsrpk; c) rskzopn; d) kropsnz; e) npzorsk.
o, z
z, k, s
n, p
7. Which of the following are true?
a) mutations causing a change in phenotype occur rarely; b) different genes mutate at different rates; c)
disruption of gene function through mutation occurs at higher frequency than restoration of function
through mutation; d) all of the above; e) none of the above.
Questions 8 and 9 pertain to the following. Four E. coli strains of genotype a+b− 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 types relative to conjugation:
either F−, F+ or Hfr with regard to a and b gene transfer.
strains
5
6
7
8
1
L
0
L
0
2
0
L
0
M
3
M
0
M
0
4
L
0
L
0
8. Which of the following can be classified as F - cells?
a) 2, 5, 7; b) 2, 3, 6; c) 1, 4, 6; d) 3, 8; e) none of the above.
9. True or false. Suppose after mixing strains 2 and 8 the culture was left to grow on medium missing the
nutrients needed by the a- and b- mutants. Virtually all of the cells arising from the minimal media plate
would be expected to have pili if examined under the electron microscope.
Five interrupted-mating experiments are performed with E. coli. Five different x+y+ z+ strs Hfr strains are
mixed in separate experiments with an x−y−z−F− strr strain. The progeny are screened on appropriate selective
media to detect recombinants. The number of minutes after the start of the experiment that the x, y and z
genes are first detected in the recombinants (i.e their time of entry) is given in the chart below; if the gene has
not been transferred, it is indicated as nt (no transfer). In the figure below, the whole chromosome has been
divided into a 100 minute map; each mark on the circle represents 5 minutes. For orientation, the origin of
replication for HfrA is located at “12 o’clock” on the map, and the genes for Hfr A will be donated in a
clockwise direction (i.e. following the arrowhead).
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
45
z+ gene
nt
5
65
nt
35
50
10
nt
5
30
50
nt
5
10. The integration site for Hfr D is how many minutes away from the integration site for Hfr E?
a) 5 minutes; b) 10 minutes; c) 25 minutes; d) 30 minutes; e) 45 minutes.