Zoo/Bot 3333
Genetics
Quiz 2
September 28, 2012
For answers to the quiz, click here:
The M and S genes encode two protein products that form a functional heterodimer. Both
protein products are needed for function, although only low threshold levels of protein
are needed and the alleles at these loci are haplosufficient. An M locus mutation (Ma),
eliminates the ability to pair with the wild type S gene product. A suppressor mutation,
however, has arisen at the S locus (Sa), that can effectively lead to heterodimer formation
with the Ma mutant, and restore function. This mutant, however, is incapable of pairing
with the wild type S gene product to form a functional heterodimer; it can only interact
with the Ma gene product.
1. In a cross between two MMaSSa heterozygotes, what fraction of the progeny will
produce functional heterodimers?
a) 14/16; b) 9/16; c) 7/16; d) 3/16; e) 1/16.
2. The F2 progeny for a dihybrid cross yields a phenotypic ratio of 9 green: 7 red. If an F1
individual was testcrossed, what would be the phenotypic ratio of the progeny?
a) all green; b) all red; c) 1 green: 1 red; d) 3 green: 1 red; e) none of the above.
Questions 3-5 pertain to the chart on the right. In
Labrador retrievers, coat color may be black, brown
(chocolate) or golden. Here are some of the many
possibilities resulting from the matings of dogs of
different coat colors:
3: True or false. The allele responsible for golden coat
color is located at the same genetic locus as the allele
for brown coat color.
4. In the cross shown in (f):
a) the first parent is heterozygous for the brown allele;
b) the second parent is heterozygous for the golden allele; c) the second parent is
heterozygous for the black allele; d) all of the above; e) none of the above.
5. The pattern of inheritance seen in these matings indicates that the golden phenotype is
an example of:
a) complementary gene action; b) duplicate genes; c) dominant epistasis; d) recessive
epistasis; e) dominant suppression.
Questions 6-8 pertain to the following. In Drosophila, Lyra (Ly)
and Stubble (Sb) are dominant mutations located on chromosome
3. A recessive mutation for bright-red eyes is discovered and
also shown to be located on chromosome 3. A female who was
heterozygous for all three mutations was crossed to a male
homozygous for the bright-red (indicated br here) mutation, with
the progeny classes produced at the right:
6. The map distance from the Stubble mutation to the Lyra
mutation is:
a) 3.4 map units; b) 4 map units; c) 13.4 map units; d) 14 map
units; e) 18 map units.
Phenotype
Ly Sb br
Ly + br
Ly + +
+++
+ Sb +
Ly Sb +
+ + br
+ Sb br
Total
Number
404
75
18
422
59
4
2
16
1000
7. True or false: For this particular data set, we see no evidence of interference between
these three genes.
8. A testcross between an individual who is trihybrid for three different
gene loci (A/a, B/b, C/c) and an individual homozygous recessive for all
three loci yields the following progeny:
Which of the following are true?
a) The A and B loci are approximately equally distant from C; b) the A
and B genes must be on the same chromosome and separated by about
25.3 map units; c) the C locus is located 12 map units from the A locus;
d) all of the above; e) none of the above.
aBC
abC
ABC
AbC
aBc
abc
ABc
Abc
214
226
26
28
30
36
218
222
Questions 9 and 10 pertain to the following. The following are ordered tetrad pairs from
a cross between strain cd and ++:
Tetrad Class
1
2
3
4
5
6
7
c+
c+
cd
+d
c+
cd
c+
c+
cd
cd
c+
++
++
+d
+d
++
++
c+
cd
cd
cd
+d
+d
++
+d
+d
++
++
1
17
41
1
5
3
1
9. The distance between the c locus and the centromere is approximately:
a) 7.3 map units; b) 14.5 map units; c) 15.9 map units; d) 20.3 map units; e) 31.8 map
units.
10. The distance between these two genes is approximately:
a) 9 map units; b) 23 map units; c) these genes do not appear to be linked to one another.