Biological molecules

The claim is incorrect.

A chain of $n$ monomers requires $n-1$ bonds, not $n$ bonds, because the number of bonds in a polymer equals the number of monomers minus one, whether the polymer is a simple unbranched chain or a branched tree structure — every monomer except the very first one added is joined by exactly one bond to the rest of the growing structure, so $n$ monomers always produce $n-1$ bonds, regardless of how those bonds are arranged into straight runs or branch points.

For 500 monomers: $500-1=499$ bonds, so hydrolysis (which uses one water molecule per bond broken) consumes 499 water molecules, not 500.

Branching makes no difference to this total: it changes WHERE the bonds are (some monomers gain a second, branching bond in addition to their main-chain bond) but not HOW MANY monomers were joined in total, and the $n-1$ rule depends only on the total monomer count, not on the specific arrangement of bonds between them.

For the polypeptide: an unbranched chain of $n$ amino acids is held together by $n-1$ peptide bonds, so $214-1=213$ peptide bonds.

Hydrolysis consumes exactly one water molecule per bond broken, so 213 water molecules are consumed.

For the triglyceride: three water molecules, one for each of the three ester bonds joining a fatty acid to the glycerol.

The reasoning differs because the $n-1$ relationship is not a separate fact but a consequence of the chain arrangement — in a chain, each monomer after the first contributes one bond, and the last has nothing beyond it to bond to.

A triglyceride is not a chain: glycerol is a single backbone forming three independent bonds, one to each fatty acid, so the count is simply the number of bonds (3), not one fewer than the number of components (4).

In both cases the rule actually being applied is the same — one water molecule per bond — and only the number of bonds has to be worked out afresh.

A is a reducing sugar (e.g. glucose) — a positive Benedict's test directly, with no need for hydrolysis first, indicates a monosaccharide (or a reducing disaccharide) already present as such.

B is a non-reducing sugar (e.g. sucrose) — the initial negative result rules out any reducing sugar being present, but a positive result only after acid hydrolysis shows that a glycosidic bond was broken by the hydrolysis, releasing reducing monosaccharides (glucose and fructose, in the case of sucrose) that were not free to react with Benedict's reagent while still joined together.

C is starch — it gives no colour change with Benedict's reagent even after hydrolysis is attempted with acid alone (starch requires more extensive hydrolysis, or is simply not tested this way), but iodine in potassium iodide is the specific test for starch, and the blue-black colour is the diagnostic positive result for its helical amylose structure.

(a) Molecule X: with a branch point roughly every 10 residues,

branch points, and (since branching approximately doubles the number of free ends at each branch point in a tree structure) around 100 free chain ends available.

Molecule Y: with a branch point roughly every 25 residues,

branch points, giving around 40 free chain ends.

(b) Molecule X would release glucose faster, because glycogen phosphorylase (the enzyme releasing glucose units) can act simultaneously at every free chain end — with roughly 2.5 times as many free ends as molecule Y ($100$ vs $40$), molecule X allows correspondingly more enzyme molecules to work in parallel, releasing glucose monomers far more rapidly when a sudden, large energy demand requires a fast response.

This is exactly the structural reasoning behind why animal glycogen is more highly branched than plant starch (amylopectin): animals typically have less predictable, more rapidly changing metabolic demands, favouring a structure optimised for fast mobilisation over one optimised purely for compact storage.

(a) Lipids are non-polar and dissolve readily in ethanol (a non-polar-enough solvent for this purpose) but are not soluble in water.

When the ethanol layer, now containing any dissolved lipid, is poured into water, the lipid immediately comes out of solution as the ethanol mixes with (and is diluted by) the water — it forms into many tiny droplets suspended throughout the mixture, an emulsion, which scatters passing light and so makes the mixture appear cloudy or milky white, rather than transparent.

(b) In increasing order of lipid content: P (0) < R (1) < Q (3) < S (4).

Sample P, scoring 0 (no cloudiness, staying clear), most likely contains no lipid at all — with no lipid present in the ethanol extract, none comes out of solution when added to water, and the mixture stays as clear as the water and ethanol alone.

(a) Triglyceride Q (saturated) is most likely solid at room temperature: its fatty acid tails, with no double bonds, are straight and can pack together closely and regularly, allowing strong collective van der Waals attraction between neighbouring molecules, giving a higher melting point.

Triglyceride R (unsaturated) is most likely liquid at room temperature: each carbon-carbon double bond introduces a kink in the fatty acid tail, preventing molecules from packing together as closely or regularly, weakening the collective intermolecular attraction between them and lowering the melting point.

(b) A phospholipid replaces one of the three fatty acid tails with a phosphate-containing group, which is hydrophilic (polar), while the remaining two fatty acid tails stay hydrophobic (non-polar) — this gives the molecule a hydrophilic head and hydrophobic tails.

