1. A. Illustrate the two polar bonds that make a water molecule. (4 pts)

 Page 112 in our text.
 
 

     B. How does the structure of water influence the following? (3 pts each)

a. hydrophobic interactions

The presence of hydrophobic molecules or regions of molecules causes water to have to cage around these regions to maintain its hydrogen bonding ability with other water molecules. This causes an unfavorable decrease in entropy. To compensate for this, the hydrophobic molecules or regions move together.
 

b. ionic bonds

Water’s dipoles attract the positive and negative ions, thus pulling them away from each other, weakening them.
 

c. solubility of polar molecules

Water’s positive dipoles attract negative dipoles on other polar molecules and water’s negative dipoles attract positive dipoles on other polar molecues.
 
 

2. The reaction between molecules A and B forming molecules C and D has a      G of –125 kcal/mole. To reach the transition state, the reactants must possess a free energy of 250 kcal/mole. Draw a graph that illustrates this chemical reaction with the progress of the reaction plotted on the x axis and the free energy on the y axis. (5 pts)
 
 

The distance between the startpoint of free energy for the reactants up to the top of the activation energy peak is 250 kcal/mole. This should have been illustrated on the graph. Similarly, the distance between the startpoint of free energy for the reactants down to the endpoint of free energy for the products is –125 kcal/mole. This should also have been illustrated.
 

3. Draw a tripeptide, indicating the side chains of the amino acids as R1, R2, and R3. Show where the regions of planarity are and indicate the phi and psi angles of rotation. (8 pts)
 

 Figure 3-4, page 134
 

4. Answer the following questions about amino acids.

A. What side chain is often found at turns in a polypeptide chain? (2 pts)

proline

B. What side chain can cause the formation of disulfide bonds? (2 pts)

cysteine

C. Circle the side chain interaction(s) that is/are not likely to be stable. (6 pts)

Hydrophobic with hydrophobic
Polar with basic
Polar with polar
Basic with acidic
Polar with acidic
Hydrophobic with polar *
Hydrophobic with acidic *
Acidic with acidic *
Basic with hydrophobic *
Basic with basic *
Hydrophobic with basic *
 

  * indicates the correct answers

5. Illustrate the following protein secondary structures in diagrammatic form. You do not need to illustrate the hydrogen bonds stabilizing the structures. (3 pts each)

A. alpha helix
 

figure 3-9, page 137
 
 
 

B. a beta sheet consisting of 2 parallel beta strands and one anti-parallel beta strand.
 
 

Two downward pointing arrows next to each other and one upward pointing arrow next to that.
 

6. Suppose there exists a metabolic pathway that creates an end-product called final. To get to final, many intermediate reactions must occur. An allosteric enzyme, finalase, catalyzes the first reaction in the pathway and is subject to both activation and inhibition. The last chemical reaction in the pathway  is: son of final + phosphate = final. Using this information, explain a way to regulate the finalase enzyme (both by activation and inhibition). Be thorough in your explanation of the mechanism. (8 pts)

Finalase exists in an active and inactive form, equally likely to be present and in equilibrium. If the concentration of final builds up, it will bind to the inactive form and stabilize it, keeping it from flipping to the active form. Active forms will still flip to inactive. Thus the total number of inactive forms increases and there is less catalysis of the first reaction in the pathway, slowing the entire pathway. This is feedback inhibition.

If the concentration of son of final builds up, that indicates the pathway is moving too slowly and not making enough final. Son of final then binds to the active form of finalase, stabilizing it and keeping it from flipping to inactive. Inactive forms can still flip to active. Therefore, there is an incease in active forms and catalysis of the first reaction increases, causing an increase in the entire pathway.
 

7. What aspects of the structure of B-DNA (Watson and Crick’s structure) explain the following pieces of evidence? (3 pts each)

A. loss of viscosity upon heating

Double-stranded (more viscous) DNA became single stranded (not viscous) due to melting of the hydrogen bonds between the nucleotide bases holding the two strands together.
 

B. .34 nm repeating features in x-ray pictures

Caused by the spacing of the nucleotides. Each is .34 nm apart from the next.

C. Chargaff’s Rules

The concentration of A and T bases is the same as is the concentration of C and G bases in DNA from many species and individuals of a species. This is due to the fact that in DNA molecules, A base pairs to T and C base pairs to G.
 

8. Answer the following questions about chromatin.

A. Outline the steps that lead to the formation of the histone octamer. (5 pts)
 

Figure 4-27, page 210
 

B. Explain how the addition of an acetyl group to a histone can be involved in regulating chromatin structure. (5 pts)

The addition of the acetyl group to lysines at the amino or carboxyl terminal of a histone interferes with the attraction of the positive lysine to the negative phosphate that is part of the DNA nucleotides. This loosens the compaction of the DNA to the histone octomer, facilitating the entry of proteins involved in activating gene expression.

9. Answer the following questions about DNA replication.

A. Why is it necessary to use the enzyme DNA primase? (4 pts)

DNA polymerase cannot make phosphodiester bonds without the presence of a pre-existing 3’OH group already present. DNA primase is can make a short RNA primer that provides this 3’OH and thus allows new DNA strands to initiate.
 
 

B. Describe the enzymatic activity that is responsible for the proofreading ability of  DNA polymerase and how it works? (6 pts)
 

This is a 3’-5’ exonuclease activity. If an incorrect nucleotide is added to the growing chain by DNA polymerase (e.g., due to a rare tautomeric shift in structure), the incorrect base pairing between it and the template is sensed by the DNA polymerase which then breaks the phosphodiester bond the the 5’ side of the incorrect nucleotide, thus removing it.

C. Draw a schematic diagram of a eukaryotic replication fork, indicating on the diagram the location of the following: helicase, topoisomerase, DNA polymerase alpha, DNA polymerase delta, and primase. (6 pts)
 

  Figure 5-28, page 254. I would also expect to see the polarities of the strands indicated.
 

10. Consider the following scenario

A. A new student begins doing research on DNA replication. He sets up a DNA replication reaction and nothing happens. He checks his materials and notices that his template sequence is ‘5GTTAACCGTAACG3’. His primer sequence is 5’ UAAC3’. Why didn’t his reaction work (assuming all other needed components were working correctly)? (5 pts)
 

His primer would base pair at the 5’ end of the template. This would leave a 5’ phosphate on the nucleotide to be extended. DNA replication can only occur in the 5’-3’ direction, using a 3’ OH. So nothing happens.
 

B. What should he look for in the refrigerator before he tries the reaction again (be specific)? (5 pts)
 

The correct primer:  5’ CGUU3’
 
 

11. This question will be given to you in class on the day of the exam. (5 pts)
 

Telomerase is an RNA-dependent DNA polymerase  because it relies on information in the form of an RNA template to direct the polymerization of a DNA polymer. This is sufficient to answer the question. But below is some additional information many of you wrote about also.

 Telomerase brings the RNA template with it as part of the telomerase enzyme. It can base pair to the 5’ end of the telomere, allowing its catalytic DNA polymerizing activity (in the protein component of telomerase) to extend the 3’ end of the opposite strand of the telomere.