Download AS.410 AS.410.601 Problem Set #1 (Johns hopkins university) and more Exams Biochemistry in PDF only on Docsity!
Selected Answer:
Correct Answer:
Response
Feedback:
[None Given]
The K a
for a weak acid (HA) is known to be 1.9 x 10
. What is the ratio of [A
] to [HA] at a
pH of 6? Explain your answer & show all mathematical work.
pH=pKa+log([A]/[HA])
6=log(1.9*10^-5)+log([A]/[HA])
6=-4.72+log([A]/[HA])
10.72=log([A]/[HA])
10^10.72=10^log([A]/[HA])
5.24*10^10 = [A]/[HA]
Since the K a
= 1.9 x 10
, then the pK a
= -log[1.9 x 10
], which is equal
to 4.72.
Using the H/H equation:
pH = pK a
]/[HA])
6 = 4.72 + log([A
]/[HA])
1.28 = log([A
]/[HA])
So: [A
]/[HA] = 19.
Question 3
Selected Answer: E.
For questions 3-6, consider the titration curve below. The chemical species being titrated by a
strong base must be a __________ , since it is capable of donating ______ proton(s).
triprotic acid; three
0.5 out of 0.5 points
Answers: A.
B.
C.
D.
E.
weak base; no
strong base; three
monoprotic acid; one
diprotic acid; two
triprotic acid; three
Question 4
Specified Answer for: Blank1 basic
Correct Answers for: Blank
Evaluation Method Correct Answer Case Sensitivity
Exact Match
basic
Response
Feedback:
The chemical group being titrated around the point labeled (C) in the graph is most likely a(n)
________________ (basic or acidic) group. [Blank1]
Since the graph indicates the pKa is approximately 12.5 at point (C), then this must
be a basic group, such as an amino group.
Question 5
Specified Answer for: Blank1 Arginine
Correct Answers for: Blank
Evaluation Method Correct Answer Case Sensitivity
Exact Match
arginine
Response
Feedback:
The titration curve shown in Question #3 is that of a free amino acid in aqueous solution.
Based on the graph, which particular amino acid is most likely represented? Simply name
the amino acid. [Blank1]
The titration curve clearly shows that the amino acid in question has three ionizable
groups, due to the three distinct "steps" on the graph. So this means that the amino
acid must be Tyr, Cys, Lys, His, Arg, Asp, or Glu (these are the only free amino acids
with ionizable side chains). The chemical group at point (A) has a pKa of a little over 2;
the chemical group at point (B) has a pKa of approximately 9; the chemical group at
point (C) has a pKa of around 12.5. The only amino acid that fits this pKa profile is
ARGININE.
Question 6
Assume that the equilibrium represented around point (A) in the titration can generically be
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1 out of 1 points
C.
D.
E.
Response Feedback:
The ionizable groups are:
The amino terminal group of His
The side chain of His
The side chain of Asp
The side chain of Tyr
The carboxy terminal group of Ser
Question 9
Selected Answer: B.
Answers: A.
B.
C.
D.
E.
Response
Feedback:
At a pH of 6, which of the following best approximates the net charge on the peptide?
At a pH of 6, which is equal to the pKa for the imidazole side chain on His,
the charges are as follows:
N-terminal group is +
His side chain is +1/
Asp side chain is -
Tyr side chain is 0
C-terminal group is -
Question 10
Selected
Answer:
Within this peptide, which amino acid residue is the most hydrophobic? Explain your
answer.
Phe is most hydrophobic as it has an aromatic R group which makes it somewhat
hydrophobic. Also the hydropathy index of Phe is 2.
0.5 out of 0.5 points
1 out of 1 points
Correct
Answer:
Response
Feedback:
[None Given]
I will accept an answer of either Met or Phe for this question. Both contain very
hydrophobic side chains; what is most important is your reasoning behind your
answer.
