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MCAT Amino Acid Synthesis, Peptide Bond and Nucleotide Practice Questions

By StudyStore · Published 2026-10-20 · 10 original questions · about 15 min

Ten free MCAT-style practice questions on the Strecker and Gabriel syntheses, imines and enamines, peptide bond chemistry, phosphodiester bonds and ATP hydrolysis, with full explanations.

Quick concept primer

Two lab routes build amino acids. In the Strecker synthesis an aldehyde reacts with ammonia to form an imine, cyanide adds to the imine carbon to give an alpha-aminonitrile, and acid hydrolysis turns the nitrile into a carboxylic acid. The aldehyde carbon becomes the alpha carbon, so the aldehyde's R group becomes the side chain. The Gabriel route alkylates the nitrogen of phthalimide (often as a phthalimidomalonate ester), then hydrolysis and decarboxylation release the amino acid. Neither method controls stereochemistry, so both give racemic products. Primary amines condense with aldehydes and ketones to give imines; secondary amines cannot form a neutral C=N and give enamines instead.

A peptide bond is an amide. Resonance gives the C-N bond partial double-bond character, so the peptide group is planar, rigid and usually trans. Each bond forms by condensation, releasing one water. Nucleic acids are linked by phosphodiester bonds joining the 3' hydroxyl of one sugar to the 5' phosphate of the next. ATP stores energy in phosphoanhydride bonds whose hydrolysis relieves charge repulsion and gives better-stabilized products.

Traps: "high-energy bond" means a large negative free energy of hydrolysis, not a strong bond; and a peptide bond is thermodynamically unstable in water but kinetically very stable.

10 practice questions

Try each question before opening the explanation. Click "Show answer and explanation" to check yourself. All questions are original and written for study practice.

Question 1

A chemist wants to prepare racemic valine, whose side chain is an isopropyl group, by the Strecker synthesis. Which aldehyde should be used?

  1. Acetaldehyde, CH3CHO
  2. 2-Methylpropanal, (CH3)2CHCHO
  3. 3-Methylbutanal, (CH3)2CHCH2CHO
  4. Propanal, CH3CH2CHO
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Answer: B. 2-Methylpropanal, (CH3)2CHCHO

In the Strecker synthesis the aldehyde carbon becomes the alpha carbon, and the group attached to the aldehyde carbon becomes the side chain. 2-Methylpropanal carries an isopropyl group, giving valine. Acetaldehyde would give alanine, 3-methylbutanal would give leucine (isobutyl side chain), and propanal would give 2-aminobutanoic acid, which has an ethyl side chain.

Question 2

Why does a Strecker synthesis starting from propanal give a racemic amino acid?

  1. The acid hydrolysis step inverts half of the alpha carbons
  2. Ammonia attacks the aldehyde from only one face
  3. Cyanide adds with equal probability to either face of the planar imine
  4. The amino acid racemizes spontaneously in water at room temperature
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Answer: C. Cyanide adds with equal probability to either face of the planar imine

The new chiral center forms when cyanide adds to the imine carbon. That carbon is trigonal planar, and with no chiral catalyst or reagent present, attack from the two faces is equally likely, so the R and S aminonitriles form in equal amounts. Hydrolysis of the nitrile does not touch the chiral center, and amino acids do not racemize quickly at room temperature.

Question 3

In the Gabriel synthesis of amines and amino acids, what is the main advantage of using phthalimide as the nitrogen source instead of ammonia?

  1. The nitrogen can be alkylated only once, which avoids a mixture of over-alkylated amine products
  2. Phthalimide is a stronger base than ammonia, so it reacts with any alkyl halide, including aryl halides
  3. Phthalimide makes the product a single enantiomer
  4. Phthalimide supplies the carboxyl group of the amino acid
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Answer: A. The nitrogen can be alkylated only once, which avoids a mixture of over-alkylated amine products

An amine formed from ammonia and an alkyl halide is itself a good nucleophile and keeps reacting, giving secondary, tertiary and quaternary products. The phthalimide anion has one acidic N-H to remove, and once alkylated the nitrogen bears no hydrogen and is flanked by two carbonyls, so it does not react further. Phthalimide is a weak base, aryl halides do not undergo SN2, and the product is racemic.

Question 4

A student makes phenylalanine (side chain -CH2C6H5) by deprotonating diethyl phthalimidomalonate, alkylating the carbanion, then hydrolyzing and decarboxylating. Which alkylating agent is needed?

  1. Bromobenzene, C6H5Br
  2. Methyl bromide, CH3Br
  3. 2-Phenylethyl bromide, C6H5CH2CH2Br
  4. Benzyl bromide, C6H5CH2Br
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Answer: D. Benzyl bromide, C6H5CH2Br

The malonate carbon becomes the alpha carbon, and the alkyl group added becomes the entire side chain. Phenylalanine needs a CH2 linking the ring to the alpha carbon, so benzyl bromide is used, and it is an excellent SN2 substrate. Bromobenzene cannot undergo SN2 at an sp2 carbon, methyl bromide would give alanine, and 2-phenylethyl bromide adds one carbon too many.

