Download 2008 exam 3 answers

C2005/F2401 ’09
Answers to Review Questions for Exam #3 – Key to Exam #3 of ‘08
Most answers and explanations were worth 2 points each.
1. A. 5. 1 for gln, 1 for trp, 2 for gly, 1 for cys. The code is highly degenerate, but because of wobble, you only need 1
tRNA to read CAA/G (gln), 2 for gly (GGX), and 1 for cys (UGU/C).
B. 16. You need to consider all possible codons for each amino acid. There are 2 for gln, 1 for trp, 4 for gly, & 2 for
cys. So the number of possible different coding sequences is 2 X 1 X 4 X 2 = 16. You need this many different
oligonucleotides – one will match the DNA, but you don’t know in advance which one it will be.
C-1. U. (See the wobble rules).
C-2. Gln yes, the others no. In your explanation, 1 pt each was awarded for explaining cases (a) & (c).
Gln is the only sure bet. The tRNA with the anticodon (3’) GUU (5’) can read both gln codons.
(a) You can’t use a tRNA with U in the wobble position to read the codon for trp (UGG) because that tRNA would
also read the stop codon UGA.
(b) You can’t use a tRNA with U in the wobble position to ready cys, because the 2 codons for cys have U & C, not A
& G, in the wobble position.
(c) You can use a tRNA with U in the wobble position to read 2 of the 4 gly codons (GGA/G), but that tRNA cannot
read the other 2 gly codons (GGU/C). So if you use the tRNA with the anticodon (3’) CCU (5’) it might work for gly, but
it may not – it will depend on what codon is present in the mRNA.
D-1. Either one. The mRNA will be either the same length (patient B) or one base shorter (patient A). Since the
mRNA is quite long, this difference is insignificant. Frameshifts usually generate nonsense codons (premature stop
codons), but these have a direct effect on translation, not on transcription. The protein made in patient A will probably be
significantly shorter, but the mRNA will be the same length.
D-2. >4 Amino acids. The mutation will cause a frame shift, and all amino acids after the codon with the insertion will
be incorrect or missing.
D-3. 0, 1, or >4 amino acids. Amino acid #122 is cys. A substitution of one base in the codon for cys can generate a
stop codon (UGU/C → UGA), a substitution (to a codon for tyr, ser, phe, gly or arg), or another codon for cys (UGU ↔
UGC). To get full credit for the explanation, you had to consider the actual codon involved, not just the general case.
E. DNA from patient A will definitely hybridize. DNA from patient B may or may not. (For this problem, credit
awarded depended entirely on your explanation; your circled answer was ignored. Explanation = 4 pts; 2 pts for each
patient.)
Why will the probe hybridize to the DNA from patient A? An insertion causes a shift in the reading frame at
translation, leading to a major change in the protein. However it has no effect on the ability of the unchanged parts of the
DNA to hybridize. The part of the DNA which hybridizes to the probe is unchanged. Whichever oligonucleotide in the
collection hybridized to the normal DNA will hybridize to the DNA of patient A. (We know the probe works – that is,
hybridizes to normal DNA – because you can use the probe to isolate the normal human gene.)
The answer for patient B depends on whether you are using the entire mixture of 16 probes, or just the one probe that
is complementary to the normal DNA sequence. If you are using just the one probe, it will not hybridize to the DNA of
patient B. Why not? A base substitution will alter the match between the correct strand of the DNA (here the template
strand) and the oligonucleotide. Base pairing will not be perfect, so the hybrid will not be stable.
If you are using the entire mix of oligonucleotides, the result will depend on the nature of the change in codon 122. If
the base in position 3 of the codon (the wobble position) is changed from C to U, or the reverse, then one of the probes in
the mix will hybridize to the DNA of patient B, but it won’t be the same probe that hybridized to the normal DNA. If any
other base in the codon is changed, or the base in position 3 is changed to A or G, then none of the probes in the mix will
hybridize to the DNA from patient B.
Note: It is possible to make probes that will not hybridize if there is a mismatch of one base pair only. However the real
probes are generally a bit longer than the ones in this problem.
