Chapter 32 — Quiz

Twenty-five questions on carbohydrate structure, reactivity, and biology. ∗ marks questions answered in the answer key.

Multiple choice

1.∗ D-glucose is an: (a) aldohexose (b) ketohexose (c) aldopentose (d) ketopentose

2.∗ D-fructose is an: (a) aldohexose (b) ketohexose (c) aldopentose (d) deoxypentose

3.∗ Open-chain glucose at equilibrium in water is approximately: (a) ~99% (b) ~50% (c) ~10% (d) ~0.02%

4.∗ The β-anomer of D-glucopyranose has its anomeric C1-OH: (a) axial (b) equatorial (c) trans-diaxial (d) at C2

5.∗ Mutarotation is: (a) racemization at C5 (b) interconversion between α and β anomers via the open-chain form (c) oxidation (d) reduction

6.∗ A glycosidic bond is: (a) a hemiacetal (b) an acetal (c) an alcohol (d) an ester

7.∗ Sucrose's glycosidic linkage is: (a) α-1,β-2 (between glucose's α-C1 and fructose's β-C2) (b) α-1,4 (like maltose) (c) β-1,4 (like cellulose) (d) α-1,6

8.∗ Cellulose's glycosidic linkage is: (a) α-1,4 (like starch) (b) β-1,4 (linear, indigestible by humans) (c) α-1,6 (branched) (d) α-2,4

9.∗ Starch's glycosidic linkage (in amylose) is: (a) α-1,4 (helical, digestible) (b) β-1,4 (linear, indigestible) (c) α-2,4 (d) random

10.∗ Biological sugars are predominantly: (a) D-configured (b) L-configured (c) racemic (d) achiral

11.∗ Why does β-D-glucopyranose dominate over α at equilibrium (64% vs 36%)? (a) all 5 substituents are equatorial in β chair (sterics) (b) electronic effect alone (c) it's the kinetic product (d) it's the only stable chair

12.∗ A reducing sugar is one that: (a) has a free anomeric C (in equilibrium with open-chain CHO; can reduce Tollens or Fehling reagents) (b) reduces all electrophiles (c) is a reducing agent in industrial processes (d) is non-reducing

13.∗ Sucrose is a non-reducing sugar because: (a) both anomeric carbons (C1 of glucose and C2 of fructose) are tied up in the glycosidic bond (b) it has no carbonyl (c) it's an ester (d) it has no aldehyde

14.∗ Lactose intolerance is caused by deficiency of: (a) α-amylase (b) lactase (β-galactosidase) (c) glucagon (d) insulin

15.∗ Glycogen is highly branched because: (a) it allows fast glucose release from many terminal points (multiple anomeric Cs available) (b) it's the most stable structure (c) it's randomly distributed (d) all of the above

16.∗ HbA1c (glycated hemoglobin) is the result of: (a) imine formation between glucose's C1 + N-terminal valine of hemoglobin, with subsequent Amadori rearrangement (b) glucose oxidation (c) hemoglobin proteolysis (d) DNA modification

17.∗ Why are blood types determined by sugars? (a) carbohydrate antigens on red cell surface glycoproteins (A, B, O differ by trisaccharide composition) (b) protein-only differences (c) DNA differences (d) lipid differences

18.∗ A pentose-furanose (5-ring form) appears in: (a) DNA (2-deoxyribose) (b) RNA (ribose) (c) both DNA and RNA (d) starch only

19.∗ Glycolytic aldolase catalyzes: (a) cleavage of fructose-1,6-bisphosphate into DHAP + G3P (a retro-aldol) (b) glucose oxidation (c) lactose hydrolysis (d) DNA synthesis

20.∗ Cellulose is the most abundant biopolymer on Earth because: (a) it's the structural material of plant cell walls (~50% of biomass) (b) it's mostly water (c) it's metabolized everywhere (d) it's inorganic

Short answer

21. Draw α-D-glucopyranose and β-D-glucopyranose in chair conformation. Explain why β is preferred (steric and electronic reasons).

22. Sketch the mechanism of glycoside formation: D-glucose + methanol + H⁺ → methyl glucoside. Show the oxocarbenium intermediate.

