1. Draw the structure of a tripeptide, designating the side chains as R1, R2, and R3. Identify the omega, psi, and phi torsion angle locations on the diagram. Also, show the region of planarity. (10 pts)

See exam key in cabinet outside 021 McKinly.
 

Explain why the region is planar. (4 pts)

Both the C-O and the C-N of the peptide bond have partial double bond character due to resonance. See diagram on exam key in cabinet.
 
 
 
 
 
 

2. What three non-covalent interactions stabilize the tertiary structure of a protein, excluding van der Waal’s forces? (6 pts)

Hydrogen bonding
Ionic interactions
Hydrophobic interactions
 
 

What categories of amino acids could interact to form each of these interactions (give one example for each)? (6 pts)

Hydrogen: polar amino acids interacting
Ionic:  basic amino acids with acidic amino acids
Hydrophobic: Two hydrophobic amino acids
 

3. Define the following terms: (4 pts each)

A. Chaperone: Protein that binds to exposed hydrophobic side chains in regions of polypeptides to keep them from aggregating until the protein has folded into its appropriate 3D structure.
 
 

B. Ubiquitination: ATP-requiring processs of attaching a polymer of ubiquitins to a lysine reside in a polypeptide, targeting it for degredation in a proteosome.
 

C. Proteosome:  Barrel-shaped structure containing  proteases that degrades ubiquitin-tagged polypeptides.

D. induced fit: Conformational change in the active site of an enzyme caused by the initial binding of the substrate. It allows a tighter fit to the substrate. Our example: glycine lid induced to close over ATP as substrates bind to the cyclic AMP dependent protein kinase.
 

4. What aspects of the structure of B-DNA explain the following:  (2 pts each)

A. 34 angstrom repeating features in the x-ray crystallographic data

1 full turn of the DNA double helix, consisting of 10 base-pairs

B. 20 angstrom diameter

hydrogen bonding between a purine and a pyrimidine holding the two DNA strands together
 

C. loss of viscosity upon heating: melting of the hydrogen bonds holding the bases together, causing the double-stranded DNA to become single-stranded
 
 

5. Draw a trinucleotide polymer of single-stranded DNA. Designate the bases by their letter abbreviations (use any). (8 pts)

See figure 4-3 in the textbook
 
 

6. Below are the results of a DNAse sensitivity assay performed on two cell types, one that makes the ethanol dehydrogenase protein (cell A) and one that does not (cell B). Knowing that the patterns seen are from Southern blot studies on DNA from either cell A or B, probed with sequences that detect the gene for ethanol dehydrogenase, circle the result that would be expected from cell A. (4 pts)

The diagram on the right is correct. See exam key in cabinet
 

  Explain your answer. (6 pts)

 Cell A, which expresses the gene, has a looser chromatin structure than cell B that does not express it. This allows the DNAse to have easier access to the DNA, making the DNA more sensitive to digestion, as seen by the quick disappearance of the expected intact band on the gel.
 
 

7. Jane’s new graduate student has just completed her first experiment. She claims that her results identify a new genetic codon for methionine in an unusual cell type she is studying. She has used the Nirenberg method. Look at her results and decide if you agree or not. Explain your answer. (6 pts)

Codon labeled amino acid counts on filter  counts in flowthrough

 GUU  alanine  no    yes
 GUU  tryptophan no     yes
 GUU  methionine no     yes
 GUU  glutamine no    yes
 GUU  valine  yes    no
 AUG  alanine  no    yes
 AUG  tryptophan no    yes
 AUG  methionine yes    no
 AUG  glutamine no    yes
 AUG  valine  no    yes

Disagree. The data show that when AUG is the codon used, the labeled methionine gives a radioactive filter, indicating association of its tRNA anticodon with the AUG codon on the ribosome. The codon GUU gives a negative result for methionine but a positive for valine, indicating GUU codes for valine. AUG is not a new codon for methionine.
 

8. Explain either in words or by diagram the chemical reaction/s that allow amino acyl tRNA synthetase to attach an amino acid to a tRNA. (8 pts)

Amino acyl tRNA synthetase binds to both ATP and to the amino acid.
ATP is hydrolyzed to AMP plus PP. PP immediately becomes 2P.
The energy released is used to attach AMP to the COOH end of the amino acid, thus activating it.
This also causes a conformational change in the amino acyl tRNA synthetase which now can bind to the tRNA.
The AMP is broken off from the amino acid and the energy released is used to attach the amino acid to the 3’terminal A of the tRNA.
 

9. eEF1-GTP acts as an important safeguard during translation elongation. Explain in detail why this is so. (10 pts)

eEF1 binds to the tRNA and this complex goes to the A site of the ribosome. The anticodon of the tRNA bonds to the codon in the mRNA. EeF1 sterically inhibits the attachment of the amino acid end of the tRNA with the ribosome and so the tRNA is in the A/T orientation. GTP is hydrolyzed to GDP, changing the shape of eEF1 which no longer can remain bound to the tRNA and the tRNA can now bind its amino acid end to the A site, using energy released by the GTP hydrolysis (A/A orientation). These events require some time to be accomplished. If the anticodon/codon interaction of the tRNA and mRNA is not sufficiently strong (three base pairs), kinetic motion will knock this tRNA out of the A site before a stable association can occur. This guarantees that an incorrect charged tRNA, not coded for by the codon at the A site, will not be used by peptidyl transferase as the next amino acid added. This is kinetic proofreading.
 

10. Phosphorylation of eIF2 (used during initiation of translation) prevents its ability to be used during the initiation of translation of the globin mRNAs when heme concentration is limited in the cell. Explain why this would make sense, remembering that heme is a prosthetic group for hemoglobin. Also, explain why eIF2 is the factor that is used in this way, remembering its role in the initiation process. (10 pts)

Heme is a necessary prosthetic group for all four subunits of hemoglobin. Without it, hemoglobin cannot bind oxygen. Therefore, if heme is insufficient, globin mRNA translation in not useful. Therefore, translation initiation is stopped by phosphorylation of eIF2, rendering it incapble of participating in its usual initiation roles. These include the interaction with the initiator tRNA carrying methionine and the small ribosomal subunit (subunit initiation complex). This must form first before the complex can bind to eIF4 at the Cap and ultimately scan to find the AUG start codon. Also, eIF2-GTP hydrolyzes the GTP to GDP and uses this released energy to associate the large ribosomal subunit with the small subunit to create a functional ribosome. Translation of globin mRNA cannot proceed if eEF2 is not able to perform these roles.
 
 
 

Problem One Question
 

Briefly explain why the amino acid changes in the alpha 1-antytripsin inhibitor cause trouble for the livers cells that make this protein?

 To gain maximum credit for this answer the following had to be discussed:

Location of synthesis and secretion of the AAT inhibitor. (E.R. and endomembrane system)

Effect upon structure of the protein induced by the point mutations causing polymerization of AAT molecules. (The greater detail here, the higher the score)

The build-up of inclusion bodies due to the above reaction and its disturbance of liver architecture and function.