Download 2009 exam 3

C2005/F2401 ’10
Review Questions for Exam #3 – Slightly Modified Version of Exam #3 of ‘09
1. Researchers have recently analyzed the DNA from a famous person (FP) who is deceased. He is long dead, but his
remains have been located, and his DNA examined. This person died of other causes, but there is a strong suspicion that
he had a genetic disease.
A. First of all, the researchers examined FP’s DNA looking for missense mutations. This means they probably examined
the DNA of (introns) (exons) (both) (either one) (neither). Explain very briefly.
B. The researchers reported that they found ‘no non-synonymous missense mutations’ in any of the relevant genes. This
implies they did find some synonymous mutations.
B-1. Synonymous mutations were probably ignored because they are expected to change
(the genotype only) (the phenotype only) (both) (neither) (beats me).
B-2. Which of the following is/are ‘non-synonymous missense mutations? (See code on last page.) Change in mRNA
could be (CUG to CUA) (AUG to AUA) (CGA to AGA) (UAC to UAA) (none of these).
B-3. Which of the following could they have found, in other words, which ARE synonymous? Change in mRNA could be
(CUG to CUA) (AUG to AUA) (CGA to AGA) (UAC to UAA) (none of these). Explain briefly.
C. Suppose one of the 4 mutations listed above occurs in the middle of a coding region (not at the start), and is the cause
of FP’s disease (whether this fits with the statement in B above or not). Which of the changes listed is most likely to be
the one that is the cause of the disease? _________________________________________________ . Explain how you
know.
2. Suppose a ribosome is translating normal mRNA from a eukaryotic gene. The second tRNA (#2) has just moved into
the P site of the ribosome. Assume codons two to four are not codons for methionine.
A. The initiator tRNA could be in (the P site) (the A site) (the E site) (A or P) (A or E) (E or P) (any of these).
B. Methionine should be attached directly to (tRNA #1) (AA #2 = amino acid #2) (tRNA #2) (AA #3)
(peptidyl transferase) (either tRNA) (tRNA or AA #2) (either AA) (none of these) (any of these).
C. The ribosomal site closest to the 5’ end of the mRNA should be (A) (E) (P) (can’t predict). No explanation required for
A to C. If you think there is any ambiguity, you can explain on back.
D. Suppose the ribosome is translating mRNA from a mutant version of the same gene. Codon #2 is changed, but
translation produces a normal peptide.
D-1. Assume you have all possible tRNAs. In that case, peptide could be normal if change in mRNA is (AAG to AAC)
(AGU to UCU) (ACU to ACC) (ACU to ACG) (none of these). Circle all correct answers.
D-2. Suppose the same tRNA #2 is in the P site as in parts A to C. In that case, peptide could be normal if change in
mRNA is (AAG to AAC) (AGU to UCU) (ACU to ACC) (ACU to ACG) (none of these). Circle all case(s) that would
produce a normal peptide, using the same tRNA.
Explain briefly how you ruled each case in or out. (Code and wobble rules were provided; look them up.)
C2005/F2401 ’10 --- Review Questions for Exam #3
3. This problem is about the production of a toxin by the bacterium S. toxis. On the next to last page there is a description
of the structure of the DNA region containing the genes involved (for part A) and the results of some genetic experiments
(for the remaining parts).
A. What is the simplest interpretation of the (structural) results described on the next to last page?
A-1. Genes 4 & 5 are structural genes (in the same operon) (in different operons) (either way).
A-2. Suppose you isolate a single mRNA molecule made from this region of the DNA. This mRNA could include
information to make (all 6 enzymes) (only one of the 6 enzymes) (enzymes 3 & 4) (enzymes 4 & 5) (can’t predict).
Explain both answers briefly.
B. Consider a del 2 strain (deletion 2 on the chromosome) that you have transformed with the plasmid described on the
next to last page.
B-1. Suppose you isolate a single mRNA molecule made from this strain. Could this mRNA contain all the
information needed to make the toxin? (yes) (no) (can’t decide).
B-2. Production of the toxin in this strain should require transcription of (only plasmid DNA)
(only chromosomal DNA) (both DNAs) (recombinant DNA) (none of these) (beats me).
Explain both parts.
B-3. The transformed cells would NOT make any toxin if the plasmid contained a deletion of (gene 1) (gene 2)
(gene 3) (gene 4) (gene 5) (gene 6) (P2) (none of these – cells would make some toxin no matter what).
B-4. These cells would make LOW levels of toxin (<10% of normal) if the plasmid contained a deletion of (gene 1)
(gene 2) (gene 3) (gene 4) (gene 5) (gene 6) (P1) (P2) (none of these – cells would make high levels of toxin no matter
what). For B-3 & B-4, circle all correct answers and explain both parts briefly.
