BioSc 231
General Genetics
Exam 2
Name __________________________________
Multiple Choice. (2 points each)
_____An autosome is ___.
A.
B.
C.
D.
E.
a non-sex determining chromosome
an alternate form of a gene
another term for epistasis
present only in males and is responsible for sex determination
found in mitochondria but not in nuclei
____ In a complementation test the number of complementation groups indicates
A.
B.
C.
D.
E.
the number of genes required for a specific phenotype
the penetrance of phenotype
the number of phenotypes for a gene
the number of chromosomes in an organism
the quantity of gene product required for a phenotype
_____ In a complementation test
A.
B.
C.
D.
mutations that complement are allelic
mutations that complement belong to the same complementation group
mutations that complement are in two different genes required for the wild-type phenotype
mutations that are allelic are required for complementation
_____ The percentage of individuals with a given genotype who exhibit the phenotype associated with that genotype is called
A.
B.
C.
D.
E.
penetrance
expressivity
incomplete dominance
co-dominance
lethality
_____ A situation where each allele produces a protein that can be detected in the heterozygote is called
A.
B.
C.
D.
E.
penetrance
expressivity
incomplete dominance
co-dominance
lethality
_____ A person who has type O blood has
A.
B.
C.
D.
A antigens on the cell surface
B antigens on the cell surface
both A and B antigens on the cell surface
no surface antigens
_____ In Drosophila the recessive alleles for brown and scarlet eyes (resulting from two independent genes) interact so that the double
homozygous recessive is white. A pure-breeding brown and pure breeding scarlet are crossed. What proportion of the F2 will be white?
A.
B.
C.
D.
E.
1/4
3/4
1/16
7/16
9/16
_____ In chickens the dominant allele Cr produces the creeper phenotype (having short legs). However, the creeper allele is lethal in
the homozygous condition. If two creepers are mated, what proportion of the living progeny will be creepers?
A.
B.
C.
D.
E.
1/4
1/2
3/4
1/3
2/3
_____ The maximum recombination frequency between two genes is
A.
B.
C.
D.
E.
100%
80%
50%
10%
1%
_____ In crossing over
A.
B.
C.
D.
Genetic exchange occurs before chromosome replication
The probability of its occurrence decreases with increasing distance between the genes exchanged
Occurs between two loci very close together
The reciprocal exchange between homologous chromosomes is random
_____ A plant of genotype C D/C D is crossed to c d/ c d and an F1 testcrossed to c d/c d. If the genes are unlinked, the percentage of
c D/c d recombinants will be
A.
B.
C.
D.
E.
10%
25%
30%
40%
50%
_____ The maize genes sh and bz are linked, 40 map units apart. If a plant sh+ bz/sh bz+ is testcrossed, what proportion of the progeny
will be sh bz/sh bz?
A.
B.
C.
D.
0.04
0.10
0.20
0.40
_____ In sweet peas, the two allelic pairs C, c and P, p are known to affect pigment formation in the flowers. The dominants, C and P,
are both necessary for colored flowers. Absence of either results in white. A dihybrid plant with colored flowers is crossed to a white
one which is heterozygous at the “c” locus. What are the genotypes of these two plants?
A.
B.
C.
D.
E.
CcPp and Ccpp
CCPP and Ccpp
ccpp and Ccpp
CcPp and ccpp
CcPp and ccPp
_____ Assume that an additional allelic pair in sweet peas also affects pigment formation in addition to the genes mentioned in the
first question. The presence of the dominant R allele is required for red flowers and the recessive r allele produces yellow flowers.
Which of the following genotypes would result in red flowers?
A.
B.
C.
D.
E.
CcPpRr
CcppRR
CcPPrr
ccPPRR
CcppRR
_____ In the human, the dominant alleles, D and E, are both required for normal development of the cochlea and the auditory nerve,
respectively. The recessive alleles, d and e, can result in deafness due to impairment of these essential parts of the ear. Which of the
following sets of parents would produce all hearing children?
A.
B.
C.
D.
