StudyStore

Free MCAT-style practice, 5D

MCAT Carbohydrate and Lipid Chemistry Practice Questions

By StudyStore · Published 2026-10-20 · 8 original questions · about 12 min

Eight free MCAT-style practice questions on epimers, anomers and mutarotation, ring forms, glycosidic bonds, reducing sugars, sugar reduction, saponification and lipid classes, with explanations.

Quick concept primer

Carbohydrate stereochemistry is a vocabulary test. Epimers differ at exactly one stereocenter (glucose and galactose differ at C4, glucose and mannose at C2). In the ring form the carbonyl carbon becomes a new stereocenter, the anomeric carbon, and its two configurations are anomers (alpha: anomeric OH on the opposite face from the CH2OH group of a D-sugar; beta: same face). In solution the anomers interconvert through the open-chain form, a process called mutarotation.

A glycosidic bond links the anomeric carbon of one sugar to an alcohol, and the alpha or beta configuration of that bond determines enzyme recognition. A sugar is reducing if it has a free anomeric carbon able to open to an aldehyde. Lipids are classified by structure: triacylglycerols and glycerophospholipids have a glycerol backbone, sphingolipids have sphingosine, and steroids have a four-ring skeleton. Base-catalyzed ester hydrolysis of fats, saponification, gives glycerol and carboxylate salts (soaps).

Traps: epimer is not the same as anomer, and a nonreducing disaccharide has both anomeric carbons tied up in the glycosidic bond.

8 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

D-glucose and D-galactose differ only in the configuration at C4. These two sugars are best described as

  1. enantiomers
  2. epimers
  3. constitutional isomers
  4. anomers
Show answer and explanation

Answer: B. epimers

Epimers are diastereomers that differ at a single stereocenter. Galactose is the C4 epimer of glucose. Anomers differ only at the anomeric carbon of the ring form, and enantiomers are mirror images that differ at every stereocenter (D-glucose and L-glucose).

Question 2

Crystalline alpha-D-glucose is dissolved in water and its optical rotation is measured over several hours. The rotation changes gradually before reaching a stable value. Which explanation is correct?

  1. Glucose racemizes completely to L-glucose
  2. The glucose is decomposing
  3. The alpha anomer reacts with water to form an acid
  4. Mutarotation through the open-chain aldehyde form converts alpha-glucose into an equilibrium mixture with the beta anomer
Show answer and explanation

Answer: D. Mutarotation through the open-chain aldehyde form converts alpha-glucose into an equilibrium mixture with the beta anomer

The hemiacetal ring opens reversibly to the aldehyde, and re-closure can give either anomer. At equilibrium, glucose in water is about 36 percent alpha and 64 percent beta (with a trace of open-chain form), and the observed rotation settles at the weighted average of the two pure anomers. The configuration at the other stereocenters does not change, so L-glucose does not form.

Question 3

Which of the following disaccharides is NOT a reducing sugar?

  1. Sucrose (alpha-glucose 1 to 2 beta-fructose)
  2. Lactose (beta-galactose 1 to 4 glucose)
  3. Maltose (alpha-glucose 1 to 4 glucose)
  4. Cellobiose (beta-glucose 1 to 4 glucose)
Show answer and explanation

Answer: A. Sucrose (alpha-glucose 1 to 2 beta-fructose)

Maltose, lactose and cellobiose each leave one glucose unit with a free anomeric carbon (a hemiacetal) that can open to an aldehyde and reduce Benedict's or Tollens' reagent. In sucrose the glycosidic bond connects the anomeric carbon of glucose with the anomeric carbon of fructose, so no free hemiacetal remains and the molecule is nonreducing.

Question 4

Humans digest starch but not cellulose, although both are polymers of glucose. What is the key structural difference?

  1. Cellulose contains peptide bonds
  2. Cellulose has alpha-1,4 linkages and starch has beta-1,4 linkages
  3. Starch has alpha glycosidic linkages that human amylase can hydrolyze, while cellulose has beta-1,4 linkages that human enzymes cannot hydrolyze
  4. Starch is made of galactose
Show answer and explanation

Answer: C. Starch has alpha glycosidic linkages that human amylase can hydrolyze, while cellulose has beta-1,4 linkages that human enzymes cannot hydrolyze

Enzymes recognize the three-dimensional shape of their substrate. Amylase cleaves alpha-1,4 bonds in starch and glycogen, while cellulose has beta-1,4 bonds that give straight, hydrogen-bonded chains that human enzymes cannot cleave. Some microorganisms produce cellulase, which is why herbivores depend on gut bacteria.

Question 5

D-glucose is treated with sodium borohydride (NaBH4) in water. What is the major product?

  1. Gluconic acid
  2. Glucose pentaacetate
  3. Glucaric acid
  4. Sorbitol (glucitol), a polyalcohol
Show answer and explanation

Answer: D. Sorbitol (glucitol), a polyalcohol

NaBH4 reduces the aldehyde of the open-chain form of glucose to a primary alcohol, giving a six-carbon alditol, sorbitol. Gluconic acid would come from mild oxidation of the aldehyde (such as with bromine water), and glucaric acid from strong oxidation (nitric acid) of both ends. Acetylation requires acetic anhydride.

Question 6

The cyclic hemiacetal form of glucose in which C1 reacts with the hydroxyl on C5 contains a ring of how many atoms?

  1. Five atoms (furanose)
  2. Six atoms (pyranose)
  3. Seven atoms
  4. Four atoms
Show answer and explanation

Answer: B. Six atoms (pyranose)

Ring closure between C1 and the oxygen of C5 creates a ring containing C1 to C5 plus the ring oxygen, which is six atoms and is called a pyranose. Fructose usually closes between C2 and the C5 hydroxyl, which gives a five-membered furanose. Five- and six-membered rings are favored because they are relatively unstrained.

Question 7

Tristearin (a triacylglycerol of stearic acid) is heated with aqueous NaOH. What are the products?

  1. Glycerol phosphate and stearic acid
  2. Glycerol and three equivalents of stearic acid
  3. Stearyl alcohol and sodium glycerate
  4. Glycerol and three equivalents of sodium stearate
Show answer and explanation

Answer: D. Glycerol and three equivalents of sodium stearate

Saponification is base-promoted ester hydrolysis. Hydroxide attacks each ester carbonyl, and the resulting carboxylic acid is deprotonated under the basic conditions, so each fatty acid ends up as its sodium carboxylate (soap). Free carboxylic acids (choice B) would only be obtained after acidification.

Question 8

Which of the following lipids does NOT have a glycerol backbone?

  1. Sphingomyelin
  2. Phosphatidylcholine
  3. Phosphatidylserine
  4. Triacylglycerol
Show answer and explanation

Answer: A. Sphingomyelin

Sphingomyelin is built on sphingosine, an amino alcohol with a long hydrocarbon chain, to which a fatty acid is attached by an amide bond and a phosphocholine group by an ester. Phosphatidylcholine and phosphatidylserine are glycerophospholipids, and triacylglycerols are fatty acid triesters of glycerol.

Keep practicing

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.