Questions 1-4:  Answer True or False. (2 pts each)

1.       Ionic bonds are weaker when water is present. ____T

2.       Entropy is the driving force behind hydrophobic interactions. ____T

3.       Alpha-glucose monomers are used to create structural polysaccharides like cellulose. ___F

4.       Two condensation reactions are used to make a triglyceride molecule. ___F

5.       Draw a peptide bond. Illustrate the planar region and the omega, psi, and phi angles. Side chains can be designated by the symbols R1 and R2.(8 pts)

See Figure 3-3

6.       Briefly describe or illustrate the structure of the following and explain how the structure is stabilized. (4 pts each)

a.       An alpha helix

Right handed helix stabilized by hydrogen bonding between peptide bonds where the C=O oxygen bonds to the NH hydrogen four peptide bonds away.  There are about 3.6 amino acids per turn of the helix. The The R groups of the amino acids project to the outside of the helix.

b.      An antiparallel beta sheet

Secondary structure formed by two or more linear strands of amino acids that run in opposite polarities (relative to the carboxy and amino termini of the polypeptide). Stabilized by hydrogen bonds between the strands’ peptide bonds where a C=O oxygen of one peptide bond of one strand bonds to the NH hydrogen of a peptide bond  the other strand. The hydrogen bonds are linear and therefore stronger than in the parallel sheets below.The R groups project above and below the plane of the sheet.

c.       A parallel beta sheet

Secondary structure similar to that described in but having the strands run in the same polarity. The hydrogen bonds are thus forced into an angle which makes these sheets less strong.

7.       Answer EITHER question A or question B. (10 pts)

A.      Why does the mechanism by which a serine protease catalyzes breaking of a peptide bond not allow the enzyme to function in the stomach?  

The catalytic activity depends upon the ability of the histidine at the active site to function as a weak base. Binding of the polypeptide chain to the enzyme causes an induced fit that brings an aspartic acid near a histidine at the active site and pulls a H from histidine towards the negatively charged aspartic acid. Histidine then pulls an H toward itself and away from the OH group of a nearby serine at the active site. This activates the serine O for attack on the peptide bond of the substrate. If in the stomach, the low pH would have histidine already in a protonated state and thus be unable to participate in the above reaction. The serine O could not then become activated.

B.      Briefly explain how lysozyme lowers the activation energy of the hydrolysis reaction that breaks a glycosidic bond. You do not need to go into detail about the organic chemistry of the reaction.

Upon binding of the polysaccharide to the enzyme, an induced fit occurs that forces the configuration of one of the sugars into a strained conformation. This situates the glycosidic bond to be broken near two acidic amino acids that participate, along with water, in the breaking of the glycosidic bond, forming a transient  intermediate (transition state) that covalently connects the enzyme to the strained sugar before the reaction is completed.

8.       Answer EITHER question A or question B. (10 pts)

A.      Describe how a chaperone protein works to help a protein fold correctly.

An oligomeric chaperone protein, bound to ATP, is helped by  another chaperone protein to bind to exposed hydrophobic regions on nascent polypeptides or unfolded polypeptides. ATP is hydrolyzed to ADP, causing the chaperone to more closely bind and allow the protein time to fold correctly. Following dissociation of the ADP and the helper chaperone, rebinding of ATP to the chaperone causes it to release polypeptide that hopefully is now correctly folded. If it is not, this process can repeat several times until the still incorrectly folded polypeptide is moved to a chaperonin or possibly targeted for destruction.

B.      Explain how a protein gets targeted for destruction in a proteosome.

The protein will have an amino acid signal for attachment of a polyubiquitin tag to one of its amino terminal lysine side chains. This can either be inherent in the amino acid sequence of short-lived proteins or be exposed due to misfolding. The attachment of the tag begins when the E1 protein interacts with ATP which is hydrolyzed to AMP and PP which then further breaks down to 2P. The energy released is used to attach the AMP to the COOH terminal of an ubiquitin polypeptide. This activates the COOH for the attachment to an E1 cysteine side chain following removal of the AMP and the release of energy. The E1 interacts with an E2E3 complex called ubiquitin ligase. The E1 transfers the ubiquitin to E2. The E3 of the complex associates specifically with the tagged protein and the ubiquitin is transferred to the lysine side chain on the protein. This process repeats multiple times to attach a large string of ubiquitins.

