AS Level Biology 9700 — 2025 Exam Study Guide

Where to spend your time

Topic frequency ranking

Tier 1 — Essential

Immunity

Topic 11 · Every paper

~12 question parts

Tier 1 — Essential

Enzymes

Topic 3 · Data and graph heavy

~10 question parts

Tier 2 — High value

Cell membranes

Topic 4 · Always in Paper 1

~9 question parts

Tier 2 — High value

Biological molecules

Topic 2 · Proteins everywhere

~8 question parts

Tier 2 — High value

Nucleic acids

Topic 6 · DNA + protein synthesis

~8 question parts

Tier 2 — High value

Cell structure

Topic 1 · Organelles + microscopy

~7 question parts

Tier 2 — High value

Infectious disease

Topic 10 · Antibiotics + pathogens

~7 question parts

Tier 3 — Solid support

Cell cycle

Topic 5 · Mitosis + stem cells

~6 question parts

Tier 3 — Solid support

Transport in mammals

Topic 8 · Bohr shift + heart

~5 question parts

Connections

Cross-topic links

Proteins — collagen (T2), haemoglobin (T2 & T8), enzymes (T3), channel proteins (T4), antibodies (T11). Strong protein knowledge pays off across half the paper.

70S vs 80S ribosomes — tested in cell structure (T1), antibiotic action (T10), and protein synthesis (T6). One fact, marks in three topics.

Phospholipid bilayer — membranes (T4), cell comparisons (T1), cholesterol in malaria (T10). Same structure, three contexts.

Mitosis — cell cycle (T5), clonal expansion in immunity (T11), tumour formation (T5). Always say mitosis, not meiosis.

Colorimeters — enzyme kinetics (T3) and ELISA for monoclonal antibodies (T11). Same principle, tested twice per year.

Specific receptors — membrane signalling (T4), phagocytosis (T11), clonal selection (T11). One concept, three question contexts.

Cell Structure & Microscopy

Tier 2

Microscopy and measurements

Question	What was asked	Key mark scheme points
m25-22-1(a)	Name the type of microscope used for a root image	Light / optical microscope
m25-22-1(b)	Calculate actual length (µm) of a root hair cell from a micrograph	Formula: image size ÷ magnification Answer: 460–464 µm Must show working and units
w25-24-2(c)(i)	Calculate magnification of a Plasmodium image (actual size = 2 µm)	Image size ÷ actual size (2 µm) Answer: ×12,000
s25-21-2(b)	Suggest a measurement to estimate microvilli surface area	Number of microvilli per unit area of cell surface

Calculation tip

Always write: formula → substitution → answer with units. Correct method earns marks even if the arithmetic is wrong. Convert mm to µm by ×1000 before dividing.

Organelles — identification and function

Question	What was asked	Key mark scheme points
s25-22-1(a)	Identify plant cell structures X, Y, Z from a diagram	X: (large/permanent) vacuole Y: tonoplast (vacuolar membrane) Z: nucleus
w25-24-1(a)(ii)	Describe features of a nucleus, excluding chromatin	Nuclear envelope: double membrane, protects DNA Nuclear pores: allow mRNA / ribosomal subunits to exit Nucleolus: site of rRNA synthesis and ribosome assembly
s25-21-2(a) / w25-22-1(a)	Compare cilia and microvilli	Cilia: microtubules in 9+2 arrangement; attach to basal body; rhythmic movement moves mucus Microvilli: contain actin microfilaments; increase surface area for absorption; no movement
s25-22-1(d)(i)	Roles of Rough ER vs Smooth ER	Rough ER: ribosomes on surface → protein/enzyme synthesis and transport Smooth ER: no ribosomes → lipid synthesis
w25-22-1(e)(ii)	Evidence from an EM image that a cell is specialised for secretion	Abundant Rough ER Many ribosomes High density of Golgi / secretory vesicles
s25-21-2(c)	Identify organelle Z and its role in epithelial cells	Mitochondrion: produces ATP for energy-requiring processes such as active transport or exocytosis
m25-22-6(b)	Define telomere, centriole, centromere	Telomere: repeated DNA sequences at chromosome ends Centriole: organises microtubules / spindle Centromere: attachment point between sister chromatids
w25-23-1(b)	How viruses travel between plant cells without crossing membranes	Via plasmodesmata — cytoplasmic strands connecting adjacent cells through cell walls

