Download C2005/F2401 `09

C2005/F2401 ’10
Key to Review Questions for Exam #3
Note: The explanations given here are generally much more lengthy & comprehensive than expected from
students on the exam.
1. A. Exons. Missense mutations are changes from one amino acid to another. Only the exons include sections
that code for amino acids; introns do not. (Note: both exons and introns are transcribed, but only exons are
translated. Introns are removed before the ribosome attaches to the mRNA)
B-1. Genotype only. B-2. AUG to AUA. B-3. CUG to CUA and CGA to AGA (1 pt for each correct choice
for B-3.) Explanations:
B-1. Synonymous mutations are changes in the code that do not change the corresponding amino acid. Since
the code is degenerate, there are multiple codons for most amino acids, so changes (especially in the 3rd position
of the codon) often do not change the resulting amino acid. See the code table. Therefore it is possible to change
the genotype (the DNA) without changing the phenotype (the function or appearance).
B-2. AUG to AUA is missense – it changes the amino acid from met to ile. UAC to UAA causes a change
but it is nonsense, not missense – it creates a premature stop codon.
B-3. See the code table. The two correct choices are synonymous, although CGA to AGA doesn’t look it at
first. (Note that the ability to use the same tRNA or a different one is not important here. That’s an issue of
wobble, and the issue here is degeneracy.)
C. UAC to UAA. This creates a premature stop, so the protein will be much shorter than normal, and is very
unlikely to work. AUG to AUA may have a serious effect, but it may not. It all depends on where the change is
in the protein and what that part of the protein does. You can be pretty sure that UAC to UAA will knock out
function, but you can’t predict about AUG to AUA -- it may or may not affect function. The mutation that is
most likely to lead to a lack of function is most likely to cause a disease. Note that AUG to AUA will not cause
a problem with initiation of protein synthesis – the mutation is in the middle of a coding region, not at the start.
2. A-C. Answers: E site; AA #2, E.
In this case we are just starting translation; the initiator tRNA has just been used to start the chain. Met was
detached from the initiator tRNA and connected to the amino end of AA #2, which was still attached to tRNA
#2. tRNA #2, with dipeptide attached, moved into the P site, and the initiator tRNA moved to the E site. Look at
class handout to see relative positions of A, P and E sites relative to end of mRNA.
D-1. AGU to UCU, ACU to ACC, and ACU to ACG.
D-2. ACU to ACC will work, but none of the others.
(1). AAG to AAC. This mutation changes the amino acid encoded. So there is no way to translate the
mRNA and get a normal peptide. Wobble won’t help at all.
(2). AGU to UCU. This mutation encodes the same amino acid, but a different tRNA is needed to do the
translation. Wobble only helps if the first two bases are the same.
(3). ACU to ACC. This mutation encodes the same amino acid, and the same tRNA can be used to do
the translation. If the base in the wobble position of the tRNA (first base of anticodon) is G or I, it will match up
with either C or U in the third position of the codon. In this case the first two bases are the same, and the wobble
rules indicate that there is at least one anticodon that can match up with either codon.
(4). ACU to ACG. This mutation encodes the same amino acid, but the same tRNA can not be used to
do the translation. There is no base in the wobble position of the tRNA (first base of anticodon) that can match
up with U or G in the third position of the codon. In this case the first two bases are the same, but the wobble
rules indicate that there is no anticodon that can match up with both codons.
3. A. In different operons; enzymes 3 & 4. The placement of promoters and terminators indicates that there are
two separate operons here, one that encodes enzymes 1-4 and a second operon that encodes enzymes 5 & 6.
There should be two separate transcripts, one starting at P1 or P2 and going to T1; the other starting at P3 and
going to T2. Each transcript is polycistronic and encodes more than one enzyme, but there is no transcript that
encodes all 6 enzymes. Enzymes 4 & 5 are encoded in separate operons and transcribed separately; only
enzymes 3 & 4 are encoded on the same operon and are part of the same transcription unit.
