1. The disease sickle-cell anemia is caused by a single amino acid change from glutamic acid to valine. The side chain of glutamic acid is –CH2CH2COOH. The side chain of valine is –CH .
H3C CH3
This substitution occurs in the hemoglobin protein. Using what you have
learned
about amino acid side chains and their influence on protein structure,
explain
what may happen to hemoglobin when this substitution occurs. (8 pts)
Valine is non-polar and glutamic acid is negatively charged (acidic). Valine can only make hydrophobic interactions with other amino acids; glutamic acid can make ionic interactions with other basic amino acids and can also interact well with polar uncharged amino acids. Therefore, the tertiary structure of hemoglobin can become significantly changed when the above substitution occurs.
2. Discuss why peptide bonds must be planar. (7 pts)
Due to resonance of the double bond in the peptide linkage, the omega angle of rotation is fixed. This resonance occurs as the C=O resonates with the C-N bond giving both C-O and C-N a partial double bond characteristic. The only free rotation in the linkage is at the psi and phi angles connected to the alpha carbons. This forces the planarity of the peptide bond.
3. Hydrogen bonding is important for protein secondary structure. Describe its role in the following: alpha helix, antiparallel beta sheet, parallel beta sheet. (6 pts).
Every peptide linkage in the alpha helix is hydrogen bonded to another peptide linkage four residues away. The H bonding is between the O of the peptide bond and H of the other peptide bond. All H bonds point in the same direction relative to the axis of the helix, giving the alpha helix a polarity.
Antiparallel beta sheets have similar H bonds but they are located between peptide linkage across the span of two beta strands that run in opposite polarities. They are relatively linear. Parallel beta sheets have similar H bonds to antiparallel but they are skewed and at an angle, making them slightly less strong. The sheets are pleated in both types due to the need to maintain peptide bond planarity.
4. What is meant by a hydrophobic interaction and why does it form in a water environment? (5 pts)
Water is driven to form intermolecular hydrogen bonds with other water molecules due to enthalpy considerations. However, in the presence of a hydrophobic molecule or part of a molecule, water must cage around the hydrophobic to form the H bonds. This caging imposes order on the system which is not favored by entropy considerations. The way to maximize the enthalpy and minimize the negative effect upon entropy is for the system to bring hydrophobic molecules or regions together and then water need form one larger cage instead of many smaller cages. This is more favorable as in its entirety the system will have less order in this type of system.
5. What differs about the binding pockets of trypsin, chymotrypsin, and elastase and how does this cause a particular peptide bond to be broken in a polypeptide.? (9 pts)
Trypsin has a binding pocket with an aspartic acid in it that can attract the basic amino acid side chains of lysine and arginine. Trypsin then cleaves the peptide bond after one of these. Chymotrypsin has a binding pocket containing serine and is fairly large. It accommodates the bulky aromatic side chains such as phenylalanine and cleaves after them. Elastase has two valine side chains in its binding pocket that are located opposite one another and project inside the pocket, leaving very little room. Only very small nonpolar side chains such as in glycine or alanine can fit there and therefore elastase cleaves after glycine.
6. Serine is part of the catalytic triad at the active site of trypsin. How is its oxygen activated to begin the attack on the peptide bond being cleaved by trypsin? (5 pts)
The acidic aspartic acid residue 102 is brought close to the histidine 57 in the active site of the enzyme due to induced fit. The aspartic acid draws a proton from the histidine. The histidine is then able to draw a proton from the serine OH side chain. This is acid-base catalysis. This activates the O on serine which attacks the C=O of the peptide bond to be broken.
Why is this a pH-dependent reaction? (3 pts)
Only when the histidine is not-protonated will the acid-base catalysis work. If the pH of the enzyme’s environment is low enough, the histidine will pick up a proton from the solution and not from the serine OH group. Thus, the O will not be activated as described above.
7. Feedback inhibition describes what happens when the final product of a metabolic pathway acts as an inhibitor of an allosteric enzyme that works early in the pathway. Explain how this inhibition happens, using what we have discussed about allosteric enzyme regulation. (6 pts)
The allosteric enzyme exists in two equally likely conformations, active and inactive. The active (correct active site shapes) can flip to inactive (incorrect active site shapes) in an equilibrium that favors neither. Each has two catalytic subunits containing the active sites and two regulatory subunits containing allosterics sites that can bind to regulatory molecules. In feedback inhibition, the product of the pathway acts as the regulatory molecule designed to slow down the entire pathway by down-regulating the allosteric enzyme. It does this by binding the allosteric sites on the regulatory subunits of the inactive forms, thus stabilizing these forms. They will then not flip back to active. But actives can still flip to inactive. Therefore the entire population of molecules is enriched for inactive forms, the equilibrium is shifted to the inactive side of the equation, and the reaction catalyzed by the allosteric enzyme is inhibited. This causes all downstream reactions then to also be downregulated.
