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 (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

exon 1 exon 2 exon 3 exon 4 & introns 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 Toxin levels

Del 1 gene 1 0

Del 2 gene 2 0

Del 3 gene 3 0

Del 4 gene 4 0

Del 5-6 genes 5 & 6 100%

Del P P2 4%

WT nothing 100%

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.)

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