name:_______________________
student ID:_____________________
Genetics L311 exam 1
September 18, 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
_______
(25 points possible)
page 4
_______
(18 points possible)
page 5
_______
(25 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 stage of the cell cycle, composed of G1, S and G2 phases, when the cell is not dividing is
called interphase .
B. 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.
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. A(n) diploid cell has its homologous chromosomes present in pairs.
E. Cytokinesis , the act of a cell dividing its cytoplasm, often accompanies mitosis.
F. The centromere serves to hold sister chromatids together and provides the point of
attachment for the spindle during cell division.
G. The reciprocal exchange of corresponding segments of homologous chromosomes is
called crossing over .
For the following, please provide a brief definition of the term given:
H. chiasmata: These are the apparent points of crossing over.
I. pseudoautosomal region: A region of DNA found on both X and Y chromosomes.
J. sex linkage: The presence of genes on the X or Y chromosomes.
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name:_______________________
student ID:_____________________
2. Polycystic kidney disease (PKD) is a progressive disease characterized by the development of cysts
on kidney and other internal disorders. The pedigree to the right shows an extended family in which the
disease is segregating.
A. What mode of inheritance does PDK show (4 points)?
Autosomal dominant
B. Obligate carriers are individuals who must carry the
mutation that produces the trait but are not affected (i.e. do
not have the disease). Please list the obligate carriers in the
pedigree at right (3 points).
There are no obligate carriers
C. What is the probability that III-3 and III-4’s next child will be affected (will have the disease, 3
points)?
1/2
D. Please provide the genotypes of the following individuals (3 points):
I-1 pp
II-2 Pp
IV-4 Pp
3. During a recent trip through the jungles of Central and South America you find 3 geographically
isolated white-flowered strains, called A, B and C, of a rare species of orchid, E. qinae, that usually
produces red flowers. You suspect that these strains each arose independently. To gain insight into the
origins of these strains, you collect individuals from the different locations and cross them. You show
that the mutation that produces white flowers is recessive in each case.
A. Your cross of strain A X strain B produces offspring with red flowers. How do you explain this result
(3 points)?
Two different genes are mutated in strains A and B. In other words, the two mutations
complement one another.
B. Your cross of strain B X strain C produces offspring with white flowers. How do you explain this (3
points)?
Strains B and C have mutations in the same gene. In other words, the two mutations fail to
complement one another.
C. What result will a cross of strain A X strain C produce? Why (3 points)?
Red flowered offspring because the mutations affect two different genes.
D. What outcome will result from a cross of the F1’s in part C? Please give phenotypes, general
genotypes and their relevant frequencies (3 points).
A-B- - these are 9/16, red flowered
So 9/16 are red flowered and 7/16 are white
A-bb – these are 3/16, white flowered
flowered
aaB- - these are 3/16, white flowered
aabb - these are 1/16, white flowered
3
name:_______________________
student ID:_____________________
4. T. bastinae is a newly discovered red-spotted poison dart frog from Costa Rica. Karyotype analysis of
T. bastinae 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 T. bastinae 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)?
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).
4
name:_______________________
student ID:_____________________
5. In your studies of C. zengae, a small fish found only in the mountains of California you find two
unusual strains, cleverly named strain 1 and strain 2. Strain 1 has dark blue scales whereas strain 2 has
lavender scales. To understand the genetics of scale color, you perform the crosses and obtain the results
shown below.
Cross 1: dark blue X lavender => ½ of the progeny have dark blue and ½ lavender
Cross 2: lavender X lavender => ¼ dark blue, ½ lavender, ¼ white
Cross 3: lavender X white (from cross 2) => ½ lavender and ½ white
A. What are the genotypes of the original strain 1 (dark blue) and strain 2 (lavender) fish (4 points)?
Strain 1: WW
Strain 2: Ww
B. What phenotype(s) would you expect from a cross of dark blue scale X white scale? Please include
the frequencies (4 points).
all lavender
C. You find another mutation in the same gene that produces dark purple scales. When you cross dark
purple by homozygous wild type you get ½ dark purple and ½ wild type. When you cross dark purple X
dark purple you get 2/3 dark purple and 1/3 wild type. How do you explain this (4 points)?
Dark purple appears to be dominant for color but also recessive lethal.
First cross is WP/w+ X w+/w+ => WP/w+ and w+/w+
Second cross is WP/w+ X WP/w+ => 2/3 WP/w+ and 1/3 w+/w+
6. A newly discovered species of flying fish, K. badgerae, 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 (5 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
5
name:_______________________
student ID:_____________________
7. A new species of lemur is found to have two true breeding strains. Strain A has red fur and short
claws, while Strain B has black fur and long claws. A cross of males from Strain A and females from
Strain B results in offspring that all have black fur and long claws. 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 clawed females
3/16 black, long clawed males
3/16 red, long clawed males
1/8 black, short clawed females
1/16 black, short clawed males
1/16 red, short clawed males
A. Fill in the genotypes of the F2 progeny (6 points).
B. What do you expect from a cross of black, long clawed female F1 X red, short clawed F2 males?
Please provide the phenotypes and relative frequencies (6 points).
LlXRXr X llXrY
LlXRXr and LlXRY ¼ long clawed, black fur
LlXrXr and LlXrY ¼ long clawed, red fur
llXRXr and llXRY
¼ short clawed, black fur
llXrXr and llXrY
¼ short clawed red fur
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