Question 11
Selected
Answer:
Correct
Answer:
Response
Feedback:
[None Given]
Calculate an approximate pI (isoelectric point) for this peptide. Please use the pK a
's listed
within Table 3-1 of Lehninger. Show and/or explain all your work; you must be very
clear about WHY you use particular pK
a
's in your calculation.
pKa histidine = (1.82+9.17)/2 = 5.
pKa serine = 2.
pI= (5.50+2.21)/
pI=3.
The approximate pI for this peptide is the average of the pK a
's for the His side
chain (6.00) and the Asp side chain (3.65), resulting in a value of 4.80. At a pH
of 4.8, the His side chain is still-protonated (the pK a
is more than one pH unit
above the pH) and therefore contributes a +1 charge. Also at a pH of 4.8, the
Asp side chain is fully deprotonated (the pK a
is more than one pH unit below the
pH) and therefore contributes a -1 charge. All other pK a
's are well away from
4.8, thereby rendering them fully protonated or deprotonated: the amino
terminus is +1; the Tyr side chain is 0; the carboxy terminus is -1. Therefore,
at a pH of 4.8, the net charge on our peptide is 0.
Question 12
Selected
Answer:
Correct
Answer:
Response
Feedback:
[None Given]
Based on your answer for Question #11, would you categorize this peptide as "acidic" or
"basic"? Explain.
We would categorize this peptide as acidic because the isoelectric point is 3.
which corresponds to an acidic pH. isoelectric point tells us the overall
acidic/basic character of a protein.
This is definitely an acidic peptide because it's pI is well below neutral pH (7).
This means that the peptide loses more than half of its dissociable protons at low
pH values, meaning it contains a majority of acidic groups.
Question 13
0 out of 1 points
1 out of 1 points
1 out of 1 points
Correct
Answer:
Response
Feedback:
[None Given]
Answers will obviously vary, based on PTM selected.
Question 16
Selected Answer: A.
Answers: A.
B.
C.
D.
E.
Disulfide linkages:
are covalent bonds.
are covalent bonds.
are a type of electrostatic interaction.
occur between methionine residues.
can be disrupted by SDS.
can only occur within the same polypeptide chain.
Question 17
Selected Answer: A.
Answers: A.
B.
C.
D.
E.
The idea that primary sequence determines tertiary structure first came from experiments in
the 1950's about:
the renaturation of RNaseA (ribonuclease A).
the renaturation of RNaseA (ribonuclease A).
the denaturation of lysozyme.
the 3D structure of lysozyme.
the role of Hsp70 in protein folding.
the role of PDI (protein disulfide isomerase) in protein folding.
Question 18
Questions #18 & 19 refer to the following information. Altered conformations of the prion
protein (PrP) are implicated in the infectivity and/or pathogenesis of Scrapie-like diseases,
such as "mad cow disease" and Creutzfeldt-Jacob Disease (CJD). Below are ribbon diagrams
of two different conformations of PrP:
0.5 out of 0.5 points
0.5 out of 0.5 points
1 out of 1 points
Selected
Answer:
Correct
Answer:
Response
Feedback:
[None Given]
Which of the conformations illustrated above (a or b) is more likely PrP
c
(the normal cellular
form of PrP) and which is more likely PrP
Sc
(the pathogenic form of PrP)? Why?
A is the PrP
c
form and B is the PrP
Sc
form.
Prions typically appear as form B, with beta-sheets, when in pathogenic form
(e.g., amyloid beta plaques found in neurodegenerative diseases). This form is
more structurally stable due to the cross beta structure which is supported by
hydrogen bonding.
The tertiary conformations of PrP
c
and PrP
Sc
differ largely in the relative
amounts of secondary structures, namely beta-sheet structures. PrP
Sc
is high in
beta-sheet structure, whereas PrP
c
has very little beta-sheet structure. As you
can see from the ribbon diagrams, the conformation on the left (a) is largely
composed of alpha-helical, turns, and random coil structures. The conformation
on the right (b) has at least two apparent beta sheet structures (two strand in
green and two strands in red). Therefore, it's more likely that (b) represents
PrP
Sc
and that (a) represents PrP
c
.
Question 19
Selected Answer: E.
Answers: A.
