OCR SUMMER 2026
PREDICTED PAPER
A Level Biology B (Advancing Biology)
H422/01 Fundamentals of biology
Time allowed: 2 hours 15 minutes
You can use:
a scientific or graphical calculator
a ruler (cm/mm)
Please write clearly in black ink. Do not write in the barcodes.
Centre number Candidate number
First name(s)
Last name
INSTRUCTIONS
• Use black ink. You can use an HB pencil, but only for graphs and diagrams.
• Write your answer to each question in the space provided. If you need extra space use
the lined pages at the end of this booklet. The question numbers must be clearly shown.
• Answer all the questions.
• Where appropriate, your answer should be supported with working. Marks might be
given for using acorrect method, even if your answer is wrong.
INFORMATION
• The total mark for this paper is 110.
• The marks for each question are shown in brackets [ ].
• Quality of extended response will be assessed in questions marked with an asterisk (*).
• This document has 40 pages.
ADVICE
• Read each question carefully before you start your answer.
Turn over
for more: tyrionpapers.com
, 2
Section A
You should spend a maximum of 40 minutes on this section.
Write your answer to each question in the box provided.
1 A SEM image shows detailed surface contours of a pollen grain. Which statement
correctly explains how this image was formed?
A Electrons passed through the specimen and hit a fluorescent screen
B Electrons were scattered off the surface in a pattern reflecting its contours, analysed by
computer to produce a 3D image
C Electromagnets focused electrons onto a photographic plate beneath the specimen
D Visible light was reflected off heavy metal stains applied to the surface
Your answer [1]
2 Why are neutrophils unable to sustain a prolonged immune response against a
persistent infection?
A They lack a nucleus and cannot produce antibodies
B They are agranulocytes and cannot perform phagocytosis repeatedly
C They cannot renew their lysosomes and die after breaking down only a few pathogens
D They are too large to leave the capillaries and reach the infection site
Your answer [1]
3 A bacterium is treated with an antibiotic that inhibits hydrogen bond formation.
Which structural consequence would directly result?
A Plasmids would fuse with the main circular DNA
B Ribosomes would increase in size to match eukaryotic cells
C Pili would detach from the cell surface membrane
D Cellulose microfibrils in the cell wall would fail to cross-link, reducing mechanical
strength
Your answer [1]
for more: tyrionpapers.com
, 3
4 A student uses a stage micrometer with 50 divisions per mm. 8 eyepiece divisions
align with 20 stage micrometer divisions. What is the actual length of one
eyepiece division?
A 25 μm
B 40 μm
C 50 μm
D 80 μm
Your answer [1]
5 A patient has a condition causing abnormally low plasma protein levels
(hypoproteinaemia). What would be the most likely consequence at the
capillaries?
A Increased urea excretion into the tissue fluid
B Reduced hydrostatic pressure forcing more fluid into cells
C Excess tissue fluid accumulation as reduced plasma proteins lower the osmotic pull
returning fluid to the blood
D Increased glucose absorption into the plasma from tissue fluid
Your answer [1]
6 Amylopectin has branching via α(1→6) bonds occurring every 24–30 glucose
units. What is the structural and functional significance of this specific branching
pattern?
A It reduces the molecular weight of amylopectin, making it easier to transport in phloem
B It creates numerous terminal end points at regular intervals, maximising the sites available
for simultaneous hydrolysis and rapid glucose release
C It converts the linear amylose helix into a soluble form suitable for temporary leaf storage
D It allows amylopectin to form hydrogen bonds with water, increasing its hydrophilic
properties
Your answer [1]
for more: tyrionpapers.com
, 4
7 A student constructs a comparison table of haemoglobin and lysozyme:
Feature Haemoglobin Lysozyme
Number of 4 1
polypeptide chains
Prosthetic group Haem (×4) None
Location Blood plasma Secretions (tears, mucus)
Levels of structure Primary–Quaternary Primary–Tertiary
Using the table, which conclusion is best supported?
A Lysozyme has a more complex structure than haemoglobin as it functions in more
locations
B Haemoglobin lacks tertiary structure as it possesses quaternary structure instead
C Both proteins have identical bonding interactions as they are both globular proteins
D Haemoglobin's quaternary structure and prosthetic groups give it a greater structural
complexity than lysozyme, consistent with its specialised role in oxygen transport
requiring cooperative binding across four subunits
Your answer [1]
8 Aspirin inhibits COX by adding an acetyl group to an amino acid very close to the
active site. Which type of enzyme inhibition does this best represent and why?
A Competitive inhibition — the acetyl group has a similar shape to the substrate
arachidonate and competes for the active site directly
B Allosteric activation — the acetyl group binds to a separate site and increases COX
activity
C Non-competitive inhibition — the acetyl group binds adjacent to but not at the active site,
permanently altering the active site shape and preventing substrate binding
D Reversible competitive inhibition — increasing arachidonate concentration would restore
COX activity
Your answer [1]
for more: tyrionpapers.com
PREDICTED PAPER
A Level Biology B (Advancing Biology)
H422/01 Fundamentals of biology
Time allowed: 2 hours 15 minutes
You can use:
a scientific or graphical calculator
a ruler (cm/mm)
Please write clearly in black ink. Do not write in the barcodes.