2. A. The probe would hybridize to the sense strand and to the primary transcript. You started with mRNA and made its
complement, the cDNA, with reverse transcriptase. Therefore the cDNA is not the same as the mRNA, primary transcript
or sense strand, which are all equivalent. Instead the cDNA is complementary to all of them. The cDNA is the same as the
template strand.
B-1. Two bands. The joints between the cloned fragment and the plasmid will be cut by PstI, giving you back one
piece = the plasmid and one piece = cloned fragment. The joints were made by joining the sticky ends of the plasmid
generated by PstI to the sticky ends of the fragment generated by Pst I. When you join complementary sticky ends, they
form a restriction site. (Since no other restriction enzymes are mentioned, it is assumed you followed standard procedure
and used the same enzyme to cut the plasmid and the human DNA.)
B-2. One radioactive band. The cDNA probe will hybridize to the cloned fragment, but not to the plasmid (without an
insert). You know the fragment contains a sequence that will hybridize to the cDNA probe because the colony containing
the recombinant plasmid hybridized to the probe.
C. 1400 BP. You got one band of DNA. The simplest explanation is that you cut the recombinant plasmid only once
or not at all. Therefore you have only one piece, and it will hybridize to the probe. This piece is the same length as the
‘isolated plasmid DNA’ = length of recombinant plasmid = plasmid plus insert = 1400 BP. It is either linear or circular,
depending on whether it was cut once, or not at all. (See note at *.)
A less likely, but possible explanation for the result (one band) is that the DNA was cut twice, forming two bands of
exactly the same length. In this case, you would get two pieces of DNA, but they would be found in the same position, at
the spot corresponding to 700 BP. The insert could be entirely on one half or part on each half, but the result would be the
same – one radioactive band.
The radioactive band will be at the position of 1400 BP (or 700 BP), not 1500 BP, because the position in the gel (&
on the blot) depends on the length of the DNA you ran on the gel, not on the length of the probe added later.
*Note: Circular DNA and linear DNA of the same length don’t run the same way on gels, and are easily distinguished. So
in real life you would know whether the plasmid had been cut once or not at all. However for this problem, we have
assumed that both linear and circular full length DNA’s would form a band in the same position as the 1400 BP fragment.
D-1. Part of at least one exon. An exonic sequence is required for the DNA in the colony to hybridize to the probe. If
the cloned DNA contained only intronic sequences, it wouldn’t hybridize to the probe. You don’t need a whole exon in
each clone – just enough to hybridize to the cDNA.
D-2. Can’t tell. There is no information here about the total number of introns or exons. You do know the gene
contains at least one intron. If the gene had no introns at all, the total length of the gene would = the sum of all the cloned
fragment lengths.
D-3. At least two sites in one intron. The probe did not detect all the colonies that contained fragments from your
gene. It detected only colonies that had a plasmid with an exonic sequence that was long enough to hybridize to the probe.
The probe didn’t detect colonies that contained plasmids with inserts derived entirely from intronic DNA. The total of the
7 fragments is too short, because some cloned fragments of the gene where entirely from introns and didn’t hybridize to
the probe. The only way to get a fragment that is entirely derived from an intron is to cut the DNA twice within an intron.
There is an alternative way to generate a fragment that won’t hybridize to the probe. It can happen if one or both of
the cuts are in the surrounding exon, very close to the exon/intron border. In that case, the cloned fragment would be
mostly from an intron, and would not have enough exonic sequences to hybridize to the cDNA probe.
3. A-1. mRNA should increase; A-2. Binding of repressor should decrease.
B-1. mRNA in culture should decrease – no more is being made and bacterial mRNA usually has a short ½ life -- the
old mRNA should be degraded.
B-2. Level of peptide B in culture should stay the same. No more is being made, but enzymes are usually stable. As
cells grow, level of peptide B per cell will decrease, but the total amount should sts.
