BioSc 231
General Genetics
Exam 2
Name __________________________________
Multiple Choice. (2 points each)
1. ____ 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 a phenotype
the number of phenotypes for a gene
the number of chromosomes in an organism
the quantity of gene product required for a phenotype
2. _____ The percentage of individuals with a given genotype who exhibit the phenotype associated with that genotype is called
A.
B.
C.
D.
penetrance
incomplete dominance
co-dominance
lethality
3. _____ A plant of genotype C D/C D is crossed to c d/ c d and the resulting F1 testcrossed to c d/c d. If the genes are unlinked, the
percentage of c D recombinants will be
A.
B.
C.
D.
E.
10%
25%
30%
40%
50%
4. _____ A situation where each allele produces a phenotype (usually a protein) that can be detected in the heterozygote is called
A.
B.
C.
D.
E.
penetrance
expressivity
incomplete dominance
co-dominance
lethality
5. _____ In Drosophila the alleles for brown and for scarlet eyes (resulting from two independent genes) interact so that the double
homozygous recessive is white. A pure-breeding brown (BBss) and pure breeding scarlet (bbSS) (P generation) 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
6. _____An autosome is ___.
A.
B.
C.
D.
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
7. _____ 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
8. _____ The maximum recombination frequency between two genes is
A.
B.
C.
D.
E.
100%
80%
50%
10%
1%
9. _____ 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
10. _____ The maize genes bl and ue are linked, 30 map units apart. If a plant bl+ ue/bl ue+ is testcrossed, what proportion of the
progeny will be bl ue/bl ue?
A.
B.
C.
D.
0.03
0.15
0.20
0.50
11. _____ 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
12. _____ Assume that an additional allelic pair in sweet peas also affects pigment formation in addition to the genes mentioned in the
previous 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
13. _____ In humans, 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
14. _____ 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
15. _____ 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
16. _____ 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 more frequently between two loci very close together
The reciprocal exchange between homologous chromosomes is random
17. _____ A dihybrid cross results in a phenotypic ratio of 12:3:1. This type of ratio most likely results from
A.
B.
C.
D.
E.
incomplete dominance
co-dominance
epistasis
co-penetrance
variable penetrance
Short Answer. (variable points)
(4) Two-point testcrosses revealed the following map results:
br__________ui 7 map units
ui___________ns 3 map units
A. Draw the two possible maps for these loci.
B. Other than a 3 point-test cross, what other cross would resolve the two possible maps and what are the possible outcomes
of that cross?
Compound Tested
Mutant
(5) The table to the right shows the results of a series of experiments to
determine the sequence of intermediates in a biochemical pathway. 4
independent auxotrophic mutants which all require compound E (an amino
acid) as a nutritional supplement were analyzed with 4 compounds that are
precursors in the synthesis of compound E. Each mutant was grown on a
minimal medium supplemented with each of the indicated compounds. +
indicates growth that is supported by the indicated precursor. Using the
diagram below, show the order of the intermediates in the pathway and
indicate which step in the pathway is catalyzed by each mutant by placing
the letter representing the appropriate compound in each box and the number
of the appropriate mutant in each circle.
A
B
C
D
E
1
+
+
--
+
+
2
--
--
--
--
+
3
+
+
--
--
+
4
--
+
--
--
+
(6) In pumpkins, jack-o-lantern shaped fruit (o) is recessive to round shaped fruit (o+). Branching vines (s) is recessive to simple
vines (s+). In the P generation, plants from two different pure-breeding lines are crossed. One variety bears round fruit and has
simple vines. The other variety has jack-o-lantern fruits and branched vines. The resulting F1 plants were testcrossed and the
following 240 progeny were obtained:
70
46
56
68
- round, simple
- jack-o-lantern, simple
- round, branched
- jack-o-lantern, branched
Calculate the chi-square and P values based on the prediction that the genes are not linked. (chart is on the last page)
(6) In corn, the genes u, m and r are linked. The data given below summarize the result of 1000 offspring from 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.
u+
u
u
u+
u+
u
u
u+
m+
m
m+
m
m
m+
m
m+
r+
r
r
r+
r
r+
r+
r
304
295
70
64
119
108
18
22
(5) Based on the complementation data below, A) how many complementation groups exist, and B) which mutations belong to each
group? (+ = complementation, -- = no complementation)
Mutation
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
20
21
30
GA1
AG
4 top
4 bottom
44
11
38
59
r 89998
DFR
5 top
5 bottom
41
50
45
34
Which marker is linked to the flowering
mutation?