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AQA A Level Biology (yr 12 content) > Exam question practise > Flashcards

Exam question practise Flashcards

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1
Q

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)

A

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;

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2
Q

Describe the mass flow hypothesis for the mechanism of translocation in plants. (4)

A
  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;
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3
Q

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

A
  1. Condensation;
  2. removal of water molecule;
  3. from amine 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;
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4
Q

Describe how molecular shape is important in explaining the way in which enzymes may be affected by inhibitors. (6)

A

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;

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5
Q

Explain how oxygen is loaded, transported and unloaded in the blood (6)

A
  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);
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6
Q

What is atheroma and how may it cause myocardial infarction? (5)

A
  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;
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7
Q

Describe how the structures of starch and cellulose molecules are related to their functions. (5)

A

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;

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8
Q

Describe the processes involved in the transport of sugars in plant stems. (5)

A
  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;
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9
Q

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)

A
  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;
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10
Q

Describe the role of the enzymes of the digestive system in the complete breakdown of starch. (5)

A 
Amylase;
(Starch) to maltose:
Maltase;
Maltose to glucose;
Hydrolysis;
(Of) glycosidic bond;
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11
Q

Scientists have investigated the effects of competitive and non-competitive inhibitors of the enzyme maltase.
Describe competitive and non-competitive inhibition of an enzyme. (5)

A
  1. Inhibitors reduce binding of enzyme to substrate / prevent formation of ES complex;
    (Competitive inhibition),
  2. Inhibitor similar shape (idea) to substrate;
  3. (binds) in to active site (of enzyme);
  4. (Inhibition) can be overcome by more substrate;
    (Non-competitive inhibition),
  5. Inhibitor binds to site on enzyme other than active site;
  6. Prevents formation of active site / changes (shape of) active site;
  7. Cannot be overcome by adding more substrate;
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12
Q

Scientists believe that it may be possible to develop vaccines that make use of microfold cells (lines 9 -10). Explain how this sort of vaccine would lead to a person developing immunity to a pathogen. (5)

A
  1. (Vaccine contains) antigen/attenuated/dead pathogen;
  2. Microfold cells take up/bind and present/transport antigen (to immune system/lymphocytes/T- cells);
  3. T-cells activate B-cells;
  4. B-cells divide/form clone/undergo mitosis;
  5. B-cells produce antibodies;
  6. Memory cells produced;
  7. More antibodies/antibodies produced faster in secondary response/on reinfection;
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13
Q

The events that take place during interphase and mitosis lead to the production of two genetically identical cells. Explain how. (4)

A
  1. DNA replicated;
  2. (Involving)
    specific/accurate/complementary
    base-pairing;
  3. (Ref to) two identical/sister
    chromatids;
  4. Each chromatid/ moves/is separated
    to(opposite) poles/ends of cell;
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14
Q

Explain how the structure of DNA is related to its functions. (6)

A
  1. Sugar-phosphate (backbone)/double
    stranded/helix so provides strength/stability
    /protects bases/protects hydrogen bonds;
  2. Long/large molecule so can store lots of
    information;
  3. Helix/coiled so compact;
  4. Base sequence allows information to be
    stored/ base sequence codes for amino
    acids/protein;
  5. Double stranded so replication can occur
    semi-conservatively/ strands can act as
    templates;
  6. Complementary base pairing / A-T and G-C
    so accurate replication/identical copies can
    be made;
  7. (Weak) hydrogen bonds for replication/
    unzipping/strand separation;
  8. Many hydrogen bonds so stable/strong;
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15
Q

Describe how DNA is replicated. (6)

A
  1. Strands separate / H-bonds break;
  2. DNA helicase (involved);
  3. Both strands/each strand act(s) as (a) template(s);
  4. (Free) nucleotides attach;
  5. Complementary/specific base pairing / AT and GC;
  6. DNA polymerase joins nucleotides (on new strand);
  7. H-bonds reform;
  8. Semi-conservative replication / new DNA molecules contain one old strand and one new strand;
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16
Q

Describe how tissue fluid is formed and how it is returned to the circulatory system. (6)

