KEY FOR EXAM ONE, SPRING 2006

1.                  Consider the following categories of amino acid side chains:

A.                 nonpolar

B.                 polar, uncharged

C.                 acidic

D.                 basic

Which categories can interact to stabilize the structure of a protein? Give all possibilities. (5 pts)

            A with A

            B with B

            C with D

             B with C

             B with D

Which categories form ions at physiological pH? (2 pts)

                        C and D

Why are cysteine and proline categorized as “special” amino acids? (6 pts)

Cysteine contains a sulfhydryl group (SH) which, under appropriate cellular conditions, can react with another cysteine to produce a S-S (disulfide bond). This is covalent and has a dramatic effect on protein structure

Proline is an imino acid. This means that there is a covalent linkage between the side chain and the amino group that is attached to the alpha carbon in the amino acid. This creates a bulky ring that is not easily accommodated in the folded structure of the protein and is usually found in turns or looped out regions.

2.                   

A.                 Draw a tripeptide, indicating the three R groups as R1, R2, and R3. (6 pts)

                                See diagram in the textbook and our pictures from the linked website that we used in class.

B.                 Illustrate on the structure you drew the regions that are planar, the psi angles, and the phi angles. (7 pts)

                            Same answer as for A

    C.    Why is the peptide bond planar? (5 pts)

 This is due to resonance. The electrons of the double bonded carbonyl group can move to the C-N bond creating a partial double-bonded structure there. To maintain this, the peptide bond region must remain planar.

3.                  What is the difference between a parallel and an anti-parallel beta pleated sheet.? (5 pts)

The sheet is made up of beta strands. If the polarity (NH2 terminal direction versus COOH teminal direction) of the strands is opposite, the strands are antiparralel. The hydrogen bonds connecting the two strands are straight and therefore stronger than in the parallel beta strands. The parallel strands run in the same direction relative to their polarity and the subsequent hydrogen bonds holding them together are angled, thus not as strong as for the antiparallel.

4.                  Explain how nascently growing peptides on ribosomes are prevented from aggregating before they assume their proper folding structure. (5 pts)

As the polypeptide chain grows during translation, hydrophobic amino acid side chains appear. Chaperone proteins, such as hsp70, interact with ATP. This opens the chaperone’s structure to expose its own hydrophobic amino acids. These bind to the new hydrophobic amino acid side chains on the growing polypeptide, preventing other hydrophobic interactions from occurring between the polypeptide and other polypeptides. This gives time for the 3D structure of the new polypeptide to form. Once formed, the ATP is hydrolyzed to ADP and the chaperone protein disengages from the polypeptide.

5.                  How does a cell target and degrade short-lived proteins? (5 pts)

An amino acid sequence near the NH2 terminal of short-lived proteins induces the attachment of a polymer of ubiquitin proteins to be attached at a lysine residue near the NH2 terminal. This is achieved by three enzymes, E1, E2, and E3 which, respectively, bind a ubiquitin, transfer it to the next enzyme, and then ligating the ubiquitin to the lysine. The process repeats over and over again, creating the polymer. Once formed, the polymer allows the protein to be brought to a proteosome, deposited inside it, where it is degraded.

6.                  How does binding of ATP to protein kinase A alter the enzyme’s tertiary

      structure? (5 pts)

When ATP enters the binding cleft created by the large and small domains of the catalytic subunits of protein kinase A (the active site), it induces a conformational change in the active site (induced fit). A glycine lid moves closer down over the ATP, preventing it from leaving the active site and bringing the polypeptide to be phosphorylated by the ATP closer to the ATP. When ATP is hydrolyzed by the enzyme to ADP, the removed gamma phosphate is covalently linked to the side chain of a serine or threonine amino acid in the active site. This causes the glycine lid to open again, releasing the ADP and the now phosphorylated peptide.

