name:_______________________
student ID:_____________________
Genetics L311 exam 1
September 19, 2014
Directions: Please read each question carefully. Answer questions as concisely as possible.
Excessively long answers, particularly if they include any inaccuracies, may result in deduction of
points. You may use the back of the pages as work sheets, but please write your answer in the space
allotted and please show all your work. Clearly define your genetic symbols. We will not make
guesses as to what a particular symbol is intended to mean. Also, don’t assume that strains are truebreeding unless this is stated in the question. Finally, show all your work. Good luck.
page 2
_______
(20 points possible)
page 3
_______
(26 points possible)
page 4
_______
(18 points possible)
page 5
_______
(24 points possible)
page 6
_______
(12 points possible)
total
_______ (of 100 points possible)
1
name:_______________________
student ID:_____________________
1. Short answers (2 points each, 20 points total)
A. The pairing of homologous chromosomes during meiosis I is referred to as synapsis .
B. Haploinsufficient is when having a single wild-type copy of a gene is not enough to
prevent a mutant phenotype.
C. Different versions of a gene are called alleles .
D. Codominance is when both alleles in a heterozygote are expressed.
E. The genetic makeup of an individual is referred to as its genotype .
F. Pleiotropy is when mutation of a single gene produces multiple phenotypes.
G. The stage of the cell cycle when a cell is not in mitosis is called interphase . It consists of
two growth phases and S phase.
For the following, please provide a brief definition of the term given:
H. pseudoautosomal: A region found on both the X and Y chromosomes.
I. nondisjunction: Failure of chromosomes to separate.
J. polar body: Cell produced during oogenesis as a means of eliminating extra chromosome
sets. This is a dead end product of meiosis.
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name:_______________________
student ID:_____________________
2. Reggaephilia, an extremely rare disorder, causes an obsession with reggae music. The disorder
appears in individuals by age 3 or 4. All members of the pedigree are adult.
A. What is the most likely mode of transmission for reggaephilia (4 points)?
X-linked dominant
B. What is the probability that III-3 is heterozygous for the
mutation (i.e. is heterozygous, 2 points)?
0 (male won’t have two copies)
C. Please provide the genotypes of the following individuals
(6 points)?
I-2 XRY
II-4 XrY
IV-7 XRXr
3. You wish to study nervous system development in Drosophila. To begin these studies, you use a
genetic screen to find mutations that alter development of a particular neuron. You find 8 different
mutations, initially numbered one through eight. You perform crosses between your mutants in all
possible pairwise combinations, producing the results shown in the table.
A. What is the name given to the general
procedure described here (2 points)?
complementation test
B. Please indicate which mutations are allelic
to one another (4 points).
gnA: 1 and 6
gnB: 2 and 5
gnC: 3
gnD: 4 and 7
gnE: 8
1
2
3
4
5
6
7
8
1
–
2
+
–
3
+
+
–
4
+
+
+
–
5
+
–
+
+
–
6
–
+
+
+
+
–
7
+
+
+
–
+
+
–
8
+
+
+
+
+
+
+
–
C. How many different genes were identified in this screen (4 points)?
5 genes
D. Using gene names of your choosing (eg. gene1+ or geneA+, etc.), please give the genotype of the
progeny of the cross of homozygous mutant 1 X homozygous mutant 2 (4 points).
gnA-/gnA+; gnB-/gnB+ or, if you prefer Mendelian notation, AaBb
3
name:_______________________
student ID:_____________________
4. During an expedition to Madagascar, you find a new species of tortoise that you name S. reifeisae. S.
reifeisae is diploid with 3 pairs of chromosomes - one long, one medium and one short. Usually tortoise
offspring are produced from the fusion of two haploid gametes, which are formed via normal meiosis.
Furthermore, you find that your favorite S. reifeisae specimen, named Sue, is heterozygous for three
genes, A, B and C, where A is on the long chromosome, B is on the medium chromosome and C is on
the short chromosome. A. Draw one of Sue's somatic cells including chromosomes and genes (6pts).
c
A
C
B
a
b
B. Show one of Sue's cells in anaphase I of meiosis. Circle one pair of homologues. Draw a box
around one pair of sister chromatids. You need not include genes (6 points).