In water, phospholipids spontaneously orient into structures (such as a bilayer) that shield their hydrophobic tails from water while exposing their hydrophilic heads to it, rather than clustering together into a compact, water-excluding droplet the way a pure triglyceride (uniformly hydrophobic, with no polar head) can.

This is exactly what makes phospholipids suited to forming membranes rather than energy stores, and triglycerides suited to the reverse.

Inhibitor X is competitive.

A competitive inhibitor has a shape similar enough to the substrate to bind reversibly at the active site itself, competing directly with substrate molecules for the same, limited number of active sites.

At a sufficiently high substrate concentration, substrate molecules vastly outnumber inhibitor molecules in any given collision with an active site, so substrate molecules outcompete the inhibitor for those sites almost all of the time — the inhibitor's effect becomes negligible, and the reaction can reach the same maximum rate as if no inhibitor were present at all, just requiring a higher substrate concentration to get there.

Inhibitor Y is non-competitive.

A non-competitive inhibitor binds at a site other than the active site (an allosteric site), changing the enzyme's tertiary structure and so the active site's shape, making it no longer complementary to the substrate.

Because this inhibitor does not compete for the active site itself, raising substrate concentration cannot displace it or restore the active site's shape — the proportion of enzyme molecules disabled by the inhibitor stays the same regardless of how much substrate is present, so the maximum rate achievable is permanently reduced below the no-inhibitor maximum, no matter how far substrate concentration is raised.

(a) The biuret test: add biuret reagent (alkaline copper(II) sulfate solution) to the original sample.

A colour change from blue to lilac/purple confirms the presence of peptide bonds, and so of protein; a sample giving no colour change (staying blue) would indicate no peptide bonds, and so no protein, was present.

(b) Chromatography: a spot of the hydrolysed mixture (now containing free, individual amino acids, since hydrolysis has broken every peptide bond) is placed near the base of a sheet of chromatography paper, and the paper is stood in a shallow layer of solvent, which rises up the paper by capillary action, carrying each amino acid with it.

Because different amino acids have different solubilities in the solvent relative to their affinity for the paper (depending on the chemical properties of their particular R group), they travel different distances and separate into distinct spots.

Each spot's identity is established either by comparing its position (specifically its $R_f$ value, distance travelled by the spot divided by distance travelled by the solvent front) against known reference $R_f$ values, or by running known amino acid standards on the same sheet and comparing spot positions directly.

Below 40 °C: raising the temperature increases the kinetic energy of both enzyme and substrate molecules, so they collide more frequently and with more energy.

More successful collisions per unit time means more enzyme–substrate complexes form per unit time, so the rate rises.

Above 40 °C: the enzyme molecules vibrate more, and this breaks the hydrogen bonds and ionic bonds between R groups that hold the tertiary structure in place.

The active site changes shape, is no longer complementary to the substrate, and the enzyme is denatured.

As progressively more enzyme molecules are denatured, fewer functional active sites remain, so the rate falls — steeply, because denaturation is not a gradual slowing of the same process but the permanent removal of enzyme molecules from it.

The reversibility question distinguishes the two effects.

At 45 °C only some enzyme molecules have denatured; the rest are unaffected, and cooling restores the kinetic-energy contribution, so the rate returns close to its previous value.

At 55–60 °C essentially every enzyme molecule has denatured; cooling reduces the vibration but the bonds that determined the original folding do not spontaneously re-form in the correct arrangement, so the active sites are not restored and the rate stays low.

The key point is that the rise is a reversible effect on molecular motion, while the fall is an effectively irreversible change to the enzyme's structure.

(a):

(b):

Ratio:

— the pH 4.0 solution has 1000 times more H⁺ ions than the pH 7.0 solution, even though the pH values themselves differ by only 3 units.

This is the direct consequence of pH being a logarithmic scale: each single pH unit represents a tenfold change in H⁺ concentration, so a 3-unit drop in pH means $10^3=1000$ times more H⁺.

(a) In double-stranded DNA, adenine pairs with thymine and cytosine pairs with guanine (Chargaff's base-pairing rules), so:

The remaining bases, cytosine and guanine, are also present in equal proportions to each other:

shared equally, giving:

(b) Uracil replaces thymine in RNA.

The mRNA's overall base composition cannot be predicted from the double-stranded DNA percentages found in part

(a) alone, because mRNA is transcribed from only ONE of the two DNA strands (the template strand) over a specific, limited region (a gene), not from the whole double-stranded molecule — the double-stranded percentages describe both strands combined, in equal (complementary) proportion, and give no information about the base sequence of any one single strand over any specific short region.

Start from what each model predicts, then compare.

Conservative replication would keep the two original heavy strands together and build a wholly new molecule from ¹⁴N nucleotides, so after one round there would be two bands — one fully heavy, one fully light — and nothing of intermediate density.

