AQA A-Level Biology 25 markers

• Many different substances enter and leave a cell by crossing its cell surface membrane. Describe how substances can cross a cell surface membrane. (5)

  1 (Simple / facilitated) diffusion from high to low

  concentration / down concentration gradient;

  2 Small / non-polar / lipid-soluble molecules pass via phospholipids / bilayer;

  OR

  Large / polar / water-soluble molecules go through proteins;

  3 Water moves by osmosis / from high water potential to low water potential / from less to more negative water potential;

  4 Active transport is movement from low to high

  concentration / against concentration gradient;

  5 Active transport / facilitated diffusion involves proteins/carriers;

  6 Active transport requires energy / ATP;

  7 Ref. to Na+ / glucose co-transport;
• Describe and explain how the lungs are adapted to allow rapid exchange of oxygen between air in the alveoli and blood in the capillaries around them. (5)

  1 Many alveoli / alveoli walls folded provide a large surface area;

  2 Many capillaries provide a large surface area;

  3 (So) fast diffusion;

  ________________________________________________

  4 Alveoli or capillary walls / epithelium / lining are thin / short distance between alveoli and blood;

  5 Flattened / squamous epithelium;

  6 (So) short diffusion distance / pathway;

  7 (So) fast diffusion;

  ________________________________________________

  8 Ventilation / circulation;

  9 Maintains a diffusion / concentration gradient;

  10 (So) fast diffusion;
• Scientists believe that it may be possible to develop vaccines that make use of microfold cells. Explain how this sort of vaccine would lead to a person developing immunity to the pathogen (5)

  1.Vaccine contains antigen/ dead pathogen

  2.Microfold cells take up/bind and present/transport antigen (to immune system)

  3.T-cells activate B-cells

  4.B cells divide / undergo mitosis

  5.B cells produce antibodies

  6.Memory cells produced

  7.More antibodies produced faster in secondary response
• Explain the role of B-lymphocytes and T-lymphocytes in the defence of the body against a virus infection. (6)

  1.B lymphocytes produce antibodies/involved in humoral response;

  2.T lymphocytes involved in cell mediated immunity;

  3.Macrophages present antigens;

  4.(specific) B lymphocytes recognise/bind to antigen;

  5.increase in numbers by mitosis;

  6.produce plasma cells (which make antibodies);

  7.antibodies bind to and clump/ agglutinate virus;

  8.memory cells produced by 1st exposure/cloned on 2nd exposure;

  9.T lymphocytes(helpers) produce 10.lymphokines/chemicals;

  11.which aid B lymphocyte cloning;

  12.encourages phagocytes to engulf clumped virus;

  13.killer T cells kill virus infected cells;
• Explain how water enters a plant root from the soil and travels through to the endodermis. (5)

  1. Water enters root hair cells;

  2. By osmosis;

  3. Because active uptake of mineral ions has crated a water potential gradient;

  4. Water moves through cortex;

  5. Down water potential gradient;

  6. Through cell vacuoles and cytoplasms (symplastic pathway);

  7. And through apoplastic pathway (cell walls);
• Root pressure is a force that is partly responsible for the movement of water through xylem in stems. Explain how the active transport of mineral ions into the xylem vessels in the roots results in water entering these vessels and then being moved up the xylem tissue. (5)


  1. Entry of ions leads to a reduced water potential;

  2. Water potential established between xylem and surrounding cells;

  3. Plasma membranes of surroudning cells are partially permeable;

  4. Water enters xylem by osmosis;

  5. Volume of water in xylem increases;

  6. Cannot move back due to gradient;

  7. Pressure in xylem increases and forces water upwards.
• The presence of an air bubble in a xylem vessel in the stem blocks the movement through that vessel. Use the cohesion-tension theory to explain why. (4)

  1. Evaporation from leaves (transpiration);

  2. Water in xylem under tension (pulled up);

  3. Because water molecules cohere (stick together) (because of hydrogen bonds);

  4. So water travels in a single column;

  5. A water bubble would break the column;
• In daylight, most of the water evaporates from the leaves but some is used by the plant. Describe the ways in which this water could be used by the plant. (6)

  1. (water is used in) the light-dependent reactions of photosynthesis;