Question 5

The free amino acids glycine, alanine and serine have molar masses of about 75.07, 89.09 and 105.09 g/mol, and water is about 18.02 g/mol. What is the approximate molar mass of the linear tripeptide Gly-Ala-Ser?

  1. 269.25 g/mol
  2. 251.23 g/mol
  3. 233.21 g/mol
  4. 215.19 g/mol
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Answer: C. 233.21 g/mol

Forming each peptide bond is a condensation that releases one water, and a linear tripeptide has two peptide bonds. The mass is 75.07 + 89.09 + 105.09 - 2(18.02) = 233.21 g/mol. Choice A forgets the water loss, choice B removes only one water, and choice D removes three, which would describe a cyclic peptide with three amide bonds.

Question 6

All of the following describe a typical peptide bond in a protein EXCEPT:

  1. The atoms of the peptide group lie in one plane because of resonance
  2. There is free rotation about the bond between the carbonyl carbon and the nitrogen
  3. The trans arrangement is much more common than cis for most residues
  4. The carbonyl carbon to nitrogen bond is shorter than a typical C-N single bond
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Answer: B. There is free rotation about the bond between the carbonyl carbon and the nitrogen

The nitrogen lone pair is delocalized into the carbonyl, which gives the C-N bond partial double-bond character. That makes the bond shorter than a normal single bond, holds the peptide group planar, and blocks free rotation, so trans and cis forms do not interconvert easily and trans is favored on steric grounds. Rotation in the backbone happens at the bonds to the alpha carbon instead.

Question 7

Hydrolysis of a peptide bond in water has a negative standard free energy change, yet proteins in neutral water at room temperature persist for years without catalysts. What best explains this?

  1. Hydrolysis is endergonic at neutral pH
  2. Peptide bonds cannot be cleaved except by enzymes
  3. Hydrolysis lowers entropy, so it is not spontaneous
  4. The reaction is favorable but has a very high activation energy, so it is extremely slow without a catalyst
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Answer: D. The reaction is favorable but has a very high activation energy, so it is extremely slow without a catalyst

Thermodynamics says whether a reaction can proceed; kinetics says how fast. The resonance-stabilized amide is a poor electrophile, so the uncatalyzed transition state is high in energy and the rate is tiny. Proteases or hot concentrated acid speed the reaction greatly, which shows that enzymes are not the only route (choice B) and that hydrolysis is favorable (ruling out A and C).

Question 8

How many phosphodiester bonds are present in a linear, double-stranded DNA fragment that is 40 base pairs long?

  1. 78
  2. 80
  3. 40
  4. 39
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Answer: A. 78

Each strand of 40 nucleotides has a phosphodiester bond between each adjacent pair, which is 40 - 1 = 39 per strand, so the duplex has 2 × 39 = 78. Choice D counts only one strand, and 80 would be correct for a circular duplex, where each strand closes into a ring with one extra bond. Hydrogen bonds between base pairs are not phosphodiester bonds.

Question 9

In a cell at 310 K, ATP hydrolysis to ADP and phosphate has a standard free energy change of about -30.5 kJ/mol. If [ATP] = 5.0 mM, [ADP] = 0.50 mM and [Pi] = 5.0 mM, what is the approximate actual free energy change? Use R = 8.314 J/(mol K) and treat the concentrations in molar units.

  1. -10.9 kJ/mol
  2. -30.5 kJ/mol
  3. -50.1 kJ/mol
  4. -19.6 kJ/mol
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Answer: C. -50.1 kJ/mol

Use ΔG = ΔG°' + RT ln Q with Q = [ADP][Pi]/[ATP] = (5.0 × 10-4)(5.0 × 10-3)/(5.0 × 10-3) = 5.0 × 10-4. RT ln Q = (0.008314)(310)(-7.60), about -19.6 kJ/mol, so ΔG is about -50.1 kJ/mol. Choice A adds the term with the wrong sign, and choice D is the correction term alone.

Question 10

Cyclohexanone is heated with pyrrolidine, a cyclic secondary amine, and a trace of acid while water is removed. What type of product forms?

  1. An enamine
  2. An imine with a C=N double bond
  3. An amide
  4. A hemiacetal
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Answer: A. An enamine

Addition of the amine and loss of water give an iminium ion. A primary amine could lose its remaining N-H proton to form a neutral imine, but a secondary amine has no N-H left, so a proton is removed from the adjacent alpha carbon instead, giving a C=C bonded to nitrogen: an enamine. Amides require an acyl derivative, and hemiacetals come from alcohols.

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Written by StudyStore from standard scientific facts as a study aid. These are not official exam questions and are not affiliated with or endorsed by any testing organization. Always check values against the reference provided with your exam or course.