Download C2005/F2401 `09

C2005/F2401 ’09
Exam #3
Questions & Answers
Note: Each answer was worth 2 pts and each explanation 2 pts unless it says otherwise. The explanations given here are
generally much more lengthy & comprehensive than expected from students on the exam.
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.
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. 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.
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. 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. Which of the changes listed is most likely to be the one that is the cause of the disease?
_________________________________________________ . Explain how you know.
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. 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, explain on back.
Problem 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. Suppose the ribosome is translating mRNA from a mutant version of the same gene. Codon #2 is changed, but the
same tRNA #2 is in the P site as above. In which of the following cases could translation produce a normal peptide?
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 cases that would produce a normal peptide, using the same tRNA. Explain briefly how you
ruled each case in or out. (Code and wobble rules are on the last page.)
ACU to ACC will work, but none of the others. (1 pt explanation of each case).
(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. 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.
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. 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.
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.)
Question 3, cont.
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.
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. 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.
Bit 1; none of these; no. (Explanation 2 pts – 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. 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.
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. 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.
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 (co-repressor) 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. 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.
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.
Question 4, cont.
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.
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. 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 DNA could you insert in your plasmid?
(1) (2) (3) (4) (5) (none of these).
A-2. If you want to use the plasmid for gene therapy in patients with hemo. B, which DNA could you insert?
(1) (2) (3) (4) (5) (none of these). For each part, circle all reasonable choices and explain.
Consider only choices 1-4. (Choice 5 was considered separately; see below.) Then the answers are
A-1. only 2 will work; A-2. 1 & 2 – both will work. (1 pt each = 3 pts total). Explanations (2 pts):
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.
Choice 5 (2 pts; only the explanation considered). 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 will work perfectly well in either
prokaryotes or eukaryotes.
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.
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.
Question 5, cont.
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.
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. 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.
4 bands max; same number in mutant and normal; 0 labeled bands in a different place. (1 pt each answer; 2 for
explanation of D-3.) Here is a picture of the DNA containing the gene for factor 9:
Restriction Sites (2 outside gene and 3 inside) = ↓
↓
↓
↓
↓
↓
unknown # of exons
& introns
exon 1
exon 2
exon 3
exon 4
last exon
-----------------XXXXXX----------XXXXXX----------XXXXXX----------XXXXXX………………………XXXX---------↑
Intron 1
intron 2
intron 3
↑
Start of Transcription
End of Transcription
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
splice 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.
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.
See text or handout for the genetic code and the wobble rules. (These were included on the exam.)