Download Exam 1 Key

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Page 2 of 6
Question 1. (15 points)
a) The base sequence of a region where DNA replication begins is shown. What is the sequence
of the primer that is synthesized complementary to the underlined bases? Write the sequence
below the underlined bases. Be sure to indicate the 5´ and 3´ ends.
5´- T C G A G A C G C A C A T C T G G C A G C G A G T C G A -3´
3´ G U A G A C C G U 5´
b) Circle the purines in your primer.
c) If DNA polymerase is added to the template strand and primer given in part a), indicate the
direction(s) of DNA synthesis with an arrow.
Question 2. (15 points)
The plant, Banana examinus, is diploid and 2n = 4. There is one long pair and one short pair of
chromosomes. Each diagram represents an anaphase stage of an individual cell during meiosis or
mitosis in a plant that is genetically dihybrid (A/a ; B/b) for genes on different chromosomes.
The lines represent chromosomes or chromatids, and the points of the Vs represent centromeres.
In each case, determine whether the diagram represents a cell in meiosis I, meiosis II, or mitosis,
OR if a diagram is "abnormal" (not correct) indicate this and very briefly explain why.
A)
B)
a
A
a
A
B
b
A
B
A
A
b
C)
a
a
a
b
B
B
B
B
b
b
b
A
a
A
a
B
B
A
a
B
B
D)
a
B
A
B
A) Meiosis I
B) Abnormal – Too many chromatids for 2n=4.
C) Meiosis II
D) Abnormal – There should be 2 “B” alleles and 2 “b” alleles.
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Page 3 of 6
Question 3. (10 points).
The year is 2080; a NASA mission returns from Mars with a bacterial-like life form. After
extensive experimentation, scientists succeed in growing the Martian bacteria and determine that
its genetic material is DNA. The scientists then conduct the Meselson-Stahl experiment on the
Martian bacteria:
generation 0: Bacteria are grown for many generations on 15N ("heavy" nitrogen).
generation 1: The cells are then shifted to 14N ("light" nitrogen) for 1 generation.
generation 2: The cells continue to grow in 4N for one generation.
After centrifugation in a CsCl gradient, the DNA molecules band according to their density. In
the test tubes below, the location of DNA is indicated after generation 0, l and 2.
Generation
0
A
1
2
3
14 N
15 N
control
a) Based on the result above, what can you say about DNA replication on Mars? Briefly, how
does this compare to DNA replication on Earth?
DNA replication is not semi-conservative, as it is on Earth. On Mars, each DNA
strand is made up of some old DNA and some new DNA. On Earth, the old DNA
strands are kept intact and a complementary strand is synthesized de novo.
b) In the empty test tube above, indicate where you expect to find DNA if the Martian bacteria
are grown for one more generation in 14N (generation 3).
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Page 4 of 6
Question 4. (20 points)
William Shakespeare writes, "What have we here? A man or a fish? Dead or alive? He smells
like a fish; a very ancient and fish-like smell . . ." to describe a character in “The Tempest.” This
man is apparently afflicted with fish odor syndrome, an extremely rare genetic disorder. Let’s
assume this man is individual 8 in the pedigree below.
1
3
2
4
5
6
7
8
a) What mode(s) of inheritance could explain the pedigree above? Circle all possible answers.
1) autosomal dominant
2) autosomal recessive
3) X-linked dominant
4) X-linked recessive
b) For each mode of inheritance you answered in part a), give the genotypes of individuals 1
through 8. Define all symbols used.
a = recessive fish odor syndrome
A = dominant, no syndrome
Y = Y chromosome
Individual
Autosomal
Dominant
Autosomal
Recessive
X-Linked
Dominant
X-linked
Recessive
1
A/A
A/A
2
a/a
a/Y
3
A/A
A/Y
4
A/a
A/a
5
A/a
A/a
6
A/a
A/a
7
A/a
A/Y
8
a/a
a/Y
Proper genetic notation (i.e. A/A, not AA) required for full credit.