The red arrows within the (b) conformation are best categorized as:
antiparallel beta-strands
alpha helices
0.5 out of 0.5 points
Selected Answer: A.
Answers: A.
B.
C.
D.
E.
Question #20: According to the gel (Step 6), which protein has the highest molecular
weight?
Protein A
Protein A
Protein B
Protein C
Protein D
YFP
Question 21
Selected
Answer:
Correct
Answer:
Response
Feedback:
[None Given]
Are the bands indicated in Step 6 (SDS-PAGE) likely visualized by western blot or by a general
protein stain (such as Coomassie or silver)? Explain your answer.
it is a general protein stain using a stain such as coomassie blue which allows us
to see proteins separated by size. For western blotting, antibodies need to be
washed through the gel after general SDS-PAGE procedure.
Since all proteins in the complex can be "seen" on the gel, then it must not be a
western, but rather a general protein stain. If we were seeing a western, then
we would only have seen one of the many bands shown, as an antibody against
just one of those proteins would have been used.
1 out of 1 points
Question 22
Selected
Answer:
Correct
Answer:
Response
Feedback:
[None Given]
Step 7 states "Excise bands and analyze by mass spectrometry."
(A) Following excision of the bands (simply cutting each band out of the gel), briefly describe
each step, in order, that must be taken prior to analysis in the mass spectrometer.
(B) What's the big experimental question that the researchers hope to answer following Step
A) we would need to first do an in-gel digestion. This procedure allows us to
remove staining from the SDS-PAGE experiment, break up disulfide bonds, break
the protein up using a proteolutic agent, and then finally extract the
peptide/protein for MS analysis.
B) MS will allow researchers to determine the identify of the proteins in the
sample and their quantity in the sample.
(A) Sequence of steps (to be done for each band independently):
- reduce (elim. disulfides) with BME or DTT
- digest with protease (like trypsin) or CNBr
- separate peptides generated in Step 2 via chromatography (reverse-phase
HPLC)
- sequence one (or several) peptides, from Step 3, by mass spec
(B) The big question: "What proteins specifically interact with YFP?" The mass
spec analysis will allow them to IDENTIFY the proteins (A, B, C, D) that associate
with YFP.
Question 23
Selected
Answer:
C.
Answers: A.
B.
C.
Following Step 5 ("competitive elution with peptide encoding the epitope"), it appears that
the YFP/protein complex has dissociated -- all proteins are drawn floating freely. Which of
the following is the most likely explanation for dissociation of the protein complex?
Treatment of the eluate (Step 5) with sodium dodecyl sulfate, in preparation
for loading onto the gel (Step 6)
Treatment of the eluate (Step 5) with beta-mercaptoethanol, in preparation for
loading onto the gel (Step 6)
Treatment of the eluate (Step 5) with dithiothreitol, in preparation for loading
onto the gel (Step 6)
1 out of 1 points
0.5 out of 0.5 points
Response
Feedback:
[None Given]
Question 25
Selected Answer: A.
Answers: A.
B.
C.
D.
E.
For the table shown in Question #24, the purification step labeled "40-60% (NH 4
SO
" is
most likely a(n):
salting in/out step.
salting in/out step.
cation exchange chromatography.
anion exchange chromatography.
hydrophobic interaction chromatography.
size exclusion chromatography.
Question 26
Selected Answer: B.
Answers: A.
B.
C.
D.
E.
For the table shown in Question #24, the purification step labeled "CM-sepharose" is most
likely a(n):
cation exchange chromatography.
salting in/out step.
cation exchange chromatography.
anion exchange chromatography.
hydrophobic interaction chromatography.
size exclusion chromatography.
Question 27
Selected Answer: E.
Answers: A.
B.
C.
For the table shown in Question #24, the purification step labeled "Gel filtration" is most
likely a(n):
size exclusion chromatography.
salting in/out step.
cation exchange chromatography.
anion exchange chromatography.
0.5 out of 0.5 points
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0.5 out of 0.5 points
Monday, April 4, 2022 6:31:14 PM EDT
D.
E.
hydrophobic interaction chromatography.
size exclusion chromatography.
← OK
