Download C2005/F2401 Key to Exam #3

C2005/F2401 Key to Exam #3
All answers and explanations are 2 pts each unless it says otherwise. If it says ‘not required’ than no points were
given for explaining that part, and nothing was deducted for failing to explain it. The explanations that were ‘not
required’ are included below fyi. If you find any errors, please email Dr. M.
1.
A. Answers: A-1. an exon; A-2. Same as in normal.
Explanations:
A-1. (not required) Exons code for the entire sequence of mRNA, including the 5’UTR and 3’UTR. Not
all the sequences encoded by exons are translated.
A-2. Since the insert is before the start of translation, ribosomes and initiator tRNA will attach at the
normal spot and translate the mRNA normally. There will be no frameshift.
B. Answers (1 pt each): B-1. either one; B-2. One base different from normal.
Explanations.
B-1. The mutation could be in an intron, because the intron will be spliced out, and so the mutation will
have no effect on the sequence of the mRNA, and no effect on phenotype. (Most mutations in introns do not
affect splicing.)
The mutation could be in an exon, because of the degeneracy of the genetic code. If the mutation
changes a codon to a different codon for the same amino acid, such as AAA to AAG, then the protein will be the
same and the function the same.
Note that it says the mutation does not change the untranslated parts of the mRNA, so the mutation
cannot be in the 5’ or 3’ UTR.
B-2. (not required) It says at the top of the page, that each mutation in this problem involves only one
base pair. So that’s the maximum change possible. If the mutation is in an intron, the mRNA will be the same
as normal. If the mutation is a synonymous change in a codon, the mRNA will be one base different.
C. Answers (1 pt each): C-1. an exon; C-2 translation; C-3. Same as normal.
Explanations:
C-1 & C-3 (not required): RNA is made and spliced as usual. Both the primary transcript and the
mRNA are the usual length, but there is a single base change in the mRNA, in the translated part.
C-2. A stop codon has been mutated to a codon for an amino acid. Therefore translation continues past
the position of the usual stop codon. Translation continues until the ribosome hits a stop codon in the 3’UTR.
This is the reverse of a nonsense mutation.
D. Answers: D-1. Sometimes; D-2. Always.
Explanations:
D-1. If you change a codon from XYG to XYC you may change the amino acid that is specified, or you
may not. (X & Y are any two bases.) In some cases XYA, XYG, XYC and XYU code for the same amino acid,
so the change is synonymous. (This is the case if XY = CC or GU for example.) However in many cases the 4
codons are not synonymous – XYU & XYC always code for one amino acid, but XYC &/or XYU often code
for a different amino acid. (This is the case if XY = UU or GA for example.) This question is about degeneracy,
not wobble. D-2 is about wobble.
D-2. No base in the wobble position of the tRNA anticodon (1st base of anticodon) can pair with both G
and C in the wobble position of the mRNA codon (3rd base of codon). There is no way to orient the bases so
that the same base (in the tRNA) can form H bonds to either G or C. Therefore the same tRNA can not be used
to read XYG and XYC in the mRNA even if both codons specify the same amino acid. (It is not sufficient to say
“the wobble rules don’t allow it.” For full credit, you have to explain why not – that it’s an issue of base
pairing.)
2.
A. #6. Note the question asks about the sites at the ends of the introns, not the sites at the ends of the
exons. That’s why the answer is #6, not #7. In kidney cells, exon 6 is excised along with introns 5 & 6 as one
big intron. As a result, exon 5 is connected to exon 7. The splicing out of the big intron connects the 3’ end of
exon 5 (which is the 5’ donor site of intron 5) to the 5’ end of exon 7 (which is the 3’ end of intron 6). Donor
and acceptor sites for RNA splicing are defined by the ends of the introns, not by the ends of the exons. (For an
explanation, it was sufficient to draw exons 5 to 7 with accompanying introns, and show where the splicing
occurs.)
B. Answers (1 pt each): B-1. 160 AA; B-2. the same length; B-3. Smaller.
Explanations:
B-1. 30 nt corresponds to 10 codons, or 10 additional amino acids. Therefore the protein in eye cells
will be 10 AA longer than the protein found in kidney cells, which is 150 AA long.