Prokaryote vs eukaryote — comparison

Question	What was asked	Key mark scheme points
s25-23-3(a)	Comparison table: prokaryote vs eukaryote features	Prokaryote: circular DNA; 70S ribosomes; 1–5 µm; no nuclear envelope Eukaryote: linear DNA + histones; 80S ribosomes; membrane-bound organelles; nuclear envelope
s25-21-2(d)	Organisation and distribution of DNA in prokaryotes vs eukaryotes	Eukaryotic: linear DNA associated with histones; enclosed in nuclear envelope Prokaryotic: circular DNA; found free in cytoplasm (nucleoid region)
w25-22-4(a)	Similarities and differences: bacterial outer membrane vs eukaryotic cell membrane	Similarities: both have phospholipid bilayer; transport proteins Differences: bacteria have lipopolysaccharides; eukaryotes have cholesterol and glycoproteins
w25-24-2(a)	Classify disease-causing organisms as prokaryotic or eukaryotic	Cholera and TB: prokaryotic Malaria (Plasmodium): eukaryotic
w25-22-1(e)(iii)	Difference between Paneth cells and stem cells	Paneth cells: differentiated / specialised Stem cells: undifferentiated

Common mistake

Students list only differences. Both prokaryotes and eukaryotes have a phospholipid bilayer and transport proteins — these similarities earn marks too.

Biological Molecules

Tier 2

Carbohydrates — structure, bonds, and tests

Question	What was asked	Key mark scheme points
s25-21-1(a)	Define "macromolecule" and "polymer"	Macromolecule: large molecule with high molecular mass Polymer: composed of many repeating monomers (e.g. α-glucose)
m25-22-2(a) / w25-22-3(d)	Complete / draw the ring structure of α-glucose	–OH below at C1 (defines α); –H above at C1 All carbons numbered correctly — examiners check C1 and C4 specifically
s25-22-3(a)	Differences between galactose and α-glucose	Positions of –H and –OH on C1 and C4 are reversed compared to α-glucose
s25-22-3(b)	Support conclusions about sugars in carrots (Benedict's / hydrolysis tests)	Fructose, glucose, galactose are reducing sugars; sucrose is non-reducing Glucose, fructose, galactose are monosaccharides; sucrose is a disaccharide
w25-21-4(a)	Identify bonds and monomers in chitin	Correct glycosidic bond circled Monomer shown with –OH at C1 and C4 after hydrolysis
w25-24-1(b)(i)	Structural differences between starch and cellulose	Starch: α-glucose; same orientation; coiled (amylose) or branched (amylopectin); no H-bonds between chains Cellulose: β-glucose; 180° alternating rotation; linear; H-bonds between chains → microfibrils → high tensile strength

Drawing tip

Label carbon numbers on all ring diagrams. Getting the –OH direction wrong on C1 loses the α/β mark. Draw C1 first, then work around the ring.

