BCH 403 - Cumulative Final Exam Study Questions Solved 100% Correct | Latest Update
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Course
BCH 403
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BCH 403
The polypeptide shown below needs to be separated from other polypeptides of similar
size (at pH 7.0). What purification technique is best suited to isolate this peptide?
A. Nickel affinity chromatography, since it contains a histidine residue that can bind
metal.
B. Size exclusion chromatograp...
BCH 403 - Cumulative Final Exam Study
Questions Solved 100% Correct | Latest Update
The polypeptide shown below needs to be separated from other polypeptides of similar
size (at pH 7.0). What purification technique is best suited to isolate this peptide?
A. Nickel affinity chromatography, since it contains a histidine residue that can bind
metal.
B. Size exclusion chromatography, since the polypeptide has some hydrophobic
residues.
C. Anion exchange chromatography, since the polypeptide has a net positive charge.
D. Anion exchange chromatography, since the polypeptide has a net negative charge. -
Answer D. Anion exchange chromatography, since the polypeptide has a net negative
charge.
The helical wheel projection represents a 14-residue α-helix that is part of a larger
protein structure. Which of the following would apply?
A. The α-helix is amphipathic with respect to hydrophobicity and therefore will be
partially buried in the protein structure.
B. The α-helix is not amphipathic at all.
C. At neutral pH, the α-helix is amphipathic with respect to charge.
D. At basic pH, the α-helix would be entirely buried in the hydrophobic core of the
structure. - Answer C. At neutral pH, the α-helix is amphipathic with respect to charge.
The small side chain of glycine allows it to do which of the following?
A. Fit into the small space at the center of β-turns.
B. Destabilize β-turns.
C. Fit into the small space between the interwound strands of fibroin.
D. Fit in between the closely packed β-sheets of α-keratin.
,E. All of the answers are correct. - Answer A. Fit into the small space at the center of β-
turns.
The classical protein folding experiment with Ribonuclease A demonstrated that:
A. Disulfide bonds play a central role in determining protein tertiary structure.
B. Urea promotes hydrogen bonding, which stabilizes tertiary structure.
C. The primary amino acid sequence contains the necessary information to fold a
protein.
D. Denatured proteins can be oxidized more easily than natively folded proteins.
E. Protein unfolding is not reversible. - Answer C. The primary amino acid sequence
contains the necessary information to fold a protein.
In an experiment, researchers have replaced the distal histidine of myoglobin with
serine, which has a shorter side chain than histidine. This has the effect of increasing the
distance from the distal residue to the iron. What would you predict will happen to the
P50 value of this mutant protein?
A. The P50 value will not change significantly.
B. The P50 value will increase because oxygen will now bind stronger.
C. The P50 value will increase because oxygen will now bind more weakly.
D. The P50 value will decrease because the protein will become unstable. - Answer C.
The P50 value will increase because oxygen will now bind more weakly.
A protein that binds a ligand is found to have a hill coefficient of 1.5. Which of the
following is TRUE regarding its cooperativity?
A. It cannot bind more than 1.5 molecules of ligand at once.
B. Binding of the first molecule of ligand increases binding for the second molecule.
C. Binding of the second molecule of ligand causes the first molecule to release.
D. The protein displays negative cooperativity.
,E. The binding curve will be hyperbolic. - Answer B. Binding of the first molecule of
ligand increases binding for the second molecule.
The basis of cooperativity in hemoglobin relies on:
A. Rearrangement of hydrogen bonds between the α1 and α2 subunits.
B. Breaking of hydrogen bonds and salt-bridge interactions between the α1 and β2
subunits.
C. Rearrangement of hydrogen bonds, salt-bridge interactions, and hydrophobic
contacts between the α1 and β2 subunits.
D. Movement of the iron to outside of the heme plane upon binding to oxygen.
E. The presence of 2,3-BPG. - Answer C. Rearrangement of hydrogen bonds, salt-bridge
interactions, and hydrophobic contacts between the α1 and β2 subunits.
Which of the following is NOT a feature of the Bohr effect?
A. Low pH protonates a histidine residue in hemoglobin.
B. A protonated (thus positively charged) histidine forms a salt-bridge interaction with
an aspartate residue.
C. The Oxy (R state) is stabilized, thus more oxygen can readily bind.
D. The Deoxy (T state) is stabilized, thus more oxygen is unloaded.
E. The Bohr effect is most likely to occur in oxygen-starved tissue. - Answer C. The Oxy (R
state) is stabilized, thus more oxygen can readily bind.
Fetal hemoglobin has a higher affinity for oxygen than adult hemoglobin because:
A. It's not subject to the Bohr effect.
B. It has a lower affinity for 2,3-BPG.
C. It contains a histidine to threonine mutation on the β1 subunit.
D. It undergoes more allostery than adult hemoglobin.
, E. It contains a negatively charged cavity that binds 2,3-BPG tighter. - Answer B. It has a
lower affinity for 2,3-BPG.
A kinetics experiment reveals that a Michaelis constant (Km) of an enzyme for its
substrate is equal to 7 μM. What can you safely conclude regarding the behavior of the
enzyme?
A. At 5 μM, the reaction is second order with respect to the enzyme.
B. At 7 μM, the reaction is zero order with respect to substrate.
C. The reaction Vmax occurs at greater than 7 μM substrate concentration.
D. The reaction is already operating at Vmax.
E. All of the answers are true. - Answer C. The reaction Vmax occurs at greater than 7
μM substrate concentration.
Under conditions where [S] >> Km, enzymatic reactions become zero order with respect
to substrate because:
A. The enzyme reaches one-half saturation.
B. The reactions are actually still first order with respect to substrate.
C. The enzyme becomes saturated with enzyme and is performing catalysis at maximum
capacity.
D. At high concentrations, substrate can allosterically modulate enzyme catalysis.
E. The enzyme has degraded. - Answer C. The enzyme becomes saturated with enzyme
and is performing catalysis at maximum capacity.
Upon adding a catalyst, the ΔΔG°‡ of a reaction becomes very large. What can you
conclude about the rate constant of the reaction?
A. It has most likely increased.
B. It has most likely decreased.
C. I cannot make any conclusions about the rate constant with this information.
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