B-2 (not required): The primary transcript is the same length in the two cell types; it’s the way it’s
spliced that’s different in kidney cells and eye cells. After splicing, 30 additional nt are included in the mRNA in
eye cells.
B-3. (not required): In kidney cells, one large intron is spliced out as a lariat. The big intron or lariat
includes introns 5 & 6 and exon 6. (See A.) In eye cells, the two introns, 5 & 6, are removed as separate
(smaller) lariats. The individual lariats are smaller in eye cells, but the number of lariats is smaller in kidney
cells.
C. Answers: C-1. both; C-2. eye cells; C-3. the mRNA (but see below); C-4. Either probe with mRNA
from eye cells.
Explanations:
C-1. Probe 5 hybridizes to exon 5 in the mRNA. This exon is found in mRNA from both cell types.
C-2. Probe 6 hybridizes to exon 6 in the mRNA. This exon is only found in mRNA from eye cells. This
exon is transcribed in both cell types, but spliced out of the primary transcript in kidney cells. (See above.)
C-3. The mRNA is separated by size on a gel first. Then the probe is added to visualize the location of
the mRNA. The size of the probe does not affect the position of the band – the probe sticks and forms a visible
band wherever the mRNA happens to be, and that depends on the size of the mRNA.
Some students chose ‘neither’ because they felt that this mRNA was too long to code for a protein that
contains only 160 amino acids. Therefore they reasoned that the RNA in the gel or blot must be the primary
transcript, not the mRNA. Fifteen hundred nt is on the long side, but mRNAs often contain very long UTRs, and
the problem says it is messenger RNA. Partial credit was given for this answer. The amount of credit (up to ¾
for answer + explanation) depended on the thoroughness of the explanation.
Answer to 2C, cont.
C-4. 4 pts for explanation. 2 pts for proper roles of probes; 2 pts for differences in mRNAs. There are 4
possible cases here – two different mRNAs with each of two different probes. Here are the patterns of bands
you will get on the blots:
Kidney mRNA
Probe 5
Probe 6
Eye Cell mRNA
Probe 5
xxxxx
Probe 6
xxxxx
xxxxx
A
B
C
D
You are looking for two cases that give the same pattern of labeled bands. You can see from the results
above that C and D are the same, but none of the other pairs are.
The bands in A and C are in different places because the two mRNAs (from the two cell types) are of a
different length. Whatever probe you use, if you compare blots of the two mRNAs with the same probe (A vs C
or B vs D) you won’t get the same pattern.
If you want to get the same pattern in different cases, you have to use the same mRNA both times. In
other words, you need to use two different probes with the same mRNA (A vs B or C vs D). If you use kidney
cell mRNA (A vs B), you’ll get one band with probe 5 and no bands with probe 6. If you use eye cell mRNA (C
vs D), you’ll get one band with either probe, and both bands will be in the same spot (since the position of the
band depends on the length of the mRNA, which is identical in both cases). The two different probes will bind
to different parts of the mRNA, but the position of the band will not change.
In this problem, many students compared the wrong blots – they compared the bands you get when
using probe 5 with the two different mRNAs (A vs C) instead of comparing the two different mRNAs with the
same probe (A vs B and C vs D).
3.
A. Answers (1 pt each): A-1. Gene X; A-2 1 promoter, 4 start codons, and 0 introns; A-3. A start codon.
Explanation: You had to draw the operon, including an operator, a promoter (before gene X) and at least one
start codon at the beginning of one of the structural genes. Some students drew introns between the structural
genes. This is incorrect because (1) this is a bacterial operon, and bacteria lack introns, and (2) even if there
were introns, the spacers between genes are not introns. Most eukaryotic genes have introns, but bacterial genes
generally do not.
B. Answers: sappy polycistronic mRNA and RNA pol. transcribing the hap gene.
Explanation: Deprepression = turn on of repressed operon. This involves removal of co-repressor, not addition
of inducer. When operon is turned on, the operon is transcribed, and new mRNA is made and translated.
C. Answers (1 pt each): bound to operator and bound to co-repressor.
Explanation: Repressor protein is made constitutively; it is always present. It’s the state it’s in that varies. Free
repressor has one shape (that doesn’t bind to the operator); the repressor-co-repressor complex has a different
shape, and sticks to the operator. When co-repressor is present, co-repressor binds to repressor, and the complex
binds to the operator, blocking transcription and turning off the operon.
