1. Describe the role in transcription initiation performed by the following: (4 pts each)

A. TFIIH
General Transcription Factor that provides the helicase activity that melts the hydrogen bonds at the TATA box region, allowing the template strand to be accessible for transcription. Also, it attaches phosphate groups to the CTD of RNA pol II, facilitating promoter clearance.

B. Mediator
A large complex of proteins that forms a bridge between specific transcription factors bound to upstream promoter elements or to enhancer/silencer sequences and general transcription factors at the TATA box. This can either activate or repress the loading of RNA polymerase II at the TATA box.

C. histone acetylase
An activating enzyme, part of complexes bound at regulatory sequences in upstream promoter proximal regions or enhancer/silencer regions that attaches acetyl groups to the amino terminal region of histone proteins, loosening the chromatin structure to facilitate transcription factor binding at the nearby TATA box

D. TATA Box
Sequence of nucleotides located approximately 35 bases upstream of the startpoint of transcription of most eukaryotic genes. Serves as the site of binding of the general transcription factors and the loading of RNA polymerase II.

E. Specific transcription factors
Proteins that bind to upstream promoter proximal regions or to enhancer/silencer regions and that are involved in either activating or repressing the assembly of the general transcription factors and RNA polymerase II at the TATA box.
 

2. Describe how the Human Immunodeficiency Virus (HIV) relieves the premature termination of transcription that causes RNA polymerase II to stall shortly after initiation. (8 pts)

After transcribing the 5’ end region of the transcript, called the TAR sequence, The HIV Tat protein binds the stem-loop in the TAR. This seeds assembly of a complex of cellular proteins that includes Cyclin T (also bound the TAR), NELF, Spt4/5, and Cdk9 (a protein kinase). The Cdk9 hyperphosphorylates the CTD of RNA polymerase II, which has stalled in its transcription. This relieves the termination and the transcription is completed.

3. Describe the 5’ Cap  on eukaryotic mRNAs. (6 pts)

This is a G nucleotide, methylated at the N7 position of the guanine. It is attached to a the first nucleotide of the RNA transcript by a 5’ to 5’ triphosphate bridge. The ribose of the first nucleotide in the transcript has a methyl group attached at the 2’ carbon and, in vertebrates, the same methylation occurs on the second nucleotide of the transcript as well.
 

4. What mechanism is used by eukaryotic cells to generate the 3’ end of mRNAs? (8 pts)
 

RNA polymerase II transcribes a poly-A signal, AAUAAA and a G/U rich sequence further down the transcript. CPSF binds to the poly A signal. CStF binds the G/U sequence and the bound CPSF, forming a loop in the RNA. CF1 and CFII bind, stabilizing the complex. Poly A polymerase binds and stimulates the cleavage at the poly A site (10-35 nucleotides downstream of the poly A signal). CStF, CFI, and CFII are released and the RNA transcript 3’ to the cleavage site is degraded eventually. Poly A polymerase slowly addes A nucleotides to the new 3’ end using ATP as a source of nucleotides and energy. After about 12 are added, PABPII binds the A polymer, stimulating rapid addition of the remaining poly A tail of about 200 nucleotides in length.
 

5. Below is a diagram of a eukaryotic gene. Open boxes indicate introns. Darkened boxes are exons. Draw the results of an R-looping experiment using this gene. (6 pts)
Consult your exam for the diagram.

Answer should show that there are 3 single standed DNA looped out regions, representing the introns. There are 4 hybrid regions showing interactions between the exons and their counterparts in the mRNA. Also, there would be a poly A tail region of single-stranded RNA, not paired to anything and attached to the mRNA 3’ end. Also, there would be a single stranded DNA region at the 5’ end of the DNA, not paired to anything, representing the TATA box and the DNA located between the TATA box and the startpoint of transcription, +1.

6. Describe in detail how the U1 snurp and U2 snurp interact with intron-exon junctions and how this helps begin the splicing process. (10 pts)

The RNA component of the U1 snurp is complementary and antiparallel to the consensus sequence located at the junction between the end of the first exon and the 5’ end of the intron. These pair, marking the appropriate junction region. The U2 snurp binds through its RNA component to the consensus sequence surrounding the branch point A in the intron. However, the A itself has no counterpart and therefore bulges out. This facilitates the spatial accessibility of the branch point A which is involved in the first transesterification reaction, where the 2’OH of the A attacks the phosphodiester bond connecting the 3’end of the exon to the 5’ G end of the intron, breaking it and forming a new bond between the 2’C of the ribose of the A nucleotide and the 5’ phosphate of the G nucleotide.
This forms a lariat structure.
 