A 
Formation
1. High blood / hydrostatic pressure /
pressure filtration;
2. Forces water / fluid out;
3. Large proteins remain in capillary;
Return
4. Low water potential in capillary /
blood;
5. Due to (plasma) proteins;
6. Water enters capillary / blood;
7. (By) osmosis;
8. Correct reference to lymph;
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17
Q

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

A

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;

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18
Q

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)

A
  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
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19
Q

Explain the role of B-lymphocytes and T-lymphocytes in the defence of the body against a virus infection. (6)

A
  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;
  10. which aid B lymphocyte cloning;
  11. encourages phagocytes to engulf clumped virus;
  12. killer T cells kill virus infected cells;
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20
Q

Explain how water enters a plant root from the soil and travels through to the endodermis. (5)

A
  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);
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21
Q

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)

A
  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.
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22
Q

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.

A

1) Sunken stomata;
water evaporation into pit creates local humidity;
increased humidity reduces gradient for water evaporation;
2) close arrangement of stomata;
diffusion shells of individual stomata overlap;
interferes with water diffusion and slows evaporation;
3) 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;

4) 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;
5) reduced leaves/spines/small surface area to volume;
less surface area for evaporation;
more distance across leaf for water to diffuse;
rolled leaves;
6) stomata enclosed in localised humidity;
increased humidity reduces gradient for water evaporation;

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23
Q

Xylem transports water through a plant. Describe and explain how the cells of xylem are adapted for this function. (5)

A

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

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24
Q

Describe the structure of a cell membrane. (5)

A
  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;
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25
Q

Describe the part played by cell surface membranes in regulating the movement of substances into and out of cells. (6)

A
  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;
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26
Q

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)

A 
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;
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27
Q

Describe how proteins are arranged in a plasma membrane and the part they play in transporting substances into and out of cells. (6)

A

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;

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28
Q

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)

A

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;

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29
Q

Describe how the regular contraction of the atria and ventricles is initiated and coordinated by the heart itself. (5)

A
  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;
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30
Q

The diet of a person can increase the risk of coronary heart disease. Explain how. (5)

A
  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.
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31
Q

Explain why the diffusion of chloride ions involves a membrane protein and the diffusion of oxygen does not. (5)

A
  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;
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32
Q

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)

A
  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;
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33
Q

Describe and explain how cell fractionation and ultracentrifugation can be used to isolate mitochondria from a suspension of animal cells. (5)

A
  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;
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34
Q

Describe the principles and the limitations of using a transmission electron microscope to investigate cell structure. (5)

A

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;

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35
Q

Use your knowledge of protein structure to explain why enzymes are specific and may be affected by non-competitive inhibitors. (5)

A

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

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36
Q

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)

A 
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;
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37
Q

Describe the processes involved in the absorption of the products of starch digestion. (5)

A 
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;
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38
Q

Describe the structure of a cellulose molecule and explain how cellulose is adapted for its function in cells. (6)

A
  1. made from β-glucose;
  2. joined by condensation / removing molecule of water / glycosidic bond; 3. 1 : 4 link specified or described;
  3. “flipping over” of alternate molecules;
  4. hydrogen bonds linking chains / long straight chains;
  5. cellulose makes cell walls strong / cellulose fibres are strong; 7. can resist turgor pressure / osmotic pressure / pulling forces; 8. bond difficult to break;
  6. resists digestion / action of microorganisms / enzymes;
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39
Q

In humans, the enzyme maltase breaks down maltose to glucose. This takes place at normal body temperature.
Explain why maltase:
-only breaks down maltose
-allows this reaction to take place at normal body temperature. (5)

A
  1. Tertiary structure / 3D shape of enzyme (means);
  2. Active site complementary to maltose / substrate / maltose fits into active site / active site and substrate fit like a lock and key;
  3. Description of induced fit;
  4. Enzyme is a catalyst;
  5. Lowers activation energy / energy required for reaction;
  6. By forming enzyme-substrate complex;
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40
Q

Explain how the heart muscle and the heart valves maintain a one-way flow of blood from the left atrium to the aorta. (5)

