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AQA Biological Molecules Mark Schemes Questions with Answers
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AQA Biological Molecules Mark Schemes Questions with Answers
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AQA Biological Molecules MarkSchemes Questions with Answers
Test for a non-reducing sugar (3) - Answer-Heat with Benedict's reagent and no colour
change (1);
boil with dilute HCl and then neutralize with NaHCO3 (1);
re-heat with Benedict's reagent and colour change from blue to brick-red (1)
Test for starch (2) - Answer-Add iodine in potassium iodide solution (1);
colour change from brown to blue-black (1)
Explain why glucose and maltose both taste sweet but starch does not. - Answer-
Glucose and maltose soluble/starch insoluble;
Active site function - Answer-1. Tertiary structure / 3D shape of enzyme (means);
Accept references to active site
2. Active site complementary to maltose / substrate / maltose fits into active site / active
site and substrate fit like a lock and key;
Idea of shapes fitting together
3. Description of induced fit;
4. Enzyme is a catalyst / lowers activation energy / energy required for reaction;
Accept "provides alternative pathway for the reaction at a lower energy level"
5. By forming enzyme-substrate complex;
Accept idea that binding stresses the bonds so more easily broken
Inhibitors - Answer-1. Inhibitors reduce binding of enzyme to substrate / prevent
formation of ES complex;
Max 3 if only one type of inhibition dealt with. Accept maltase and maltose as examples
of enzyme and substrate (and others)
Only once, for either inhibitor
(Competitive inhibition),
2. Inhibitor similar shape (idea) to substrate;
3. (Binds) in to active site (of enzyme);
Accept allows max rate of reaction to be reached / max product will eventually be
formed
Accept complementary to active site
4. (Inhibition) can be overcome by more substrate;
(Non-competitive inhibition),
5. Inhibitor binds to site on enzyme other than active site;
6. Prevents formation of active site / changes (shape of) active site;
Accept does not allow max rate of reaction to be reached / max product will not be
formed
7. Cannot be overcome by adding more substrate;
, Saccharin, cyclamates and sucrose are chemically different but they all taste sweet.
Suggest why. - Answer-Have similar molecular shape/structure / similarly positioned
chemical groups;
so bind to/fit receptors;
Describe how a peptide bond is formed between two amino acids to form a dipeptide -
Answer-1. Condensation (reaction) / loss of water; Accept each marking point if shown
clearly in diagram.
2. Between amine / NH2 and carboxyl / COOH; Accept between amino (group) and
carboxylic / acid (group)
The secondary structure of a polypeptide is produced by bonds between amino acids.
Describe how. - Answer-Hydrogen bonds; Accept as a diagram Reject N - - - C / ionic /
disulfide bridge / peptide bond
2. Between NH (group of one amino acid) and C=O (group); OR Forming β pleated
sheets / α helix;
Two proteins have the same number and type of amino acids but different tertiary
structures. Explain why. - Answer-1. Different sequence of amino acids OR Different
primary structure; If candidate assumes proteins are the same, accept effect of different
pH/ temperature
2. Forms ionic / hydrogen / disulfide bonds in different places;
The chilling requirement of seeds of certain plant species is considered to be an
adaptation for survival in countries with seasonal changes in environmental conditions.
Suggest how this adaptation may enable these plant species to survive and respond to
seasonal changes. - Answer-Seeds/embryo remain dormant/inactive in winter/cold OR
Growth/development of seed/embryo during winter/cold; Ignore: hibernate.
Accept: 'seed survives winter/cold'.
Reject: plant develops or seed germinates during winter/cold.
2. Seeds/plants develop in spring/summer OR Seeds/plants develop when
temperature/light increases;
Accept: seeds/plants develop when more light or when temperature is higher.
Accept: seed germinates/'sprouts' during spring/summer or when temp/light increases.
3. Plant photosynthesise (in spring/when warm);
4. Produce (more) seeds/offspring in spring/growing season;
A high concentration of galactose slows down the breakdown of lactose by lactase. Use
your knowledge of competitive inhibition to suggest why. - Answer-1. Galactose is a
similar shape / structure to lactose / both complementary; Q Reject: Same shape /
structure
2. (Inhibitor / Galactose) fits into / enters / binds with active site (of enzyme); Accept
blocks active site
3. Prevents / less substrate fitting into / binding with (active site) / fewer or no E-S
complexes;
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