Download Sample Exam 3 answer key

Bcor101
Sample questions
Midterm 3
1. The maps of the sites for restriction enzyme EcoR1 (R1) in the wild type and mutated
cystic fibrosis genes are shown below:
Wild Type
CF probe
Mutant
R1
12 kb
R1 4 kb R1
_|_________________________|__________|__
____________________________
R1
10 kb R1 2 kb R1 4 kb R1
_|___________________|______|__________|__
Samples of DNA obtained from a fetus (F) and her parents (M and P) were analyzed by
gel electrophoresis followed by the Southern blot technique and hybridization with the
radioactively labeled probed designated "CF probe" in the above figure. The
autoradiographic results are shown in the following figure:
___P
F
M___
___ ___ ___
___
___
___ ___ ___
___
___
________________
Given that cystic fibrosis is a recessive mutation, will the fetus be affected? Explain.
Both parents are heterozygous – contain 4 EcoR1 fragments (one of which (4 kb) is
common between the wild type and mutant gene). The fetus has this common fragment
and one other fragment (12 kb) that corresponds to the size found in the wt gene.
Therefore the fetus is homozygous wt and will not be affected.
2. Several different haploid and diploid genotypes for the lac operon are given below.
For each genotype, determine whether b-galactosidase and permease will be produced
under conditions of noninduction (no lactose present) and induction (lactose present).
Fill in the table with a (+) if the enzyme is produced and a (-) if the enzyme is not
produced.
Permease
no lactose lactose
b-galactosidase
no lactose lactose
Genotype
(a)
i+ p- o+ z+ y+
-----
(b)
i+ p+ oc z+ y -
+
(c)
i+ p+ oc z+ y - /
i- p+ o+ z- y+
(d)
is p+ o+ z- y+/
i+ p + o + z + y -
---
(e)
i+ p+ oc z+y+/
is p+ o+ z- y -
+
+
---- ---+
+
-------
----
----
+
---
-----
---
+
+
+
3) A linear DNA molecule 1000 base-pairs long is digested with the following restriction
enzymes with the following results:
EcoRI
BglII
EcoRI + BglII
400 bp, 600 bp
250 bp, 750 bp
250, 350, 400 bp
EcoR1
400bp
BglII
350 bp
250 bp
Determine the restriction map of this DNA
G
A
__
T
C
__
(top of gel)
__
__
__
__
__
__
__
__
4. Identify the open-reading frame
represented in this DNA sequence
__
__
__
__
__
__
__
__
__
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__
__
__
__
__
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__
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__
__
__
__
__
__
__
__
(bottom of gel)
DNA sequence:
5’CTGACTGCTTACCTAGGTACCGCATGGCATGCGATTTGTACATA3’
met-ala-cys-asp-leu-tyr-ileu
5) A short DNA fragment with the sequence 3’-CTGATAAGGCTTTG-5’ is used as a
template for sequencing by the dideoxynucleotide method. Each of the four reaction
vessels contains DNA polymerase, a primer with the sequence 5’-GACTA-3’, labelled
deoxynucleotide triphosphates and a small amount of either ddATP, ddTTP, ddCTP,
ddGTP. Draw the dideoxy sequencing gel pattern expected for this template.
G
A
T
C
___
___
___
___
___
___
___
___
bottom of gel
6. You are studying Neurospora mutants that are proline auxotrophs (unable to grow in
the absence of proline). One of your students isolates a revertant which is able to grow
on minimal media. You cross the revertant with a wild type strain of Neurospora in
order to determine the nature of the mutation in the revertant. What proportion of the
progeny from this cross would be proline independent if
a) the reversion precisely reversed the original change that produced the pro - mutant
allele?
100%
b) the revertant phenotype was produced by a suppressor mutation in a second gene
located on a different chromosome?
75% - Will get four classes of genotypes in equal frequency:
pro-sup- -same genotype as the revertant parent – proline independent
pro-sup+ - recombinant – doesn’t contain the suppressor mutation – proline dependent
pro+ supindependent
recombinant – sup- mutation has no phenotype on its own – proline
pro+sup+ - same as wild type parent – proline independent
c) the suppressor mutation occurred in a second gene located 15 map units from the pro locus on the same chromosome?
92.5%
42.5%: pro-sup- -same genotype as the revertant parent – proline independent
7.5%: pro-sup+ - recombinant – doesn’t contain the suppressor mutation – proline
dependent
7.5%: pro+ supproline independent
recombinant – sup- mutation has no phenotype on its own –
42,5%: pro+sup+ - same as wild type parent – proline independent
7. A plant was transformed using Ti plasmid which contains a kanamycin-resistance
gene between the left and right borders of the T-DNA region. Two kanamycin-resistant
plants were identified for further study. The plants were allowed to self, and the results
were as follows:
Progeny from selfing plant A:
3/4 progeny were resistant to kanamycin and 1/4 progeny were sensitive to
kanamycin
Progeny from selfing plant B:
15/16 of the progeny were resistant to kanamycin and 1/16 of the progeny were
sensitive to kanamycin.
a) Explain the molecular and genetic basis for these two different ratios.
Plant A: A single insertion event of the T-DNA into a plant chromosome – the parent
plant is hemizygous for the insertion. When this plant is selfed, you will get a 1:2:1
segregation for kan-resistance, since this is a dominant marker. Therefore, 3/4 will be
kanamycin-resistant and 1/4 will be sensitive to kanamycin
Plant B: Two independent insertions of the T-DNA occurred on different chromosomes.
Therefore the two T-DNA insertions are unlinked. When plant B is selfed, the two loci
will segregate 9:3:3:1, with 15 individuals that are kanamycin resistant and 1 individual
out of 16 that is kanamycin sensistive.
b) what ratios would you expect from these two plants if they were used in a test cross?
Plant A: 50% kanamycin resistant;50% kanamycin sensitive
Plant B: 75% kanamycin resistant; 25% kanamycin sensitive