A
  1. Atrium has higher pressure than ventricle (due to filling/contraction);
  2. Atrioventricular valve opens;
  3. Ventricle has higher pressure than atrium (due to filling/contraction);
  4. Atrioventricular valve closes;
  5. Ventricle has higher pressure than aorta;
  6. Semilunar valve opens;
  7. Higher pressure in aorta than ventricle (as heart relaxes);
  8. Semilunar valve closes;
  9. (Muscle/atrial/ventricular) contraction causes increase in pressure;
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41
Q

A mutation can lead to the production of a non-functional enzyme. Explain how. (6)

A 
1. Change/mutation in base/nucleotide
sequence (of DNA/gene);
2. Change in amino acid
sequence/primary structure (of
enzyme);
3. Change in hydrogen/ionic/disulfide
bonds;
4. Change in the tertiary
structure/shape;
5. Change in active site;
6. Substrate not complementary/cannot
bind (to enzyme/active site) / no
enzyme-substrate complexes form;
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42
Q

Some substances can cross the cell-surface membrane of a cell by simple diffusion through the phospholipid bilayer.
Describe other ways by which substances cross this membrane. (5)

A

By osmosis (no mark)
1. From a high water potential to a low water potential/down a water potential gradient;
2. Through aquaporins/water channels;
By facilitated diffusion (no mark)
3. Channel/carrier protein;
4. Down concentration gradient;
By active transport (no mark)
5. Carrier protein/protein pumps;
6. Against concentration gradient;
7. Using ATP/energy (from respiration);
By phagocytosis/endocytosis (no mark)
8. Engulfing by cell surface membrane to form vesicle/vacuole;
By exocytosis/role of Golgi vesicles (no mark)
9. Fusion of vesicle with cell surface membrane;

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43
Q

Atheroma formation increases a person’s risk of dying. Explain how. (5)

A
  1. Atheroma is fatty material/cholesterol/foam cells/plaque/calcium deposits/LDL;
  2. In wall of artery;
  3. (Higher risk of) aneurysm/described;
  4. (Higher risk of) thrombus formation/blood clot;
  5. Blocks coronary artery;
  6. Less oxygen/glucose to heart muscle/cells/tissue;
  7. Reduces/prevents respiration;
  8. Causing myocardial infarction/heart attack;
  9. Blocks artery to brain;
  10. Causes stroke/stroke described;
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44
Q

When HIV infects a human cell, the following events occur.
• A single-stranded length of HIV DNA is made.
• The human cell then makes a complementary strand to the HIV DNA.
The complementary strand is made in the same way as a new complementary
strand is made during semi-conservative replication of human DNA.
Describe how the complementary strand of HIV DNA is made.
[3 marks]

A
  1. (Complementary) nucleotides/bases pair
    OR
    A to T and C to G;
  2. DNA polymerase;
  3. Nucleotides join together (to form new
    strand)/phosphodiester bonds form;
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45
Q

Contrast the structures of DNA and mRNA molecules to give three differences.
[3 marks]

A 
1. DNA double stranded/double helix and
mRNA single-stranded;
2. DNA (very) long and RNA short;
3. Thymine/T in DNA and uracil/U in RNA;
4. Deoxyribose in DNA and ribose in RNA;
5. DNA has base pairing and mRNA doesn’t/
DNA has hydrogen bonding and mRNA
doesn’t;
6. DNA has introns/non-coding sequences
and mRNA doesn’t;
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46
Q

Describe the difference between the structure of a triglyceride molecule and the structure of a phospholipid molecule. (1)

A
  1. In phospholipid, one fatty acid

replaced by a phosphate;

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47
Q

Describe how you would test for the presence of a lipid in a sample of food.
[2 marks]

A
  1. Add ethanol, then add water;

2. White (emulsion shows lipid);

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48
Q

Describe how a saturated fatty acid is different from an unsaturated fatty acid.
[1 mark]

A 
Saturated single/no double bonds
(between carbons)
OR
Unsaturated has (at least one) double
bond (between carbons);
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49
Q

This fat substitute cannot be digested in the gut by lipase.
Suggest why.
[2 marks]

A 
1. (Fat substitute) is a different/wrong
shape/not complementary;
OR
Bond between glycerol/fatty acid
and propylene glycol different (to
that between glycerol and fatty
acid)/no ester bond;
2. Unable to fit/bind to (active site of)
lipase/no ES complex formed;
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50
Q

