name:_______________________
student ID:_____________________
Genetics L311 exam 1
February 5, 2016
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
_______
(24 points possible)
page 4
_______
(14 points possible)
page 5
_______
(22 points possible)
page 6
_______
(20 points possible)
total
_______ (of 100 points possible)
1
name:_______________________
student ID:_____________________
1. Short answers (2 points each, 20 points total)
A. Co-dominance is when both alleles in a heterozygote are expressed.
B. A cross of an individual of unknown genotype by homozygous recessive is called a(n) test
cross .
C. The failure of proper chromosome segregation, called nondisjunction , can lead to offspring
with abnormal chromosome complements.
D. The centromere serves to hold sister chromatids together and provides the point of
attachment for the spindle during cell division.
E. The pairing of homologous chromosomes during meiosis is called synapsis .
F. The physical manifestations of crossing over are called the
chiasmata .
G. The genetic makeup of an individual is referred to as its genotype .
For the following, please provide a brief definition of the term given:
H. heterozygous: An individual with two different versions/alleles of a gene.
I. alleles: Different versions of a gene.
J. pleiotropy: When mutation in a single gene produces multiple phenotypes.
2
name:_______________________
student ID:_____________________
2. Wiskott-Aldrich syndrome (WAS) is a very rare disorder affecting about 1/500,000, characterized by
eczema, low platelet count in the blood and immune deficiency. The pedigree below shows a family in
which WAS is segregating.
A. What mode of inheritance does WAS show (3 points)?
X-linked recessive
B. Obligate carriers are individuals who must carry the
mutation that produces the trait but who are not affected
(i.e. do not have the disease). Please list the obligate
carriers in the pedigree at right (3 points).
I-1, II-2, III-5, IV-3, IV-5, IV-7
C. What is the probability that III-3 and III-4’s next child will be affected (will have the disease, 3
points)?
Effectively 0
D. Please provide the genotypes of the following individuals (3 points):
I-1 XWXw
II-3 XWXW
IV-4 XWY
3. A new species of fruit fly, J. gavadeae, is found to have two true breeding strains. Strain A has a blue
thorax and short wings, while strain B has a black thorax and long wings. A cross of strain A males with
strain B females results in offspring that all have black thoraxes and long wings. Crossing the F1 animals
results in the following:
S–XBXB or b
S–XBYb
S–XbY
ssXBXB or b
ssXBY
ssX bY
752 black thorax, long winged females
373 black thorax, long winged males
377 blue thorax, long winged males
249 black thorax, short winged females
123 black thorax, short winged males
127 blue thorax, short winged males
A. Fill in the genotypes of the F2 progeny (6 points).
B. What do you expect from a cross of black thorax, long winged female F1 X blue thorax, short winged
F2 males? Please provide the genotypes, phenotypes and relative frequencies (6 points).
SsXBXb X ssXbY=> 1/8 each:
SsXBXb black, long, female
SsXBY black, long, male
SsXbXb blue, long, female
SsXbY blue, long, male
B
b
ssX X
black, short, female
ssXBY black, short, male
b
b
b
ssX X blue, short, female
ssX Y blue, short, male
3
name:_______________________
student ID:_____________________
4. During an expedition to Madagascar, you find a new species of singing sea robin that you name A.
millerae. A. millerae is diploid with 3 pairs of chromosomes - one long, one medium and one short.
Usually singing sea robin offspring are produced from the fusion of two haploid gametes, which are
formed via normal meiosis. Furthermore, you find that your favorite A. millerae specimen, named Sarah,
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 Sarah's somatic cells in prophase of
mitosis including chromosomes and genes (4 points).
c
A
C
B
a
b
B. Show one of Sarah'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 (4 points).
A
a
B
b
C c
C. Occasionally in A. millerae, 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 was 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. In your studies of the unusual flying fish species A. farlowae you find two true breeding strains. Strain
1 has light blue skin and light blue fins. The second strain has lavender-hued skin with lavender-hued
fins. A cross of strain 1 females with strain 2 males produces F1 that all have dark blue skin and dark
blue fins. Crossing F1s produces:
BB (or bb) 183 light blue skin and light blue fins
bb (or BB) 187 lavender-hued skin and lavender-hued fins
Bb
371 dark blue skin and dark blue fins
A. Please give the genotypes of the F2s on the lines above (6 points).
B. From a cross of light blue skinned, light blue finned X dark blue skinned, dark blue finned animals,
what is the probability of obtaining animals with dark blue skin and dark blue fins (4 points)?
BB X Bb => BB and Bb so 1/2
6. 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
5
2/3(1/2) + 1/3(0) = 1/3
name:_______________________
student ID:_____________________
7. 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
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
6