Enzyme M binds with substrate A to produce product B. Compound C is an allosteric activator of enzyme M. Scientists made three mutated forms of M by introducing an amino acid change at residue 57 which normally is valine. They then measured the enzymatic activity of M and M mutants with and without the addition of C. The following data summarize the results. Use these to answer the questions below.
Enzyme Activity
Alteration Minus C Plus C
none 10.3 51.4
valine to serine 10.5 30.2
valine to glutamate 10.2 11.1
valine to alanine 10.1 49.5
a. Is valine more likely to be part of the allosteric site or the active site of the enzyme? Explain your reasoning. (6 pts)
Allosteric site. Minus C shows activities due only to normal active
site interaction. Little change with any mutant. Plus C indicates significant
loss of activity with the mutants suggesting that the alteration interferes
with the ability of C to activate. Most likely due to allosteric site changes.
b. Explain why substituting serine, glutamate, and alanine resulted in the different effects on enzyme activity. Discuss this in terms of their chemical properties. (6 pts)
Alanine is nonpolar like valine therefore alters the enzyme structure
the least. Serine is polar and therefore different from the nonpolar valine
and will disrupt any hydrophobic interactions and perhaps interact with
other polar or charged groups nearby, changing the structure in that region.
Glutamate is acidic, therefore negatively charged. This would dramatically
interfere with hydrophobicity in the region and might also attract basic
side chains in the region as well as polar groups, showing the most dramatic
structural changes of the mutants.
2. Draw a tripeptide with the side chains indicated as R1, R2, and R3. On the diagram, indicate the following: (10 pts)
· All planar regions
· All psi bonds
· All phi bonds
· All omega bonds
Consult the website linked to our course webpage for the picture.
3. Why are non-polar amino acids usually found in the interior of globular proteins? Explain this in terms of energy considerations. (5 pts)
Driven by entropy. Water molecules must cage around hydrophobic groups
in order to form hydrogen bonds with one another. This puts order into
the system and therefore causes a decrease in entropy (disorder). This
is energetically unfavorable. If hydrophobic groups move together, the
amount of caging of water will be less than if, individually, each individual
hydrophobic region had to be caged. Therefore, disorder is increased. This
is energetically favorable.
4. Answer A and B about beta-sheets.
A. Why is the sheet pleated? (4 pts)
The resonance of the peptide bonds causes the atoms within the peptide
bonds to be required to lie in the same plane. Thus, when the H bonds form
between beta strands, the sheet is forced to be pleated to maintain the
planarity.
B. What is the major difference between a parallel beta sheet and an
antiparallel beta sheet? (4 pts)
The polarities, relative to the amino terminal end and the carboxy
terminal end of the strands participate in forming the sheets are different.
Each strand runs in the same polarity for the parallel beta sheet and in
opposite polarities for the antiparallel beta sheet.
5. Describe the biological role played by a chaperone and explain the
mechanism that it uses to play that role. (8 pts)
Chaperone proteins bind to exposed hydrophobic side chains on growing
peptides as they are synthesized in the cell. This prevents clumping and
aggregation due to hydrophobic interactions with other hydrophobic side
chains. This allows the protein to begin to fold accurately. They accomplish
this by binding to ATP and taking on an open conformation that allows a
hydrophobic group in the chaperone to interact with the hydrophobic side
chain. When ATP is hydrolyzed, the chaperone goes back to a closed conformation,
burying the hydrophobic region and must therefore release the side chain
in the now, hopefully, properly folded protein.
6. Answer the following questions about cyclic AMP dependent protein kinase.
A. How does it become active when it is inactive? (5 pts)
The enzyme has two regulatory subunits and two catalytic subunits
that are in contact with each other when the enzyme is inactive (the active
sites on the catalytic subunits can are blocked by the regulatory subunits).
When cAMP levels rise in the cell, cAMP binds to the regulatory subunits,
changing their shape. The regulatory subunits no longer can remain bound
to the catalytic subunits, which are released. The active sites are no
able to bind substrate.