Cells constantly hydrolyse ATP to provide energy.
Describe how ATP is resynthesised in cells.
[2 marks]

A
  1. From ADP and phosphate;
  2. By ATP synthase;
  3. During respiration/photosynthesis;
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51
Q

This fat substitute is a lipid. Despite being a lipid, it cannot cross the cell-surface
membranes of cells lining the gut.
Suggest why it cannot cross cell-surface membranes.
[1 mark]

A

It is hydrophilic/is polar/is too large/is

too big;

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52
Q

Give two ways in which the hydrolysis of ATP is used in cells.
[2 marks]

A 
1. To provide energy for other
reactions/named process; (Active transport, respiration)
2. To add phosphate to other
substances and make them more
reactive/change their shape;
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53
Q

Y is a protein. One function of Y is to transport cellulose molecules across the
phospholipid bilayer.
Using information from Figure 3, describe the other function of Y.
[2 marks]

A 
1. (Y is) an enzyme/has active
site/forms ES complex;
2. That makes cellulose/attaches
substrate to cellulose/joins β
glucose;
OR
3. Makes cellulose/forms glycosidic
bonds;
4. From β glucose;
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54
Q

Describe the induced-fit model of enzyme action.

[2 marks]

A 
1. (before reaction) active site not
complementary to/does not fit
substrate;
2. Shape of active site changes as
substrate binds/as enzymesubstrate complex forms;
3. Stressing/distorting/bending bonds
(in substrate leading to reaction);
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55
Q

A quantitative Benedict’s test produces a colour whose intensity depends on the concentration of reducing sugar in a solution. A colorimeter can be used to measure the intensity of this colour.
The scientist used quantitative Benedict’s tests to produce a calibration curve of colorimeter reading against concentration of maltose.
Describe how the scientist would have produced the calibration curve and used it to obtain the results in Figure 4.
Do not include details of how to perform a Benedict’s test in your answer.
[3 marks]

A 
1. Make/use maltose solutions of
known/different concentrations
(and carry out quantitative
Benedict’s test on each);
2. (Use colorimeter to) measure
colour/colorimeter value of each
solution and plot calibration
curve/graph described;
3. Find concentration of sample from
calibration curve;
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56
Q

Human papilloma virus (HPV) is the main cause of cervical cancer. A vaccine
has been developed to protect girls and women from HPV.
Describe how giving this vaccine leads to production of antibody against HPV.
[4 marks]

A 
1. Vaccine/it contains antigen (from
HPV);
2. Displayed on antigen-presenting cells;
3. Specific helper T cell (detects antigen
and) stimulates specific B cell;
4. B cell divides/goes through
mitosis/forms clone to give plasma
cells;
5. B cell/plasma cell produces antibody;
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57
Q

There is genetic diversity within HPV.
Give two ways doctors could use base sequences to compare different types of
HPV.
[2 marks]

A
  1. Compare (base sequences of) DNA;
  2. Look for mutations/named mutations
    (that change the base sequence);
  3. Compare (base sequences of)
    (m)RNA;
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58
Q

Endopeptidases and exopeptidases are involved in the hydrolysis of proteins.
Name the other type of enzyme required for the complete hydrolysis of proteins
to amino acids.
[1 mark]

A

Dipeptidase/s;

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59
Q

Suggest and explain why the combined actions of endopeptidases and
exopeptidases are more efficient than exopeptidases on their own.
[2 marks]

A
  1. Endopeptidases hydrolyse internal (peptide
    bonds)
    OR
    Exopeptidases remove amino
    acids/hydrolyse (bonds) at end(s);
  2. More ends or increase in surface area (for
    exopeptidases);
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60
Q

The addition of a respiratory inhibitor stops the absorption of amino acids.
Use Figure 1 to explain why.
[3 marks]

A 
1. No/less ATP produced
OR
No active transport;
2. Sodium (ions) not moved (into/out of cell);
3. No diffusion gradient for sodium (to move
 into cell with amino acid)
OR
No concentration gradient for sodium (to
move into cell with amino acid);
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61
Q

Give two reasons why it was important that the student counted the number of
stomata in several parts of each piece of leaf tissue.
[2 marks]