C. The plasmid used in these experiments was made by genetic engineering. Suppose bacterial DNA was cut up with the
restriction enzyme Bit1. A single piece of DNA containing the entire sequence from P1 to T2 (see next to last page) was
isolated from the digest. Also suppose the original plasmid used as a cloning vector had one site each for Bit1 and Bit2.
The two enzymes produce different sticky ends. The vector also has one gene for resistance to the drug bubimycin, and
the site for Bit1 is in that gene.
C-1. To make the final plasmid, the genetic engineers probably cut up the vector with (Bit1) (Bit2) (either one) (both).
C-2. Now suppose you have the final recombinant plasmid, and you use it transform a del 5-6 strain. Which of the
following will allow you to detect transformed cells? (growth on medium containing toxin) (ability to produce toxin)
(growth on medium without toxin) (ability to break down toxin) (growth on medium containing bubimycin)
(replica plating to medium with bubimycin) (none of these). Circle all that apply.
C-3. Suppose you transformed a del 2 strain instead (with the same plasmid). Would you answer be the same?
(yes) (no) (maybe) (beats me). Explain your answers briefly.
C2005/F2401 ’10 --- Review Questions for Exam #3
4. This problem is about the regulation of anthrose production. Anthrose is an unusual sugar that is found only in the
bacterium that causes anthrax (B. anthracis), and a few closely related other bacteria that are equally dangerous. Many
techniques for detecting the anthrax bacillus depend on detecting anthrose.
When B. anthracis cells are actively growing, the cells contain no anthrose. When the bacteria stop growing and form
spores, they produce anthrose. The spores are dormant forms of the bacteria encased in a resistant coat. (These are
prokaryotic spores, not the same as spores formed by eukaryotes.) Only sporulating cells produce anthrose. Consider the
anthrose operon – the operon that codes for the enzymes needed to make anthrose. (You have to figure out below if it is
inducible or repressible.)
A. Suppose you compare sporulating bacteria and actively growing bacteria. Which of the following would you expect to
find in sporulating bacteria, but NOT in actively growing ones? (mRNA from the structural genes of the operon)
(mRNA from the repressor gene for the operon) (enzymes needed to synthesize anthrose) (enzymes needed to degrade
anthrose) (genes for the enzymes needed to produce anthrose) (none of these). Circle all reasonable possibilities and
explain briefly.
B. Suppose you compare sporulating bacteria, and actively growing bacteria. This time you look at a different set of
things.
B-1. Which of the following would you expect to find in one state, but NOT in the other? (repressor) (co-repressor)
(inducer) (RNA polymerase) (promoter of operon) (repressor gene of operon) (operator of operon) (promoter of repressor
gene) (operator of repressor gene). Circle all reasonable possibilities.
B-2. Suppose at least one of the items on the list in B-1 is found in sporulating bacteria, but not in actively growing
bacteria. In that case, the operon is probably (inducible) (repressible) (constitutive) (inducible or repressible) (any of these
-- can’t tell from information given). Explain your answers.
C. Cells with extra copies of the operon make very high levels of the enzymes of anthrose synthesis, but they make
normal amounts of anthrose. Given all the information so far, what is the role of anthrose in this system? Anthrose is
probably an (inducer) (co-repressor) (feedback inhibitor of anthrose synthesis) (activator of anthrose synthesis) (none of
these) (can’t predict). Circle one or more choices and explain.
D. You discover a mutant strain of B. anthracis. The mutant bacteria make anthrose even when they are actively growing.
All other aspects of bacterial function are normal. The mutation turns out to be a deletion of about 12 base pairs. Which of
the following is a reasonable site for the mutation? The (promoter of the operon) (operator of the operon) (gene for RNA
polymerase) (promoter of the repressor gene of the operon) (one of the structural genes of the operon) (operator of the
repressor gene). Circle all reasonable possibilities and explain.
C2005/F2401 ’10 --- Review Questions for Exam #3
5. Hemophilia B is caused by mutations in the gene (F9) that codes for factor 9 (required for blood clotting). In some
cases, the mutation is in an intron and alters splicing – it changes the position of the splice point. As a consequence of the
mutation, the mRNA contains 2 nucleotides that are normally part of an intron. All the questions on this page are about
this mutation.
A. You want to get DNA coding for factor 9, insert it into a plasmid, and use the plasmid for gene therapy or to make lots
of factor 9. You have a choice of 5 DNAs (All details spelled out on the next to last page.)
(1 -- normal F9 gene) (2 -- normal cDNA) (3 – hemoph. cDNA) (4 – hemoph. F9 gene) (5 – intronless hemoph. F9 gene)
A-1. If you want bacteria to make normal factor 9, which of the following DNAs could you insert in your plasmid?
