"Biomolecules" — Practice

Question 1

A carbohydrate does not reduce Fehling's solution, but on hydrolysis with dilute acid it gives two hexoses, both of which reduce it. Identify the carbohydrate and give the structural reason for both observations.

Accepted Answer

The carbohydrate is sucrose. It fails the test because its glycosidic linkage joins C1 of glucose to C2 of fructose, and both of those are anomeric carbons, so no ring in the molecule can open to a carbonyl. Hydrolysis breaks that linkage and releases free glucose and free fructose. Glucose now has a free anomeric carbon and reduces the reagent directly. Fructose does so indirectly, since Fehling's solution is alkaline and in base fructose isomerises through an enediol to glucose and mannose, which are oxidised.

Question 2

State the reaction that proves each of the following about glucose: an unbranched six-carbon chain, the presence of a carbonyl group, that the carbonyl is an aldehyde, the presence of five hydroxyl groups, and a primary alcohol at the far end of the chain.

Accepted Answer

Prolonged heating with hydroiodic acid gives n-hexane, proving six carbons in an unbranched chain. Formation of an oxime with hydroxylamine, or a cyanohydrin with hydrogen cyanide, proves a carbonyl group. Oxidation by bromine water to gluconic acid of the same carbon count proves the carbonyl is an aldehyde, since bromine water does not oxidise ketones. Acetylation with acetic anhydride to a penta-acetate proves five hydroxyl groups. Oxidation by nitric acid to saccharic acid, a dicarboxylic acid with the same carbon count, proves that the terminal carbon opposite the aldehyde carries a primary alcohol.

Question 3

Pure alpha-D-glucose has a specific rotation of +111 degrees and pure beta-D-glucose +19.7 degrees. A freshly prepared solution of either, left to stand, settles at +52.7 degrees. Name the phenomenon and explain it.

Accepted Answer

The phenomenon is mutarotation. Each anomer differs only in the configuration at the anomeric carbon C1, created when the C5 hydroxyl closed onto the C1 aldehyde to give the cyclic hemiacetal. In solution the ring opens back to the small open-chain form and recloses, and it can reclose either way. The two anomers therefore interconvert until they reach a fixed equilibrium mixture, whose rotation is +52.7 degrees. That both starting materials reach the same value from opposite directions is what identifies it as an equilibrium rather than decomposition.

Question 4

Glucose forms an oxime with hydroxylamine, yet it gives a negative Schiff's test and forms no addition product with sodium hydrogensulphite. Its penta-acetate reacts with hydroxylamine not at all. Reconcile these four observations.

Accepted Answer

Glucose exists almost entirely as the cyclic hemiacetal, with roughly 0.02 per cent present as the open chain at any moment. Oxime and cyanohydrin formation are effectively irreversible, so they drain that small fraction and the ring keeps opening to replace it, letting the reaction run to completion. The Schiff and bisulphite reactions are readily reversible and require a real standing concentration of free aldehyde, which does not exist, so both fail. The penta-acetate is decisive: with C1 capped as an ester the ring cannot open at all, so no aldehyde is ever available and hydroxylamine finds nothing to react with.

Question 5

Name the constituent monosaccharides and the linkage in maltose and in lactose, and state why both are reducing sugars.

Accepted Answer

Maltose is two units of alpha-D-glucose joined by an alpha 1 to 4 glycosidic linkage. Lactose is beta-D-galactose joined to beta-D-glucose by a beta 1 to 4 linkage. In each case the linkage uses the anomeric carbon of one unit and an ordinary C4 hydroxyl of the other, so one anomeric carbon is left free. That free end can open to an aldehyde, so both sugars reduce Tollens and Fehling reagents.

Question 6

Amino acids melt at high temperatures, dissolve well in water and dissolve poorly in benzene. Explain, and define the isoelectric point.

