AQA A LEVEL BIOLOGY PAPER 2
7402-2 MARK SCHEME MS ACTUAL
EXAM PAPER 2026 QUESTIONS WITH
ANSWERS GRADED A+
◍ AMP-activated protein kinase (AMPK) is an enzyme that regulates a
number of cellular processes. Exercise leads to activation of AMP
K. Figure 1 shows one effect of activation of AMPK during exercise.CPT1
is a channel protein that transports fatty acids into mitochondria.Using
Figure 1, explain the benefit of activation of AMPK during exercise.[3
marks].
Answer: nan
◍ Dengue is a serious disease that is caused by a virus. The virus is carried
from one person to another by a mosquito, Aedes aegypti. One method used
to try to reduce transmission of this disease is the Sterile Insect Technique
(SIT). This involves releasing large numbers of sterile (infertile) male
A. aegypti into the habitat. These males have been made infertile by using
radiation.Explain how using the SIT could reduce transmission of dengue.
[2 marks].
Answer: nan
◍ Hydrolysis.
Answer: Breaks a chemical bonds between two molecules; Using water;
◍ Glycogen Structure.
Answer: Polysaccharide of α-glucose; (Joined by) glycosidic bonds;
Branched structure;
◍ Describe how the mark-release-recapture method could be used to determine
, the population of
A. aegypti at the start of the investigation.[3 marks].
Answer: nan
◍ Cellulose vs Glycogen.
Answer: Cellulose is made up of β-glucose (monomers) and glycogen is
made up of α-glucose (monomers); Cellulose molecule has straight chain
and glycogen is branched; Cellulose molecule has straight chain and
glycogen is coiled; Glycogen has 1,4- and 1,6- glycosidic bonds and
cellulose has only 1,4- glycosidic bonds;
◍ Suggest why the scientists released more transgenic males every week.[1
mark].
Answer: nan
◍ The release of transgenic males proved successful in reducing the number of
A. aegypti.Describe how the results in Figure 2 support this conclusion.[2
marks].
Answer: nan
◍ Glycogen Structure Related to Function.
Answer: Insoluble (in water), so doesn't affect water potential; Branched /
coiled / (α-)helix, so makes molecule compact; Polymer of (α-)glucose so
provides glucose for respiration; Branched / more ends for fast breakdown /
enzyme action; Large (molecule), so can't cross the cell membrane;
◍ The scientists then compared the length of time that the control mice and the
trained mice could carry out prolonged exercise. The trained mice were able
to exercise for a longer time period than control mice.Explain why.[3
marks].
Answer: nan
◍ Starch Properties Related to Function.
Answer: Insoluble; Don't affect water potential; Helical; Compact; Large
molecule; Cannot leave cell;
◍ The scientists also compared the diameter of samples of muscle fibres taken
, from young mice and adult mice.Some of their results are shown in Figure
4.Describe two differences between these samples of muscle fibres.[2
marks].
Answer: nan
◍ Phospholipids vs Triglycerides.
Answer: Both contain ester bonds (between glycerol and fatty acid); Both
contain glycerol; Fatty acids on both may be saturated or unsaturated; Both
are insoluble in water; Both contain C, H and O but phospholipids also
contain P; Triglyceride has three fatty acids and phospholipid has two fatty
acids plus phosphate group; Triglycerides are hydrophobic/non-polar and
phospholipids have hydrophilic/polar and hydrophobic/polar region;
Phospholipids form monolayer (on surface)/micelle/bilayer (in water) but
triglycerides don't;
◍ Explain why the student set up Tube 1.(1 cm3 of solution without
chloroplasts and 9 cm3 of DCPIP solution in light)[2 marks].
Answer: nan
◍ Protein Structure.
Answer: Polymer of amino acids; Joined by peptide bonds; Formed by
condensation; Primary structure is order of amino acids; Secondary structure
is folding of polypeptide chain due to hydrogen bonding; Tertiary structure
is 3-D folding due to hydrogen bonding and ionic/disulfide bonds;
Quaternary structure is two or more polypeptide chains;
◍ Explain the results in Tube 3.(1 cm3 of chloroplast suspension and 9 cm3 of
DCPIP solution in light)[2 marks].
Answer: nan
◍ Explain the advantage of the student using the IC50 in this investigation.[1
mark].
Answer: nan
◍ Explain how chemicals which inhibit the decolourisation of DCPIP could
slow the growth of weeds.[2 marks].
, Answer: nan
◍ Arbuscular mycorrhiza fungi (AMF) are fungi which grow on, and into, the
roots of plants. AMF can increase the uptake of inorganic ions such as
phosphate.Suggest one way in which an increase in the uptake of phosphate
could increase plant growth.[1 mark].
Answer: nan
◍ Suggest one way in which AMF may benefit from their association with
plants.[1 mark].
Answer: nan
◍ Explain why an increase in shoot biomass can be taken as a measurement of
net primary productivity.[2 marks].
Answer: nan
◍ Induced Fit Model.
Answer: (before reaction) active site not complementary to/does not fit
substrate; Shape of active site changes as substrate binds/as
enzyme-substrate complex forms; Stressing/distorting/bending bonds (in
substrate leading to reaction);
◍ Increased Temperature and Reaction Rate.
Answer: Particles have more kinetic energy; therefore they move more; so
there are more collisions between substrates and active sites; so more ES
complexes form;
◍ Using the data from Figure 5, evaluate the effect on plant productivity of
adding AMF species and adding phosphate to the soil.[4 marks].
Answer: nan
◍ Denaturation.