Proteins — amino acids, structure levels, and specific proteins

Question	What was asked	Key mark scheme points
s25-23-2(a)(ii)	Draw the general structure of an amino acid	Central carbon bonded to –COOH (carboxyl) and –NH₂ (amine group), plus –H and variable R-group
m25-22-2(b)(i)	Identify covalent bonds C and D in a glycoprotein	C: peptide bond (between amino acids) D: glycosidic bond (between carbohydrate units)
s25-23-2(b)	Match descriptions to protein structure levels	Primary: sequence of amino acids, joined by peptide bonds Secondary: H-bonds → α-helix or β-sheet; proline disrupts this Tertiary: 3D fold; H-bonds, ionic, disulfide bonds, hydrophobic interactions between R-groups Quaternary: 2+ polypeptide chains associated together
m25-22-2(c)	Why β-casein lacks secondary and tertiary structure	High proline content prevents H-bonding / helix formation (disrupts secondary) Little cysteine → few/no disulfide bonds; little serine → fewer H-bonds (disrupts tertiary)
w25-23-2(a)	Structure and assembly of collagen	3 polypeptides wound into a triple helix Molecules arranged parallel and staggered, held by covalent bonds between R-groups
w25-23-2(c)(ii)	Importance of glycine in collagen	Every third amino acid; smallest R-group (–H) → allows tight packing → high tensile strength
s25-22-2(a) / w25-22-6(b)	Haemoglobin as a globular protein; quaternary structure and cooperative binding	Hydrophilic R-groups on exterior → soluble Quaternary: 2α + 2β chains, each with haem group containing Fe²⁺ First O₂ binding causes conformational change → increases affinity for subsequent O₂

Lipids and glycoproteins

Question	What was asked	Key mark scheme points
w25-21-1(c)	Effect of unsaturated fatty acids on membrane fluidity	Double bonds cause kinks/bends in hydrocarbon tails → prevent close packing → weakens hydrophobic interactions → increases fluidity
w25-24-2(e)	Role of cholesterol in Plasmodium membranes	Maintains / regulates fluidity and stability Prevents membrane becoming too fluid at high temp or too rigid at low temp
m25-22-2(b)(ii)	Role of glycoproteins in cell surface membranes	Act as receptors on cell surface; bind specific ligands — cell recognition and signalling

Cell Membranes & Transport

Tier 2

Fluid mosaic model — structure and components

Question	What was asked	Key mark scheme points
w25-21-1(a)	Identify components A, B, C of the fluid mosaic model	A: glycolipid B: cholesterol C: glycoprotein
w25-21-1(d)	Distribution of R-groups in a membrane channel protein	Facing fatty acid tails: hydrophobic R-groups Lining the pore: hydrophilic R-groups (allow polar molecules through)
w25-21-1(c) / w25-24-2(e)	Effect of unsaturated fatty acids on fluidity; role of cholesterol	Unsaturated: kinks prevent close packing → increases fluidity Cholesterol: regulates fluidity and stability across temperature changes

Transport mechanisms

Question	What was asked	Key mark scheme points
w25-23-1(a)	Match substances to transport mechanisms	O₂ / CO₂: simple diffusion through phospholipid bilayer; no protein; no ATP Glucose / ions (down gradient): facilitated diffusion via channel or carrier protein; no ATP Ions against gradient: active transport via carrier protein + ATP
m25-22-1(d)	Transport of Na⁺ and K⁺ into root hair cells	Na⁺: facilitated diffusion — down concentration gradient; channel protein; no ATP K⁺: active transport — against gradient; carrier protein; ATP required
s25-22-1(d)(ii)	Mechanisms for secreting enzymes and lipids from cells	Enzymes (large molecules): exocytosis — vesicles fuse with membrane Lipids (small, hydrophobic): simple diffusion through bilayer
w25-24-2(f)	Role of flippase, floppase, and scramblase	Flippase / floppase: active / energy-requiring; maintain membrane asymmetry Scramblase: random movement; activated by Ca²⁺; disrupts asymmetry

Common mistake

Both facilitated diffusion and active transport use carrier/channel proteins. The distinction is: with or against concentration gradient, and whether ATP is used. State both clearly.

