BIO 304
Fall 2003
Exam II: October 23, 2003
Please print your name:____________________________________________________
1. This exam consists of 8 numbered pages. Be sure that your exam includes all 8 pages.
2. Answer the questions in the space provided. You may use the back sides of each sheet if
required. Be sure to show ALL work. Credit will only be given for answers that have been
derived. Partial credit will be given where appropriate.
3. The point value of each question is given. Read the questions carefully and be sure to
give what is specifically requested. The exam is worth 100 points.
Points:
Page 2
________/20
Page 3
________/15
Page 4
________/15
Page 5
________/13
Page 6
________/15
Page 7
________/8
Page 8
________/14
TOTAL
________
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Part 1: Fill-in 20pts (2 pts each)
1. _____HETEROGAMETIC____is a term applicable to the sex that produces gametes
bearing structurally different sex chromosomes.
2. _____SEXUAL___________is a form of reproduction involving the union of haploid
gametes to form a diploid zygote.
3. In humans, males are ___HEMIZYGOUS____for any gene on the X chromosome, in
other words any gene on the X chromosome is present in only a single dose.
4. _DOSAGE COMPENSATION____, a genetic mechanism that regulates the levels of
gene products on the X chromosome is achieved via X inactivation.
5. Seedless watermelons are ____POLYPLOID___________, they contain more than 2
complete sets of chromosomes.
6. Any variation in chromosome number that does not involve complete sets of
chromosomes is referred to as ___ANEUPLOIDY_________________.
7.__TRANSLOCATION_______describes the movement of a chromosomal segment to
a new location.
8. Cri-du-chat syndrome (46, 5p-) is an example of this type of chromosomal abnormality
___DELETION_________________________.
9. Two genes that demonstrate a recombination frequency of 16.5% are
___16.5_________map units apart.
10. The type of phenotypic variation that is associated with quantitative traits is
__CONTINUOUS__________variation.
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Part 2: Short Answer 30 pts (5pts each)
1. During one lab/discussion section, each of you were assigned a model genetic organism.
From that discussion we compiled a list of “what makes a model organism”. List five
items/components from that list and briefly (in one or two sentences) describe the
significance of each. Any 5 of the following:
1. Cheap and easy to maintain – less $ spent on maintenance = more $ on research
2. Rapid life cycle – the quicker the life cycle, the sooner you can analyze progeny
3. Easy to manipulate for crosses – can perform what ever crosses you are interested in.
4. Genome sequenced – the more researchers know about the exact sequence of genes,
the more that can get accomplished
5. Small genome size – researchers can get work with a “minimum” amount of DNA;
not a lot of “junk” to search through
6. Homology to humans or other important species – ease of relating what we’ve
learned to something “more important”
7. Lots of offspring – the more offspring there are, the more you can analyze.
8. Mutational analyses are easily performed
9. Lots of genetic information around prior to any DNA sequencing
2. Briefly describe the XX-XO and the XX-XY modes of sex determination. How do they
differ from one another?
XX-XO (also known as Protenor mode) – mode of sex determination in which once sex
has two identical sex chromosomes (XX for example) and the other sex has only one
sex chromosome (i.e. – has JUST one X). There is no differentiation of sex
chromosomes in this mode.
XX-XY (also known as Lygaeus mode) – mode of sex determination in which there are
two distinct sex chromosomes. One sex is homogametic – produces gametes bearing
chromosomes which are structurally the same (i.e.- XX) where as the other sex is
heterogametic – produces gametes bearing structurally different sex chromosomes
(i.e.- XY)
3. During meiosis in humans, the X and Y align as “homologous” chromosomes. Briefly
compare/contrast these chromosomes.
Both chromosomes have two regions – 1) homologous region necessary for synapsis,
this region contains sequences in common an 2) differential region – that does not pair
in synapsis and contains different genes.
The X chromosome is very similar to an autosome. It contains >2300 genes, most
unrelated to sex.
The Y contains very few genes, including the SRY (sex determining region) which
contains a gene that is involved in the development of testes from undifferentiated
gonadal tissue.
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4. Briefly describe the mechanism involved in X inactivation.
There is a region of the X chromosome known as the X-inactivation Center (XIC). The
XIC produces an RNA molecule known as XIST. The XIST RNA is thought to coat the
X chromosome (from which it was synthesized) which in turn silences the genes on the
X. (In other words XIST “coats” the X chromosome, making it non-functional).
5. Describe and diagram how nondisjunction of homologous chromosomes results in
aneuploidy.
*Nondisjunction of homologous
chromosomes occurs during the 1st meiotic
division, resulting in one cell that contains
both homologous chromosomes and one
that does not contain any.
*Assuming normal disjunction at the
2nd meiotic division, the end result will
be 2 gametes that contain 2 “red”
chromosomes and 2 gametes that don’t
contain any “red” chromosomes.
*If either of the gametes that contain 2
“red” chromosomes combines with a
normal gamete, the resulting offspring
will be trisomic for the “red”
chromosome, wherease if either of the
gametes that don’t have a “red”
chromosome combines with a normal
gamete, the resulting offspring will be
monosomic for the “red” chromosome.
6. List 5 enzymes/proteins necessary for DNA replication in E.coli. What are the
functions of each of these enzymes?
