AQA A-level Biology · 3.1

Biological molecules

50 revision questions with answers for 3.1 Biological molecules in AQA A-level Biology (spec 7402). Practise with flashcards, a topic quiz, a random quiz across the course, 5 crosswords, or read every question and answer below. Not official AQA material.

All 50 questions and answers

Question 1 · Spec 3.1.1 · Monomers and polymers · Recall
What are monomers and polymers? Give three biological examples.
Answer:
A monomer is a small building unit. A polymer contains many joined monomers. Monosaccharides form polysaccharides, amino acids form polypeptides, and nucleotides form polynucleotides.
Question 2 · Spec 3.1.1 · Biological reactions · Application
How do condensation and hydrolysis reactions differ?
Answer:
Condensation joins molecules by forming a covalent bond and releasing water. Hydrolysis uses water to break a covalent bond. Forming a linear chain of n monomers requires n - 1 condensation reactions.
Question 3 · Spec 3.1.2 · Disaccharides · Recall
Which monosaccharides form maltose, sucrose and lactose, and which bond joins them?
Answer:
Maltose contains glucose and glucose. Sucrose contains glucose and fructose. Lactose contains glucose and galactose. Each pair joins by condensation, forming a glycosidic bond.
Question 4 · Spec 3.1.2 · Glucose isomers · Recall
How do alpha-glucose and beta-glucose differ?
Answer:
They have the same molecular formula but differ in the position of the hydroxyl group on carbon 1. In the standard ring diagrams, this group is below the ring in alpha-glucose and above it in beta-glucose.
Question 5 · Spec 3.1.2 · Starch · Application
How is starch adapted for storage in plant cells?
Answer:
Starch contains alpha-glucose. Amylose is coiled, making it compact. Amylopectin is branched, providing ends for enzyme action. Starch is insoluble, so it has little osmotic effect and does not readily diffuse out of cells.
Question 6 · Spec 3.1.2 · Glycogen · Application
How does glycogen's structure suit its role in animals?
Answer:
Glycogen is an insoluble, compact polymer of alpha-glucose. It is more highly branched than amylopectin, providing many ends for enzymes to release glucose rapidly for respiration.
Question 7 · Spec 3.1.2 · Cellulose · Application
How does cellulose strengthen plant cell walls?
Answer:
Beta-glucose units form straight, unbranched chains, with alternate units inverted. Many hydrogen bonds form between parallel chains. Bundles form microfibrils with high tensile strength, helping the wall resist stretching under turgor pressure.
Question 8 · Spec 3.1.2 · Reducing sugars · Practical
How do you test for reducing sugars?
Answer:
Add Benedict's solution to the sample and heat in a hot water bath. A positive result changes blue solution to a green, yellow, orange or brick-red precipitate, depending on concentration. Compare samples under the same conditions.
Question 9 · Spec 3.1.2 · Non-reducing sugars · Practical
How do you test for a non-reducing sugar after a negative Benedict's test?
Answer:
Use a fresh sample. Heat with dilute hydrochloric acid to hydrolyse the sugar, then cool and neutralise with sodium hydrogencarbonate. Repeat Benedict's test. A positive result now supports the presence of a non-reducing sugar.
Question 10 · Spec 3.1.2 · Starch test · Practical
What is the test for starch, and how should a negative result be described?
Answer:
Add iodine in potassium iodide solution. Starch produces a blue-black colour. A negative result remains yellow-brown. Use a known starch sample and water as positive and negative controls.
Question 11 · Spec 3.1.3 · Triglycerides · Recall
How is a triglyceride formed?
Answer:
One glycerol molecule joins to three fatty acids by three condensation reactions. Three ester bonds form and three water molecules are released. The fatty acid R groups may differ.
Question 12 · Spec 3.1.3 · Fatty acids · Application
How can saturated and unsaturated fatty acids be distinguished in a diagram?
Answer:
A saturated fatty acid has no carbon-carbon double bonds in its hydrocarbon chain. An unsaturated fatty acid has at least one. A cis double bond creates a kink, reducing close packing of neighbouring molecules.
Question 13 · Spec 3.1.3 · Lipid storage · Application
Why are triglycerides effective energy stores?
Answer:
Their many carbon-hydrogen bonds provide a large energy yield when oxidised. They are insoluble, so storage has little effect on water potential. Their compact storage provides a high energy content per unit mass.
