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BIOCHEM 210 FINAL COMPREHENSIVE EXAM QUESTIONS WITH VERIFIED ANSWERS

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BIOCHEM 210 FINAL COMPREHENSIVE EXAM QUESTIONS WITH VERIFIED ANSWERS

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BIOCHEM 210 FINAL COMPREHENSIVE
EXAM QUESTIONS WITH VERIFIED
ANSWERS



1. What is the net charge of the pentapeptide Asp-Glu-Gly-Lys-Arg at pH 7.0?

A. -1


B. +2


C. +1


D. 0


Answer: D


Conceptual Explanation: At pH 7.0, the side chains of Asp and Glu are negatively charged

(-1 each), and the side chains of Lys and Arg are positively charged (+1 each). The N-

terminus is positive (+1) and the C-terminus is negative (-1). Calculation: (-1) + (-1) + (0) +

(+1) + (+1) + (+1) + (-1) = 0.


2. Which of the following describes the ‘Hydrophobic Effect’ in protein folding?

A. It is driven by the enthalpy of hydrogen bonding between nonpolar side chains.


B. It decreases the overall stability of the protein’s native state.

,C. It is caused by the attraction of hydrophobic groups for each other via Van der Waals

forces.


D. It is an entropy-driven process resulting from the release of ordered water molecules.


Answer: D


Conceptual Explanation: The hydrophobic effect is primarily driven by the increase in

entropy of the surrounding water molecules as they are released from the structured

‘clathrate’ cages around nonpolar solutes when those solutes aggregate.


3. In the Henderson-Hasselbalch equation, what happens when the concentration of the

conjugate base equals the concentration of the weak acid?

A. The pH is equal to the pKa.


B. The pH is 7.0.


C. The solution is at its lowest buffering capacity.


D. The pKa is equal to 0.


Answer: A


Conceptual Explanation: When [A-] = [HA], the log([A-]/[HA]) term becomes log(1),

which is 0. Thus, pH = pKa.


4. Which amino acid is most likely to be found in the interior of a globular protein?

A. Serine


B. Lysine

, C. Isoleucine


D. Glutamate


Answer: C


Conceptual Explanation: Isoleucine has a nonpolar, branched aliphatic side chain, making

it highly hydrophobic and likely to be buried in the protein core to avoid water.


5. What is the structural basis for the ‘Bohr Effect’ in hemoglobin?

A. Protonation of His146 leads to the formation of salt bridges that stabilize the T-state.


B. Carbon dioxide binds to the heme iron, displacing oxygen.


C. 2,3-BPG binds to the center of the tetramer, increasing oxygen affinity.


D. High pH shifts the equilibrium toward the T-state.


Answer: A


Conceptual Explanation: Lower pH (higher [H+]) results in protonation of specific

residues like His146, which form ion pairs (salt bridges) that stabilize the T-state

(deoxyhemoglobin), promoting oxygen release.


6. How does 2,3-bisphosphoglycerate (2,3-BPG) affect hemoglobin’s affinity for oxygen?

A. It decreases affinity by stabilizing the T-state via the central cavity.


B. It decreases affinity by binding to the heme group.


C. It increases affinity by stabilizing the R-state.

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