401 exam 1 key spring 2009

Key for exam 1, BISC401 Honors, Spring 2009

1. 1. Use the above figure to answer the following questions:

A. Which number represents the omega angle? __2_____ (2 pts)

B. Which number represents the phi angle? ____3____(2 pts)

C. Which number represents the psi angle? ___1____(2 pts)

Below, draw the resonance structure of the peptide bond. (4 pts)

See page 74 of the textbook

2. 2. We discussed three secondary structure motifs in class.

A. What is a secondary structure motif? (3 pts)

A combination of secondary structures frequently seen in proteins that perform a similar activity of some type, (for example DNA binding, or calcium binding, etc.)

B. Choose one of the three motifs we discussed and describe its structure and the role that the motif plays in the proteins that contain it. (5 pts)

See figure 3-9 on page 70 of the textbook for either the coiled-coil, the EF hand/helix-loop-helix, or zinc finger motifs.

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

A. Molecular Chaperone: member of a group of proteins (e.g. hsp 70) that interact with exposed hydrophobic side chains on nascent or denatured proteins in order to prevent aggregation of the proteins and facilitate their normal folding structure.

B. Chaperonin: Barrel-like protein structure that helps many proteins fold properly by allowing them to enter the barrel and have needed time to fold properly.

C. Proteosome: Barrel-like structure which receives proteins designated by ubiquitination tags for degradation and uses proteases within it to break the proteins into peptides.

Choose one of the three terms you defined above and describe the mechanism by which it performs its function. (6 pts)

Molecular Chaperones: See figure3-16; Chaperonins: see figure 3-17; proteosomes: see figure 3-29

4. 4. Describe the structure of eukaryotic DNA. Include all details about the B-DNA structure defined by Watson and Crick and also what constitutes interphase chromatin structure. (10 pts)

Your answer should include: Right-handed double helix of two DNA strands running antiparallel to one another. The sugar-phosphate backbones are located to the outside of the double helix with nucleotide bases projecting inwards perpendicular to the axis of the helix. Bases are hydrogen bonded between the two strands with A bonding to T with two hydrogen bonds and C bonding to G with three hydrogen bonds. There are roughly 10 base pairs per turn of the double helix represented by 3.4 nm and .34 nm between the base pairs. Spaces between the strands on the outside of the double helix produce major and minor grooves. The diameter of the helix is about 20 nm.

Interphase chromatin consists of nucleosomes connected by linker regions and further compacted into 30-nm fibers. The nucleosomes are 147 base pairs of ds DNA wrapped one and ¾ times around a core of histones consisting of 2 copies of H2A, H2B, H3, and H4. Linker regions are of variable lengths from about 540-200 base pairs long. Histone H1 sits at the junctions between the nucleosomes and the linkers and facilitates the stacking of the histones and formation of a 30-nm diameter higher folding structure called a cylinder, fiber, or solenoid. (See figure 6-30 and page 249).

5. 5. Describe how the side-chain specificity binding pockets of trypsin, chymotrypsin and elastase determine the specificity of these enzymes. (4 pts each)

Trypsin binds basic amino acid side chains and the pocket has a centrally located aspartic acid whose negative charge facilitates this. Chymotrypsin binds large aromatic side chains and has a binding pocket that is wider and contains a serine. Elastase binds small hydrophobic side chains like glycine or alanine and has two inwardly projecting valines whose branched hydrophobic side chains prevent larger amino acid side chains from entering the pocket. (See figure 3-25b).

6. 6. Answer the following questions about the chemical mechanism of trypsin digestion.

A. What three amino acids make up the catalytic triad? (3 pts)

Aspartic acid, histidine, and serine

B. How is this mechanism an example of acid-base catalysis? (6 pts)

Histidine has a proton drawn toward aspartic acid and this allows histidine to draw the proton from the hydroxyl group on the serine side chain. This activates the oxygen for the attack on the peptide bond and formation of the first tetrahedral intermediate in the reaction. Later, histidine gives a proton back to form the NH2 terminal of the first peptide released. It again attracts a proton from water as the second tetrahedral intermediate is formed and later gives it serine to reform the OH side chain.

C. How many transition states occur during the reaction? (2 pts)

7. 7. Explain the accuracy of the DNA polymerase enzyme. In other words, what mechanism is used to help insure no errors in DNA replication occur by using this enzyme? (6 pts)

First, the incoming nucleotide is base-paired to the template to check for correct hydrogen-bonding before the phosphodiester bond formation is catalyzed. Second, there is a 3’-5’ exonuclease activity associated with DNA polymerase delta. When the polymerase detects a mispaired base it releases the 3’ end of the growing strand and the end is brought to the 3’-5- exo site of the enzyme. The incorrect nucleotide is removed (cuts the phosphodiester bond to the 5’ side) and then the 3’ end is returned to the polymerizing site for addition of the correct nucleotide. See figure 4-34)

8. 8. Match the following terms to the correct statement. The terms can be used more than once. (2 pts each)

DNA polymerase alpha DNA polymerase delta DNA ligase

Primase topoisomerase PCNA

Lagging strand Leading strand telomerase

A. Does not require a template to catalyze formation of a phosphodiester bond. _Either primase or DNA ligase __

B. Relieves torsional strain as the replication the fork progresses _topoisomerase_

C. Increase the processivity of continuous DNA replication _PCNA___

D. Uses a 3’-5’ exonuclease activity __DNA polymerase delta____

E. Provides the 3’OH needed to begin the synthesis of new DNA strands__primase___

F. Copied from a template strand that runs 3’-5’ relative to the movement of the replication fork___leading strand__

G. An RNA-dependent DNA polymerase. _____telomerase____

9. 9. Outline the mechanism by which two Okazaki fragments are connected by a phosphodiester bond. (6 pts)

First, ATP is hydrolyzed to AMP plus pyrophosphate. Pyrophosphate breaks down to 2Ps and the energy released is used by DNA ligase to attach the AMP to the 5’P of one of the Okazaki fragments to be joined. Then ligase breaks that bond and transfers the energy to the formation of phosphodiester bond between the 5’P and the 3O of the other Okazaki fragment (a condensation reaction).

10. 10. This question will be given to you in class on the day of the exam. (8 pts)

The lagging strand at the telomere cannot be completed because there is insufficient room for the priming and synthesis of another Okazaki fragment before reaching the end of the template. This produces an overhang of about 15 nucleotides in length where the strand ending with a 3’OH (that played the role of the template for the lagging strand synthesis) overhangs the 5’ end of the strand that was the lagging strand. If these two strands then would be used for synthesis of new DNA strands in another round of replication, eventually there would be continuing shortening of the DNA at the telomere that could impact genetic information. (See figure 6-48).
Telomerase is a reverse transcriptase that contains both a protein component and a small RNA component that is complementary in sequence to the telomere sequence at the end of the 3’ strand. Telomerase can base-pair there, forming a DNA-RNA hybrid and then use the RNA template to add DNA nucleotides to the 3’ end of that telomere strand, lengthening it. It repeats the lengthening again and again, adding enough nucleotides to extend the 3’ end of the telomere sufficiently for additional Okazaki fragments to be able to form on a lagging strand that would use the elongated strand as a template during DNA replication. (see figure 6-49)