8. Either draw or describe in words the structure of B-DNA. Be very detailed in your description. ( 8 pts)
Two right handed alpha helices wrap around one another in antiparallel directions, running 5’-3’ on one strand of the DNA and 3’-5’ on the other. The backbone of both strands is phosphodiester bonds between the individual nucleotides of the strand (sugar-phosphate backbone). The bases of the nucleotides extend across the inside of the double-helix, perpendicular to the axis of the double helix and connect to each other by hydrogen bonds. C bases form 3 hydrogen bonds with G bases and A bases form 2 hydrogen bonds with T bases. The diameter of the double helix is about 2.0 nm. Each turn of the double helix contains approximately 10 basepairs with a total of about 3.4 nm per turn. Thus the basepairs are about .34 nm apart. The spaces between the helices that form during the winding of the two strands are one larger space (major groove) and one smaller space (minor groove) that alternate.
Describe a nucleosome. (4 pts)
About 147 basepairs of double stranded DNA wrap 1 2/3 times around a core of histones consisting of 2 each of H2A, H2B, H3, and H4. These nucleosomes are separated on the DNA by linker regions without nucleosomes. Histone H1 is placed at the junctions between nucleosomes and linkers and is used to direct higher folding into the 30 nm fiber.
9. A new graduate student is tasked with performing a DNA replication experiment in the laboratory. He opens the freezer and finds four possible reagents to use. These are: alpha-P* dATP, alpha-P* UTP, gamma-P*GTP, gamma-P* dGTP. Which should he/she choose? Explain why the others are incorrect and your choice is correct. (6 pts)
To introduce * into the DNA molecule being made, the student needed to choose a reagent labeled at the alpha P. The reaction that adds nucleotides to a growing DNA strand cleaves PP off the nucleotide and attaches the alpha P to the 3’OH of another nucleotide. Thus the gamma-P* answers are incorrect. From the potential alpha-P* choices, only dATP is a DNA nucleotide. Alpha dUTP* is an RNA nucleotide and would not incorporate into DNA.
Note: If you chose alpha dUTP* for incorporation into the RNA primer, that would be accepted as a correct choice also.
10. The following enzymatic activities are required for DNA replication. Describe how and when they work in the process. (12 pts)
5’-3’ exonuclease activity:
Removes the RNA primers by breaking phosphodiester bonds from a 5’ end to a 3’ direction.
3’-5’ exonuclease activity:
This is the proofreading activity of the DNA polymerase delta enyzyme. If an incorrect base is incorporated during replication, the polymerase will transfer the 3’ end of the new strand to the 3’-5’ exo site, break the phosphodiester bond just made, and return the 3’ end to the polymerizing site for addition of the correct nucleotide.
Primase:
An RNA polymerase that synthesizes a short 10-15 nucleotide long RNA polymer to start all new DNA polymers by providing a preexisting 3’OH for the DNA polymerases to use.
DNA ligase:
Creates the
phosphodiester bonds
that connect
Topoisomerase:
Relieves torsional strain (due to supercoiling) of the double-stranded DNA upstream from the opening replication fork. It does this by transiently bonding covalently to a 5’P in one strand of the DNA, thus creating an opening in that strand through which the other strand can swivel and relieve the strain. It then releases and reseals the broken phosphodiester bone.
Helicase:
Melts the hydrogen bonds that hold the two DNA strands together, thus creating the templates for new DNA synthesis. Requires ATP to do this.
11. What roles in DNA replication are played by the following? (4 pts)
PCNA
Increases the processivitiy of DNA polymerase delta by clamping onto it, with assistance from Rfc, holding it to the single stranded template.
RPA
Non-specifically binds to the single stranded DNA created by helicase opening up the template strands. This keep the two strands separate until the replication enzymes can move through.
12. On the following diagram, which DNA strand, A or B, represents the template for the leading strand? ____A_ ( 2 pts)
Which DNA strand is copied discontinuously? ___B_ (2 pts)
13. Describe the telomerase enzyme that works to lengthen telomeres in chromosomal DNA. You do not need to describe the mechanism. (6 pts)
It contains a protein component that acts as a reverse transcriptase (RNA dependent DNA polymerase) to lengthen the 3’ end of the template for the lagging strand which overhangs the 5’ end of the lagging strand. To do this it also possesses an RNA component that is complementary to the telomere repeat sequence. It acts as a template for addition of complementary DNA nucleotides.
14. Outline one DNA repair mechanism that we discussed in class. (7 pts)
See figure 4-36 (base excision repair) or figure 4-39 (nucleotide excision repair).