1.       1. Name the DNA repair process that would be used to repair the following: (2 pts each)

a.       A pyrimidine dimer __nucleotide excision repair

b.      A misincorporated nucleotide detected during DNA replication but missed by the 3’-5’exonuclease activity _mismatch repair

c.       A deaminated cytosine __base excision repair

d.      A double-stranded DNA break _either homologous or non-homologous recombination

e.      A depurination _base excision repair

2.      2. What is the difference between a general transcription factor and a specific transcription factor? (6 pts)

General trancscriptions bind to the promoter, usually the TATA box, and help load the RNA polymerase on so it can begin transcribing at the appropriate location. They are required for all genes to be transcribed. Specific transcription factors are unique as regulators for a specific gene or cell type and bind to upstream regulatory sequences (usually 50 to 200 nucleotides upstream from the startpoint of transcription) or to enhancer/silencer regions which can be located kilobases away from the transcription startpoint on either side. They are regulatory and either enhance or inhibit the ability of the general transcription factors to bind.

3.      3.    Name two methods that a cell uses to decompact chromatin structure in preparation for the initiation of transcription. (8 pts)

Possible answers:

Chromatin remodeling complexes that loosen the connection between the DNA and the histone core in chromatin, allowing the nucleosome cores to slide.

Histone acetylases that attach acetyl groups to the amino terminal regions of the histones thus loosening the attraction between the negatively charged DNA and the histone core.

Histone chaperones that bind to histone cores once they have been loosened from the DNA and escort them out to other locations or back again.

4.       4. Give two reasons why the phosphorylation of the CTD of RNA polymerase II is necessary. (6  pts)

It causes the DNA and the general transcription factors at the TATA box to loosen their attraction to the RNA polymerase, allowing it to clear the promoter region and continue transcription.

It uses additional ser phosphorylations to bind factors involved in the capping reaction and later in the splicing reactions and the cleavage and polyadenylation reactions.

It is involved in “stalling” of the polymerase and then in its restarting to complete transcription, often used as a regulatory mechanism (HIV; hsp70 examples)

5.       5. Answer one of the following questions. (10 pts)

A.      Describe the modification that occurs on the 5’ end of eukaryotic mRNAs and explain all steps in the process that creates the modification.

B.      Outline the steps by which RNA polymerase II terminates the transcription process and how the 3’ end gets modified in eukaryotes.

A.       See Figures 6-22 and 6-24

B.       See Figures 6-37 and 6-38

6.      6. The following  have a role to play in determining that an RNA transcript is correctly spliced. Describe that role. (4 pts each)

a.       U1

Binds to RNA sequences at the 3’exon/5’intron splice junction using its complementary, antiparallel snurp RNA component. This marks the junction region.

b.      U2

Binds to RNA sequences surrounding the branchpoint A in the intron through complementary, antiparallel basepairing to its snurp RNA component. But it cannot bind to the A itself, causing it to bulge out. This helps with the first transesterification reaction.

c.       Branch point A

Located in the 3’ third of the intron it begins the first transesterification reaction by attacking the first exon/intron junction with its 2’OH. A phospodiester bond is formed between the 2’OH and the 5’Phosphate of the first nucleotide of the intron (always a G). This forms the lariat.

d.      SR proteins

Bind to exonic enhancer sequences within exons and facilitate loading of the snurps on either side within the introns. Specifically, they help U2AF load onto the pyrimidine rich region in the 5’ side intron (U2AF helps the U2 snurp load on). And they help U1 snurp load onto its binding site on the 3’ side intron. Therefore, this helps prevent cryptic sites within the exons from being used as introns. They “mark” the exons in this way and also facilitate correct splicing.

7.     7.  Describe in detail the molecular switch that allows the developing B cell to become an antibody secreting plasma cell. (10 pts)

See Figure 7-99

8.      8. A mutation in the tra protein in Drosophila embryos distorts the sex determination of the flies. Explain why.  Include all of the molecular details in your answer. (8 pts)

See figure 7-98 for the molecular details, concentrating on the regulation of the doublesex gene splicing, which is acted on by the tra protein complex acting as an SR-like complex in females only. In the answer, you were also required to address the ultimate result expected if tra is mutated, therefore not being able to function. This would cause all doublesex transcripts to be spliced as if they were in the male pathway with the result that all the embryos would be have a male phenotype.

9.       9. Define the term wobble and explain how it relates to third base degeneracy of the genetic code. (6 pts)

Wobble is non-Watson-Crick base pairing between the first nucleotide base of the anticodon and the third nucleotide base of the codon. This interaction is stable enough to allow a tRNA that participates in such a wobble relationship to be used during translation. Since the amino acid carried on such a tRNA will be incorporated into the polypeptide during translation if the interaction between the codon and anticodon is stable, it is essential that any amino acid that is specified by such codons always be the same amino acid that is coded for and that is carried on the tRNA. This requires that the third base of the codon be irrelevant for specificying the amino acid. In this way, the wobble interaction no matter what it is will not influence the amino acid being incorporated in the growing chain. This leads to the degeneracy of the genetic code where many possible third bases are allowed.

10.  10. How does the cell guarantee that only completely processed mRNAs enter the cytoplasm? (6 pts)

Special SR proteins associate with the mRNA as it is being processed only if they are phosphorylated. Following poly-A addition a phosphatase is activated that removes the phosphates from these SR proteins, causing them to leave the mRNA. This allows access to the mRNA by the nuclear export receptor which facilitates movement of the mRNA out of the nuclear pores.

11.   11. If a new planet was discovered where a quadruple genetic code was the norm and there were only 5 possible amino acids called 1, 2, 3, 4, and 5, describe the results of a Nirenberg-like experiment if the codon AUGC actually coded for amino acid 4. (6 pts)

A filter binding assay would be performed. 5 test tubes containing all translation components, the AUGC codon, all tRNAs, all amino acids, and one radioactive amino acid would be filtered. Only complexes between a tRNA carrying an amino acid and the codon would be retained on the filter. All other components not part of the complex would flow through. The filter that is radioactive would be from the test tube containing the radioactive form of the amino acid coded for by the codon. In this example, the test tube had radioactive amino acid 4 would result in a radioactive filter.

12.  12. This question is to be determined and will be given to you at the start of the exam. (8 pts)

Describe the charging reaction.

Amino acyl tRNA synthetase binds the amino acid and ATP, It hydrolyzes ATP to AMP and pyrophosphate. Pyrophosphate breaks down to 2Pi. The energy released is used by the enzyme to attach the AMP to the amino acid COOH group. The enzyme may use its editing region to check out the amino acid at this time, releasing this complex if the amino acid is incorrect. Or, the second part of the mechanism may happen immediately whereby the AMP is removed from the amino acid and the energy released used to attach the amino acid to the 2’ or 3’OH of the nucleotide at the 3’ end of the tRNA, an A in the conserved sequence CCA. If attached to the 3’OH the amino acid will be then transferred to the 2’OH.  The amino acid is again checked in the editing region and if incorrect this complex is removed.

Bonus Question:  (4 pts)

Draw the result you would expect from an R-looping procedure performed with a gene containing 4 introns.

The drawing would have four single-stranded DNA loops representing the 4 introns. The exons would be hybrids between one strand of DNA and one strand of mRNA (complementary and antiparallel).