A
  1. Distribution may not be uniform
    OR
    So it is a representative sample;
  2. To obtain a (reliable) mean;
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62
Q

Suggest two reasons why the rate of water uptake by a plant might not be the
same as the rate of transpiration.
[2 marks]

A 
1. Water used for
support/turgidity;
2. Water used in photosynthesis;
3. Water used in hydrolysis;
4. Water produced during
respiration;
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63
Q

Species of tubifex worm that live in ponds, lakes and rivers cannot survive in
seawater.
Use your knowledge of water potential to explain why they cannot survive in
seawater.
[2 marks]

A 
1. Water potential higher in worm
OR
Lower water potential in seawater;
2. Water leaves by osmosis (and
worm dies);
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64
Q

On the intensive farms, the farmers had removed hedges to increase land for
grazing. This resulted in a decrease in the diversity of birds on these farms.
Explain why the removal of hedges caused a decrease in the diversity of birds.
[3 marks]

A
  1. Removes species/types of
    plant/insect;
  2. Fewer food sources;
  3. Fewer habitats/niches;
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65
Q

The scientists concluded that an increase in phosphate in the embryo was linked
to growth of the embryo.
Suggest two reasons why an increase in phosphate can be linked to growth of
the embryo.
[2 marks]

A
  1. (Phosphate required) to make RNA;
  2. (Phosphate required) to make DNA;
  3. (Phosphate required) to make
    ATP/ADP;
  4. (Phosphate required) to make
    membranes;
  5. (Phosphates required) for
    phosphorylation;
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66
Q

Scientists determined the mean FEV1 value of 25-year-olds in the population.
Suggest two precautions that should have been taken to ensure that this mean
FEV1 value was reliable.
[2 marks]

A
  1. Large sample size;
  2. Individuals chosen at random;
  3. Are healthy;
  4. Equal number of males and
    females;
  5. Repeat readings;
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67
Q

The mean FEV1 value of non-smokers decreases after the age of 30.
Use your knowledge of ventilation to suggest why.
[1 mark]

A
  1. Internal intercostal muscle(s) less
    effective
    OR
    Less elasticity (of lung tissue);
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68
Q

One of the severe disabilities that results from emphysema is that walking
upstairs becomes difficult.
Explain how a low FEV1 value could cause this disability.
[3 marks]

A
  1. Less carbon dioxide removed;
  2. Less oxygen (uptake/in blood);
  3. Less (aerobic) respiration/ATP
    OR
    (More) anaerobic respiration;
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69
Q

HIV attaches to a specific protein receptor on helper T cells. A low percentage
of people have a mutation of the CCR5 gene which codes for this protein
receptor. This mutation results in a non-functional protein receptor.
Explain how this mutation can result in the production of a non-functional protein
receptor.
[4 marks]

A
  1. Change in DNA base/nucleotide
    (sequence);
  2. Change in amino acid
    (sequence)/primary structure;
  3. Alters (position of)
    hydrogen/ionic/disulfide bonds;
  4. Change in tertiary structure (of
    receptor);
70
Q

Some people produce a much higher ventricular blood pressure than normal.
This can cause tissue fluid to build up outside the blood capillaries of these
people.
Explain why.
[2 marks]

A 
1. More fluid forced/filtered out of
capillary/blood (due to high
pressure);
2. Less return of fluid (into
capillary/blood) due to pressure
OR
Lymph(atic) (system) cannot drain
away all excess fluid;
71
Q

Suggest how widening of blood vessels can reduce ventricular blood pressure.
[2 marks]

A
  1. Larger lumen/volume (of blood
    vessels) ;
  2. Reduces (blood) pressure (in blood
    vessels) ;
  3. Less friction/resistance (in blood
    vessels) ;
72
Q

High blood pressure leads to an accumulation of tissue fluid. Explain how (3 marks)

A
  1. High blood pressure = high hydrostatic pressure;
  2. Increases outward pressure from (arterial) end of capillary / reduces inward pressure at (venule) end of capillary;
  3. (So) more tissue fluid formed / less tissue fluid is reabsorbed;
73
Q

Describe and explain the mechanism that causes forced expiration (4 marks)

