Download 06BIO201 Exam 3 KEY

Name _________KEY________________
Section ______________
Biology 201 (Genetics)
Exam #3 120 points
20 November 2006
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Happy thanksgiving!
1
10 pts.
1. Gene expression in eukaryotes can be regulated at five different levels. List five levels at which
regulation can occur.
Chromatin remodeling
Transcription
RNA processing
Translation
Posttranslational modification
Question was directly from Figure 18.1
12 pts.
2. Based on the genotypes provided, indicate the level of transcription of the lacZ gene for each culture
condition listed using the legend below.
Legend:
+ = high transcription
- = low or very, very low transcription
Normal E. coli
lacI mutant
lacO mutant
cap mutant
lacI mutant /pLacI+
lacO mutant /pLacO+
cap mutant/pCAP
lacO & lacZ mutant/ pLacO+ lacZ+
Glycerol only
(no lactose or glucose)
+
+
+
+ (or -)
lacZ transcription level in media with:
Lactose only
Glucose only
Lactose and Glucose
+
+
+
+
+
+
+
Question was based upon in class activity of 10/27
2
-
-
12 pts.
3. The tol gene in bacteria encodes a protein that breaks down toluene. When toluene is present, the tol
gene is transcribed. When toluene is absent, the tol gene is NOT transcribed. Mutants have been
isolated that change the normal regulation of tol transcription. A mutation in a gene called regA gene
(no functional RegA protein) results in a bacterium that transcribes the tol gene in the presence AND
in the absence of toluene. A mutation in a gene called regB (no functional RegB protein) results in a
bacterium that NEVER transcribes the tol gene under any condition.
Is the RegA protein likely to be a positive or negative regulator of tol gene expression? Negative
Is the RegB protein likely to be a positive or negative regulator of tol gene expression? Positive
Is toluene an inducer or a repressor of tol gene expression? inducer
Propose a model for how toluene might affect regulation of tol gene expression.
When toluene is absent, RegA repressor/negative regulator protein can bind to the tol promoter and
prevent transcription. When toluene is present, it binds the RegA repressor/negative regulator
protein and inactivates it so that RegA can no longer bind to the tol promoter and prevent
transcription.
Question was based upon ideas in the problem of the day from 10/30/06, Group discussion problem
1-2 from Ch 17, and old exam question.
12 pts.
4. Briefly describe three ways in which the regulation of prokaryotic and eukaryotic gene expression is
similar.
A variety of answers were possible. Question was based upon in class activity of 11/03, reading in
Ch 17-Ch 18, and Ch 18 conceptual review question 1.
Briefly describe three ways in which the regulation of prokaryotic and eukaryotic gene expression is
different.
A variety of answers were possible. Question was based upon page 394-5 of the textbook, in class
activity of 11/03, reading in Ch 17-Ch 18, and Ch 18 conceptual review question 1.
3
12 pts.
5. The entire genome of the Richmond spider (a species of spider) has been sequenced. In the
Richmond spider genome, there is a gene that is similar to a web spinning gene (called the web gene)
in other spider species. You want to clone the web gene from Richmond spiders. Since the genome
has been sequenced, you have the DNA sequence of the Richmond spider web gene and flanking
DNA sequence. List the steps you would use to clone the Richmond spider web gene so that it can
be expressed in bacteria. Your starting material is one Richmond spider.
Method #1: Isolated spider cDNA, PCR just web gene, and clone into plasmid.
1. Using reverse transcriptase, make cDNA from mRNA isolated from Richmond spider cells that spin
web.
2. Use PCR to make copies of just the cDNA that corresponds to the web gene (you have the DNA
sequence and so you can use primers that just copy the web cDNA
3. Isolate plasmid DNA for vector from bacteria.
4. Digest amplified cDNA corresponding to the web gene and plasmid DNA with the same restriction
enzymes.
5. Ligate the sticky ends of the restriction enzyme digested cDNA from spider cells and plasmid DNA
together using DNA ligase.