(1) (2) (3) (4) (none of these).
A-2. If you want to use the plasmid for gene therapy in patients with hemo. B, which of the following DNAs could you
insert? (1) (2) (3) (4) (none of these).
Explain your choices.
A-3. Suppose you insert DNA 5 in your plasmid, and use the recombinant plasmid for gene therapy. Will the person be
okay? Explain why or why not.
B. The mutation described here is a single base change from A to G. The mutation is in intron 3, 2 bases before the 3’ end
of the intron. This creates a new splice site 2 bases away from the normal intron/exon boundary at the 3’ end of the intron.
B-1. The primary transcript in the mutant should be (longer than) (shorter than) (the same length as) normal.
B-2. The lariat released upon removal of intron 3 in the mutant should be (longer than) (shorter than) (the same length as)
normal. Explain briefly.
C. Suppose translation starts in the middle of exon 2, which is 200 bases long. When the ribosome translates the mutant
mRNA, which of the following should be different, or in a different position?
C-1.The (start codon) (a stop codon) (both) (one or the other) (neither) should be different from normal.
C-2. The part of the peptide corresponding to (exon 3) (exon 4) (one or the other) (both) (neither) should be different from
normal. Explain briefly.
D. Suppose there are restriction sites for EcoR1 in exons 2 & 4 of the gene for factor 9; there are 2 sites in exon 2 and one
site in exon 4. (No other EcoR1 sites in the gene.) You cut up both normal and mutant genomic DNA with EcoR1 and do
a Southern blot. (Separate DNA by gel electrophoresis, blot to nitrocellulose, hybridize blot to your probe.) Your labeled
probe consists of a full length cDNA from a normal person.
D-1. The maximum number of labeled bands on your blot with normal DNA will be _______________.
D-2. The maximum number of labeled bands with mutant DNA will be (higher than) (lower than) (the same number as)
normal.
D-3. On the mutant blot, what is the maximum number of labeled bands that could be in a different position from normal?
(0) (1) (>1, but not all) (all) (can’t predict). Assume that differences in length as small as a single base can be detected.
Explain your prediction.
Information for problem 3
This problem is about the production of a toxin by the bacterium Streptococcus toxis (S. toxis). Assume the toxin has no
effect on S. toxis but kills other organisms. Six linked genes have been identified that might code for the enzymes needed
to make the toxin. Below is a description of the structure of the DNA containing the genes (for part A) and the results of
some genetic experiments (for the remaining parts).
Structural Information
Researchers identified 6 genes near each other on the bacterial chromosome. These are collectively called genes 1-6 in
order of location (going 5’ to 3’ on the sense strand). All these genes code for enzymes that might be involved in toxin
synthesis, and all 6 genes are transcribed in the same direction (relative to the start of gene 1). The region of the DNA
containing all 6 genes was sequenced, and compared to known sequences. The sense strand of the DNA looks like this:
5’… P1 P2
Gene 1 Gene 2
Gene 3
Gene 4 T1
P3 Gene 5
Gene 6
T2…. 3’
P1, P2 & P3 are sequences similar to those found in other promoters.
T1 & T2 are sequences similar to those found in other transcriptional terminators.
Genetic Information
You can make strains with various deletions in this region. You can also make a plasmid carrying a normal copy of this
DNA region.
The mutations are called del 1, del 2, del 3, del 4, del 5-6, or del P, and have deletions of genes 1, 2, 3, 4, both 5 & 6, or
P2 respectively.
The plasmid has a normal copy of the entire DNA region shown above.
You grow each strain and measure the level of toxin. A table summarizing the results is shown below. If you transform
the cells with the plasmid described above, all mutant cells that take up the plasmid make high levels of toxin (> 100%).
Strain What’s deleted
Del 1
gene 1
Del 2
gene 2
Del 3
gene 3
Del 4
gene 4
Del 5-6
genes 5 & 6
Del P
P2
Toxin levels
0
0
0
0
100%
4%
WT
100%
nothing
Information for problem 5:
Here are the choices of DNA.
1 – normal F9 gene -- F9 gene from normal person
2 – normal cDNA -- cDNA from normal person
3 -- hemophilia cDNA – cDNA from a person with hemophilia B
4 – hemophilia F9 gene – F9 gene from a person with hemophilia B
5 – intronless hemophilia F9 gene – F9 gene from a person, with hemophilia B, introns (of normal length) removed by
genetic engineering.
All people with hemophilia B have the splicing mutation described in problem 5.
You can insert any of these DNA’s into a plasmid with the right origins, promoters etc. and get proper transcription in
either bacteria or eukaryotes, as needed.