Answer to 3D, cont.
D. Answers: D-1. no (1 pt); D-2 UAA and UGA (1 pt each); D-3 translation (1 pt); D-4 hap (2 pts).
Explanation: You had to draw the stop codons (that end translation of hap) and show how they could overlap
the start codon (for translation of nar). There is no possible overlap between the stop codon UAG and the start
codon AUG. However UAA and UGA can overlap AUG. For UAA, the last A in UAA can overlap the first A in
AUG. For UGA, there are two possible overlaps. Either the last A in UGA can overlap the first base of AUG or
the last two bases of AUG can overlap the first two bases of UGA. For all the possible combinations, if the stop
codon is used to stop translation of hap, the start codon can then be used to start translation of nar. The hap
protein will be released when the stop codon is in the A site of the ribosome. The mRNA and ribosome can then
move relative to each other until the AUG is in the P site, and synthesis of the nar chain can begin.
4. Answers: A. 0; B. 2; C-1. All of these; C-2. lys. Explanations:
A & B (not required) – If the strep resistance gene works, then it was not cut up by Hind III, and nothing
was inserted into it. It is intact. The pME plasmid had one Hind III site. When it was cut and the fragment
inserted, each end of the fragment joined one half of the original Hind III site, making two Hind III sites, one at
each end of the insert. (See handouts or texts.)
C-1. The recipient bacteria must be grown on medium containing strep to eliminate bacteria that did not
get a plasmid. However, the medium must also contain lys and B12, as these are required for growth of this
bacterial strain. A few of the bacteria will get fragments that convey the ability to make lys or B12, and these
recipients will grow without added lys or B12, but most recipients will still require lys or B12 in order to grow.
C-2 (not required). You want to select for bacteria that can make their own lysine and eliminate bacteria
that cannot make lysine. In other words, you want bacteria that can grow without added lysine. You would need
replica plating if you wanted to select the bacteria that can’t make lysine.
D. Answers: D-1. A B. examinus promotor; D-2, on the plasmid.
Explanations (2 pts each).
D-1. The same gene cannot be transcribed (successfully) from either strand – only one strand is ‘sense’
and only the ‘other’ strand can successfully act as template. The orientation of the gene relative to the promoter
determines which strand is transcribed. If the enzyme Z gene can be transcribed (successfully) in either
orientation, then the fragment itself must contain the promoter of the Z gene as well as the coding region for the
enzyme. Therefore, the gene is always in the same orientation to its promoter no matter which way the fragment
is inserted. In other words, the insert in the plasmid contains a section of B. examinus DNA that includes the
promoter of the enzyme Z gene.
Some students suggested that the plasmid contained two promoters, one at each end of the insert. That
wouldn’t work – if they both were used, you’d get two complementary mRNAs. They would hybridize (like
sense and antisense mRNA) and no enzyme Z would be made.
D-2. A promoter must be next to the structural genes that it controls – a promoter can only promote
transcription of genes on the same DNA molecule. If DNA polymerase binds to a promoter, it won’t help
transcription on a different DNA molecule. Since the gene being transcribed is on the plasmid, the promoter
must be on the plasmid too. We know that the plasmids are not integrated, since it says all the recipients can
make lysine – integration, a rare event, is not needed to allow transcription.
E. Answer: Wham I.
Explanation (5 pts):
Hind III cuts only at the ends of the fragment (in the recombinant plasmid). However the fragment is
inserted, digestion with Hind III will give the same two pieces – the plasmid and the fragment. (2 pts for why
Hind III won’t work.)
In the recombinant plasmid, Wham I cuts once inside the fragment and once in the plasmid, outside the
fragment. The result will be 2 pieces. The longer piece, containing most of the plasmid, will include either 1/3 of
the fragment or 2/3 of it, depending on the orientation of insertion. The shorter piece will consist of a short
section of the plasmid plus either 2/3 or 1/3 of the fragment. So the two pieces produced by Wham digestion
will be different lengths, depending on the orientation of insertion of the plasmid. (3 pts for why Wham will
work.)