  2. electrons from water enable ATP production / H+ are used to reduce NADP / produces O2;

  3. (water can be used in) hydrolysis reactions within the plant;

  4. to create turgor;

  5. as a solvent for transport;

  6. as a medium for chemical reactions;

  7. component of cells / cytoplasm;
• Describe two features you would expect in the leaves of a tree adapted to a dry environment. Explain how each feature helps the tree's survival. (6)

  Sunken stomata;

  water evaporation into pit creates local humidity;

  increased humidity reduces gradient for water evaporation;

  close arrangement of stomata;

  diffusion shells of individual stomata overlap;

  interferes with water diffusion and slows evaporation;

  restriction of stomata to lower side of leaf;

  rate of air movement below leaf less/ heating effect of sun less;

  gradient for water evaporation reduced/ water molecules have less

  kinetic energy;

  
  

  thick cuticle/wax/suberin (on upper surface);

  (wax/suberin )waterproof;

  water unable to diffuse onto surface to evaporate,

  presence of trichomes/ hairs;

  surface traps water close to leaf surface;

  increased humidity reduces gradient for water evaporation;

  reduced leaves/spines/small surface area to volume;

  less surface area for evaporation;

  more distance across leaf for water to diffuse;

  rolled leaves;

  stomata enclosed in localised humidity;

  increased humidity reduces gradient for water evaporation;
• Xylem transports water through a plant. Describe and explain how the cells of xylem are adapted for this function. (5)

  Thick cell walls;

  Withstand tension / negative pressure;

  Lignin in cell walls;

  Walls waterproof / withstand tension / negative pressure;

  Xylem cells have no end walls / tubular (not hollow);

  So a continuous column of water;

  Xylem vessels are stacked on top of each other;

  So a continuous column of water;

  Have no cytoplasm / hollow;

  Reduces resistance to flow of water / so a continuous column of water;

  Xylem cells have pores / pits (in side walls);

  Enable sideways flow / by-pass blockages / allows entry or exit of water;

  Narrow tubes;

  Allows capillarity / increased surface area for adhesion;

  (Molecules in) cell walls;

  Allows adhesion
• Explain how the structure of the endodermis affects the passage of water by this (apoplast) pathway. (4)

  1. Casparian bands;

  2. Which are waterproof (impermeable);

  3. Lower water potential of endodermis cell;

  4. Water enters symplastic pathway;

  5. By osmosis;
• Describe the mass flow hypothesis for the mechanism of translocation in plants. (4)

  1. In the leaf sugars are actively transported into phloem;

  2. By companion cells;

  3. Lowers water potential of sieve tubes and water enters by osmosis;

  4. Increase in pressure causes mass movement (towards roots);

  5. Sugars used (converted) in root for respiration/for storage;
• A laboratory has oat plants containing the resistance gene and a supply of plasmids. Describe how bacteria may be produced which have the resistance gene in their plasmids. (6)


  1. Cut desired gene (from DNA) of oat plant;

  2. Using restriction endonuclease (restriction enzyme);

  3. Cut plasmid open;

  4. With (same) restriction endonuclease;

  5. Sticky ends/unpaired bases attached;

  6. Use (DNA) ligase to join;

  7. Return plasmid to (bacterial) cells;

  8. Use of Ca2+/calcium salts/electric shock;
• The polymerase chain reaction (PCR) can be used to produce large quantities of DNA. Describe how the PCR is carried out. (6)

  1. DNA heated to 90-95°C;

  2. Strands separate;

  3. Cooled to temperature below 70°C;

  4. Primers bind;

  5. Nucleotides attach;

  6. By complementary base pairing;

  7. Temperature 70-75°C;

  8. DNA Polymerase joins nucleotides together;

  9. Cycle repeated;
• Describe how an action potential is produced in an axon. (6)

  1. A stimulus causes a temporary reversal of charge and inside of axon becomes positive (depolarisation);

  2. Energy of stimulus causes some sodium channels to open;

  3. So sodium diffuses into axon;

  4. Once an action potential of +40mV has been established, channels close and there is no more influx of sodium;

  5. Potassium channels open and potassium ions diffuse out, causing repolarisation of axon;

  6. Axon becomes more negative than usual, so potassium gates close and sodium-potassium pump again actively transports sodium ions out and potassium ions in;
• Describe the sequence of events which allows information to pass from one neurone to the next neurone across a cholinergic synapse. (6)