Centre number Candidate number
First name(s)
Last name
INSTRUCTIONS
• Use black ink. You can use an HB pencil, but only for graphs and diagrams.
• Write your answer to each question in the space provided. If you need extra space use
the lined pages at the end of this booklet. The question numbers must be clearly shown.
• Answer all the questions.
• Where appropriate, your answer should be supported with working. Marks might be
given for using acorrect method, even if your answer is wrong.
INFORMATION
• The total mark for this paper is 110.
• The marks for each question are shown in brackets [ ].
• Quality of extended response will be assessed in questions marked with an asterisk (*).
• This document has 40 pages.
ADVICE
• Read each question carefully before you start your answer.
Turn over
for more: tyrionpapers.com
, 2
Section A
You should spend a maximum of 40 minutes on this section.
Write your answer to each question in the box provided.
1 A SEM image shows detailed surface contours of a pollen grain. Which statement
correctly explains how this image was formed?
A Electrons passed through the specimen and hit a fluorescent screen
B Electrons were scattered off the surface in a pattern reflecting its contours, analysed by
computer to produce a 3D image
C Electromagnets focused electrons onto a photographic plate beneath the specimen
D Visible light was reflected off heavy metal stains applied to the surface
Your answer [1]
2 Why are neutrophils unable to sustain a prolonged immune response against a
persistent infection?
A They lack a nucleus and cannot produce antibodies
B They are agranulocytes and cannot perform phagocytosis repeatedly
C They cannot renew their lysosomes and die after breaking down only a few pathogens
D They are too large to leave the capillaries and reach the infection site
Your answer [1]
3 A bacterium is treated with an antibiotic that inhibits hydrogen bond formation.
Which structural consequence would directly result?
A Plasmids would fuse with the main circular DNA
B Ribosomes would increase in size to match eukaryotic cells
C Pili would detach from the cell surface membrane
D Cellulose microfibrils in the cell wall would fail to cross-link, reducing mechanical
strength
Your answer [1]
for more: tyrionpapers.com
, 3
4 A student uses a stage micrometer with 50 divisions per mm. 8 eyepiece divisions
align with 20 stage micrometer divisions. What is the actual length of one
eyepiece division?
A 25 μm
B 40 μm
C 50 μm
D 80 μm
Your answer [1]
5 A patient has a condition causing abnormally low plasma protein levels
(hypoproteinaemia). What would be the most likely consequence at the
capillaries?
A Increased urea excretion into the tissue fluid
B Reduced hydrostatic pressure forcing more fluid into cells
C Excess tissue fluid accumulation as reduced plasma proteins lower the osmotic pull
returning fluid to the blood
D Increased glucose absorption into the plasma from tissue fluid
Your answer [1]
6 Amylopectin has branching via α(1→6) bonds occurring every 24–30 glucose
units. What is the structural and functional significance of this specific branching
pattern?
A It reduces the molecular weight of amylopectin, making it easier to transport in phloem
B It creates numerous terminal end points at regular intervals, maximising the sites available
for simultaneous hydrolysis and rapid glucose release
C It converts the linear amylose helix into a soluble form suitable for temporary leaf storage
D It allows amylopectin to form hydrogen bonds with water, increasing its hydrophilic
properties
Your answer [1]
for more: tyrionpapers.com
, 4
7 A student constructs a comparison table of haemoglobin and lysozyme:
Feature Haemoglobin Lysozyme
Number of 4 1
polypeptide chains
Prosthetic group Haem (×4) None
Location Blood plasma Secretions (tears, mucus)
Levels of structure Primary–Quaternary Primary–Tertiary
Using the table, which conclusion is best supported?
A Lysozyme has a more complex structure than haemoglobin as it functions in more
locations
B Haemoglobin lacks tertiary structure as it possesses quaternary structure instead
C Both proteins have identical bonding interactions as they are both globular proteins
D Haemoglobin's quaternary structure and prosthetic groups give it a greater structural
complexity than lysozyme, consistent with its specialised role in oxygen transport
requiring cooperative binding across four subunits
Your answer [1]
8 Aspirin inhibits COX by adding an acetyl group to an amino acid very close to the
active site. Which type of enzyme inhibition does this best represent and why?
A Competitive inhibition — the acetyl group has a similar shape to the substrate
arachidonate and competes for the active site directly
B Allosteric activation — the acetyl group binds to a separate site and increases COX
activity
C Non-competitive inhibition — the acetyl group binds adjacent to but not at the active site,
permanently altering the active site shape and preventing substrate binding
D Reversible competitive inhibition — increasing arachidonate concentration would restore
COX activity
Your answer [1]
for more: tyrionpapers.com