C-1. Shorter chains of peptide C. The ribosome reads the mRNA from 5’ to 3’. The ribosome closest to the 5’ end of
the coding sequence for peptide C has the shortest chain – it has not gone as far down the mRNA. Each succeeding
ribosome in the peptide C section (that is, closer to the 3’ end) has already translated more of the mRNA for peptide C and
should have a longer chain.
C-2 Peptide A or B. There is a single mRNA for the whole operon (a polycistronic mRNA). Ribosomes on that
mRNA could be making peptides A, B, C, or D. The order of genes in the operon is ABCD. Therefore, going 5’ to 3’ on
the mRNA, the coding sequences for the peptides should be in the same order. A ribosome reading 5’ to 3’ could get to
the coding sequences for A & B before it gets to C. But it can only get to the sequences coding for D after it finished
making peptide C.
Note: Directions are always given relative to the order on the sense strand = order on the mRNA. So if it says the order of
genes is ABCD, it means the genes are in the order A-D, reading along the sense strand. That means they are transcribed
and translated A to D, not D to A. (Only 1 pt was deducted for getting it backwards.)
D-1. Low; D-2 stay the same. The production of the fusion protein is under the control of the regulatory elements of
the spv operon. The presence or absence of lactose makes no difference. There is no lac operator controlling the regulation
of the lacZ gene. So there should be no response to the lactose, even if the lac repressor is present in the cell. When the
spv operon is induced, the fusion protein will be made and β galactosidase activity per cell will be high. That doesn’t
happen until log phase.
E. Not virulent. The peptide B gene has been disrupted by the insertion of the lacZ gene in the middle. It is unlikely
that the fusion protein will carry out the function usually performed by peptide B. (If the lacZ gene were inserted near one
end of the B gene, the B peptide might work. But the problem says the lacZ gene is inserted in the middle.)
4. A-1 & A-2. RPX is acting like a repressor; its effector is acting like a co-repressor. If RPX is missing, too much
enzyme is made. Therefore the normal job of RPX is to reduce transcription and enzyme production. This is what a
repressor does. When the effector is missing, enzyme levels are also too high, indicating too much enzyme production.
Therefore the job of the effector is to act as a co-repressor and allow RPX to bind to the DNA and reduce transcription. If
the effector were acting as an inducer, and it were missing, then there should be unusually low enzyme levels, not
unusually high levels.
Note: This operon must have more than one repressor-like protein – or one protein that responds differently to several
effectors.
A-3. Higher than normal. The spv operon produces a polycistronic message that codes for all 4 peptides. If you have
an unusually high level of one peptide, you’ll have unusually high levels of the others as well.
B. Higher than normal. If there is no effector (co-repressor), then RPX (acting like a repressor protein) can’t bind to
the DNA and slow down transcription. In this case, lacking either regulator protein or its effector should have the same
effect. (If an effector is an inducer, lacking the regulator protein and lacking the inducer will have opposite effects.)
C-1. Transformation (adding DNA = transformation).
C-2. (a) Both. (1 pt for each.) You need each gene to make its product. Therefore each gene must be transcribed, and
each plasmid must have a promoter. You need each plasmid to replicate so that all the descendants of the transformed
cells will get copies. You have to grow the cells up to stationary phase, and get many descendants with the plasmid, in
order to detect high levels of peptide A. Therefore you need an origin of replication.
(b) Both. (Only answer that received pts). To restore proper regulation you need both the regulatory protein and
its effector. Therefore you need both plasmids in each cell. One plasmid to code for RPX and one for the enzyme to make
the effector. The only way to ensure this is to select for cells that received both plasmids. That requires the addition of
both drugs.
C-3. Complementation. This set up is equivalent to complementation. There are 3 pieces of DNA – the bacterial
chromosome and two plasmids. Each piece of genetic material codes for proteins the others can’t make, and all the
proteins are needed to restore normal regulation. Recombination does not affect the result here – there are no mutant
pieces of DNA that have to be cut and rejoined to make a normal gene or restore function. All the genes needed are
already in good shape, but they have to be put into the same cell to supplement (complement) each other and restore
function.