  1. Impulse causes calcium ions to enter axon;

  2 Vesicles move to/fuse with membrane;

  3. Acetylcholine is released;

  4. Diffuses across synaptic cleft/synapse;

  5. Binds with receptors on postsynaptic neurone;

  6. Sodium ions enter postsynaptic neurone;

  7. Depolarisation of postsynaptic membrane;

  8. If above threshold, nerve impulse/action potential is produced;
• How does temperature affect the speed of a nerve impulse? (5)

  1. The higher the temperature the faster the nerve impulse;

  2. Sodium-potassium pump uses active transport;

  3. Which requires energy from ATP from respiration;

  4. Respiration is enzyme-controlled;

  5. More kinetic energy so more enzyme-substrate complexes formed and faster diffusion
• Explain how nervous control in a human can cause increased cardiac output during exercise. (4)

  1. Coordination via medulla (of brain) / cardiac centre;

  2. (Increased) impulses along sympathetic (/ cardiac accelerator) nerve;

  3. To S.A. node / pacemaker;

  4. Release of noradrenalin;

  5. More impulses sent from / increased rate of discharge of S.A. node / pacemaker;

  6. Increased heart rate / increased stroke volume;
• Describe the structure of a cell membrane. (5)

  1. Double layer of phospholipid molecules;

  2. Detail of arrangement of phospholipids;

  3. Intrinsic proteins/protein molecules passing right through;

  4. Some with channels/pores;

  5. Extrinsic proteins/proteins only in one layer/on surface;

  6. Molecules can move in membrane/dynamic/membrane contains cholesterol;

  7. Glycocalyx/carbohydrates attached to lipids/proteins;
• Describe the part played by cell surface membranes in regulating the movement of substances into and out of cells. (6)

  1. Non-polar/lipid soluble molecules move through phospholipid layer/bilayer;

  2. Small molecules/water/gases move through phospholipid layer/bilayer;

  3. Ions/water soluble substances move through channels in proteins;

  4. Some proteins are gated;

  5. Reference to diffusion;

  6. Carriers identified as proteins;

  7. Carriers associated with facilitated diffusion;

  8. Carriers associated with active transport/transport with ATP/pumps;

  9. Different cells have different proteins;

  10. Correct reference to cytosis;
• Explain how amino acid molecules may be linked to form a polypeptide chain which is folded into a specific tertiary shape. (6)


  1. Condensation;

  2. removal of water molecule;

  3. from amino and carboxyl groups;

  4. forming peptide bonds;

  5. same amino acids in same sequence;

  6. bonds form between R-groups/side chains;

  7. e.g. sulphur-containing amino acids / ionic bonds / hydrogen bonds;

  8. bonds form in same place;
• Describe how molecular shape is important in explaining the way in which enzymes may be affected by inhibitors. (6)

  1 Active site (of enzyme) has particular shape;

  2 (Into which) substrate molecule fits / binds;

  3 Appropriate reference linking induced fit and shape;

  4 (Competitive inhibitor) has similar shape to substrate;

  5 Also fits active sites;

  6 Prevents substrate access;

  7 (Non-competitive inhibitor) fits at site other than active site;

  8 Distorting shape of active site / enzyme;

  6 Prevents substrate access; (award once only)

  9 Two types identified as competitive and non-competitive;
• The bacteria in the intestine are prokaryotic cells. The epithelial cells which line the small intestine are eukaryotic cells. Describe the ways in which prokaryotic cells and eukaryotic cells differ. (6)


  1 Prokaryotic cells do not have a nucleus / have genetic material

  in cytoplasm;

  2 DNA in loop / ring;

  3 Not associated with proteins / do not have chromosomes /

  chromatin / do not divide by mitosis;

  4 Smaller ribosomes;

  5 No membrane-bound organelles;

  6 Such as mitochondria / lysosomes / endoplasmic reticulum /

  Golgi / chloroplasts;