Dna A, B and C are helicases that unwind the double helix
SSBP’s (Single strand binding proteins) prevent closure of the open double helix
DNA gyrase reduces tension from supercoiling
Primase synthesizes short RNA primer
DNA synthesis via DNA polymerase III
Primer removal and replacement with DNA by DNA polymerase I
Ligase closes up the gaps between Okazaki fragments
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Part 3: Problems 50pts
1. (28pts) In Drosophila the following three genes are located on chromosome 2:
Gene
wing conformation
body color
antennae length
Alleles
B = straight wings; b = bent wings
Y = white; y = yellow
A = long antennae; a = short antennae
A true breeding female with bent wings, white body and long antennae was crossed to a true
breeding male with straight wings, yellow body and short antennae. All of the F1 were
white, had straight wings and long antennae.
(a) (5pts) Given this information, what was the genotype of the two parents and the F1 with
respect to these three linked genes?
P:
b Y A/ b Y A
X
B y a/ B y a
Genotype of the female parent
Genotype of the male parent
F1:
b Y A/ B y a
Genotype of the F1 progeny
(b) (8pts) F1 females were testcrossed to true breeding males with bent wings, yellow color
and short antennae. Five-hundred progeny were examined and classified based on
phenotype. In the table below, fill in genotypes on chromosome 2 inherited from the F1
parent, with the correct locus in the middle. Also fill in the correct progeny class
corresponding to each genotype (i.e. parental, SCO between ___ and ___, DCO)
Number Genotype of chromosome 2
Phenotype
of
inherited from F1 parent. Put
Progeny Class
progeny the correct locus in the middle.
Straight,
19
BYA
SCO b/w B and Y
white, long
Bent, white,
201
bYA
PARENTAL
long
Straight,
2
BYa
DCO
white, short
Straight,
31
ByA
SCO b/w Y and A
yellow, long
Bent, white,
30
bYa
SCO b/w Y and A
short
Bent, yellow,
1
byA
DCO
long
Straight,
198
Bya
PARENTAL
yellow, short
Bent, yellow,
18
bya
SCO b/w B and Y
short
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PROBLEM CONTINUED FROM PREVIOUS PAGE
(c) (3pts) Draw a linkage map for these three loci with the correct locus in the middle. Use
the symbols B, Y and A for the genes.
B
8 MU
Y
12.8 MU
A
(d) (3pts) Determine the map unit distance between the gene locus you placed on the right
and the gene locus you placed in the center (c).
31 + 30 + 1 + 2 = 64 = 0.128 X 100% = 12.8 MAP UNITS B/W Y AND A
500
500
(e) (3pts) Determine the map unit distance between the gene locus you placed on the left and
the gene locus you placed in the center (c).
19 + 18 + 1 + 2 = 40 = 0.08 x 100% = 8 MAP UNITS B/W B AND Y
500
500
(f) (3pts) Determine the map unit distance between the gene locus you placed on the right
and the gene locus you placed on the left in (c).
8 MU + 12.8 MU = 20.8 MAP UNITS
(f) (3pts) Is there any evidence of chromosome interference in this interval of the Drosophila
genome? Solve and explain. YOU CAN EARN POINTS EVEN IF YOUR MAP
DISTANCES ARE INCORRECT IF YOU SHOW THAT YOU UNDERSTAND THIS
GENETIC CONCEPT.
EXPECTED FREQUENCY OF DCO’s = (0.08)(0.128) = 0.01024
(0.01024)(500) = 5.12 DCO INDIVIDUALS EXPECTED
C = # OBS DCO’S = 3 = 0.59
# EXP DCO’S
5.12
I = 1-C = 1-0.59 = 0.41 OR 41% INTERFERENCE
*Interference - When a crossover event in one region of the chromosome inhibits a
second event in nearby regions
In other words, something is preventing (or interfering) cross-over
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2. (8pts) A bacterial culture was grown an 15N containing medium for several generations.
The culture was then transferred to an 14N containing medium and allowed to replicate.
Below is a “picture” of a replicating chromosome isolated from a bacterium exposed to the
scenario above. Smooth lines indicate parental or “old” DNA, whereas squiggles indicate
newly synthesized DNA. For each experiment below, the DNA was chopped up into smaller
fragments and subjected to density dependent centrifugation.
Assuming semiconservative replication, draw the banding patterns of DNA found in each
band in the gradients below.
14
15
N
N
(a) To in 15N containing media
(b) 0.5 Generations in 14N media
(c) 0.5 Generations in 14N media, and
DNA strands were denatured (pulled apart
from one another) prior to centrifugation
(d) 1 Generation in 14 media
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3. (5pts) How many Barr bodies are expected in interphase cells of the following
individuals?
(a) Klienfelter syndrome, 47, XXY ____1_____
(b) Turner syndrome, 45, X
____0_____
(c) 47, XYY
____0_____
(d) 47, XXX
____2_____
(e) 48, XXXX
____3_____
4. (4pts) Three genes are located on the same chromosome, in the following order: A B C
Your instructor has all ready performed a number of genetic crosses and has determined that
the recombination frequency between genes A and B is 12% and the recombination
frequency between B and C is 20%. Determine the probability of a double cross-over
(DCO) event between all three genes.
(0.12)(0.20) = 0.024 or 2.4%
5. (5pts) In Drosophila, black body color (b) is recessive to the normal wild-type body color
(b+). Cinnabar eye (cn) is recessive to the normal wild-type eye color (cn+). A homozygous
wild-type fly was crossed to a fly with black body and cinnabar eyes. The resulting
heterozygous F1 fly was mated to a fly with black body and cinnabar eyes. These were the
results of the offspring:
90 wild-type
92 black body and cinnabar eyes
9 black body and wild-type eyes
9 wild-type body and cinnabar eyes
What is the map distance between the gene for black body and the gene for cinnabar eyes?
9+9
90 + 92 + 9 + 9
= 18 = 0.09 X 100% = 9 map units
200
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