1. Using the free energy equation, G = H – TS, explain why hydrophobic interactions are such a powerful influence in aqueous environments. (5 pts)

Negative delta G values indicate a favorable reaction. The enthalpy component is favored by hydrogen bonding between water molecules. In the presence of hydrophobic molecules, however, water must form a cage to hydrogen bond. This increases the order of the system and is unfavorable due to a decrease in entropy. If hydrophobic molecules interact with one another, this decreases the amount of caging and thus the increase in order. Entropy thus increases and the delta G is more negative (favorable).
 

2. A single amino acid change of valine for glutamic acid can lead to the formation of dysfunctional proteins that causes serious diseases like sickle-cell anemia. What effect upon protein structure could result from this change that could explain this? (5 pts)

Non-covalent interactions between amino acid side-chains are the most significant factor in determing the 3-dimensional shape of proteins. This shape, (tertiary structure) is the main reason that proteins function properly. Valine is hydrophobic. Glutamic acid is acidic and carries negative charge. Therefore, glu can form ionic interactions with basic side chains and interact with polar side chains of other amino acids near it. Valine favors interactions with other hydrophobic side chains and cannot interact with polar or charged side chains. Therefore, these changes in interactions can significantly influence tertiary structure and function of proteins if they occur.
 
 

3. Draw a tripeptide, designating the side chains as R1, R2, and R3 and label the regions of planarity and the angles of free rotation, psi and phi. (10 pts)
 

See the link on our course website to get this picture.
 

4. Use the following diagram to answer questions b and c. (4 pts)
 

Consult your exam for the diagram.
 

a. How many parallel beta strands are present? 2

b. How many antiparallel beta strands are present? 3
 

5. List the 4 major non-covalent bonds or interactions that determine protein tertiary structure. (8 pts)

Hydrophobic interactions, ionic interactions, hydrogen bonds, and van der Waal’s forces.
 

6. Define these terms: (3 pts each)

A. Chaperone A protein that assists in the proper folding of a polypeptide as it is being made by binding to exposed hydrophobic side chains, preventing aggregation, until the protein can finally assemble into the correct tertiary structure.
 

B. Ubiquitination A process that tags proteins for destruction in a proteosome. The tag consists of the addition of a ubiquitin polymer to a lysine side chain near the amino terminal of the protein to be degraded.
 

C. Proteosome A cylindrical structure lined with proteases that receives other protein tagged with ubiquitin polymers and degrades them.
 
 

7. Use the axes below to draw what a hypothetical exergonic chemical reaction that is not catalyzed would look like, labeling the activation energy peak and the overall change in free energy. (6 pts)

Consult your notes. The activation energy is the distance between the G of the reactants and the G of the transition state. The delta G is the distance between the G of the products of the reaction and that of the reactants.

8. How do the conformational changes that are induced by the binding of ATP to protein kinase A demonstrate the concept of induced fit? (6 pts)

Induced fit refers to a change in the conformation of the active site of an enzyme caused by the binding of its substrate. When ATP enters the binding site of protein kinase A, it induces a conformational change that moves the glycine lid over the binding site in a closed conformation, holding the ATP there until the transfer of its terminal phosphate to the amino acid side chain of its target protein. When this reaction is completed, the glycine lid moves to the open conformation once again.
 

9. Describe the structure of B-DNA in detail. (10 pts)

Points that needed to be shown or described:

Double-right handed helix.

The two strands wrap around each other, running antiparallel (3’ to 5’ and 5’ to 3’).

Sugar-phosphate backbone on the exterior of the double-helix, interacting with water.

Interior region is hydrophobic due to the base pairs extending into the interior, perpendicular to the axis of the helix, and hydrogen bonding with one another.

A bonds T (2 H bond) and C bond G (three H bonds).

Approximately 10 base pairs per turn of the helix.

Major grooves and minor grooves form on the surface of the helix with a major followed by a minor followed by a major, etc. One opposing sides, a major is opposite a minor.
 

10. If a gene is transcribed in cell A but not in cell B, which cell would exhibit degredation of the DNA containing that gene in the presence of DNAse? Explain why. (5 pts)

Cell A because the chromatin structure of transcribed genes is more open and accessible to DNAse due to the need for the transcription factors to bind in order to transcribe the gene.
 

11. Match the following terms to the best statement. (2 pts each)

Topoisomerase
Helicase
 DNA polymerase alpha
 DNA ligase
 Primase
 DNA polymerase delta
 Telomerase
 Single-stranded DNA binding protein
 

a. Connects Okazaki fragments __DNA ligase__________________
b. Synthesizes the leading strand __DNA polymerase delta (I also accepted alpha since it begins the synthesis.__________________
c. Melts hydrogen bonds between DNA strands __Helicase________________
d. Relieves tension as the replication fork progresses ___Topoisomerase_____________
e. An RNA polymerase __Primase_______________________
f. Contains a 3’-5’ exonuclease activity __DNA polymerase delta_____________
g. Prevents shortening of chromosomes in some cells ___Telomerase_____________
h. Prevents melted hydrogen bonds from reforming ____single stranded DNA binding protein____________
 

12. Below is a list of 4 single-stranded nucleic acid primers available to a graduate student who is trying to copy a DNA molecule into its complementary strand using DNA polymerase. The sequence of the template he is trying to copy reads 5’CTTAA……GCCGCA3’ Circle the correct primer he/she should use (5 pts) and explain why this one is correct but the other choices are not. (5 pts)

A. 5’GAATT3’
B. 5’TGCGG3’ the correct choice
C. 5’TTAAG3’
D. 5’GGCGT3’
 
 
 

Explanation for Q 12 choice:

This primer would base pair correctly (complementary bases and antiparallel) to the template at its 3’ end. That is required because DNA polymerase requires a free 3’OH be present for it to extend the polymer.
 

13. Outline how thymine dimers are removed from double-stranded DNA. (6 pts)

Consult the textbook for this mechanism. All factors needed to be correctly named and their activity described for full credit.