  7 Prokaryotic cells may have mesosomes;

  8 Prokaryotic cells smaller;

  9 May be enclosed by capsule;
• Describe how proteins are arranged in a plasma membrane and the part they play in transporting substances into and out of cells. (6)


  1 Some proteins pass right through membrane;

  2 Some proteins associated with one layer;

  3 Involved in facilitated diffusion;

  4 Involved in active transport;

  5 Proteins act as carriers;

  6 Carrier changes shape / position;

  7 Proteins form channels / pores;

  8 Protein allows passage of water soluble molecules / charged particles / correct named example;
• Skin cells may be studied with a transmission electron microscope or an optical microscope. Explain the advantages and limitations of using a transmission electron microscope to study cells. (6)

  1 TEM uses (beam of) electrons;

  2 These have short wavelength;

  3 Allow high resolution/greater resolution/Allow more detail to be seen/greater useful magnification;

  4 Electrons scattered (by molecules in air);

  5 Vacuum established;

  6 Cannot examine living cells;

  7 Lots of preparation/procedures used in preparing specimens/ fixing/staining/sectioning;

  8 May alter appearance/result in artefacts;
• Explain how oxygen is loaded, transported and unloaded in the blood (6)

  1. Haemoglobin carries oxygen / has a high affinity for oxygen / oxyhaemoglobin;

  2. In red blood cells;

  3. Loading / uptake/association in lungs;

  at high p.O2;

  4. Unloads / dissociates / releases to respiring cells / tissues;

  5. at low p.O2;

  6. Unloading linked to higher carbon dioxide (concentration);
• Describe how the regular contraction of the atria and ventricles is initiated and coordinated by the heart itself. (5)

  1. (cardiac) muscle is myogenic;

  2. sinoatrial node/SAN;

  3. wave of depolarisation/ impulses /electrical activity (across atria);

  4. initiates contraction of atria

  atrioventricular node/AVN;

  5. bundle of His/purkyne tissue spreads impulse across ventricles;

  6. ventricles contract after atria/time delay enables ventricles to fill;
• What is atheroma and how may it cause myocardial infarction? (5)

  1. Cholesterol/ plaque / lipoprotein / LDL / fatty material / cells;

  2. In artery wall / under lining / endothelium of artery / blood vessel;

  3. Atheroma linked with blood clotting / thrombosis;

  4. (Blocks) coronary artery / artery supporting heart muscle / tissue / cells;

  5. Reduces oxygen / glucose supply (to heart muscle / tissues / cells);

  6. (Heart muscle / tissue / cells) unable to respire / dies;
• The diet of a person can increase the risk of coronary heart disease. Explain how. (5)

  1. Too much saturated fat / cholesterol in diet;

  2. Increase in LDL / cholesterol in blood;

  3. Atheroma / fatty deposits / plaques in artery walls;

  4. Reduces diameter of / blocks coronary arteries;

  5. Less oxygen / glucose to heart muscles / tissues / cells;

  6. Increase in blood pressure;

  7. (Increased risk of) clot / thrombosis / embolism / aneurysm.
• Describe how the structures of starch and cellulose molecules are related to their functions. (5)


  Starch (max 3)

  1. Helical/ spiral shape so compact;

  2. Large (molecule)/insoluble so osmotically inactive;

  3. Branched so glucose is (easily) released for respiration;

  4. Large (molecule) so cannot leave cell/cross cell-surface membrane;

  Cellulose (max 3)

  5. Long, straight/unbranched chains of β glucose;

  6. Joined by hydrogen bonding;

  7. To form (micro/macro)fibrils;

  8. Provides rigidity/strength;
• Describe the processes involved in the transport of sugars in plant stems. (5)

  1. (At source) sucrose is actively (transported) into the phloem/sieve element/tube;

  2. By companion/transfer cells;

  3. Lowers water potential in phloem/sieve element/tube and water enters by osmosis;

  4. (Produces) high (hydrostatic) pressure;

  5. Mass flow/transport towards sink/roots/storage tissue;

  6. At sink/roots sugars are removed/unloaded;
• Explain why the diffusion of chloride ions involves a membrane protein and the diffusion of oxygen does not. (5)

