Chemistry of biomolecules sample question 1

Resonance gives the C-N bond partial double-bond character, so the atoms around it lie in one plane, usually in the trans arrangement.

Chemistry of biomolecules sample question 2

Stearic acid > oleic acid > linoleic acid

Chemistry of biomolecules sample question 3

Drug P in the aqueous interior and drug Q in the hydrocarbon core of the bilayer

Chemistry of biomolecules sample question 5

About −20°, because the strongly levorotatory fructose outweighs glucose in the equimolar mixture.

MCAT · Biological and Biochemical Foundations · General and Organic Chemistry

Chemistry of biomolecules

Chemistry of biomolecules is part of General and Organic Chemistry in the MCAT Biological and Biochemical Foundations section. The questions below are written in the style of the real test, each with the answer and a worked explanation.

Question types you will see here include sugar stereochemistry and anomers, glycosidic bonds and reducing sugars, fatty acid and triglyceride properties and amphipathic molecules and micelles.

Common traps: confusing epimers with enantiomers; mixing alpha and beta by the OH orientation relative to the CH2OH group; reading mutarotation as a change in chemical identity.

Sample questions

Question 1
Which statement best explains why the peptide bond is planar and rotation about it is restricted?
- A.
  The C-N bond is a pure single bond that rotates freely, and planarity comes only from hydrogen bonds.
- B.
  Resonance gives the C-N bond partial double-bond character, so the atoms around it lie in one plane, usually in the trans arrangement.
- C.
  The C-N bond is an ionic bond between the carbonyl carbon and the nitrogen, which keeps the atoms in line.
- D.
  Steric hindrance forces every peptide bond into the cis arrangement, which is the only planar form.
Answer: B. Resonance gives the C-N bond partial double-bond character, so the atoms around it lie in one plane, usually in the trans arrangement.
B is correct: donation of the nitrogen lone pair into the carbonyl gives the C-N bond partial double-bond character. A wrongly says rotation is free; C calls the bond ionic, but it is covalent; D reverses the usual preference, since trans is favored for most residues (cis occurs mainly before proline).
Question 2
Stearic acid (18:0), oleic acid (18:1, cis) and linoleic acid (18:2, cis,cis) all have 18 carbons. Which ranking of melting points, highest to lowest, is correct?
- A.
  Stearic acid > oleic acid > linoleic acid
- B.
  Linoleic acid > oleic acid > stearic acid
- C.
  Oleic acid > stearic acid > linoleic acid
- D.
  All three have the same melting point, because they have the same chain length
Answer: A. Stearic acid > oleic acid > linoleic acid
Each cis double bond introduces a kink that disrupts close packing and weakens dispersion forces, so melting point falls as the number of cis bonds rises. B reverses the trend; C places the one-kink molecule above the straight saturated chain; D ignores that packing and not chain length differs here.
Question 3
Researchers load liposomes with two drugs: drug P, a polar compound with a charged group, and drug Q, a nonpolar compound. Where are they most likely to be found?
- A.
  Drug P in the aqueous interior and drug Q in the hydrocarbon core of the bilayer
- B.
  Drug P in the hydrocarbon core of the bilayer and drug Q in the aqueous interior of the vesicle
- C.
  Both drugs in the aqueous interior, because a liposome is a water-filled sac
- D.
  Both drugs among the polar head groups on the outer surface of the vesicle
Answer: A. Drug P in the aqueous interior and drug Q in the hydrocarbon core of the bilayer
A liposome is a bilayer vesicle enclosing water: polar, charged drugs dissolve in the aqueous lumen, whereas nonpolar drugs partition into the hydrophobic tail region between the leaflets. The reverse arrangement (B) puts a charged drug in the nonpolar core, a nonpolar drug is poorly soluble in water (C), and head groups do not retain both drugs (D).
Question 4
The imidazole side chain of a histidine residue has a pKa of 6.0. At pH 7.0, what is the approximate ratio of the neutral (deprotonated) form to the positively charged (protonated) form?
- A.
  1:1
- B.
  100:1
- C.
  10:1
- D.
  1:10
Answer: C. 10:1
By the Henderson–Hasselbalch equation, log([A]/[HA]) = pH − pKa = 1, so the neutral form outnumbers the protonated form by 10 to 1. Getting 1:10 comes from reversing the equation; the side chain is mostly protonated only when the pH is below its pKa.
Question 5
Sucrose has a specific rotation of +66°, glucose +52.7° and fructose −92°. What is the expected rotation after complete hydrolysis of sucrose to an equimolar mixture of the two sugars?
- A.
  +66°, because the mass is conserved during hydrolysis, and optical rotation depends only on the amount of material and not on its structure.
- B.
  +145°, because the rotations of the two products add.
- C.
  0°, because the products form a racemic mixture.
- D.
  About −20°, because the strongly levorotatory fructose outweighs glucose in the equimolar mixture.
Answer: D. About −20°, because the strongly levorotatory fructose outweighs glucose in the equimolar mixture.
Average of the two products = (52.7 − 92)/2 ≈ −20°, so the rotation changes sign (inversion), which is why the product is called invert sugar. A ignores the change in composition, B adds rotations rather than averaging, and C wrongly treats glucose and fructose as enantiomers.

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