"Biomolecules" — Practice

Question 1

Explain why D-glucose, D-fructose and D-mannose all give the same osazone, and predict whether D-galactose would give it too.

Accepted Answer

Osazone formation reacts at carbon one and then oxidises and reacts at carbon two, so any difference confined to those two carbons is destroyed.

Glucose and fructose differ only in whether the carbonyl is at carbon one or two. Glucose and mannose differ only in configuration at carbon two.

All three therefore converge on a single osazone.

D-Galactose is the **carbon-four** epimer of glucose, and carbon four is untouched by the reaction. Its osazone is therefore different.

Question 2

A pure sample of $\beta$-D-glucose has a specific rotation of $+19^{\circ}$, which rises to $+52.7^{\circ}$ on standing. Explain, and calculate the percentage of each anomer at equilibrium.

Accepted Answer

The ring opens to the free aldehyde and recloses either way, so the pure anomer converts partly to the other until equilibrium is reached. This is mutarotation.

Let the fraction of $\alpha$ be $x$: $112x+19\left(1-xight)=52.7$

$93x=33.7\qquad\Rightarrow\qquad x=0.362$

So about $36\%$ $\alpha$ and $64\%$ $\beta$ at equilibrium, the $\beta$ form being favoured because its anomeric hydroxyl is equatorial.

Question 3

Classify each pair: $\alpha$-D-galactose and $\beta$-D-galactose; D-glucose and D-galactose; D-fructose and D-glucose.

Accepted Answer

$\alpha$- and $\beta$-D-galactose differ only at the anomeric carbon: **anomers**.

D-Glucose and D-galactose differ at carbon four alone: **epimers**.

D-Fructose and D-glucose differ in the position of the carbonyl group itself, one being a ketose and the other an aldose. They are **functional isomers**, not stereoisomers of any kind.

The last pair is the trap, since they share a molecular formula and give the same osazone but are not stereoisomers at all.

Question 4

Explain why maltose reduces Fehling's solution but its methyl glycoside does not.

Accepted Answer

In maltose the glycosidic bond uses the anomeric carbon of only one glucose unit, leaving the other free as a hemiacetal.

That ring can open to a free aldehyde, which reduces the copper(II) in Fehling's solution.

Converting the free anomeric hydroxyl to a methyl glycoside makes it an **acetal**, which is stable to the alkaline conditions of the test and cannot open.

With no route to a free carbonyl, the glycoside is non-reducing and also fails to show mutarotation.

Question 5

Sucrose has a specific rotation of $+66.5^{\circ}$. Explain what happens on hydrolysis and calculate the final rotation, given glucose at $+52.7^{\circ}$ and fructose at $-92^{\circ}$.

Accepted Answer

Hydrolysis breaks the glycosidic bond, giving equal amounts of glucose and fructose.

The rotation of the mixture is the average: $\dfrac{52.7+\left(-92ight)}{2}=-19.7^{\circ}$

The sign has changed from positive to negative, which is why the product is called **invert sugar**.

Fructose's large negative rotation more than cancels glucose's positive one, even though the two are present in equal amounts.

Question 6

Starch and cellulose are both polymers of D-glucose. Explain why one is a digestible food and the other an indigestible fibre.

Accepted Answer

Starch uses $\alpha$-$1,4$ linkages, which bend the chain at every unit so that it coils into a loosely packed helix.

Cellulose uses $\beta$-$1,4$ linkages, which let successive units sit rotated by half a turn so that the chain runs straight. Straight chains hydrogen bond extensively into rigid fibres.

Human digestive enzymes are specific to the $\alpha$-$1,4$ bond and cannot hydrolyse the $\beta$ form.

The monomer is identical in both; only the stereochemistry at one bond differs, and it changes both the physical form and the biological fate completely.

Question 7

Compare amylose, amylopectin and glycogen in terms of branching, and explain why animals store glucose as the most branched of the three.

Accepted Answer

Amylose is unbranched, with $\alpha$-$1,4$ bonds only. Amylopectin has $\alpha$-$1,6$ branches every twenty to twenty-five units. Glycogen has them every eight to twelve, making it the most branched.

Enzymes release glucose from the **non-reducing ends** of a chain, so a highly branched polymer offers far more points of attack.

Glycogen can therefore be mobilised very rapidly when energy is needed.

Plants, being sessile and slower in their demands, store the less branched starch instead.

Question 8

Alanine has $pK_a$ values of $2.35$ and $9.69$. Find its isoelectric point and describe its behaviour in electrophoresis at pH $2$, pH $6$ and pH $11$.

Accepted Answer

$pI=\dfrac{2.35+9.69}{2}=6.02$

At pH $2$, well below $pI$, the carboxyl is protonated and the molecule carries a net **positive** charge, migrating to the cathode.

At pH $6$, essentially at $pI$, it exists as the zwitterion with no net charge and does **not migrate**.

At pH $11$, well above $pI$, the amino group is deprotonated and the molecule is **negative**, migrating to the anode.

Question 9

Glutamic acid has $pK_a$ values of $2.19$, $4.25$ and $9.67$, the middle one belonging to its side chain. Find its isoelectric point.

Accepted Answer

The side chain is acidic, so at low pH the molecule is a cation with both carboxyls protonated.

The neutral zwitterion is the species existing after the first carboxyl has lost its proton but before the side chain does.

The two values bracketing it are $2.19$ and $4.25$.

$pI=\dfrac{2.19+4.25}{2}=3.22$, well below $7$ as expected for an acidic amino acid.

Question 10

Explain why the peptide bond is planar and why that matters for protein structure.

Accepted Answer

The nitrogen lone pair is delocalised onto the carbonyl oxygen, giving the carbon to nitrogen bond substantial double-bond character.

A double bond cannot rotate freely, so the six atoms of the amide unit are held in one plane.

That rigidity removes rotational freedom from the backbone, leaving only two rotatable bonds per residue.

Regular repeating structures such as the $\alpha$-helix and $\beta$-sheet are possible only because of that restriction, since a freely rotating backbone could not maintain a fixed geometry over many residues.

Question 11

A DNA sample contains $31\%$ thymine. Find the percentage of the other three bases and predict how its melting temperature compares with a sample containing $22\%$ thymine.

Accepted Answer

Adenine equals thymine, so adenine is also $31\%$, accounting for $62\%$ between them.

The remaining $38\%$ is shared equally by guanine and cytosine: $19\%$ each.

The second sample has $22\%$ each of adenine and thymine, leaving $28\%$ each of guanine and cytosine.

The second sample is richer in guanine and cytosine, whose pairs have three hydrogen bonds against two, so it melts at the **higher** temperature.

Question 12

Give three structural differences between DNA and RNA and state the functional consequence of one of them.

Accepted Answer

**Sugar:** DNA has $2$-deoxyribose, RNA has ribose with a hydroxyl at carbon two.

**Bases:** DNA uses thymine, RNA uses uracil in its place.

**Strands:** DNA is a double helix, RNA is usually single stranded.

The missing hydroxyl in DNA is the important one functionally: it makes the backbone far less prone to hydrolysis, which is why DNA is stable enough to store information over a lifetime while RNA is a short-lived working copy.

Practice — "Biomolecules"

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