Download 2008 exam 3

C2005/F2401 ’09
Review Questions for Exam #3 = Exam #3 of ‘08
A copy of the genetic code, and the wobble rules, were provided. REMEMBER YOU SHOULD BE ABLE TO
EXPLAIN EVERYTHING!!
1. You have a protein that includes the sequence gln-trp-gly-cys. The protein has only one peptide chain, 432 amino acids
long, and the sequence shown represents amino acids 119-122.
A. Suppose you are carrying out translation in a test tube. You are translating the mRNA for the entire protein, and the
ribosome has reached the section that codes for the sequence shown above. How many different tRNA’s must you have
present in your test tube to be certain you can translate this section of the mRNA?____________ ______
B. You wish to make an oligonucleotide probe for the section of the gene that codes for the sequence above. If you
make a collection of oligo probes for this section of the gene, how many different probes would you have to make to be
sure that one of the oligos in the collection will hybridize to the DNA? ___________ (These oligo probes are a bit short,
to keep part B manageable, but assume the probes are long enough to form a stable hybrid.)
C. Suppose you have a tRNA that recognizes either A or G in the wobble position.
C-1. For this tRNA, the base in the wobble position of the anticodon should be
(A) (I) (G) (C) (U) (T) (A or G) (A or I) (U or C) (U, C or A) (U, C, A or G) (can’t tell from info given).
C-2. You wish to use this tRNA in your test tube to add one of the amino acids shown above to the growing chain.
You want to be sure this is the only tRNA you need to read the codon for the respective amino acid. If the other bases in
the tRNA are correct, you could use this tRNA to add (gln) (trp) (gly) (cys) (none of these). Circle all correct answers and
explain your answers to C-2. (No explanation needed for C-1.)
D. You have 2 patients, A & B, with mutations in the gene for the protein under discussion. Patient A has an insertion
of 1 base pair in the codon for amino acid (AA) #115. Patient B has a substitution of 1 base in the codon for AA #122.
(The section shown at the top of the page consists of amino acids #119-#122). Assume that neither of these mutations
affects RNA splicing.
D-1. The mRNA is likely to be about the normal length if it comes from (A) (B) (either one) (neither) (can’t
predict).
D-2.The protein from A is likely to differ from normal by (0) (0 or 1) (1) (1 or 2) (2) (3) (4) (>4) (0, 1 or >4) AA.
D-3.The protein from B is likely to differ from normal by (0) (0 or 1) (1) (1 or 2) (2) (3) (4) (>4) (0, 1 or >4) AA.
Remember to consult the genetic code on the last page. Explain D-3 only.
E. You use the oligo probe(s) made in B to isolate the normal human gene for your protein. Then you attempt to
hybridize the same oligo probe(s) to the DNA from your 2 patients described in part D. You use sufficiently stringent
conditions – if the match is not perfect, the probe will not hybridize. The probe should hybridize to the DNA of
(patient A) (patient B) (both) (neither) (can’t predict).
2. Suppose that you can isolate mRNA for human protein Q. Protein Q contains one peptide chain. You make a probe for
gene Q, using the mRNA and reverse transcriptase. After you finish with the reverse transcriptase, you add an enzyme to
degrade the mRNA and purify the single stranded cDNA that remains.
A. Suppose you hybridize the cDNA to nucleic acids from humans. Your cDNA should hybridize to
(the sense strand of gene Q) (the template strand of gene Q) (the primary transcript from gene Q) (none of these). Circle
all correct answers and explain very briefly.
C2005/F2401 ’09
Review Questions for Exam #3 = Exam #3 of ‘08
Problem 2, cont.
You have a library of human DNA made by cutting up the DNA with PstI and cloning it in plasmid PU5. Using colony
hybridization, you find 7 colonies (clones) in your library that hybridize to the cDNA probe for gene Q. The rest of this
question is about the library.
B. You grow up the cells from one of the 7 colonies that hybridizes to your probe. You isolate the total DNA from the
cells you have grown up. You separate the plasmids from the rest of the DNA, and treat the isolated plasmid DNA with
PstI. You run a gel, and blot the DNA to nitrocellulose. You first examine the blot with a technique that visualizes all
DNA.
B-1. How many DNA bands will you see on the blot? (0) (1) (2) (3) (4) (5) (>5) (can’t predict).
B-2. Now you hybridize the blot to your cDNA probe, which is radioactive. How many radioactive bands should
you be able to detect? (0) (1) (2) (3) (4) (5) (>5) (can’t predict).
C. You do another experiment. You grow up the cells from a different colony that hybridized to your probe. You treat
the isolated plasmid DNA (from this one clone) with EcoRI instead of PstI. Then you run a gel, do a blot. If you visualize
all the DNA, you get one DNA band on your blot. Then you hybridize the blot to the radioactive cDNA probe. Suppose
the cDNA is 1500 base pairs (BP) long, PU5 (without an insert) is 1200 BP, and the isolated plasmid DNA is 1400 BP.
For comparison, you have run a gel and made a blot with DNA fragments of 200, 1200, 1400 and 1500 BP. When you
look at the hybridized blot, you should see radioactive band(s) that correspond to the position of fragments of length -- in
BP -- of (<200) (200) (1200) (1400) (1500) (none of these). Circle all correct answers.
D. You go back to your library, and you isolate the cloned fragments from the 7 colonies that hybridized to your
probe. All 7 sequences are different, and the 7 together add up to less than the length of the human gene. (Length of gene
= length of transcribed DNA). No other colonies hybridize, and you are sure your library includes all the DNA sequences
from the human genome. You can conclude that:
D-1. Each of the 7 fragments includes (one complete exon) (part of at least one exon) (no exons) (any of these)
(none of these) (can’t tell from info given).