Water potential, osmosis, and cell signalling

Question	What was asked	Key mark scheme points
m25-22-1(e)	Explain water movement between cells A and B with different water potentials	Water moves by osmosis From higher water potential (less negative) to lower water potential (more negative) e.g. –120 kPa → –350 kPa
m25-22-1(c)	Role of aquaporins in root hair cells	Channel proteins providing specific channels for water molecules Increase the rate of water uptake — do not change the direction
w25-22-1(d)	Sequence of events in cell signalling by GLP-1	GLP-1 secreted → transported in blood Binds to specific / complementary receptor on target cell surface membrane Triggers intracellular response — enzyme cascade / gene activation

Enzymes

Tier 1 — Essential

Mode of action — induced-fit and lock-and-key

Question	What was asked	Key mark scheme points
w25-21-3(a)	Similarities and differences between lock-and-key and induced-fit	Similarities: substrate binds to active site; ESC forms; activation energy lowered Induced-fit only: active site initially only partially complementary; conformational change occurs to become fully complementary
w25-23-3(a)	Describe the induced-fit hypothesis	Active site and substrate are initially only partially complementary Substrate binding causes enzyme to undergo a conformational change Active site becomes fully complementary → ESC forms → activation energy lowered
w25-24-3(a)(ii)	Explain active site specificity for different enzymes (Drosha/Dicer)	Active sites are specific to a particular site on the substrate Only a complementary substrate shape fits the active site

Common mistake

"Conformational change" is the required phrase for the induced-fit difference mark. Without it, full marks are not possible. Similarities (ESC forms, activation energy lowered) are also worth marks — do not skip them.

Inhibition — competitive and non-competitive

Question	What was asked	Key mark scheme points
s25-22-6(a)	Draw a graph for non-competitive inhibition on a Michaelis–Menten curve	Same general curve shape but lower plateau (lower Vmax) Same Km (0.055–0.06 mmol dm⁻³)
w25-21-3(c)	Use data to support that caffeine is a non-competitive inhibitor	Activity is reduced at all substrate concentrations Vmax/plateau is lower Km value is unchanged — same substrate concentration for half-Vmax
w25-23-5(c)	Mechanism of rifampicin as an enzyme inhibitor	Binds to active site of RNA polymerase (competitive); OR binds allosteric site → changes active site shape (non-competitive) Prevents transcription / phosphodiester bond formation
s25-23-4(b)(ii)	Effect of lowering pH on enzyme activity	H⁺ ions break ionic bonds and H-bonds in tertiary structure Active site changes shape → becomes less complementary to substrate → fewer ESCs

Three-point evidence rule for non-competitive inhibition

Cite all three from the data: (1) activity lower at every substrate concentration, (2) Vmax/plateau is lower, (3) Km is unchanged. All three = full marks.

Km, Vmax, and immobilised enzymes

Question	What was asked	Key mark scheme points
s25-21-3(b)(i)	Calculate the Michaelis–Menten constant (Km)	Read Vmax from graph → halve it → read corresponding substrate concentration from x-axis e.g. Vmax = 80 → half-Vmax = 40 → Km = 0.014 mmol dm⁻³
s25-21-3(b)(ii)	Advantage of an enzyme with a lower Km	Lower substrate concentration needed to reach half-Vmax Enzyme has higher affinity for substrate; works efficiently when substrate is scarce
s25-21-3(a)	Advantages of using a colorimeter to measure enzyme activity	Provides quantitative / numerical results Removes subjective / eye-based bias Can detect very small differences in colour intensity
w25-22-3(e)(ii)	Advantages of immobilised enzymes (e.g. lactase)	Enzyme can be recovered and reused → greater productivity More tolerant of fluctuations in temperature and pH More heat stable — less denaturation at high temperatures

The Mitotic Cell Cycle

Tier 3

Interphase, mitosis stages, and chromosome behaviour

Question	What was asked	Key mark scheme points
m25-22-6(a)(i)	Identify the correct sequence of the mitotic cell cycle from photomicrographs	Correct order: B → E → F → D → A → C
w25-22-1(e)(i)	Name the four stages of mitosis in sequence	Prophase → Metaphase → Anaphase → Telophase
w25-24-1(a)(i)	Explain the difference in chromatin between G1 and G2 phase	DNA replication (S phase) has occurred between G1 and G2 G2 has double the quantity of DNA / chromatin Sister chromatids have formed — two identical molecules joined at centromere
w25-24-5(a)(i)	Identify the component that makes up spindle fibres	Microtubules
w25-24-5(a)(ii)	Describe the events of telophase	Chromosomes at poles begin to decondense / uncoil Nuclear envelope reforms around each set Nucleolus reappears; spindle breaks down
w25-21-5(c)	Effects of endoreplication (multiple S phases without mitosis)	Nucleus contains more than two copies of each chromosome (polyploidy) Cell is larger; contains more organelles