23. Compare the structures and biological consequences of starch (α-1,4) vs. cellulose (β-1,4). Why are they so different in physical properties and digestibility?

24. Sketch the chemistry of HbA1c formation: glucose's C1 + N-terminal valine of hemoglobin → Schiff base → Amadori rearrangement product. What is the clinical use of HbA1c measurement?

25. Explain why most of glucose in water (~99.98%) is in the cyclic hemiacetal form rather than open-chain. Connect to the chemistry of Chapter 25.

Answer key

  1. a — D-glucose is an aldohexose.
  2. b — D-fructose is a ketohexose.
  3. d — ~0.02% open-chain in water.
  4. b — β-anomer is equatorial.
  5. b — Mutarotation is α/β interconversion.
  6. b — Glycosidic bond is an acetal.
  7. a — Sucrose is α-1,β-2 (unique among common disaccharides).
  8. b — Cellulose is β-1,4.
  9. a — Starch is α-1,4.
  10. a — Most natural sugars are D-configured.
  11. a — All-equatorial chair for β.
  12. a — Free anomeric C makes a reducing sugar.
  13. a — Both anomeric Cs tied up in glycosidic bond.
  14. b — Lactase deficiency causes lactose intolerance.
  15. a — Multiple terminal Cs allow fast glucose release.
  16. a — Imine formation + Amadori rearrangement.
  17. a — Carbohydrate antigens on RBCs.
  18. c — Both DNA and RNA have pentose-furanose rings.
  19. a — Aldolase performs retro-aldol.
  20. a — Cellulose is the structural material of plants.

21. β-D-glucopyranose chair: all 5 substituents (C2, C3, C4, C5-CH₂OH, ring O) are equatorial. The C1-OH is also equatorial. This is the all-equatorial chair, the most stable possible. α-D-glucopyranose chair: C1-OH is axial; the other 5 substituents are equatorial. The axial C1-OH has 1,3-diaxial interactions with axial Hs at C3 and C5. Energy difference: β is ~0.7 kcal/mol more stable, giving β:α = 64:36 (consistent with experiment).

22. Step 1: H⁺ protonates the C1-OH of glucose's hemiacetal form. Step 2: water leaves, forming an oxocarbenium ion at C1 (the carbon stabilized by the ring oxygen lone pair). Step 3: methanol attacks the C1 (the oxocarbenium carbon). Step 4: deprotonation of the new C1-OMe gives the methyl glucoside (an acetal). The reaction is reversible and stereospecific based on which face of the oxocarbenium is attacked.

23. Starch (α-1,4): the α-glycosidic bond at each glucose forces the chain into a helical conformation. Helical structure is digestible by α-amylase, cleaved into glucose for energy. Cellulose (β-1,4): the β-glycosidic bond forces the chain to alternate orientation at each glucose, giving a flat ribbon-like polymer. β-1,4 chains pack tightly via hydrogen bonds, forming microfibrils — the structural material of plant cell walls. Mammalian intestines lack β-1,4 glucosidase, so cellulose is indigestible to humans (it acts as dietary fiber).

24. Step 1: Schiff base formation between the aldehyde of open-chain glucose (the 0.02% form) and the α-amine of valine: glucose-CHO + H₂N-Val → glucose-CH=N-Val + H₂O. Step 2: Amadori rearrangement: the imine tautomerizes to a 1-amino-1-deoxy-fructose (an enaminol), which closes back to a stable ketosamine. Net: glucose + amine → 1-amino-1-deoxy-2-keto sugar. The clinical use of HbA1c: it accumulates over the lifetime of the red blood cell (~120 days), giving an average measure of blood glucose over ~3 months. Used for diabetes diagnosis (HbA1c > 6.5% = diabetes) and management.

25. The cyclic hemiacetal is favored because: (1) intramolecular cyclization is entropically favored (no need to bring two separate molecules together; the C5-OH and C1=O are in the same molecule). (2) The 6-membered ring chair conformation places all substituents equatorial — the most stable conformation possible. (3) Hemiacetal formation converts the C=O to a stronger C-O σ bond plus a new C-O σ bond from the OH. Net: ~99.98% cyclic, 0.02% open-chain in water at equilibrium. The chemistry is exactly the intramolecular hemiacetal formation discussed in Section 25.3.