BCH 403 – Cumulative Exam Questions|
Already Answered| GRADED A+
T/F: The thermodynamics of a biochemical reaction can be driven by either favorable changes in
enthalpy or favorable changes in entropy, but not by both. - ANSWER-False
T/F: Hydrogen bonds are weak electrostatic interactions that are made possible by the presence of
partial charges on H-bond donors and acceptors. - ANSWER-True
T/F: Carbonic acid has a pKa of 6.37; acetic acid has a pKa of 4.76; therefore, carbonic acid is the
stronger acid. - ANSWER-False
T/F: Small proteins elute earliest in a gel filtration column, but travel fastest on SDS-PAGE. - ANSWER-
False
T/F: Peptide bonds are planar due to the partial double bond character of neighboring R-groups. -
ANSWER-False
T/F: Formation of a new peptide bond produces one molecule of water and one hydroxide ion. -
ANSWER-False
T/F: Protein folding is made energetically favorable because of the combined contributions of the
hydrophobic effect, van der Waals interactions, hydrogen-bonding, salt bridge interactions, and disulfide
bonds. - ANSWER-True
T/F: In protein folding, the hydrophobic effect and van der Waals interactions contribute favorably to
the entropy component of the Gibbs free energy equation. - ANSWER-False
T/F: Alpha-keratins have a pseudo-repeat motif of (abcdef)g, in which a and d are negatively charged
residues. - ANSWER-False
,T/F: The sigmoidal oxygen-binding curve of hemoglobin is an indicator of cooperativity between the
hemoglobin subunits. - ANSWER-True
Upon running SDS-PAGE on a protein complex that consists of 5 subunits, of which 3 subunits are the
same size and two have distinct sizes (different from each other and different from the other three),
how many bands should you expect to see on the gel?
A. one band
B. two bands
C. three bands
D. five bands - ANSWER-C. three bands
A reaction with a ΔG equal to zero can best be described as:
A. A system where the reactants and products are equal.
B. A system that has reached an equilibrium.
C. A system that requires ATP as a thermodynamic coupler in order to reach equilibrium.
D. A system where the change in entropy is the principle driving force.
E. A system where the change in enthalpy is the principle driving force. - ANSWER-B. A system that has
reached an equilibrium.
Use the chart to answer this question. Which of the following sets of residues would most likely be
found on a protein surface?
A. glycine, arginine, histidine, glutamate
B. glycine, proline, phenylalanine, isoleucine
C. valine, alanine, tyrosine, glutamate
D. serine, cysteine, leucine, valine - ANSWER-A. glycine, arginine, histidine, glutamate
The following primary sequence of residues folds into a structure that is mostly α-helical. How many
separate α-helices are there in the protein fold?
GMTRAVLSRDAPGTRSPYVWIARKCHTWSGPGVIDEEDPTA
A. 4
B. 3
,C. 2
D. 1 - ANSWER-B. 3
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.