Telomeres, stem cells, and tumour formation

Question	What was asked	Key mark scheme points
s25-21-4(b) / s25-23-5(a)	Location and role of telomeres	Located at the ends of chromosomes Repeated non-coding DNA sequences Prevents loss of coding sequences / genes during replication; allows continuous cell division
s25-22-5(b)	Advantages of using bone marrow stem cells to manufacture red blood cells	Haematopoietic — specialise into blood cell types Produce genetically identical cells Can divide continuously by mitosis (self-renewal)
s25-23-1(a)(ii)	Role of meristematic / procambial cells in plants	Divide continuously by mitosis Cells differentiate into vascular tissues (xylem / phloem)
s25-21-4(c)	How a tumour may form from a melanocyte	Mutation in proto-oncogene → oncogene Leads to uncontrolled mitosis Normal cell cycle checkpoints fail Mass of abnormal / non-functional cells (tumour) forms

Nucleic Acids & Protein Synthesis

Tier 2

DNA and RNA structure

Question	What was asked	Key mark scheme points
w25-23-4(a)	Define DNA monomers; compare DNA nucleotides with ATP	Monomer: repeating unit (nucleotide) used to build a polynucleotide ATP vs DNA nucleotide: ATP has ribose (–OH at C2); DNA has deoxyribose (–H at C2)
w25-23-4(b)	Identify bonds in the DNA double helix from a diagram	A–T: 2 hydrogen bonds; C–G: 3 hydrogen bonds Bond X: phosphodiester bond (linking nucleotides within each strand)
w25-24-3(a)	Complementary base pairing in double-stranded RNA	Complementary bases: A–U and G–C, held by hydrogen bonds
w25-24-3(b)	Structural differences between mRNA and siRNA	mRNA: single-stranded; siRNA: double-stranded siRNA has paired complementary nucleotides; mRNA does not mRNA is longer and codes for amino acids

Transcription and RNA processing

Question	What was asked	Key mark scheme points
m25-22-5(b)	Basic transcription mechanism and role of RNA polymerase	RNA polymerase catalyses transcription using template strand Complementary base pairing: DNA A → RNA U; T→A; G→C; C→G
w25-22-3(b)	Relationship between template strand, primary transcript, and mature mRNA	RNA polymerase → complementary primary transcript Splicing: introns removed; exons joined → mature mRNA
s25-21-4(a)	Functions of the mRNA 5' cap and poly(A) tail	Protect mRNA from degradation by nucleases Assist passage through nuclear pores Required for ribosome attachment to begin translation
s25-21-4(a)(iii)	Describe the process of RNA splicing	Introns (non-coding sequences) are removed Exons (coding sequences) are joined together → shorter mature mRNA

Translation, codons, and gene mutations

Question	What was asked	Key mark scheme points
s25-22-6(b)	Definitions of start and stop codons	Start codon (AUG): begins translation; codes for methionine Stop codon (UAA/UAG/UGA): no amino acid; polypeptide detaches from ribosome
s25-22-3(e)	Translation in viruses	Viral RNA acts directly as mRNA; translated by host ribosomes
w25-24-3(c)	Effects of cleaving mRNA via siRNA	Prevents translation No protein formed, or abnormal / shortened polypeptide that fails to fold correctly
w25-22-3(c)	Identify the mutation type where one nucleotide differs	One nucleotide replaced = base substitution
w25-23-2(c)(iii)	Compare deletion and substitution mutations	Deletion (frameshift): changes entire amino acid sequence from mutation point onwards Substitution: changes only one amino acid; rest unaffected (may be silent)

