1. Illustrate a peptide bond and show the omega, phi, and psi angles. (8 pts)
Consult course website for the link to peptide geometry. The omega angle is the peptide bond itself. The phi angle is the alpha carbon to peptide bond nitrogen and the psi angle is the alpha carbon to the peptide bond carbonyl carbon.
2. A. List three features of alpha-helices in polypeptides. (6 pts)
Some choices: right handed alpha helix, 3.6 residues per turn of helix, hydrogen bonds between peptide linkages 4 linkages apart, hydrogen bonds are between the carbonyl oxgen and the amide hydrogen of the peptide bonds spaced as above, R groups project outward from the helix, tipped slightly towards the amino terminus. Other possible answers are also possible.
B. What is an EF hand? Choose your answer from the list below: (3 pts)
a. structural domain
b. secondary structural motif
c. quaternary structure
B is the correct answer
3. Alzheimer’s disease may involve an improperly folded protein. Explain why a person with a mutated form of heat shock protein 70 could have an increased chance of developing the disease. (6 pts)
Hsp70, a chaperone, can bind to ATP and bind to hydrophobic residues on nascently growing peptides or denatured proteins to prevent their aggregation. When ATP is hydrolyzed, with the help of a second chaperone, hsp70 bends around the protein and allows it to complete its folding. Then a third protein assists in displacing ADP and allowing ATP to replace it, returning the hsp70 to its normal shape and the correctly folded protein is released. If the hsp70 protein is mutated it may not be capable of performing this protective role. This could contribute to the formation of the improperly folded Alzheimer-related misfolded protein.
4. Illustrate how three serine proteases distinguish their substrates (how do their binding pockets differ)? (9 pts)
Consult figure 3-25 in our text and the accompanying explanation of the figure.
5. Why is pH important for the catalytic functioning of the serine proteases? Be very specific in your answer. (8 pts)
Histidine is the central player in the charge relay that initiates formation of both transition states in the breaking of the peptide bond by the serine proteases. It must be non-protonated to be able to draw the H off of the serine OH side chain, causing the activation of the O and the attack on the C=O of the peptide bone to be broken. It also must be able to draw the H off the water molecule in the second part of the reaction mechanism that leads to the formation of the second tetrahedral intermediate. His can only do this at a pH that does not cause its weak basic nature to draw an H from the surrounding solution instead, therefore negating its ability to perform the above role in the reactions.
See Figure 3-26 and accompanying text.
6. Consider the following: Molecule X is the precursor in the synthesis of molecule Z. Z is the end product of a metabolic pathway. When cells undergo stress, Z concentrations fall. Molecule X has a binding site on an allosteric enzyme that works early in the pathway that makes Z. Describe how X is likely to work in this scenario to help mediate the loss of Z. (8 pts)
In this scenario, molecule X only builds in concentration when the product of the reaction pathway, Z is not being made rapidly. This indicates that the pathway needs to speed up. X will act as an activator of the pathway by binding to the regulatory binding site on the allosteric enzyme that catalyzes a reaction near the beginning of the pathway. By binding to the active configuration of the enzyme, it stabilizes that shape, preventing it from flipping to an inactive shape as would happen in the absence of any regulatory molecule being bound. In other words, the binding of X shifts the equilibrium of the two forms of the enzyme towards the active form. Now more active enzymes are found relative to inactive ones in the population of enzymes and the reaction rate increases. Z can then be formed more rapidly.
7. p53 is a tightly regulated protein with a very short half life. Describe a pathway that explains how the cell can insure this protein only remains there for a short time after synthesis. (8 pts)
p53 will be target for degredation by the proteosome pathway that utilizes uniquitin. E1, ubiquitin activating enzyme, hydrolyzes ATP to pyrophosphate and AMP and uses this energy released to attach to an ubiquitin molecule. Then E1 transfers the ubiquitin to E2, the ubiquitin conjugating enzyme. E2, with the help of E3, the ubiquitin ligase, attaches the ubiquitin to a lysine on the targeted protein. These steps repeat over and over until a large number of such ubiquitins get attached to the protein. This signals the cell to bring the protein to a proteosome.The cap on the structure stimulates ATP hydrolysis, releasing the ubiquitins, unfolding the protein, and pumping the protein to the interior of the proteosome where it is degraded by proteases that line its interior. Short peptides are then released.
8. Illustrate a phosphodiester bond between two nucleotides. You may designate the bases as A, G, C, or T. You do not need to illustrate the base’s structure. All other elements of the nucleotides and the bond must be shown. (7 pts)
See Figure 4-2
9. Answer these questions about DNA structure. (2 pts each)
a. Where are the bases found?