Question 14 · Spec 3.1.3 · Phospholipids · Application
How does a phospholipid's structure lead to bilayer formation?
Answer:
A phospholipid has glycerol, two fatty acids and a phosphate-containing group. Its head is hydrophilic and its tails are hydrophobic. In water, heads face the aqueous surroundings while tails face inward, forming a bilayer.
Question 15 · Spec 3.1.3 · Emulsion test · Practical
How is the emulsion test carried out and interpreted?
Answer:
Shake the sample with ethanol to dissolve lipids, then add water. A white, cloudy emulsion indicates lipid. A negative result remains clear. Keep ethanol away from naked flames.
Question 16 · Spec 3.1.4.1 · Amino acids · Recall
What is the general structure of an amino acid?
Answer:
A central carbon is bonded to an amine group, a carboxyl group, a hydrogen atom and a variable R group. The R group differs among amino acids and influences their properties.
Question 17 · Spec 3.1.4.1 · Peptide bonds · Recall
How are dipeptides and polypeptides formed?
Answer:
A peptide bond forms by condensation between the carboxyl group of one amino acid and the amine group of another. Two amino acids form a dipeptide. Many joined amino acids form a polypeptide.
Question 18 · Spec 3.1.4.1 · Protein structure · Recall
Distinguish primary, secondary, tertiary and quaternary structure.
Answer:
Primary structure is the amino acid sequence. Secondary structure includes alpha-helices and beta-pleated sheets formed by hydrogen bonding. Tertiary structure is a polypeptide's overall folding. Quaternary structure is the association of multiple polypeptide chains in a functional protein.
Question 19 · Spec 3.1.4.1 · Protein bonding · Recall
Which interactions stabilise protein structure?
Answer:
Hydrogen bonds stabilise secondary structure and contribute to tertiary structure. Ionic bonds form between charged R groups. Covalent disulfide bridges form between cysteine R groups. These interactions help maintain a specific three-dimensional shape.
Question 20 · Spec 3.1.4.1 · Structure and function · Application
How can changing one amino acid alter protein function?
Answer:
The new R group may change bonding and folding, altering tertiary structure. An enzyme's active site or a receptor's binding site may change shape. The effect depends on the position and properties of the replacement amino acid.
Question 21 · Spec 3.1.4.1 · Biuret test · Practical
How is the biuret test used to detect protein?
Answer:
Add sodium hydroxide solution followed by dilute copper sulfate solution, or use prepared biuret reagent. A lilac or purple colour indicates peptide bonds and therefore protein. A negative result remains blue.
Question 22 · Spec 3.1.4.2 · Enzyme action · Recall
How do enzymes increase reaction rate?
Answer:
Enzymes lower activation energy by providing an alternative reaction pathway. Binding at the active site forms enzyme-substrate complexes. The enzyme is regenerated after products leave and can catalyse further reactions.
Question 23 · Spec 3.1.4.2 · Induced fit · Recall
Explain the induced-fit model of enzyme action.
Answer:
A substrate binds to the active site and induces a change in its shape. The active site becomes more complementary to the substrate, stressing bonds or positioning reactants so the reaction requires less activation energy.
Question 24 · Spec 3.1.4.2 · Enzyme models · Analysis
How does induced fit explain specificity and improve on a rigid active-site model?
Answer:
Specificity depends on the active site's tertiary structure and chemical properties. Only suitable substrates bind effectively. Induced fit allows the active site to change shape during binding, explaining evidence that enzymes are flexible molecules.
Question 25 · Spec 3.1.4.2 · Temperature · Application
Explain the shape of an enzyme rate-temperature curve.
Answer:
Initially, higher temperature increases kinetic energy and successful collisions. Beyond the optimum, bonds maintaining tertiary structure are disrupted, altering the active site. Fewer enzyme-substrate complexes form and rate falls. Low temperature usually slows activity without denaturing the enzyme.
Question 26 · Spec 3.1.4.2, MS 0.5 · pH · Calculation
How does pH affect enzymes? What is pH when hydrogen ion concentration is 1.0 x 10⁻⁴ mol dm⁻³?
Answer:
Changes in hydrogen ion concentration can disrupt ionic and hydrogen bonding, altering active-site shape and charge. An enzyme has an optimum pH. Using pH = -log10[H+], the stated concentration gives pH 4.