Infectious Disease & Immunity

Tier 1 — Essential

Pathogens, classification, and transmission

Question	What was asked	Key mark scheme points
s25-22-4(a)	Name the main bacterium causing TB	Mycobacterium tuberculosis — must be spelled correctly and italicised
w25-22-2(a)	Name the bacterium that causes cholera	Vibrio cholerae — must be spelled correctly and italicised
s25-22-4(b)	Transmission of TB	Infected person coughs / sneezes; uninfected person inhales airborne droplets
w25-24-2(c)(ii)	Transmission of malaria	Transmitted by vector — female Anopheles mosquito Takes a blood meal and injects Plasmodium along with saliva / anticoagulant
w25-22-2(b)	Prevention of cholera through hygiene	Provide safe drinking water; hand washing with soap Break the faecal-oral route — use toilets; separate sewage from water supply
w25-21-4(b)	Why cloth filters reduce but do not eliminate cholera	Filters remove copepods (which carry bacteria on surface) Some bacteria live free in water and pass through the fabric

Spelling — zero tolerance

Incorrect spelling of species names = no mark. Capital first letter (Genus), lowercase second (species), italicised. Underline if handwriting.

Phagocytosis and the non-specific immune response

Question	What was asked	Key mark scheme points
s25-23-2(d)	Describe the process of phagocytosis	Phagocyte has specific surface receptors that bind to molecules on pathogen surface Pseudopodia surround and engulf → endocytosis Phagosome (phagocytic vacuole) pinches off inside cell Lysosomes fuse; release hydrolytic enzymes → digest contents
w25-24-2(d)	Why phagocytes destroy infected red blood cells in malaria	Phagocytes have specific receptors for molecules (e.g. phosphatidylserine) appearing on infected cell membranes Binding triggers engulfment
s25-22-4(e)	Formation and effect of granulomas in TB	Benefit: confines / isolates pathogen Harm: damages lung tissue → reduces surface area for gas exchange → fatigue, difficulty breathing

Key term — phagosome

Do not write "bubble", "bag", or "sac". The correct term is phagosome or phagocytic vacuole. This is a standalone mark point in every phagocytosis question.

Primary and secondary immune response

Question	What was asked	Key mark scheme points
w25-23-6(d)	Sequence of events in a primary immune response leading to long-term immunity	Antigen presentation by macrophage Clonal selection: lymphocyte with complementary receptor binds antigen T-helper cells release cytokines → stimulate lymphocyte proliferation Clonal expansion: lymphocytes divide by mitosis Differentiate into plasma cells (secrete antibodies) and memory B-cells
m25-22-4(a)	Why antibody concentration is higher after a second injection of the same antigen	First injection produces memory B-cells that persist in circulation Second exposure: memory cells rapidly activated → response is faster Produces a higher concentration of antibodies
s25-21-4(d)	How cytokines and cancer cell antigens stimulate the immune system	Cytokines: stimulate clonal expansion of lymphocytes; activate macrophages Antigens: stimulate clonal selection → plasma cells → antibodies Antibodies mark cells for T-killer cells → release perforin / granzymes

Secondary response — must say both

"Faster AND greater/higher concentration" are two separate marks. Writing only "more" or only "faster" typically loses one mark.

Types of immunity and vaccination

Question	What was asked	Key mark scheme points
m25-22-4(b)	Features of passive immunity	Involves injection of ready-made antibodies / immunoglobulins Provides immediate protection; effect is temporary / short-term Does not produce memory cells
w25-22-2(c)(i)	Compare an oral vaccine vs passive immunisation	Vaccine: contains antigens; artificial active immunity; stimulates memory cells; longer-lasting Passive: contains antibodies; artificial passive immunity; no memory cells; shorter duration
w25-21-4(c)(iii)	Advantage of vaccines over antibiotics in preventing resistance	Vaccines contain multiple antigens → requires many simultaneous mutations for immune evasion Stimulate secondary immunity → kills pathogens before new mutations arise