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Page 5 of 6
c) Individual 6 and individual 7 have a second child that conclusively determines the mode of
inheritance. What is the mode of inheritance and what can you say about this second child?
Show your reasoning.
Autosomal recessive. The second child is an affected female.
The gene cannot be X-linked recessive because an unaffected father will pass on a
dominant “A” to all his daughters.
The gene can be autosomal recessive because two “A/a” parents can produce “a/a”
progeny of either sex.
Question 5. (15 points)
a) A pure-breeding albino, obese mouse is crossed to a pure-breeding non-albino, non-obese
mouse. All the F1 progeny are non-abino, non-obese mice. When the F1 mice are crossed with
each other, what proportion of the F2 progeny will be non-abino, obese mice? Assume the genes
are unlinked. Define all symbols used.
a = recessive albino
b = recessive obese
A = dominant non-albino
B = dominant non-obese
P
a/a; b/b
x
F1
A/a; B/b
A/a; B/b
x A/a; B/b
F2
A/A; B/B
A/-; b/b
¾ x ¼ = 3/16 non-albino, obese progeny
b) What genotypes are represented among the non-albino, obese F2 mice? In what proportions?
A/a; b/b
A/A; b/b
½ x ¼ = 1/8
¼ x ¼ = 1/16
2/3 are A/a; b/b
1/3 are AA; b/b
c) How would you determine the genotype of a non-albino, obese F2 mouse?
Do a testcross with a homozygous recessive a/a animal.
If half the progeny are abino, the F2 parent is A/a; b/b.
If all the progeny are non-albine, the F2 parent is homozygous A/A; b/b..
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Page 6 of 6
Questions 6. (25 points)
Female Drosophila melanogaster heterozygous for three recessive alleles q, r, and t are crossed
to homozygous qrt males. The phenotypes of 10,000 progeny are scored as follows:
qRt
QrT
qRT
Qrt
qrt
QRT
qrT
QRt
3770
3770
980
980
230
230
20
20
a) Using proper genetic notation, give the genotype of the mothers.
qRT/QrT
b) Based on the observed results, what is the map order? Show your reasoning.
(4 pts) R-Q-T or T-Q-R (1 pt) Least common genotypes represent double-crossover
classes (1 pt); most common represent parental classes (1 pt); gene that differs is in the
middle (1 pt). Full credit if calculated map distances to determine.
c) Using all of the relevant data, calculate each of the three two-factor recombination
frequencies. Show your work below or on the previous page. Express all frequencies as
percentages.
(6 pts)
Recombination between
rq
qt
rt
qRt
3770
QrT 3770
qRT 980
980
980
Qrt
980
980
980
qrt
230
230
230
QRT 230
230
230
*double recombinant
qrT
20
20
20
2(20)*
count 2X for outside
QRt 20
20
20
2(20)*
genes
500
2000
2500
(may also add values for rq and qt)
recomb. freq. between r + q = Rrq = 500/10,000 = 5% (2 pts); if missed DCO's (1 pt)
recomb. freq. between q + t = Rqt = 2000/10,000 = 20% (2 pts); if missed DCO's (1 pt)
recomb. freq. between r + t = Rrt = 2500/10,000 = 25%(2 pts); if missed DCO's (1 pt)
d) What is the observed frequency of double recombination? Show your work.
(5 pts)
Double recombinants are:
qrT
20
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QRt
Page 7 of 6
20
Frequency = (20+20)/10,000 = 40/10,000 = 0.4%
If used 2 x DCO's and calculated 0.8% (2 pts); if calculated predicted rather than
observed (1%, answer for next part) (1 pt).
e) What frequency of double recombination would you predict if the data showed no
evidence of interference?
(5 pts)
Rrq x Rqt = 5% x 20% = 1%
Use of absolute number '100' (3 pts); definition of interference (1 pt).