AQA A-Level Biology Year 1 Complete Mark Scheme & Fact Recall(2)

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AQA A-LEVEL BIOLOGY COMPLETE MARK SCHEME BANK
YEAR 1 SPECIFICATION CODE 3.1 – 3.4




Specification
Question Mark Scheme
Code


3.1.1
1. Breaks a chemical bonds between two molecules;
Monomers and What is hydrolysis?
2. Using water;
Polymers



1. Polysaccharide of α-glucose;
Glycogen - Glycogen Structure (3) 2. (Joined by) glycosidic bonds;
3. Branched structure;
3.1.2
Carbohydrates

1. Cellulose is made up of β-glucose (monomers) and glycogen is made up of α-glucose (monomers);
Glycogen - Glycogen compared 2. Cellulose molecule has straight chain and glycogen is branched;
with cellulose (4) 3. Cellulose molecule has straight chain and glycogen is coiled;
4. Glycogen has 1,4- and 1,6- glycosidic bonds and cellulose hxas only 1,4- glycosidic bonds;

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1. Insoluble (in water), so doesn’t affect water potential;
2. Branched / coiled / (α-)helix, so makes molecule compact;
Glycogen - Glycogen structure
related to function (5) 3. Polymer of (α-)glucose so provides glucose for respiration;
4. Branched / more ends for fast breakdown / enzyme action;
5. Large (molecule), so can’t cross the cell membrane


1. Insoluble;
2. Don’t affect water potential;

Starch – Relate 3 properties to its 3. Helical;
function (6) 4. Compact;

5. Large molecule;
6. Cannot leave cell;


1. Both contain ester bonds (between glycerol and fatty acid);
2. Both contain glycerol;
3. Fatty acids on both may be saturated or unsaturated;
4. Both are insoluble in water;
Phospholipids compared with
3.1.3 Lipids 5. Both contain C, H and O but phospholipids also contain P;
Triglycerides (8)
6. Triglyceride has three fatty acids and phospholipid has two fatty acids plus phosphate group;
7. Triglycerides are hydrophobic/non-polar and phospholipids have hydrophilic/polar and hydrophobic/polar
region;
8. Phospholipids form monolayer (on surface)/micelle/bilayer (in water) but triglycerides don’t;


1. Polymer of amino acids;
3.1.4 Proteins 2. Joined by peptide bonds;
Protein - Protein Structure (7) 3. Formed by condensation;
4. Primary structure is order of amino acids;
5. Secondary structure is folding of polypeptide chain due to hydrogen bonding;

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6. Tertiary structure is 3-D folding due to hydrogen bonding and ionic/disulfide bonds;
7. Quaternary structure is two or more polypeptide chains;


1. (before reaction) active site not complementary to/does not fit substrate;
Enzymes – “Induced Fit” Model (3) 2. Shape of active site changes as substrate binds/as enzyme-substrate complex forms;
3. Stressing/distorting/bending bonds (in substrate leading to reaction);


1. particles have more kinetic energy
Enzymes – Increased temperature 2. therefore they move more
and reaction rate (4) 3. so there are more collisions between substrates and active sites
4. so more ES complexes form


1. Heat above the optimum breaks hydrogen bonds
2. this causes the tertiary structure to unfold
Enzymes – Denaturation (5) 3. so the active site changes shape
4. substrate can no longer bind to the active site, as it’s no longer complementary
5. so fewer ES complexes form


1. Ionic bonds holding tertiary structure break
Enzymes – Effect of Changes in pH 2. active site distorts and substrate no longer binds to active site
(4) 3. charges on amino acids in active site affected
4. fewer ES complexes form


1. (Rate of) increase in concentration of product slows as substrate is used up OR High initial rate as plenty of
substrate/more E-S complexes;
Enzymes – Concentration of
2. No increase after 25 minutes/at end/levels off because no substrate left;
Substrate (2)
Reject ref. to enzyme being used up

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1. Initial rate of reaction faster at 37 °C (than 25 °C);
Enzymes – Describe and explain 2. Because more kinetic energy;
the temperature graph of enzyme 3. So more E–S collisions/more E–S complexes formed;
rate (5) 4. Graph reaches plateau /levels off at 37 °C;
5. Because all substrate used up;


1. Competitive inhibitor binds to active sites of enzyme but non-competitive inhibitor binds at allosteric site/away
from active site;
2. (Binding of) competitive inhibitor does not cause change in shape of active site but (binding of) non-competitive
Enzymes – Comparison of
does (cause change in size of active site);
Competitive and Non Competitive
3. So with competitive inhibitor, at high substrate concentrations (active) enzyme still available but with non-
Inhibition (4)
competitive inhibitor (active) enzymes no longer available;
4. At higher substrate concentrations likelihood of enzyme-substrate collisions increases with competitive inhibitor
but this is not possible with non-competitive inhibitor;


(As temperature increases,)

1. Enzyme activity reduced/(some) enzymes denatured;
Enzymes – Effect of temperature 2. Less photosynthesis, so fewer sugars formed (plants only);
linked to plant growth (6) 3. Less (complex) biological molecules/organic substances made (that add to mass);
4. Less respiration;
5. Less energy/ATP for growth;
6. Less energy for named function associated with growth (eg mitosis)


1. Helicase unzips DNA double helix
3.1.5 Nucleic 2. by breaking hydrogen bonds
Acids are DNA Replication (8) 3. free DNA nucleotides in the nucleus complementary base pair OR A to T and C to G;
Important 4. with exposed bases on both strands
Information- 5. hydrogen bonds between complementary base pairs reform
6. both strands act as a template