Antibiotics — mode of action and resistance

Question	What was asked	Key mark scheme points
w25-23-5(b)	Mode of action of penicillin	Inhibits transpeptidase → prevents cross-links in peptidoglycan cell wall Cell wall weakens → osmotic lysis Human cells have no cell wall / no peptidoglycan
w25-23-5(a)	Why erythromycin affects bacteria but not humans	Binds to bacterial 70S ribosomes → inhibits translation Human ribosomes are 80S — binding site is absent
w25-23-5(c)	Mode of action of rifampicin	Binds to active site of RNA polymerase → blocks transcription
w25-21-4(c)	Mechanism of antibiotic resistance in bacteria	Efflux pumps — membrane proteins that use ATP to pump antibiotics out of the cell
s25-23-3(d)	Steps to reduce antibiotic resistance	Complete the full course Prescribe only when necessary — not for viral infections Use multiple antibiotics simultaneously

Exam tip — "why not humans" questions

Name the specific structural difference. "Humans are different" scores nothing. Required answers: 70S vs 80S ribosomes; or peptidoglycan is absent in human cells.

Monoclonal antibodies — hybridoma method and ELISA

Question	What was asked	Key mark scheme points
s25-23-6(a)	Outline the hybridoma method for monoclonal antibody production	Inject specific antigen into a mouse Remove B-lymphocytes (splenocytes) from spleen Fuse with myeloma (cancer) cells → hybridoma cells Screen; clone selected hybridomas
w25-22-2(c)(ii)	Why both B-lymphocytes and myeloma cells are needed	B-lymphocytes: provide antibody specificity; cannot survive long in culture Myeloma cells: cancer cells; can divide continuously / indefinitely (immortal)
w25-24-1(b)(ii)	Why multiple mAbs are needed to study complex molecules like pectin	mAbs are highly specific to one epitope / antigenic determinant Complex molecules have different shapes → require different mAbs for each region
s25-22-6(c)	Use of colorimeters in a sandwich ELISA	Measures absorbance of light from coloured product Absorbance is proportional to antigen concentration Removes subjective / eye-based judgement

Common mistake

Students state "B-cells cannot survive" but forget why myeloma cells are needed. Both halves are required: B-cells = specificity; myeloma cells = immortality / continuous division.

Transport in Mammals & Gas Exchange

Tier 3

Haemoglobin, gas transport, and the Bohr shift

Question	What was asked	Key mark scheme points
s25-21-6(a)	Features of haemoglobin structure	Contains haem group with Fe²⁺ that binds one O₂ molecule Quaternary structure: four polypeptide chains (2α + 2β)
s25-21-6(b)	Description and advantage of the Bohr shift	Description: curve shifts to the right as pCO₂ increases; lower % saturation at same pO₂ Mechanism: ↑CO₂ → carbonic acid → H⁺ → H⁺ binds Hb → haemoglobinic acid forms → O₂ released Advantage: more O₂ delivered to actively respiring tissues
w25-21-2(b)	Effect of lower CO₂ (hyperventilation) on O₂ release from haemoglobin	Less CO₂ → less carbonic acid → fewer H⁺ ions Less haemoglobinic acid formed → Hb retains O₂ → decreased O₂ release
s25-22-2(b)	Match substances to their roles in gas transport	Chloride ion: chloride shift Carbaminohaemoglobin: Hb bound to CO₂ Hydrogencarbonate ion: main CO₂ transport form in plasma Haemoglobinic acid: Hb bound to H⁺

Common mistake

"The curve shifts right" alone scores very little. The full causal chain is required: CO₂ → carbonic acid → H⁺ → binds Hb → haemoglobinic acid → O₂ released. Each arrow is a mark.