9.       Why is the interior of the DNA double helix hydrophobic? (5 pts)

All polarity associated with the nitrogenous bases that project into the interior of the double helix is negated when the bases hydrogen bond to one another.

10.   Draw a phosphodiester bond. You do not need to show the base, just designate it as B. (8 pts)

See Figure 2-28

11.   Describe the structure of a nucleosome. (5 pts)

Consists of an octameric core of histone proteins (2 each of H2A, H2B, H3, and H4) that has 147 base pairs of double-stranded DNA wrapped around it 1.7 times.

12.   What is the function of the following during DNA replication? (4 pts each)

a.       Helicase

Melts the hydrogen bonds holding the two strands of DNA together in order to expose the single stranded templates.

b.      DNA Ligase

Creates the final phosphodiester bonds that connects two Okazaki fragments together.

c.       Topoisomerase

Nicks one strand of the double helix upstream of the replication fork, providing a swivel point for the other strand in order to relieve strain and supercoiling that occurs when the double helix is opened by helicase during replication. The topoisomerase attaches covalently to a 5’ P thus breaking the phosphodiester bond. Following the relief of tension, the enzyme releases and reforms the phosphodiester bond.

d.      Sliding Clamp (PCNA in eukaryotes)

This is a protein that associates with the DNA double helix and with DNA polymerase delta to help keep the polymerase attached to the template strand for long stretches of DNA synthesis, thus increasing its processivity.

13.   The two primary DNA polymerases in DNA replication in eukaryotes are called alpha and delta. Indicate whether the activities listed below are found in or associated with one, both, or none of these. If they are found in only one, identify which. (2 pts each)

a.       Synthesis of phosphodiester bonds. ___________________both

b.      Close association with the primase. ____________________alpha

c.       3’-5’ exonuclease activity. _________________________delta

d.      Connection of two nucleotide triphosphates to begin a new DNA strand. ______neither

e.      Close association with the sliding clamp (PCNA). __________________delta

f.        High processivity. ___________________delta

14.   Answer EITHER question A or B. (6 pts)

A.      Describe the mechanism by which DNA ligase attaches two nucleotides together.

Ligase hydrolyzes ATP to AMP plus PP. PP breaks down further to 2P. The energy released attaches the AMP to the 5’ phosphate end of one Okazaki fragment. The enzyme then catalyzes the release of the AMP and the energy released by this is used to make a phosphodiester bond between the 5’ phosphate and the 3’ end of the next Okazaki fragment.

B.      Describe the role played by the RNA component of the telomerase enzyme.

The base sequence of the telomerase RNA serves as a template for the RNA-dependent DNA polymerase activity of telomerase. The RNA sequence is antiparallel and complementary to the sequence of DNA in the telomere repeat at the 3’ end of the overhang at the telomere. This DNA strand serves as the lagging strand template during DNA replication but is too short for the final Okazaki fragments to be made for the lagging strand. The telomerase extends the 3’ end of this template DNA strand by reading its annealed  RNA template and inserting complementary DNA nucleotides. The template RNA dissociates and reanneals multiple times, allowing a long extension of repeats to be made on the DNA template strand. This allows additional Okazaki fragment synthesis on the lagging strand

Bonus Question:

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. (4 pts)

A.                  5’GAAUU3’

B.                  5’UGCGG3’

C.                  5’UUAAG3’

D.                  5’GGCGU3’

The correct answer is B. This is the sequence that is antiparallel and complementary to the 3’ end of the DNA being copied. This allows the new strand being made to be extended in a 5’to 3’ direction when DNA polymerase uses the RNA template as a source of the 3’OH required to make the phosphodiester bond.