  1. Chloride ions water soluble/charged/polar;

  2. Cannot cross (lipid) bilayer (of membrane);

  3. Chloride ions transported by facilitated diffusion OR diffusion involving channel/carrier protein;

  4. Oxygen not charged/non-polar;

  5. (Oxygen) soluble in/can diffuse across (lipid) bilayer;
• Glucose is absorbed from the lumen of the small intestine into epithelial cells. Explain how the transport of sodium ions is involved in the absorption of glucose by epithelial cells. (5)

  1. Na+ ions leave epithelial cell and enter blood;

  2. (Transport out is by) active transport / pump / via carrier protein using ATP;

  3. So, Na+ conc. in cell is lower than in lumen (of gut);

  4. Sodium/Na+ ions enter by facilitated diffusion;

  5. Glucose absorbed with Na+ ions against their concentration/diffusion gradient / glucose absorbed down an electrochemical gradient;
• Blood leaving the kidney eventually returns to the kidney.

  Describe the pattern of blood circulation in a mammal that causes blood to return to the kidney. (6)

  1. (blood flows from kidney along) renal vein to vena cava;

  2. (along) vena cava to right atrium/side of heart;

  3. (along) pulmonary artery to lungs;

  4. (along) capillaries to pulmonary vein;

  5. (along) pulmonary vein to left atrium/side of heart;

  6. (along) aorta to renal artery (to kidney);

  7. Blood may pass through several complete circuits before returning to kidney;
• Describe and explain how cell fractionation and ultracentrifugation can be used to isolate mitochondria from a suspension of animal cells. (5)

  1. Cell homogenisation to break open cells;

  2. Filter to remove (large) debris/whole cells;

  3. Use isotonic solution to prevent damage to mitochondria/organelles;

  4. Keep cold to prevent/reduce damage by enzymes / use buffer to prevent protein/enzyme denaturation;

  5. Centrifuge (at lower speed/1000 g) to separate nuclei/cell fragments/ heavy organelles;

  6. Re-spin (supernatant / after nuclei/pellet removed) at higher speed to get mitochondria in pellet/at bottom;
• Describe the principles and the limitations of using a transmission electron microscope to investigate cell structure. (5)

  Principles:

  1. Electrons pass through/enter (thin) specimen;

  2. Denser parts absorb more electrons;

  3. (So) denser parts appear darker;

  4. Electrons have short wavelength so give high resolution;

  Limitations:

  5. Cannot look at living material / Must be in a vacuum;

  6. Specimen must be (very) thin;

  7. Artefacts present;

  8. Complex staining method / complex/long preparation time;

  9. Image not in 3D / only 2D images produced;
• Use your knowledge of protein structure to explain why enzymes are specific and may be affected by non-competitive inhibitors. (5)


  1 each enzyme / protein has specific primary structure / amino acid sequence;

  2 folds in a particular way / has particular tertiary structure giving an active site with a unique structure;

  3 shape of active site complementary to / will only fit that of substrate; maximum of three marks for inhibition, points 5 - 8

  4 inhibitor fits at site on the enzyme other than active site;

  5 distorts active site;

  6 so substrate will no longer fit / form enzyme-substrate complex
• Skin cells may be studied with a transmission electron microscope or an optical microscope. Explain the advantages and limitations of using a transmission electron

  microscope to study cells. (6)

  Advantages:

  1 Small objects can be seen;

  2 TEM has high resolution as wavelength of electrons shorter;

  Accept better

  Limitations:

  3 Cannot look at living cells as cells must be in a vacuum;

  4 must cut section / thin specimen;

  5 Preparation may create artefact

  6 Does not produce colour image;
• Describe the role of the enzymes of the digestive system in the complete breakdown of starch. (5)

  Amylase;

  (Starch) to maltose:

  Maltase;

  Maltose to glucose;

  Hydrolysis;

  (Of) glycosidic bond;
• Describe the processes involved in the absorption of the products of starch digestion. (5)

  Glucose moves in with sodium (into epithelial cell);

  Via (carrier / channel) protein / symport;

  Sodium removed (from epithelial cell) by active transport / sodium- potassium pump;

  Into blood;

  Maintaining low concentration of sodium (in epithelial cell) / maintaining sodium

  concentration gradient (between lumen and epithelial cell);

  Glucose moves into blood;

  By (facilitated) diffusion;