A
a
B
b
C
c
C. Occasionally in S. reifeisae, unfertilized egg cells can develop into viable offspring. The
offspring of this process, called parthenogenesis, are diploid. Diagram the four steps of meiosis
II in the atypical meiosis that could create these offspring. Assume that meiosis I is normal.
Please label each step. You need not include genes (6 points).
prophase II
anaphase II
metaphase II
telophase II
4
name:_______________________
student ID:_____________________
5. You find two populations of mice inhabiting a local park. Some mice have red fur and big ears.
Others have orange fur and small ears. You cross a true-breeding red-furred big-eared male mouse with
an orange furred small-eared female. You find that the F1s are all orange with big ears. You then cross
the F1s and see the following phenotypes in the offspring:
243 males are orange with big ears
245 females are orange with big ears
79 males are red with big ears
81 females are red with big ears
82 males are orange with small ears
78 females are orange with small ears
26 males are red with small ears
27 females are red with small ears
O–E–
O–E–
ooE–
ooE–
O–ee
O–ee
ooee
ooee
A. Please fill in the genotypes of the F2 mice on the lines above (8 points).
B. What is the probability of obtaining a red, small-eared mouse from a cross of red, big-eared F2 male
mice with red, small-eared F2 female (4 points)?
ooE– X ooee => ?? ooee
1/3 ooEE
=> all ooEe
2/3 ooEe
=> ½ ooEe and ½ ooee
2/3(1/2) + 1/3(0) = 1/3
6. In your studies of the unusual ant species K. hunterae you find two true breeding strains. Strain 1 is
yellow-skinned and has oval shaped abdomens. Strain 2 is blue-skinned with triangular abdomens.
Crossing a strain 1 female with a strain 2 male results in the following:
BbXTXt All females are yellow-skinned with triangle shaped abdomens
BbXtY All males are yellow-skinned with oval shaped abdomens
A. Please give the genotypes of the F1s on the lines above (4 points).
B. What is the probability of obtaining a female with blue skin and a triangular abdomen from a cross of
F1 males X F1 females? What is the probability of obtaining males of the same phenotype (4 points)?
Cross is BbXTXt X BbXtY
Probability of blue skinned, triangle abdomen female is ¼ X ¼ = 1/16
Probability of blue skinned, triangle abdomen male is ¼ X ¼ = 1/16
C. What genotypes and phenotypes do you expect from a cross of strain 1 males X strain 2 females (4
points)?
BBXtY X bbXTXT gives ½ BbXTXt and ½ BbXTY, which all have yellow skin and triangular
abdomens.
5
name:_______________________
student ID:_____________________
7. In your studies of the unusual flying fish species T. nicholasae you find two strains. Strain 1 has light
blue scales and long fins. The second strain has purple scales and short fins. A cross of strain 1 females
with strain 2 males produces F1 that all have lavender scales. Half of the F1’s have long fins and half
have short fins. You cross the lavender scaled, short finned F1s and find:
PPtt 1/12 purple scales, long fins
Pptt 1/6 lavender scales, long fins
pptt 1/12 light blue scales, long fins
PPTt 1/6 purple scales, short fins
PpTt 1/3 lavender scales, short fins
ppTt 1/6 light blue scales, short fins
Note: Short fins are dominant but
recessive lethal. Color is
semidominant.
A. Please give the genotypes of the F2s on the lines above (6 points).
8. Consider a cross between two tropical frogs with the following genotype:
AABbCcddEeFfGg X aaBbCCddEeFfGg
A. What is the probability of producing AaBbCCddEeffgg offspring (2 points)?
1 X ½ X ½ X 1 X ½ X ¼ X ¼ = 1/128
B. What is the probability that a child will NOT have the AaBBCCddeeffgg genotype (2 points)?
1 – (1 X ¼ X ½ X 1 X ¼ X ¼ X ¼) = 1 – 1/512 = 511/512
C. What is the probability of producing offspring with the phenotype ABCdEfg, where A is the
phenotype produced by AA or Aa and a is the phenotype produced by aa, etc (2 points).
1 X ¾ X 1 X 1 X ¾ X ¼ X ¼ = 9/256
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