B. Its catalytic mechanism is an example of induced fit. Describe the early events in this catalysis that illustrate induced fit. (5 pts)
As ATP binds in the active site of the enzyme (which is in an open
conformation) and the region of the peptide to be phosphorylated also binds,
a change in shape of the enzyme occurs which brings it into a closed
conformation. A glycine lid moves over the binding cleft, preventing ATP
from leaving the active site and facilitating the interaction between it
and the peptide side-chain. Once the reaction is completed, the glycine
lid moves to the open form again, releasing ADP and the now phosphorylated
peptide.
7. Answer the following questions about B-DNA
A. Why does the DNA lose viscosity when heated? (3 pts)
Heat melts the hydrogen bonds holding the two DNA polymers together in the interior of the double-helix. This reduces the density of the molecule and therefore the viscosity.
B. If the amount of adenine makes up 37% of the bases of a DNA molecule, what are the %s of the other three bases in that molecule? (3 pts)
37% thymine, 13% cytosine, 13% guanine
C. Illustrate or explain in words the secondary structure of B-DNA showing the following: (You do not need to show the actual nucleotide chemistry) (8 pts)
· Location of nucleotide bases
· Polarity of the strands
· Location of major and minor grooves
· Orientation of the sugar-phosphate backbone
Consult the diagrams in your textbook
8. If the DNA fragment from cell type A containing a gene that was expressed
in that cell type was subjected to increasing amounts of DNAase and analyzed
by gel electrophoresis and Southern blotting, how would the result compare
with a similar analysis using cell type B that did not express the gene?
Explain why. (8 pts)
The restriction fragment containing the gene sequences would be degraded
by increasing DNAase concentrations in the DNA from the cell that expresses
the gene but not from the cell that does not express the gene because the
chromatin structure is more loosely packed in the expressed gene and therefore
more accessible to the DNAase.
9. A graduate student is trying to replicate a strand of DNA using a
purified system in the laboratory. He has all necessary components and
knows they are working. He uses a primer that has the sequence 5’AAUUGGCCAUCG3’
His experiment is unsuccessful. He looks in the lab refrigerator and realizes
he grabbed the wrong primer. The sequence of his DNA strand was 5’TTAACCGGTAGC………………….TTAACCGGTAGC3’
Explain what happened. Why did the DNA polymerase not replicate the
DNA strand with that primer? (5 pts)
He chose the primer that had the opposite polarity to the one he
needed to use. It would not bind to the template because it was not antiparallel.
Another answer that is acceptable is that the correct primer would bind
to the 3’ end of the template strand with an entirely different sequence,
5’GCTACCGGTTAA3’. DNA polymerase could extend this primer because it would
provide the 3’OH it needed.
10. Define the role played by the following in DNA replication. (2 pts each)
A. helicase
Melts the hydrogen bonds holding the two DNA strands together to provide for the single-stranded template.
B. topoisomerase
Relieves torsional strain upstream of the advancing replication fork, caused by opening up the double helix. It nicks one strand, allowing the other to swivel through the nick, relieving the tension and then reseals the nick.
C. primase
Synthesizes the RNA primer needed to start the new DNA strands.
D. telomerase
Completes the synthesis of what was the lagging strand at the telomere regions of the chromosomes, preventing their continued shortening during rounds of DNA replication.
E. PCNA
Associates with DNA polymerase delta to increase its processivity as it synthesizes the leading strand.
11. Which of the following is associated with DNA polymerase alpha, DNA polymerase delta, with both or with neither? (2 pts each)
A. 3’-5’ exonuclease activity __delta________________________________
B. 5’-3’ polymerizing activity ___both______________________________
C. Ability to make a phosphodiester bond without a template strand _neither____
D. begins synthesis of all new DNA strands __alpha (will also accept neither)___________________
E. associates with primase ___alpha________________________