A single intermediate band, with the heavy band gone, is inconsistent with that, so conservative replication is eliminated at round one.

Dispersive replication is not eliminated at round one, because scattering old and new material evenly through both strands would also give a uniformly intermediate density.

It is eliminated at round two.

Dispersive replication predicts that every molecule always has the same density as every other, so round two would give one band, now uniformly one-quarter heavy, sitting between intermediate and light.

Two discrete bands were observed instead, which requires the original heavy material to have stayed together as intact whole strands.

What survives both tests is semi-conservative replication: each molecule has one parental and one new strand.

For the calculation, note that the number of original heavy strands never changes: there are two of them, and each stays intact in one molecule forever.

After $n$ rounds there are $2^n$ molecules, of which exactly 2 contain a heavy strand.

After four rounds: $2^4=16$ molecules, of which 2 contain a ¹⁵N strand, giving:

(a):

ATP molecules hydrolysed per second.

(b) ATP releases a small, manageable, and precisely quantised amount of energy each time a single terminal phosphate bond is hydrolysed (ATP → ADP + Pᵢ), which can be coupled directly and efficiently to one specific energy-requiring reaction, such as one cycle of an active transport pump.

Glucose, by contrast, releases a much larger amount of energy all at once when fully oxidised, and that energy is released across many separate enzyme-catalysed steps rather than in one immediately usable, appropriately sized packet — using glucose directly to power a single transport event would release far more energy than that one event needs, with no efficient way to capture and reuse the excess.

ATP acts as a universal, intermediate energy currency precisely because its hydrolysis delivers energy in a size and form that individual cellular processes, including active transport, can use directly and immediately.

Glucose stores a large amount of chemical energy, but that energy can only be released by breaking it down through a specific, multi-step metabolic pathway — it cannot be released directly and instantly at the exact time and place a particular reaction needs it, and even if it could, the quantity released from one glucose molecule is far larger than almost any single cellular reaction requires.

ATP solves this mismatch: the energy released by respiration is used, across those same metabolic pathways, to phosphorylate ADP to ATP — effectively repackaging a large, awkwardly-timed release of energy into many small, immediately usable, appropriately sized units.

Hydrolysing one ATP molecule back to ADP and Pᵢ then releases a small, manageable quantity of energy, which can be coupled directly and efficiently to one specific energy-requiring reaction, exactly when and where that reaction needs it.

(a) Water has a high latent heat of vaporisation — a relatively large quantity of energy is required to convert liquid water into water vapour, since many hydrogen bonds between water molecules must be broken for molecules to escape into the gas phase.

When sweat evaporates from the skin, it draws this energy from the body as heat, producing a significant cooling effect from the evaporation of only a small volume of water.

(b) Water has a relatively high specific heat capacity — a relatively large amount of energy is needed to raise the temperature of a given mass of water by a given amount, again because much of the absorbed energy goes into disrupting hydrogen bonds rather than immediately increasing kinetic energy.

Dry soil, lacking this extensive hydrogen bonding network, has a lower specific heat capacity, so the same energy input produces a much larger temperature change in soil than in an equivalent mass of water.

Both observations arise from the strong cohesion between water molecules — the attraction between neighbouring water molecules caused by hydrogen bonding.

At the water's surface, water molecules are attracted more strongly to each other than to the air above, producing an inward net force that behaves like a thin 'skin' under tension (surface tension); this tension is strong enough to support the weight of a sufficiently light, appropriately-shaped organism such as a water strider.

In a capillary tube, water molecules are also attracted to the tube's inner surface (adhesion); this adhesion, combined with cohesion between water molecules, draws water up the narrow tube against gravity.

Both phenomena are therefore different visible consequences of the same underlying property — cohesion between water molecules due to hydrogen bonding.

(a) A fall in blood pH means a rise in hydrogen ion (H⁺) concentration.

Excess H⁺ ions can interact with the charged R groups of amino acids within an enzyme's structure, disrupting ionic interactions and hydrogen bonds that stabilise its tertiary structure.

This changes the enzyme's active site shape, reducing how well it complements its substrate — enzyme activity falls, and may denature entirely if severe.

(b) Each haemoglobin molecule contains four iron-containing haem groups, and it is specifically the iron ion (Fe²⁺) within each haem group that reversibly binds one oxygen molecule.

An iron deficiency reduces the amount of functional haemoglobin the body can produce, directly reducing the blood's overall oxygen-carrying capacity — the basis of iron-deficiency anaemia.

(a) A 100-fold increase in $[{\text{H}}^{+}]$ corresponds to a change in pH of:

pH units.

Solution B's pH:

(b) Solution B's enzyme activity would be lower — most enzymes in the body have an optimum pH close to 7.4; a pH of 5.4 is a significant shift toward more acidic conditions, sufficient to disrupt the bonds maintaining tertiary structure and active site shape.