Question 27 · Spec 3.1.4.2 · Enzyme concentration · Application
Why does increasing enzyme concentration sometimes increase rate proportionally?
Answer:
With substrate in excess, more enzyme molecules provide more available active sites and more enzyme-substrate complexes form per second. If substrate becomes limiting, further increases in enzyme concentration produce a smaller effect.
Question 28 · Spec 3.1.4.2 · Substrate concentration · Application
Why does enzyme reaction rate approach a maximum as substrate concentration rises?
Answer:
Initially, more substrate increases successful collisions and enzyme-substrate complex formation. At high substrate concentration, almost all active sites are occupied. Enzyme concentration becomes limiting, so additional substrate has little effect.
Question 29 · Spec 3.1.4.2 · Competitive inhibition · Recall
How does a competitive inhibitor work, and how can its effect be reduced?
Answer:
It binds to the active site, competing with substrate and reducing enzyme-substrate complex formation. Increasing substrate concentration increases the chance of substrate binding and can reduce the inhibitor's effect.
Question 30 · Spec 3.1.4.2 · Non-competitive inhibition · Recall
How does a non-competitive inhibitor reduce enzyme activity?
Answer:
It binds at a site away from the active site and changes the enzyme's structure, reducing catalytic activity. Adding more substrate cannot restore the original maximum rate because some enzyme molecules remain inhibited.
Question 31 · Spec 3.1.4.2, RP1 · Required practical 1 · Practical
How could you investigate the effect of pH on amylase activity?
Answer:
Mix fixed volumes and concentrations of amylase and starch with a buffer of known pH. Keep temperature constant. Test samples with iodine at regular intervals until starch disappears. Repeat at several pH values and calculate mean rates using 1/time.
Question 32 · Spec 3.1.4.2, RP1, MS 3.6 · Required practical 1 · Calculation
How is an initial rate obtained from a product-time graph? A tangent rises by 1.8 cm³ in 30 s.
Answer:
Draw a tangent at the start of the curve. Use two well-separated points on the tangent and calculate change in product divided by change in time. Here the initial rate is 1.8/30 = 0.060 cm³ s⁻¹.
Question 33 · Spec 3.1.4.2, RP1, MS 1.11 · Required practical 1 · Calculation
How do repeats and measurement uncertainty affect evaluation? A 20.0 s time has uncertainty +/-0.5 s.
Answer:
Repeats reveal variability and allow a mean to be calculated. Percentage uncertainty here is (0.5/20.0) x 100 = 2.5%. Repeats reduce the effect of random variation on the mean, while a systematic error needs a method or calibration change.
Question 34 · Spec 3.1.5.1 · Nucleotides · Recall
What are the components of DNA and RNA nucleotides?
Answer:
Each nucleotide contains a pentose sugar, a phosphate group and a nitrogen-containing base. DNA uses deoxyribose and A, T, C or G. RNA uses ribose and A, U, C or G. Nucleotides join through phosphodiester bonds.
Question 35 · Spec 3.1.5.1 · DNA structure · Recall
Describe the bonds and arrangement in a DNA molecule.
Answer:
Two antiparallel polynucleotide strands form a double helix. Phosphodiester bonds join nucleotides within each sugar-phosphate backbone. Hydrogen bonds join complementary bases between strands. Adenine pairs with thymine and cytosine pairs with guanine.
Question 36 · Spec 3.1.5.1 · RNA structure · Recall
How does RNA differ structurally from DNA, and what are ribosomes made from?
Answer:
RNA is usually a relatively short, single polynucleotide strand containing ribose and uracil. DNA contains deoxyribose and thymine and normally forms a double helix. Ribosomes contain ribosomal RNA and proteins.
Question 37 · Spec 3.1.5.1, MS 0.3 · Base percentages · Calculation
A double-stranded DNA sample contains 18% adenine. Calculate the other base percentages.
Answer:
Complementary pairing means thymine is also 18%. Adenine plus thymine total 36%, leaving 64% for cytosine and guanine. These are equal, so cytosine is 32% and guanine is 32%. These equalities apply to the complete double-stranded molecule.
Question 38 · Spec 3.1.5.2 · DNA replication · Recall
Describe the roles of helicase and DNA polymerase in replication.
