name:_______________________
student ID:_____________________
Genetics L311 exam 1
February 6, 2015
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
_______
(16 points possible)
page 5
_______
(22 points possible)
page 6
_______
(16 points possible)
total
_______ (of 100 points possible)
1
name:_______________________
student ID:_____________________
1. Short answers (2 points each, 20 points total)
A. The situation where having one wild type gene copy is not enough to prevent any mutant
phenotype from resulting is haploinsufficient .
B. The somatic cells make up most of the body. These are all the cells except for the germ
cells.
C. Phenotype refers to observable properties of an individual.
D. The chromosomal point of attachment for the mitotic spindle is the centromere .
E. The region found on both the X and Y chromosomes is called the pseudoautosomal
region .
F. An individual in which both copies of a particular gene are identical is said to
be homozygous for that gene.
G. Nondisjunction
was viewed by many as proof of the chromosome theory of inheritance.
For the following, please provide a brief definition of the term given:
H. test cross: A cross of unknown genotype by homozygous recessive.
I. polar body: Dead end products of oogenic meiosis, these cells are used as a means of
disposing of extra chromosomes.
J. hemizygous: Having only one copy of a region of a chromosome, for example in males
with their single X chromosome.
2
name:_______________________
student ID:_____________________
2. Alkaptonuria (aka black urine disease) is a rare inherited genetic disorder that causes the urine of
affected individuals to turn brown or black after prolonged exposure to air. It results from a defect in
tyrosine metabolism. Affected individuals show the trait from birth. The pedigree to the right shows an
extended family including two individuals affected by the trait. Assume those who marry in don’t have
the mutation.
A. What mode of inheritance does alkaptonuria show
(4 points)?
autosomal recessive
B. Obligate carriers are individuals who must carry
the mutation that produces the trait. Please list the
obligate carriers in the pedigree at right (4 points).
I-1, I-2, II-4, III-2, III-3, III-5, IV-2, IV-3
C. What is the probability that II-3 carries the
mutation (i.e. is heterozygous, 3 points)?
2/3
D. What is the probability that IV-2 and IV-3’s next child will be a boy who is affected by the disease (3
points)?
½ X ½ X ½ = 1/8
3. The mythical singing field mouse, K. corneliusae, is found in isolated populations with a few
characteristic differences. Some have long hair whereas others have short hair and some have long legs
whereas others have short legs. Presented below are the results of several crosses (12 points).
Parental
1
2
3
4
Long hair, short leg X short hair, short leg
Long hair, long leg X short hair, short leg
Long hair, long leg X short hair, long leg
Short hair, long leg X short hair, long leg
Long hair,
long legs
0
103
451
0
Long hair,
short legs
181
98
0
0
Please provide the genotypes of the parental mice for each cross.
Long hair, short leg X short hair, short leg
Llss X llss
Long hair, long leg X short hair, short leg
LlSs X llss
Long hair, long leg X short hair, long leg
LlSS X LlSS
Short hair, long leg X short hair, long leg
llSs X llSs
3
Short hair,
long legs
0
101
158
358
Short hair,
short legs
178
99
0
121
name:_______________________
student ID:_____________________
4. A species of deep-sea yellow sponge has somatic cells containing 3 pairs of chromosomes, one long,
one medium and one short. The gene conferring yellow color (Y) is on the long chromosome, a gene (B)
that controls behavior leading to some individual’s predilection for wearing square pants is on the
medium chromosome and a gene (L) causing obsessive laughter is on the short chromosome.
A. A single somatic cell from an individual, named Bob, who is heterozygous for the yellow and
laughter genes and homozygous dominant for B undergoes mitosis. Please draw the products of this
division, including the genes (6 points).
L
B
Y
B
y
B
l
L
B
Y
y
l
B. Please diagram one of Bob’s cell in metaphase II. Include the genes and circle one pair of sister
chromatids. Assume no cross overs (5 points).
L
C. Rarely meiosis can go awry. Please show the products of meiosis I and meiosis II (you need not
include all the steps of each division, just the products) where nondisjunction of the long chromosome
during meiosis I has occurred. Meiosis II was normal. Show all chromosomes. You do not need to
include the genes in your answer (5 points).
4
name:_______________________
student ID:_____________________
5. A newly discovered species of deep-sea squid, M. troyerae, is found to have two true breeding strains.
Strain A has yellow skin and long tentacles, while Strain B has blue skin and short tentacles. A cross of
males from Strain A and females from Strain B results in offspring that all have blue skin and medium
tentacles. Crossing the F1 animals results in the following:
252 blue, medium tentacled females XBXB or b Tt
122 blue, medium tentacled males
XBY Tt
126 yellow, medium tentacled males XbY Tt
128 blue, long tentacled females
XBXB or b TT
60 blue, long tentacled males
XBY TT
65 yellow, long tentacled males
XbY TT
120 blue, short tentacled females
XBXB or b tt
63 blue, short tentacled males
XBY tt
66 yellow, short tentacled males
XbY tt
1002
A. Fill in the genotypes of the F2 progeny (9 points).
B. If you cross blue skinned, long tentacled males by F1 females, what fraction of male offspring will
have blue skin and medium tentacles (4 points)?
XBY TT X XBXb Tt => ½ will have blue skin and ½ of those will have medium tentacles to
answer is ¼
6. You discover a population of griffins living on a remote Pacific island. The griffins display a variety
of beautifully colored feathers. You decide to study the genetic basis for the variation in color. You
discover that all of the variations in color are the result of naturally occurring mutations that alter the
normal brown feather color. You perform a complementation test between 7 of the differently colored
animals and find the results shown in the table.
A. How many genes do mutations 1 through 7
represent (6 points)?
3
B. Mutations 1, 2, 4 and 5 all produce different
feather colors. How can you explain this
phenomenon (3 points)?
Different alleles of the gene produce
different colors.
5
1
2
3
4
5
6
7
1
–
2
–
–
3
+
+
–
4
–
–
+
–
5
–
–
+
–
–
6
+
+
+
+
+
–
7
+
+
+
+
+
–
–
name:_______________________
student ID:_____________________
7. 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:
448 are orange with big ears
151 are red with big ears
148 are orange with small ears
52 are red with small ears
O-BooBO-bb
oobb
A. Please fill in the genotypes of the F2 mice on the lines above (6 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)?
ooB- X oobb
ooBB
=>ooBb
1/3(0) + 2/3(1/2) = 1/3
ooBb
=>ooBb and oobb
8A. In a population of rare frogs, you find a single mutation in the drk gene that affects skin color. The
frogs normally appear all white. A recessive mutation in drk causes the frogs to appear dark. You cross
homozygous drk mutant frogs and find the results shown below (i.e. the frogs produced by
homozygous mutant parents are either dark or are various shades of gray.) No other genes are mutant in
these animals.
How do you explain these results (3 points)?
variable expressivity
B. In studying a different gene, ebn, you find a similarly puzzling result. When you cross homozygous
ebn mutants (again, wild type is white), you find the results shown below (i.e. the offspring of frogs
homozygous mutant for ebn are either dark or white.) No other genes are mutant in these animals.
How do you explain these results (3 points)?
incomplete penetrance
6