7. Describe the secondary structure of B-DNA in detail. (8 pts)

A right-handed double helix composed of two DNA polymers running in opposite polarities relative to their 5’ and 3’ ends. The sugar-phosphate backbones are to the exterior of the double helix, the nitrogenous bases are in the interior, perpendicular to the axis of the helix and slightly tilted. The bases interact across the diameter of the helix such that A forms two hydrogen bonds with T and C forms three hydrogen bonds with G. The topography of the helix creates major and minor grooves that alternate along the two helices and oppose one another on opposite sides. There are approximately 10.6 base pairs per turn of the helix (3.4 nm) and .34 nm distance between base pairs.

Which feature of the structure explains the following: (3 pts each)

A.                 loss of viscosity upon heating

The hydrogen bonds between the nitrogenous bases in the interior of the double helix are broken when heated, decreasing the density.

B.                 Chargaff’s rules

The concentrations of A and T are equal and those of C and G are equal because A bonds to T and C bonds to G.

C.           3.4 and .34 nm repeating features in x-ray crystallography patterns.

3.4 nm represents one turn of the helix. .34 nm represents the distances between base pairs.

8. DNAse sensitivity is often used to probe chromatin structure. Which pattern below, illustrating the results of such an assay, suggests an actively transcribed gene? (4 pts)

In our picture, gel B shows the actively transcribed gene.

            Explain your answer (5 pts)

Actively transcribed genes have a more loosely packed chromatin structure. Therefore, they are more easily degraded by DNAse. In the experiment, when increasing concentrations of DNAse are used, the susceptible gene will begin to degrade. The non-susceptible, unexpressed gene will not degrade, even at high concentrations of the DNAse.

9. What role in DNA replication is played by the following? (3 pts each)
   

A.  helicase

Using ATP, it breaks the hydrogen bonds holding the two DNA strands together, providing the DNA template for replication.

B. topoisomerase

It nicks one strand of the supercoiled double helical region upstream from the progressing replication fork to allow for the other strand to swivel through, relieving the supercoiling that was caused by the melting of the hydrogen bonds between the bases as replication proceeds.
 

C.     primase

This DNA dependant RNA polymerase enzyme lays down an RNA primer to begin all new DNA polymers during replication, in order to provide a 3’OH for DNA polymerase to extend.

D.     5’-3’ exonuclease activity

This removes the RNA primers, one nucleotide at a time, in the 5’-3’ direction.

E.   3’-5- exonuclease activity

This is a proofreading activity of DNA polymerase delta that senses the just added nucleotide is incorrect, flips it out, and cuts the phosphodiester bond that was just made (backs up, so to speak, in a 3’-5’ direction).

10. Below are three replication forks. On each, indicate which strand would be replicated discontinuously. (3 pts)

The 5’ to 3’ strand on each is the template for the lagging strand (replicated discontinuously).

11. What is the mechanism by which DNA ligase attaches two Okazaki fragments together? (5 pts)

First, ATP is hydrolyzed to AMP and PP (pyrophosphate). PP breaks down to 2P.This provides energy for the covalent attachment of the AMP to the 5’ phosphate of one of  the Okazaki fragments. That activates the phosphate. The AMP is then removed and the energy released from breaking that bond is used to created a bond between the 5’ phosphate and the 3’OH of the other Okazaki fragment, thus joining them.

Extra Credit:  Briefly explain how telomerase prevents the ends of chromosomes from shortening with each round of DNA replication.

At the telomere there is a single-stranded overhang of the DNA strand with a 3’OH end. It contains repeating sequences. Telomerase contains an RNA template that is antiparallel and complementary to that sequence. It also has a catalytic protein component that has reverse transcriptase activity that can use an RNA template and make from it a DNA polymer. When acting, the RNA component base pairs to the repeat sequences and telomerase’s catalytic activity uses the RNA information to extend the 3’OH end of the overhanging DNA strand. This extends to nucleotide number 35 on the RNA template, releases, and slides the extended strand off the RNA template, allowing it to reanneal closer to the beginning of the DNA strand again. The end is extended again up to nucleotide number 35, released, and the template reanneals again. This process is repeated over and over until sufficient telomeric repeats have been added to the 3’ overhang such that the other DNA strand (5’ phosphate end), could then have more opportunity for Okazaki fragment synthesis to occur and therefore that strand would not continue to shorten.