The heart — structure and cardiac cycle coordination

Question	What was asked	Key mark scheme points
m25-22-3(a)	Identify valves and describe blood flow in the heart	Valve V = atrioventricular (bicuspid / mitral) valve Blood flows from atria to ventricles when atrial pressure exceeds ventricular pressure
w25-23-6(a)	Adaptations of the left side of the heart	Thick muscular wall of left ventricle → pumps blood at high pressure to systemic circulation Valves prevent backflow; chordae tendineae hold valves during systole
w25-22-6(a)	Sequence of events in the left side during the cardiac cycle	Blood enters via pulmonary vein → atrial systole forces blood through bicuspid valve Ventricular systole → closes bicuspid; opens aortic semilunar valve → blood into aorta
w25-23-6(b)	Coordination of the heartbeat by SAN and AVN	SAN generates impulse → spreads across atria → atria contract Delayed at AVN (~0.1s) → allows ventricles to fill AVN → Purkyne tissue → apex → ventricles contract base-upwards

Blood vessels and tissue fluid

Question	What was asked	Key mark scheme points
w25-21-2(c)(i)	Adaptations of the tunica media in the aorta	High proportion of elastic fibres Stretch during systole → recoil during diastole to maintain high blood pressure
w25-24-5(b)	Relationship between vein structure and function	Valves: prevent backflow of low-pressure blood Large lumen: minimises resistance to flow Tunica externa (collagen): prevents collapse
w25-23-6(c)	Formation and functions of tissue fluid	Formation: high hydrostatic pressure at arteriole end → forces plasma out through endothelial pores (ultrafiltration) Function: aqueous medium for exchange of O₂, glucose, CO₂, urea between blood and cells

Exam technique

How to answer each question type

Describe / outline

e.g. "Describe phagocytosis" · "Outline the hybridoma method"

Use a numbered sequence — each step is typically 1 mark.
Name structures correctly: phagosome not "bubble"; Purkyne tissue not "fibres".
Do not explain why unless asked.
Count the marks and match the number of points you write.

Explain

e.g. "Explain the Bohr shift" · "Explain why the secondary response is greater"

Give a full cause → effect chain: "because… therefore… as a result…"
Quantify: "faster AND greater" — not just "more".
Link mechanism to consequence.
Avoid circular answers.

Compare

e.g. "Compare active and passive immunity" · "Compare cilia and microvilli"

Use parallel structure: "X has… whereas Y has…"
Cover both items — mentioning only one side loses the comparison mark.
Include similarity AND difference unless specified otherwise.
Match comparisons to marks available.

Graph / data analysis

e.g. "Identify the inhibitor type" · "Suggest why activity decreases at pH 4"

Observe: state what you see with figures.
Conclude: name the pattern or phenomenon.
Explain: give the mechanism. All three steps are usually needed.

Calculation

e.g. "Calculate actual size" · "Calculate magnification"

Write the formula first.
Show the substitution with numbers.
State the answer with correct units.
Convert units before calculating (mm → µm = ×1000).

Drawing

e.g. "Draw α-glucose ring structure" · "Sketch the inhibition curve"

Label carbon numbers — examiners check C1 and C4.
Show –OH and –H explicitly at each carbon.
For curves: label axes with units; draw smoothly; show Vmax clearly.
Use a sharp pencil for easy correction.

Zero tolerance

Critical spellings

Incorrect spelling = no mark, regardless of the rest of the answer.

✓

Mycobacterium tuberculosisCapital M, lowercase t · must be italicised

✓

Vibrio choleraeCapital V, lowercase c · must be italicised

✓

Phagosome / phagocytic vacuoleNot "bubble", "bag", or "pocket"

✓

Hydrogencarbonate ionNot "bicarbonate" — outdated term

✓

Purkyne tissueSome mark schemes accept Purkinje — check yours

✓

Peptide bond vs glycosidic bondAmino acids ≠ sugars. Never swap these.

✓

Enzyme-substrate complex (ESC)Not "compound" or "bond"

✓

Mitosis (not meiosis)Clonal expansion uses mitosis — common slip under pressure

Biology Exam Study Guide

Cell Structure & Microscopy

Biological Molecules

Cell Membranes & Transport

Enzymes

The Mitotic Cell Cycle

Nucleic Acids & Protein Synthesis

Infectious Disease & Immunity

Transport in Mammals & Gas Exchange

Describe / outline

Explain

Compare

Graph / data analysis

Calculation

Drawing