B-1 & B-2. Yes (as long as it is from the plasmid); only plasmid DNA. A transcript made from the first
operon on the plasmid contains all the information needed to make the toxin. A transcript from the chromosome
won’t help – it won’t be able to encode working enzyme 2. No recombination is needed here between the
plasmid and the chromosome. (Part credit was given if you took ‘recombinant DNA’ to refer to the plasmid, or
‘this strain’ to refer to the deletion strain without the plasmid.)
B-2 & B-3. Gene 2; P2.
B-3. If both plasmid and chromosomal DNA contained deletions of gene 2, there would be no
information available to make enzyme 2, and the toxin could not be made. (The two DNAs would NOT
complement each other; recombination wouldn’t help either since the mutations overlap.)
B-4. We know from the table that when you delete P2 you get a lower level of toxin production. The
simplest explanation is that deleting P2 cuts down on transcription, and thus on the level of enzymes. When P2
is deleted, transcription continues at a lower level, presumably using P1. (P1 must be a weaker promotor than
P2.) So if you delete P2 on the plasmid, you will get low levels of transcription and low levels of all enzymes
made from the plasmid. Therefore you will have low levels of enzyme 2, and should have low levels of toxin
synthesis.
The chromosome contains a good copy of P2, but a promoter on one DNA cannot affect transcription on a
different DNA. A good copy of say, gene 2, on one DNA can compensate for a bad copy on the other DNA, but
a good copy of P2 on the chromosome cannot compensate for a bad one on the plasmid.
C. Bit 1; none of these; no. (Explanation– it was sufficient to explain any 2 of the following points.)
(1). You need the sticky ends of the plasmid (cut once) and the sticky ends of the bacterial DNA to
match up. Therefore the same enzyme has to be used to cut them both. Since the bacterial DNA was originally
cut with Bit 1, you have to cut the plasmid with the same enzyme.
(2). If you cut the plasmid with Bit 1, and insert the bacterial DNA, you will interrupt the gene encoding
resistance to bubimycin. Therefore the cells that get the plasmid will not be resistant to the drug. Note that cells
that do or do not get the plasmid will have the same phenotype for drug resistance – both will be sensitive.
Replica plating is no help here.
(3). A del 5-6 strain makes toxin anyway – enzymes 5 & 6 are not needed for toxin production. So
production of toxin will not distinguish a transformed del 5-6 strain from one that did not get the plasmid. A del
2 strain does not make toxin, so you can tell if it got the plasmid if it starts to make toxin.
Note that it says in the details that toxin does not inhibit growth of these bacteria, so you don’t expect any
differences in growth with or without toxin, or any differences in ability to break down toxin.
4. A. Reasonable possibilities: mRNA from the structural genes; enzymes need to synthesize anthrose. When
bacteria are sporulating the operon is on, the operon is transcribed, and the mRNA is translated to make the
enzymes. When bacteria are actively growing the operon is off, and is not transcribed (or translated). The
enzyme levels are controlled by the level of transcription of the operon.
You did not need to explain the other (unreasonable) possibilities, but here is how they are ruled out: The
repressor protein is made constitutively, so mRNA from the repressor gene should be found in both states.
Genes for the enzymes are present in both states; the difference is whether they are transcribed or not. There is
no degradation of anthrose going on in either state – it is made, not degraded, by sporulating bacteria. It is a
sugar, but it is not used as an energy source.
B. Inducer or co-repressor; inducible.
B-1. You know the operon is turned on in sporulating cells, but you don’t know (at this point) if the turn
on process involves induction or release of repression. Either the operon is inducible, and there is something
special (inducer) present in sporulating cultures, or the operon is repressible and there is something special (corepressor) in actively growing cells.
B-2. The rationale: If there is something extra in sporulating bacteria, that something extra must be the
inducer. Therefore the system must be induced, not de-repressed, during sporulation. How it works: The inducer
binds to the repressor protein, interfering with binding of repressor to the operator. Repressor protein comes off
the operator and allows transcription of the operon – RNA polymerase can now bind to promoter and initiate
transcription. (Explanation of either rationale or mechanism is sufficient.)