Answer:
Helicase unwinds the double helix and breaks hydrogen bonds between complementary bases. Each exposed strand acts as a template. Free DNA nucleotides pair with complementary bases. DNA polymerase joins adjacent nucleotides by condensation, forming phosphodiester bonds.
Question 39 · Spec 3.1.5.2 · Semi-conservati ve replication · Recall
Why is DNA replication described as semi-conservative?
Answer:
Each daughter DNA molecule contains one strand from the original molecule and one newly synthesised strand. Complementary base pairing preserves the base sequence, allowing genetic information to pass accurately to daughter cells.
Question 40 · Spec 3.1.5.2 · Replication evidence · Analysis
What results distinguish semi-conservative replication from conservative replication in a nitrogen-isotope experiment?
Answer:
After heavy-nitrogen DNA is transferred to light nitrogen, one replication gives only intermediate-density DNA with semi-conservative replication. Conservative replication predicts separate heavy and light bands. After two replications, intermediate and light bands support semi-conservative replication and distinguish it from dispersive replication.
Question 41 · Spec 3.1.6 · ATP structure · Recall
What is the structure of ATP?
Answer:
Adenosine triphosphate is a nucleotide derivative containing adenine, ribose and three phosphate groups. It acts as an immediate energy source for many cellular processes.
Question 42 · Spec 3.1.6 · ATP hydrolysis · Recall
How does ATP hydrolysis support cellular reactions?
Answer:
ATP hydrolase catalyses ATP + water to ADP + inorganic phosphate. The overall reaction releases energy that can be coupled to energy-requiring processes. Transferring phosphate to another molecule can make that molecule more reactive.
Question 43 · Spec 3.1.6 · ATP resynthesis · Application
How is ATP resynthesised, and why must it be continually recycled?
Answer:
ADP and inorganic phosphate combine by condensation. ATP synthase catalyses this in respiration and photosynthesis, using energy supplied by these processes. Cells use ATP rapidly, so continual resynthesis maintains a usable supply.
Question 44 · Spec 3.1.7 · Water as a metabolite · Recall
Give examples of water's role as a metabolite.
Answer:
Water is consumed in hydrolysis, including digestion of biological polymers. Water is produced in condensation when monomers join. It is also a reactant in photosynthesis. A metabolite participates directly in a metabolic reaction.
Question 45 · Spec 3.1.7 · Water as a solvent · Application
Why is water a useful solvent in organisms?
Answer:
Water is polar. Its partial charges interact with ions and other polar molecules, allowing many substances to dissolve. This supports transport in blood and tissue fluid and provides a medium in which metabolic reactions occur.
Question 46 · Spec 3.1.7 · Heat capacity · Application
How does water's high heat capacity benefit organisms?
Answer:
A relatively large energy transfer is required to change water's temperature. Water therefore buffers temperature fluctuations in cells, body fluids and aquatic habitats, helping maintain conditions suitable for enzyme activity.
Question 47 · Spec 3.1.7 · Evaporative cooling · Application
Why does water provide effective cooling through evaporation?
Answer:
Water has a high latent heat of vaporisation because energy is needed to overcome hydrogen bonding. Evaporation removes substantial energy from a surface, allowing cooling through sweating or transpiration with a relatively small water loss.
Question 48 · Spec 3.1.7 · Cohesion · Application
How do hydrogen bonds between water molecules support transport in plants?
Answer:
Hydrogen bonding produces cohesion, helping maintain a continuous water column in xylem under tension. It also contributes to surface tension at an air-water boundary.
Question 49 · Spec 3.1.8 · Inorganic ions · Recall
State a role for hydrogen, iron, sodium and phosphate ions.
Answer:
Hydrogen ion concentration determines pH. Iron ions occur in haem groups of haemoglobin. Sodium ions provide the gradient for co-transport of glucose and amino acids. Phosphate groups are components of DNA nucleotides and ATP.
Question 50 · Spec 3.1.4.2, 3.1.5.1, 3.2.4 · Synoptic link · Synoptic
How does molecular structure explain specificity in enzymes, antibodies and DNA pairing?
Answer:
Enzymes have active sites complementary to suitable substrates. Antibody variable regions have binding sites complementary to specific antigens. DNA bases pair in complementary combinations through hydrogen bonding. In each case, molecular arrangement determines interactions and biological function.

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