Answer: Heat above the optimum breaks hydrogen bonds; this causes the
tertiary structure to unfold; so the active site changes shape; substrate can no
longer bind to the active site, as it's no longer complementary; so fewer ES
complexes form;
7402-2 MARK SCHEME MS ACTUAL
EXAM PAPER 2026 QUESTIONS WITH
ANSWERS GRADED A+
◍ AMP-activated protein kinase (AMPK) is an enzyme that regulates a
number of cellular processes. Exercise leads to activation of AMP
K. Figure 1 shows one effect of activation of AMPK during exercise.CPT1
is a channel protein that transports fatty acids into mitochondria.Using
Figure 1, explain the benefit of activation of AMPK during exercise.[3
marks].
Answer: nan
◍ Dengue is a serious disease that is caused by a virus. The virus is carried
from one person to another by a mosquito, Aedes aegypti. One method used
to try to reduce transmission of this disease is the Sterile Insect Technique
(SIT). This involves releasing large numbers of sterile (infertile) male
A. aegypti into the habitat. These males have been made infertile by using
radiation.Explain how using the SIT could reduce transmission of dengue.
[2 marks].
Answer: nan
◍ Hydrolysis.
Answer: Breaks a chemical bonds between two molecules; Using water;
◍ Glycogen Structure.
Answer: Polysaccharide of α-glucose; (Joined by) glycosidic bonds;
Branched structure;
◍ Describe how the mark-release-recapture method could be used to determine
, the population of
A. aegypti at the start of the investigation.[3 marks].
Answer: nan
◍ Cellulose vs Glycogen.
Answer: Cellulose is made up of β-glucose (monomers) and glycogen is
made up of α-glucose (monomers); Cellulose molecule has straight chain
and glycogen is branched; Cellulose molecule has straight chain and
glycogen is coiled; Glycogen has 1,4- and 1,6- glycosidic bonds and
cellulose has only 1,4- glycosidic bonds;
◍ Suggest why the scientists released more transgenic males every week.[1
mark].
Answer: nan
◍ The release of transgenic males proved successful in reducing the number of
A. aegypti.Describe how the results in Figure 2 support this conclusion.[2
marks].
Answer: nan
◍ Glycogen Structure Related to Function.
Answer: Insoluble (in water), so doesn't affect water potential; Branched /
coiled / (α-)helix, so makes molecule compact; Polymer of (α-)glucose so
provides glucose for respiration; Branched / more ends for fast breakdown /
enzyme action; Large (molecule), so can't cross the cell membrane;
◍ The scientists then compared the length of time that the control mice and the
trained mice could carry out prolonged exercise. The trained mice were able
to exercise for a longer time period than control mice.Explain why.[3
marks].
Answer: nan
◍ Starch Properties Related to Function.
Answer: Insoluble; Don't affect water potential; Helical; Compact; Large
molecule; Cannot leave cell;
◍ The scientists also compared the diameter of samples of muscle fibres taken
, from young mice and adult mice.Some of their results are shown in Figure
4.Describe two differences between these samples of muscle fibres.[2
marks].
Answer: nan
◍ Phospholipids vs Triglycerides.
Answer: Both contain ester bonds (between glycerol and fatty acid); Both
contain glycerol; Fatty acids on both may be saturated or unsaturated; Both
are insoluble in water; Both contain C, H and O but phospholipids also
contain P; Triglyceride has three fatty acids and phospholipid has two fatty
acids plus phosphate group; Triglycerides are hydrophobic/non-polar and
phospholipids have hydrophilic/polar and hydrophobic/polar region;
Phospholipids form monolayer (on surface)/micelle/bilayer (in water) but
triglycerides don't;
◍ Explain why the student set up Tube 1.(1 cm3 of solution without
chloroplasts and 9 cm3 of DCPIP solution in light)[2 marks].
Answer: nan
◍ Protein Structure.
Answer: Polymer of amino acids; Joined by peptide bonds; Formed by
condensation; Primary structure is order of amino acids; Secondary structure
is folding of polypeptide chain due to hydrogen bonding; Tertiary structure
is 3-D folding due to hydrogen bonding and ionic/disulfide bonds;
Quaternary structure is two or more polypeptide chains;
◍ Explain the results in Tube 3.(1 cm3 of chloroplast suspension and 9 cm3 of
DCPIP solution in light)[2 marks].
Answer: nan
◍ Explain the advantage of the student using the IC50 in this investigation.[1
mark].
Answer: nan
◍ Explain how chemicals which inhibit the decolourisation of DCPIP could
slow the growth of weeds.[2 marks].
, Answer: nan
◍ Arbuscular mycorrhiza fungi (AMF) are fungi which grow on, and into, the
roots of plants. AMF can increase the uptake of inorganic ions such as
phosphate.Suggest one way in which an increase in the uptake of phosphate
could increase plant growth.[1 mark].
Answer: nan
◍ Suggest one way in which AMF may benefit from their association with
plants.[1 mark].
Answer: nan
◍ Explain why an increase in shoot biomass can be taken as a measurement of
net primary productivity.[2 marks].
Answer: nan
◍ Induced Fit Model.
Answer: (before reaction) active site not complementary to/does not fit
substrate; Shape of active site changes as substrate binds/as
enzyme-substrate complex forms; Stressing/distorting/bending bonds (in
substrate leading to reaction);
◍ Increased Temperature and Reaction Rate.
Answer: Particles have more kinetic energy; therefore they move more; so
there are more collisions between substrates and active sites; so more ES
complexes form;
◍ Using the data from Figure 5, evaluate the effect on plant productivity of
adding AMF species and adding phosphate to the soil.[4 marks].
Answer: nan
◍ Denaturation.
Answer: Heat above the optimum breaks hydrogen bonds; this causes the
tertiary structure to unfold; so the active site changes shape; substrate can no
longer bind to the active site, as it's no longer complementary; so fewer ES
complexes form;