C. Feedback inhibitor of anthrose synthesis. Repression/induction controls the levels of enzyme synthesis
and therefore the levels of enzyme present. Feedback inhibition (or activation) controls the activity of the
molecules of enzyme that are already there. It doesn’t change the amount of enzyme (Eo) but it does change the
properties of the enzyme molecules (turnover number and/ or Km). See class handout 15A. Since enzyme levels
remain high, but the amount of synthesis of anthrose is not, then the anthrose (the end product) must be
inhibiting its own synthesis through feedback.
D. Operator of the operon; promoter of the repressor gene. These bacteria produce anthrose (& the enzymes
to make it) constitutively – they can’t turn off the synthesis of the enzymes. The problem must be that the
repressor cannot stick to the operator to turn the anthrose operon off. The problem could be either with the
repressor protein or with the operator of the operon. One or the other is defective. Either the promoter of the
repressor gene has a deletion (and can’t bind RNA polymerase) or the operator has a deletion (and can’t bind
repressor). As a consequence, either there is no transcription of the repressor gene (& no repressor), or repressor
is made, but can’t bind to the operator and turn the operon off.
5. A-1. Only 2 will work; A-2. 1 & 2 – both will work. A-3. Yes. Explanations:
For A-1, only 2 will work because bacteria can’t remove introns; they have no splicing enzymes (no
spliceosomes).
For A-2, either 1 or 2 will work, because eukaryotes can splice out introns – either one will code for
working factor 9.
For A-3: If you remove the sequence corresponding to the normal intron from the hemophiliac’s F9 gene,
the DNA will be the same as the normal cDNA. Therefore it should work perfectly well in either prokaryotes or
eukaryotes.
Answer cont. on next page.
B. Primary transcript the same length as normal; lariat shorter than normal. The mutation here is a substitution –
it does not change the length of the DNA or the length of the transcription unit. Before any splicing occurs, the
RNA (the primary transcript) will be the normal length. However, when splicing does occur, it will occur in the
wrong place, and the mRNA (and the part that is spliced out) will NOT be the normal length. The part that is
removed from intron 3 is the part that forms the lariat. This part will be 2 bases shorter than usual, and the
mRNA will be 2 bases longer than usual, because the intron 3/exon 4 boundary has been moved 2 bases toward
the 5’ end of the mRNA.
C. Stop codon will be in a different place; the peptide encoded in exon 4 will be different. Mutation has no
effect on the start codon, which is well before the mutation. However the mutation causes a frameshift in the
exon following the misplaced splice site. The frameshift will probably generate a stop codon well before the
normal one; in any case the frameshift will alter the position of the first stop codon.
Why exon 4? The first intron comes after the first exon – introns are sequences in between exons, not
sequences in between genes. So intron 3 comes after exon 3 and before exon 4. Exon 3 will not be affected as it
comes before the mutation; exon 4 will not be translated properly because of the frameshift caused by the
splicing mutation.
D. 4 bands max; same number in mutant and normal; 0 labeled bands in a different place. Here is a picture of the
DNA containing the gene for factor 9:
Restriction Sites (2 outside gene and 3 inside) = ↓
↓
↓
↓
↓
↓
exon 1
exon 2
exon 3
exon 4
exon
-----------------XXXXXX----------XXXXXX----------XXXXXX---------XXXXXX………………………XXXX---------↑
Intron 1
intron 2
intron 3
Start of Transcription
Transcription
unknown # of exons
& introns
last
↑
End of
Pieces that result from cutting with restriction enzyme:
---------------------------------------------------------------------------------------------------------
The DNA will be cut up the same way into 4 pieces whether it is mutant or normal (unless the substitution
causes a new restriction site in intron 3). One of the pieces will have a single difference in the base sequence (in
intron 3), but the lengths of the pieces will be the same, whether they are from normal or mutant. The single
base difference will not affect hybridization to the probe, because the cDNA probe hybridizes only to the exon
sequences, not to the introns.
Note that the DNA is being cut up here, not the mRNA. The mRNA from the mutant is 2 bases longer, but
the DNA is the same length in mutant and normal.
Each piece of DNA contains at least part of one exon. Any section with an exon will hybridize to the cDNA
probe.
The bands on the gel correspond to the lengths of the pieces of DNA, not to the length of the cDNA probe.