DDee x ddEE
DdEe x DdEe
Ddee x DdEe
DdEe x DDEe
_____ In poultry, the shape of the comb varies greatly and involves at least two pairs of alleles. The allele R can result in rose shaped
comb and the allele P can result in pea-shaped comb. If both of these dominants are present together, genic interaction produces a
walnut comb. When a bird is carrying both recessive alleles in the homozygous condition, single comb types result. Which of the
following crosses produces offspring at the ratio of 1 Walnut: 1 Rose: 1 Pea: 1 Single?
A.
B.
C.
D.
E.
rrPP x RRpp
RrPp x RrPp
RrPp x rrpp
RrPp x rrPP
RrPp x RRpp
_____ . Based on the above phenotype descriptions, which of the following crosses produces offspring at the ratio of 9 Walnut: 3
Rose: 3 Pea: 1 Single?
A.
B.
C.
D.
E.
rrPP x RRpp
RrPp x RrPp
RrPp x rrpp
RrPp x rrPP
RrPp x RRpp
Short Answer. (variable points)
(4) Two-point testcrosses revealed the following map results:
g___________m 9 map units
m___________h 12 map units
A. Draw the two possible maps for these loci.
B. What cross would resolve the two possible maps and what are the possible outcomes of that cross?
(6) In the fruit fly, the allele for purple eye color (pr) is recessive to its allele for red (pr+). The allele for vestigial (short) wings (vg) is
recessive to the wild allele for normal wings (vg+). The two genes are autosomally linked. Females from a purple stock (pure
breeding line) are crossed to males from a vestigial stock (pure breeding line). The F1 flies are all wild (red eyes and normal wings).
F1 females are testcrossed with the following results:
210
40
215
35
- purple
- wild
- vestigial
- purple and vestigial
In the F1 individuals, are the vg and pr genes in the cis or trans configuration?
Write the genotypes of each class in the testcross.
What is the recombination frequency between the vg and pr genes?
(5) In tomatoes, round fruit (o+) is dominant to long (o). Simple flowering shoot (s+) is dominant to branching flowering shoot (s).
Plants from two different pure-breeding varieties are crossed. One variety bears round fruit and has branched flowering shoots. The
other variety has long fruits and simple flowering shoots. F1 plants were testcrossed and the following progeny were obtained:
38
78
80
44
- round, simple
- long, simple
- round, branched
- long, branched
Calculate the chi-square and P values based on the prediction that the genes are not linked.
(5) In corn, the genes v, b and l are linked. The data given below summarize the result of a three-point testcross. From the data,
construct a map showing the genes in the correct order and indicating the distances between each pair of genes.
v+
v+
v
v
v+
v+
v
v
b+
b
b
b+
b
b+
b+
b
l
l
l
l+
l+
l+
l+
l+
304
119
18
70
64
22
108
295
(4) Based on the complementation data below, A) how many complementation groups exist?, and B) which mutations belong to each
group? (+ = complementation, -- = no complementation)
mutation number
Mutation
1
2
3
4
5
6
7
8
9
1
--
2
+
--
3
--
+
--
4
+
+
+
--
5
+
+
+
--
--
6
+
--
+
+
+
--
7
+
+
+
+
+
+
--
8
--
+
--
+
+
+
+
--
9
+
+
+
+
+
+
--
+
--
10
+
--
+
+
+
--
+
+
+
10
--
Bonus Question (4 pts) An Arabidopsis thaliana flowering mutation has been generated in the Columbia (Col) line. The mutant line
was then crossed with a wild-type Landsberg erectus (Ler) line to generate the F1 generation. The F1 generation was allowed to self
to produce the F2 generation. F2 plants that displayed the mutant phenotype were assayed using the CAPS system to identify a
molecular marker that is linked to the mutant flowering gene. Two markers from each of the five Arabidopsis thaliana chromosomes
were tested. The results of those tests were as follows.
Marker Name
Chromosome
# with Ler
Markers
# with Col
Markers
m 235
m 305
1 top
1 bottom
43
27
51
27
PhylB/hy3
m 429
2 top
2 bottom
38
39
36
29
g 4711
BGL1
3 top
3 bottom
29
13
21
57
GA1
AG
4 top
4 bottom
44
20
38
30
r 89998
DFR
5 top
5 bottom
41
50
45
34
Which marker is linked to the flowering
mutation?