7. Sex determination in fruit flies involves a cascade of proteins whose transcription is regulated differently in males and females. Outline this protein cascade and explain how the two gender phenotypes arise from it. (10 pts)

Early in embyogenesis, the female fruitfly uses an early promoter of the early sex lethal gene and makes the early sexlethal protein. The male is unable to use this promoter and does not make the protein. Later, both male and female embryos use the late promoter and make the late sexlethal RNA transcript. The female early sexlethal protein occludes the recognition of one of the 3’splice junctions in an intron and alternative splicing results, causing a protein to be made from the resulting mRNA (an in frame stop codon gets removed ). This is the late sexlethal protein. In the male no such alternative splicing occurs (no early sexlethal protein) and the premature stop codon remains. No viable late sexlethal protein is made. The late sexlethal protein in females regulates continuing production of  more late sexletahal protein in the identical manner and also regulates alternative splicing of another gene, transformer. The late sexlethal protein once again occuludes recognition of a 3’ splice junction in an intron leading to an mRNA that does not contain a premature stop codon and the protein made is a function transformer (tra) protein. Males keep the stop codon and make a dysfunctional protein. Regulation of the doublesex transcript results. Females, with tra functional, can make a complex with two other proteins that ACTIVATES a 3’ splice junction that is otherwise too weak to be used. The resulting mRNA gives rise to the female form of doublesex. This protein represses transcription of male-specific genes, giving rise to the female phenotype. In males, there is not activating complex to alter the splicing and a male form of the doublesex protein results. It represses transcription of the female specific genes resulting in the male phenotype.
 

8. Below is a table representing a filter binding experiment, similar to Nirenberg’s. In this case, however, one of 3 possible codons was being tested. What do you conclude? (6 pts)

Radioactive filter?   Codon   labeled amino acid

 No    CUU   phenylalanine
 No    CGA   proline
 Yes    CUU   leucine
 No    CCC   arginine
 Yes    CGA   arginine
 Yes    CCC   proline
 No    CUU   arginine
 No    CGA   leucine
 No    CCC   leucine

CUU codes for leucine; CGA codes for arginine; CCC codes for proline.
 
 

9. Describe the 2-step chemical reaction that “charges” a tRNA with its appropriate amino acid. (8 pts)
 

First, amino acyl tRNA synthetase binds to the amino acid at its active site and to ATP. ATP is hydrolyzed to AMP and PP. PP breaks down to 2P. The energy released is used to attach the AMP to the COOH end of the amino acid. This causes a shape change at the enzyme active site. This now allows the appropriate tRNA to bind there and causes the breaking off of the AMP from the amino acid, releasing energy. This energy is used to attach the amino acid to the 3’ end of the tRNA.
 

10. One regulatory mechanism in cells that make hemoglobin is to prevent the initiation of translation of the globin genes if heme levels are low. This is done by keeping eIF2 in a GDP bound form, preventing GTP from binding. How could this control translation initiation? (8 pts)

EIF2-GTP is required to form a functional preinitiation complex with the initiator tRNA carrying methionine and the small ribosomal subunit. Once the complex forms it binds to eIF4 at the Cap on the 5’ end of the mRNA, allowing for ribosome scanning to occur. This sets up the preinitiaion complex at the first AUG start codon following the Cap, setting the reading frame. The large ribosomal subunit now binds, and eIF2 hydrolyzes GTP to GDP. A functional ribosome results and translational initiation is complete.
 

11. Describe all steps by which a peptide bond is formed during translational elongation, beginning with an empty A site and continuing up until the translocation step. You do not need to discuss translocation. Caution: be very thorough in your explanation including the names of all interacting factors and their importance to the process. (10 pts)
 

In the cytoplasm tRNAs interact with eEF1alpha-GTP. These move into the open A site of the ribosome and the anticodon on the tRNA binds to the codon at the A site. If this binding is sufficiently strong (3 complementary stable relationships), the tRNA complex will remain there long enough for GTP to be hydrolyzed to GDP by the eEF1 alpha. This changes the shape of the eEF1alpha, forcing it to release the tRNA, and it leaves. The energy released when GTP is hydrolyzed is used to form a stable attachment of the tRNA to the A site. Since the eEF1alpha complex is gone, the steric hindrance is removed. Now peptidyl transferase is activated by the stable binding of the tRNA to the A site and it breaks the bond between the tRNA at the P site and the amino acid attached to it. A new peptide bond is formed between that newly released amino acid and the amino acid carried by the tRNA at the A site. Translocation will follow.