Accepted Answer

These are the properties of an ionic solid rather than a small covalent molecule. The carboxyl group protonates the amino group within the same molecule, giving a zwitterion that carries a positive and a negative charge although its net charge is zero. The solid is therefore an internal salt held by strong electrostatic forces, which raises the melting point, and its charges are solvated well by water and badly by benzene. Because it carries both an acidic and a basic centre, an amino acid is amphoteric. The isoelectric point is the pH at which the two charges balance exactly, so the molecule has no net charge and does not migrate in an electric field.

Question 7

A protein is heated until it coagulates and loses all biological activity. State precisely which levels of its structure are destroyed and which survive, and explain why.

Accepted Answer

The secondary, tertiary and quaternary structures are destroyed. All three are held by weak interactions, namely hydrogen bonds between backbone carbonyl and amide groups, ionic attractions, van der Waals forces and disulphide bridges, and heat supplies enough energy to disrupt them. The primary structure survives, because it consists of peptide bonds, which are covalent amide bonds that heating alone does not break. The denatured protein therefore has the same amino acid sequence and the same molar mass as before, but no longer has the shape that gave it its function.

Question 8

How many peptide bonds are present in a decapeptide, and how many distinct tripeptides can be built from a pool of all twenty protein amino acids?

Accepted Answer

A decapeptide has ten residues joined in a single line, and the number of links is one fewer than the number of residues, so there are nine peptide bonds. For the tripeptides, each of the three positions can independently hold any of the twenty amino acids, and order matters because a peptide chain has a direction. The count is therefore twenty raised to the third power, which is 8000 distinct tripeptides.

Question 9

Enzymes lower activation energy like any catalyst, yet they are far more selective and far more fragile than an inorganic catalyst. Explain both differences from protein structure.

Accepted Answer

An enzyme is a globular protein, and its tertiary structure folds the chain into a compact shape carrying an active site with a definite geometry and a definite arrangement of polar and non-polar groups. Only a substrate that matches that site can bind and react, which is the origin of the specificity. The same fact explains the fragility, since the active site exists only as long as the fold does. Heating, extremes of pH or heavy metal ions denature the protein, the fold collapses, the active site is destroyed, and catalytic activity is lost even though the peptide chain is intact.

Question 10

Vitamin A can accumulate to toxic levels in the body, while vitamin C must be supplied almost daily. Explain from structure, and name the one exception to the pattern.

Accepted Answer

Vitamin A is fat-soluble. Its structure is largely hydrocarbon, so it dissolves in the lipids of the liver and adipose tissue and is retained, which allows it to build up if intake is excessive. Vitamin C is water-soluble, carrying several hydroxyl groups, so it dissolves in body fluids and is excreted in the urine rather than stored, and the supply must be renewed regularly. The general rule is that fat-soluble A, D, E and K are stored while the water-soluble B group and C are not. The exception is vitamin B12, which is water-soluble yet is stored in the liver.

Question 11

Distinguish a nucleoside from a nucleotide, and name the linkage that joins successive nucleotides in a nucleic acid chain.

Accepted Answer

A nucleoside consists of a nitrogen base joined to a pentose sugar through a beta-N-glycosidic bond at the C1 of the sugar. A nucleotide is a nucleoside that additionally carries a phosphate group esterified at the C5 hydroxyl of the sugar. Successive nucleotides are joined by a phosphodiester linkage, in which the phosphate at C5 of one nucleotide is also esterified to the C3 hydroxyl of the next, giving the sugar-phosphate backbone that runs along the outside of the chain.

Question 12

A double-stranded DNA sample contains 22 per cent adenine by base count. Calculate the percentage of the other three bases, and predict how its thermal stability compares with a sample containing 35 per cent adenine.

Accepted Answer

Adenine pairs only with thymine, so thymine is also 22 per cent and the two together account for 44 per cent. The remaining 56 per cent is shared equally between guanine and cytosine, since they pair only with each other, giving 28 per cent of each. This sample is therefore relatively rich in guanine and cytosine, whose pairs are held by three hydrogen bonds against the two that hold an adenine-thymine pair. It will need a higher temperature to separate its strands than the 35 per cent adenine sample, which has only 15 per cent each of guanine and cytosine and correspondingly fewer hydrogen bonds per base pair.

Practice — "Biomolecules"

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