6. Transform the recombinant DNA molecules into E. coli.
7. Plate the transformed DNA onto agar plates containing the antibiotic to which the plasmid confers
resistance.
Method #2: Construct a plasmid library containing spider cDNA and screen the library for the Richmond
spider web gene with the cloned gene from other spiders.
1. Using reverse transcriptase, make cDNA from mRNA isolated from Richmond spider cells that spin
web.
2. Isolate plasmid DNA for vector from bacteria.
3. Digest cDNA and plasmid DNA with the same restriction enzymes.
4. Ligate the sticky ends of the restriction enzyme digested cDNA from spider cells and plasmid DNA
together using DNA ligase.
5. Transform the recombinant DNA molecules into E. coli.
6. Plate the transformed DNA onto agar plates containing the antibiotic to which the plasmid confers
resistance.
7. Transfer the colonies to filter paper.
8. Incubate the filter with a radioactive web gene from another spider species.
9. Detect the radioactive spot using autoradiography (exposing a piece of X-ray film).
10. The colony that corresponds to the radioactive spot is the one with the cloned Richmond spider web
gene.
Question was based upon Fig 19.1, Ch 19 conceptual review questions 1 and 3, and what you have been
doing in lab for the past 4 weeks.
4
9 pts.
6. For each organism listed below, indicate how you can get DNA into the cells of the organism
Bacteria: transformation
Plants: use Agrobacterium to deliver DNA on an artificially engineered Ti plasmid
Humans: use retroviruses to deliver an artificially engineered viral genome
Question was based upon Ch 19 conceptual review question 2
List two reasons why might you want to put a particular piece of DNA in plants?
Multiple possible answers
List one reason why might you want to put a particular piece of DNA in human cells?
Gene therapy: To correct a genetic disease by introducing a normal allele of the defective gene to
cells of an affected individual.
5
11 pts.
7. Below is a DNA fingerprint of five people made using four VNTR loci. The DNA fingerprints were
generated in a paternity case which seeks to determine which potential male is the father of the child.
child
*
*
(a) Which (if any) males can be eliminated as being the father of the child?
Male 1 and Male 3 (rationale: Consider largest band (a particular VNTR allele) in child’s lane on gel.
This allele could NOT have come from Mom since she does not carry it; thus, the allele MUST have
come from dad. Male 1 and 3 do NOT have this allele and thus can not be the father. There are several
other VNTR alleles that fall under this criteria on the gel).
(b) Which (if any) males can not be eliminated as being the father of the child. In other words, which (if
any) males could be the father of the child?
Male 2 (rationale: Consider largest band (a particular VNTR allele) in child’s lane on gel. This allele
could NOT have come from Mom since she does not carry it; thus, the allele MUST have come from
dad. There are several other VNTR alleles that fall under this criteria on the gel and Male 2 has each of
the alleles.
(c) The wells of the gel where the sample is loaded is indicated with the arrow on the gel above. Put a
star next to the DNA fragments that are the smallest.
See above
Question was based upon the problem of the day from 11/15, explanation in class, and old exam
question.
6
12 pts.
8. You want to perform an experiment to examine the effect of Pepsi-Cola on gene expression in
intestinal cells. In your experiment, you treat one set of intestinal cells with Pepsi-Cola and leave the
other set of intestinal cells untreated (control). mRNA from the intestinal cells treated with PepsiCola is converted into cDNA and labeled with red fluorescent nucleotides; mRNA from the intestinal
cells that were not treated is converted into cDNA and labeled with green fluorescent nucleotides.
The cDNAs are mixed together and hybridized to a microarray/ gene chip containing all the genes in
the human genome (22,000), where each spot is a difference gene sequence. When you analyze the
array, you observe the following results:
11,000 spots are black on the chip.
10,990 spots on the chip are yellow.
7 spots on the chip are green.
3 spots on the chip are red.
a. Which color spot correspond to genes whose expression is induced by exposure to Pepsi?
Red
b. Which color spot correspond to genes whose expression is repressed by exposure to Pepsi?
