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name:_______________________
student ID:_____________________
Genetics L311 exam 1
February 3, 2017
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
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(20 points possible)
page 3
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(26 points possible)
page 4
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(18 points possible)
page 5
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(22 points possible)
page 6
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(14 points possible)
total
_______ (of 100 points possible)
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name:_______________________
student ID:_____________________
1. Short answers (2 points each, 20 points total)
A. Polar body is a "dead end" product of meiosis. It is used as a means of disposing of extra
chromosomes during oogenesis. For this reason, it receives very little of the cytoplasm that was
present in the mother cell.
B. Phenotype refers to observable properties of an individual.
C. Genes present in only one copy in otherwise diploid organisms, such as X-linked genes in
human males, are said to be hemizygous .
D. Nondisjunction was viewed by many as proof of the chromosome theory of inheritance.
E. The stage of the cell cycle, composed of G1, S and G2 phases, when the cell is not dividing is
called interphase .
F. The somatic cells make up most of the body. These are all the cells except for the germ
cells.
G. The centromere serves to hold sister chromatids together and provides the point of
attachment for the spindle during cell division.
For the following, please provide a brief definition of the term given:
H. pseudoautosomal region: A region found on both the X and Y chromosomes.
I. haploinsufficient: Where having only one wild type copy of a gene is not enough to prevent
a mutant phenotype.
J. homozygous: When all alleles of a gene are the same.
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name:_______________________
student ID:_____________________
2A. Goltz syndrome produce skin anomalies (yellow-pink bumps on the skin) and short stature. What is
the most likely mode of transmission of Goltz syndrome, shown in the pedigree below (4 points)?
X-linked dominant
B. If III-3 and III-4 have another child, what is the
probability that he or she will be affected (4 points)?
1/2
C. Please provide the genotypes of the following (4
points):
I-1 XgXg
II-5 XgY
III-7 XGY
IV-7 XGXg
3. A newly discovered species of flying fish, A. schlenzae, is found in two naturally occurring strains.
Strain A has blue skin and long gills. Strain B has red skin and short gills. A cross of strain A females
with strain B males produces all red skinned, long gilled progeny. A cross of the F1’s produces:
895 red skinned, long gilled females
905 red skinned, long gilled males
305 red skinned, short gilled females
295 red skinned, short gilled males
310 blue skinned, long gilled females
302 blue skinned, long gilled males
100 blue skinned, short gilled females
102 blue skinned, short gilled males
B-SB-SB-ss
B-ss
bbSbbSbbss
bbss
A. Please fill in the genotypes of the F2 fish on the blanks provided (8 points).
B. What is the probability of obtaining blue skinned, long gilled fish from a cross of red skinned, long
gilled F2 females with blue skinned, short gilled F2 males (6 points)?
B-S- X bbss
1/9 BBSS X bbss => BbSs
2/9 BbSS X bbss => BbSs and bbSs
2/9 BBSs X bbss => BbSs and Bbss
4/9 BbSs X bbss => BbSs, Bbss, bbSs and bbss
Probability of getting bbSs is 1/9(0) + 2/9(1/2) + 2/9(0) + 4/9(1/4) = 2/9
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name:_______________________
student ID:_____________________
4. D. webbae is a newly discovered red-spotted poison dart frog from Costa Rica. Karyotype analysis of
D. webbae indicates that the species is diploid with two pairs of chromosomes, one long and one short.
Simple genetic analysis indicates that the gene(R) that specifies the red spots is located on the long
chromosome, and a gene (L) that specifies body length resides on the short chromosome.
A. Show the products at the completion of a single mitosis of a D. webbae cell, including the
genes. Assume the parent cell is heterozygous for both genes. How many chromosomes are
present in each cell? How many chromatids are present in each cell (6 points)?
4 chromosomes, 4 chromatids.
L
R
l
L
R
l
r
r
B. Diagram the appearance of the chromosomes at metaphase of meiosis I, including the genes.
Draw a circle around a pair of homologous chromosomes and a box around a pair of sister
chromatids. Assume the parent cell is heterozygous for both genes (6 points).
R
L
r
l
C. In one of the meioses you examine, nondisjunction of the large chromosome occurs during meiosis I.
Show the final products of that meiosis. You need not include genes (6 points).
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name:_______________________
student ID:_____________________
5. A new species of field cricket, J. gavadae, has been found in two strains, A and B. Strain A has a red
shell and short legs, while Strain B has a black shell and long legs. A cross of males from Strain A and
females from Strain B results in offspring that all have black shells and long legs. Crossing the F1
animals results in the following:
L-XRXR or r
L-XRY
L-XrY
llXRXR or r
llXRY
llXrY
3/8 black, long legged females
3/16 black, long legged males
3/16 red, long legged males
1/8 black, short legged females
1/16 black, short legged males
1/16 red, short legged males
A. Fill in the genotypes of the F2 progeny (6 points).
B. What do you expect from a cross of black, long legged female F1 X red, short legged F2 males?
Please provide the phenotypes and relative frequencies (6 points).
LlXRXr X llXrY
LlXRXr and LlXRY ¼ long leg, black
LlXrXr and LlXrY ¼ long leg, red
llXRXr and llXRY
¼ short leg, black
r
r
r
llX X and llX Y
¼ short leg, red
6. During your studies of the unusual leaf hopper species A. farlowae you find two true breeding strains.
Strain 1 is pale green with pale green legs. The second strain is black with black legs. A cross of strain 1
females with strain 2 males produces F1 that all are dark green with dark green legs. Crossing F1s
produces:
gg or (GG) 512 pale green with pale green legs
GG or (gg) 498 black black with black legs
Gg 1003 dark green with dark green legs
A. Please give the genotypes of the F2s on the lines above (6 points).
B. From a cross of pale green, pale green legged X black, black legged animals, what is the probability
of obtaining animals that are dark green dark green legs (4 points)?
gg X GG => Gg, so all are dark green with dark green legs.
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name:_______________________
student ID:_____________________
7. Around the world, many animals have moved from areas with day and night into the perpetual
darkness of caves. Usually these animals quickly lose their eyes because these are of no use in the dark.
You are interested in the genetics of eye degeneration so you obtain related fish from different parts of
the world. Assume that each strain has lost its eyes due to a single gene mutation. All mutations are
recessive. Strains from different countries have lost eyes independent of one another. Please include the
genotypes of parents and offspring.
A. You cross cavefish from Mexico X cavefish from France and find all the progeny have eyes. How do
you explain this result (4 points)?
They have eyes because the different fish have mutations in different genes required for eye
development. Cross is aaBB X AAbb =>AaBb
B. You cross cavefish from France (the same fish as in part A) X cavefish from Borneo and all the
progeny lack eyes. How do you explain this (4 points)?
The two strains have mutations in the same gene. Cross is bb X bb => bb
C. When you cross cavefish from Mexico (those in part A) X cavefish from Borneo (those in part B),
what outcome do you expect? Why (3 points)?
French and Bornean fish have mutations in the same gene, which I’ve called B. We know
from A that Mexican fish have a mutation in another gene, which I called A. Therefore the cross is
aaBB X AAbb => AaBb and the fish will have eyes.
D. What outcome will result from a cross of the F1’s in part C? Please give phenotypes and their
relevant frequencies (3 points).
Cross isAaBb X AaBb, which will produce:
9/16 A-B3/16 A-bb
3/16 aaB1/16 aabb
The first (i.e. A-B-) have eyes the rest do not. Therefore the
ratio of fish with eyes:eyeless is 9:7.
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