D-2. The number of introns in gene Q is (0) (>0 but <6) (6) (7) (8) (>8) (can’t tell from info given).
D-3. There are two or more PstI1 sites in (at least one intron) (at least one exon) (the cDNA) (at least one intron
or exon) (at least one exon or the cDNA) (any of these) (none of these) (can’t tell from info given).
Explain how you got your answers.
3. The genes that cause some strains of Salmonella to be virulent are on a plasmid. The genes that are responsible for
virulence (ability to cause disease) are part of the spv operon. (SPV operon = Salmonella plasmid virulence operon). This
operon is induced when the bacteria enter stationary phase. (The exact trigger for induction is unknown.) Since these
bacteria cause serious disease, several groups are studying induction of the operon.
The normal structural genes of the spv operon are ABCD, in that order. The operon codes for polypeptides A, B, C & D.
A. In a bacterium with a normal spv operon, what should happen when the cells enter stationary phase?
A-1. In each cell, the level of mRNA from the structural genes of the spv operon should
(increase) (decrease) (sts = stay the same) (sts at zero).
A-2. Binding of the repressor of the spv operon to the operator should (increase) (decrease) (sts) (sts at zero).
(No explanation required for A.)
B. Now suppose some stationary phase cells are put into fresh medium, and start to grow. Then what should happen to
the level of mRNA and protein from the structural genes of the spv operon? In the entire culture,
B-1. The level of mRNA -- in the entire culture -- should (increase) (decrease) (sts) (sts at zero).
B-2. The level of peptide B -- in the entire culture -- should (increase) (decrease) (sts) (sts at zero).
C. Suppose a ribosome is part of a polyribosome, and is making peptide C.
C-1. At the same time, the ribosomes closer to the 5’ end of the mRNA could currently be making (longer chains
of peptide C) (shorter chains of peptide C) (chains of peptide C of the same length) (none of these) AND
C-2. Some of the ribosomes closer to the 5’ end of the mRNA could currently be making (peptide A) (peptide B)
(peptide D) (none of these). For each part, circle all correct answers.
C2005/F2401 ’09
Review Questions for Exam #3 = Exam #3 of ‘08
Problem 3, cont.
D. To measure regulation of the operon, lacZ (the structural gene for β galactosidase) has been inserted into the
middle of the spvB gene, using genetic engineering. The result is that the lacZ/spvB recombinant gene produces a
lacZ/spvB recombinant protein that has β galactosidase activity, which is easily measured. Assume that there is no other
copy of the lac Z gene in the bacteria, and the spv operon is normal except for the lacZ insertion..
D-1. When the bacteria carrying the recombinant gene are growing in log phase, without lactose, the amount of β
galactosidase measurable per cell should be (low) (high) (can’t predict).
D-2. Adding lactose to the medium should cause the level of β galactosidase measurable per cell to (increase)
(decrease) (stay the same) (can’t predict).
E. All 4 peptides, A-D, are needed for virulence. The genetically engineered bacteria described in part D would
probably be (virulent) (not virulent) (can’t predict).
4. This problem is about the spv operon, which is also the subject of problem 3. For this problem, all you need to know
about the operon is that
1. The spv operon is an inducible operon coding for peptides A, B, C & D.
2. The operon is induced when the bacteria reach stationary phase.
Cells contain multiple regulatory proteins that affect the spv operon. One of the regulatory proteins is called RPX. It is an
allosteric protein, and its effector is a small molecule. RPX and its effector have many different effects on cells. In
bacteria with the spv operon that have no RPX, or no effector, the amount of peptide A in stationary cells is much higher
than normal. (Lack of RPX, or its effector, has no effect on the amount of peptide A in log phase.)
A-1. In normal bacteria, RPX is acting most like a (repressor) (co-repressor) (inducer) (operator) (inducer or corepressor) of the spv operon.
A-2. In normal bacteria, the effector of RPX is acting most like a (repressor) (co-repressor) (inducer) (operator)
(inducer or co-repressor) of the spv operon.
A-3. In cells with no RPX, the levels of peptide D in stationary phase should be (normal) (higher than normal)
(lower than normal) (can’t predict).
B. There are cells that lack gene E, which is needed to synthesize the effector. In these cells the levels of peptide A in
stationary cells should be (normal) (higher than normal) (lower than normal) (can’t predict).
C. There are doubly deficient cells that contain the spv operon but that lack both the gene for RPX and gene E
(needed to synthesize the effector). In these cells the level of peptide A is abnormally high in stationary phase. You have
two different recombinant plasmids.
Plasmid #1 has the gene for RPX and a functional gene for ampicillin resistance.
Plasmid #2 has gene E and a functional gene for tetracycline resistance.
You add one or both recombinant plasmids to medium containing the doubly deficient cells. (You have to decide what
drugs – if any -- to put in the medium; see below.) You grow up the surviving cells to stationary phase. Then you check
out the levels of peptide A in the stationary phase cultures. You hope that in some cases the levels of peptide A will be
normal.
C-1. In this experiment, the doubly deficient cells acquire their recombinant plasmids by (transformation)
(transduction) (conjugation) (can’t tell from info given). No explanation required for this one.
C-2. In order for this to work to restore the levels of peptide A to normal,
(a) Each plasmid must contain (an origin of replication) (a promoter) (both) (neither) (can’t predict) AND
(b) You should add to the medium (ampicillin) (tetracycline) (either one) (both) (neither).
C-3. In this experiment, you are hoping to restore the levels of peptide A to normal by
(complementation) (recombination) (both) (neither) (either way). Explain your answers to C-2 & C-3.