Green
c. Which color spot correspond to genes that are not expressed in either set of cells?
Black
d. Which color spot correspond to genes that are expressed in both sets of cells?
Yellow
e. Briefly explain what might you use the information from the microarray experiment for?
This question was used as a bonus (+1 or +2 added to final score). Various answers were accepted.
Question was based upon ideas in yeast microarray animation from class on 11/15, the problem of
the day from 11/17 and in class activity from 11/17.
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Multiple choice section: (30 points total – 6 points per question)
Write your answer in the blank provided to the left. If you want to explain your answer, you can do so next to
the question.
D but
Accepted E
because one
could argue
that there
would be a
tRNA in P
site without
aa attached
E
B
1. Spiderdyne, a newly developed antibiotic, inhibits translocation of the prokaryotic ribosome. If spiderdyne
were added to an actively translating ribosome, one would observe a stalled (non-moving) ribosome with:
a. two tRNAs, one carrying a long polypeptide chain, another carrying a single amino acid.
b. two tRNAs, neither carrying any amino acids.
Figure 16.5 –
c. two tRNAs, each carrying polypeptide chain of equal lengths
step 3 never
d. one tRNA in the A site and holding a polypeptide chain.
occurs
e. None of the above.
2. What does it mean to say that a genetic marker (for example, an RFLP) and a disease gene are closely linked?
a. The marker lies within the coding region for the disease gene.
Page 411-412
b. The sequence of the marker and the sequence of the disease gene are extremely similar.
c. The marker and the disease gene are on different chromosomes.
d. The marker and the disease gene are on the same chromosome.
e. The marker and the disease gene are in close physical proximity on the same chromosome and tend to be
inherited together.
3. Fundamentally, what makes one cell different from another in a multicellular eukaryote?
a. The different cells contain different sets of enhancers and promoter-proximal elements.
b. The different cells contain different sets of regulatory proteins.
(Similar to problem of the day
c. The different cells contain different sets of cell-type-specific genes.
from 11/3 and Ch 18 content
d. Answers A and B apply.
review question 4)
e. All of the above answers apply.
B but
4. Plasmids are used as vectors in both plant and bacterial genetic engineering. However, there is a major
difference in the fate of genes introduced into bacteria on most bacterial plasmids and into plants on Ti
accepted any
answer since B
plasmids. What is this difference?
could be argued
a. Gene expression tends decrease rapidly and unpredictably in bacteria; gene expression is much more
as wrong too
stable in plants.
since entire
plasmid does
b. Bacterial plasmids and the genes they carry usually are not integrated into the chromosome; Ti plasmids
not go into
and the genes they carry are integrated into the chromosome.
plant cell
c. Bacterial plasmids are circular DNAs; Ti plasmid DNA is linear.
d. In bacteria, genes are stably expressed; in plants, gene expression is quickly lost.
e. Bacterial plasmids typically do not modify the growth of the host cells; Ti plasmids modified for genetic
engineering produce plant galls.
A
5. You want to express the gene wings, which is only expressed in cells of the back of the fly, in cells from the
back of a spider (to see if we can make flying spiders!). To do this, you need high levels of gene expression;
thus, you would: (Similar to problem of the day from 11/3)
a. attach an enhancer and a basal transcription factor binding site (promoter) from a gene expressed in spider
back cells onto the wing gene and then introducing this artificially engineered gene into a culture of spider
back cells
b. attach an enhancer and a basal transcription factor binding site (promoter) from a gene expressed in fly
back cells onto the wing gene and then introducing this artificially engineered gene into a culture of spider
back cells
c. attach a basal transcription factor binding site (promoter) from a gene expressed in spider back cells onto
the wing gene and then introducing this artificially engineered gene into a culture of spider back cells
d. attach a basal transcription factor binding site (promoter) from a gene expressed in fly back cells onto the
wing gene and then introducing this artificially engineered gene into a culture of spider back cells
e. none of the above scenarios will work.
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