In the interior of the double-helix
b. How many base pairs are there per turn of the helix?
Approximately 10
c. What does antiparallel refer to?
The two strands opposite in polarity to one another, one moving in one direction from the 5’phosphate end to the 3’OH end and the other running in the opposite direction from its 5’ phosphate end to its 3’OH end.
d. What are major and minor grooves?
Spaces formed as the two helices wind around each other have a larger width (major groove) or smaller width (minor groove). These serve as access points for proteins to interact with the bases in the interior of the helix. The major and minor grooves appear sequentially: major, minor, major minor on each strand.
e. What is a nucleosome?
Approximately 146-147 base pairs of double stranded DNA wound 1.5 to 2 times around a core of histone proteins consisting of 2 each of H2A, H2B, H3, and H4. These are separated by linker region of the DNA. H1 binds at the junctions between nucleosomes and linkers and participates in drawing the nucleosomes together to form a 30 nm fiber.
10. Describe the role played by the following in DNA replication. (2 pt each)
a. helicase
Melts the hydrogen bonds holding the nucleotide bases together to create the single-stranded template.
b. topoisomerase
Operates ahead of the opening replication fork to relieve supercoiling stress imposed on the DNA due to the activity of helicase. It nicks one strand, providing a swivel point for the other strand to move through and relieve the supercoil. It then reseals the broken bond.
c. primase
An RNA polymerase that connects RNA nucleotides to form a primer for further extension by DNA polymerase. It must work every time a new DNA strand is to be initiated.
d. PCNA
Protein that interacts with RPC to clamp DNA polymerase delta more firmly to the template strand, thus increasing the processivity of the polymerase.
e. 5’-3- exonuclease
Breaks phosphodiester bonds to the 3’ side of the RNA primer, one nucleotide at a time, thus removing the primer.
f. 3’-5’ exonuclease
A DNA polymerase delta activity that breaks a phosphodiester bond to the 5’ side of the growing daughter strand when the polymerase senses it has incorporated an incorrect nucleotide during DNA replication. It causes the removal of the incorrect nucleotide and allows the polymerase to then add the correct nucleotide. It is a proofreading function.
g. DNA ligase
Uses the energy provided by ATP to connect
phosphodiester
bonds between
11. You are a scientist who needs to introduce a radioactive nucleotide in to a growing DNA polymer in a laboratory. You have available several possible reagents to use, each radioactively labeled at a particular phosphate group. Choose the correct nucleotide to use from the following list and explain your choice. * refers to the radioactive phosphate.(6 pts)
*gamma ATP
*gamma UTP
*alpha GTP
*alpha UTP
Explanation:
Alpha GTP is chosen. Assuming all of the above are deoxy NTPs, then only the G or A would be found in a DNA polymer. The alpha would need to be chosen because the chemical reaction that connects nucleotides in a growing DNA polymer attaches the new nucleotide through its alpha phosphate. dNTP is hydrolyzed to dNMP and PP. The dNMP is added to the growing 3’ end . Therefore, onlyt the alpha GTP would fulfill these two criteria and introduce radioactivity in to the polymer.
12. Describe ONE of the following mechanisms: (7 pts)
A. How telomerase extends the ends of chromosomes
B. How the cell repairs thymine dimers in DNA
C. How the cells repairs deaminated nucleotides in DNA.
We did not discuss B.
A.
Telomerase extends the 3’ overhang at the
telomeres.
The strand extended is the template for the lagging strand. Telomerase
uses its
RNA component to base pair to the tandem repeats at the telomere ends
and then
introduces DNA nucleotides up to nucleotide number 35 in its RNA
primer. The
extended strand releases from the primer and the primer reestablishes
the
interaction again on the tandem repeat region of the extended template,
lengthening it further to reside 35 of the primer again. Release,
reestablishment of primer-template interaction, and further lengthening
repeats
over and over. The 3’ template is then extended sufficiently to allow
more
C.
A mismatched
base pair is sensed by a DNA glycolyase. The incorrect base is flipped
out and
the N-glycosyl linkage broken, relasing only the base from
the nucleotide. Then
an APE1 endonuclease cuts the phosphodiester bond to the 5’ side of the
sugar
phosphate backbone containing the missing base and an
AP lyase (part of DNA
polymerase beta) cuts the opposite side. The polymerase can then use
the 3’OH
to add the correct nucleotide. DNA ligase seals the
final phosphodiester bond.
See figure 4-36.