A
  1. Contraction of internal intercostal muscles;
  2. Relaxation of diaphragm muscles / of external intercostal muscles;
  3. Causes decrease in volume of chest / thoracic cavity;
  4. Air pushed down pressure gradient.
74
Q

HSV infects nerve cells in the face (line 1). Explain why it infects only nerve cells. (3)

A
  1. Outside of virus has antigens
  2. which are complementary shape to receptor
  3. which is only found on the membrane of nerve cells
75
Q

Explain why this virus can be described as inactive (2)

A
  1. No more cells are infected

2. replication of the virus slows down or stops

76
Q

Apart from increase in the amount of DNA, give one process which occurs during interphase which enable nuclear division to occur (1)

A

ATP production/ protein synthesis/ replication of centriole

77
Q

Describe how the student could use an eyepiece graticule to determine the mean diameter of the stomata (3)

A
  1. Measure stomata using eyepiece graticule
  2. callibrate eyepiece graticule against stage micrometer
  3. Take a number of measurments to conclude a mean
78
Q

Explain the role of the heart in the formation of tissue fluid

A
  1. Contraction of ventricle causes high pressure

2. forces water and dissolved substances out of the capillaries.

79
Q

Starting with mRNA in the cytoplasm, describe how translation leads to the production of a polypeptide.

Do NOT include descriptions of transcription and splicing in your answer.

A
  1. mRNA associates with a ribosome / ribosome
    attaches to mRNA;
  2. Ribosome moves to / finds the start codon /
    AUG;
  3. tRNA brings / carries (appropriate / specific)
    amino acid;
  4. Anticodon (on tRNA complementary) to codon
    (on mRNA);
  5. Ribosome moves along to next codon;
    OR
    Ribosome ‘fits’ around two codons / can fit two
    tRNAs;
  6. (Process repeated and) amino acids join by
    peptide bonds / condensation reaction (to form
    polypeptide);
    OR
    (Process repeated and) amino acids joined
    using (energy from) ATP (to form polypeptide);
80
Q

Species richness and an index of diversity can be used to measure biodiversity within a
community.
(a) What is the difference between these two measures of biodiversity? (1 mark)

A

Species richness measures only number of (different) species / does not measure
number of individuals.

81
Q

The ecologists captured insects from a number of sites on the island. Suggest how
they decided where to take their samples.

A
  1. Draw grid over (map of) area;

2. Select squares / coordinates at random.

82
Q

What two measurements are needed to calculate an index of diversity? (2 marks)

A
  1. Number of (individuals of) each species;
    Accept: ‘population’ for ‘number’
    Total number of individuals / number of species;
    Accept: ‘species richness’
    MP2 allows for other types of diversity index
83
Q

The Amazonian forest today contains a very high diversity of bird species.
• Over the last 2 000 000 years, long periods of dry climate caused this forest to
separate into a number of smaller forests.
• Different plant communities developed in each of these smaller forests.
• Each time the climate became wetter again, the smaller forests grew in size and
merged to reform the Amazonian forest.
(a) Use the information provided to explain how a very high diversity of bird species
has developed in the Amazonian forest.
(5 marks)

A
  1. No interbreeding / gene pools are separate / geographic(al) isolation;
  2. Mutation;
  3. Different selection pressures / different foods / niches / habitats;
    Neutral: different climates
  4. Adapted organisms survive and breed / differential reproductive success;
  5. Change / increase in allele frequency / frequencies;
84
Q

Aspartic acid and proline are both amino acids. Describe how two amino acids differ
from one another. You may use a diagram to help your description. (1)

A
  1. Have different ‘R’ group
85
Q

Deletion of the sixth base (G) in the sequence shown in the diagram above would
change the nature of the protein produced but substitution of the same base would
not. Use the information in the table and your own knowledge to explain why. (3)

A
  1. Substitution would result in CCA / CCC / CCU;
  2. (All) code for same amino acid / proline;
  3. Deletion would cause frame shift / chang
86
Q

Messenger RNA (mRNA) is used during translation to form polypeptides.
Describe how mRNA is produced in the nucleus of a cell.
(6 marks)

A
  1. Helicase;
  2. Breaks hydrogen bonds;
  3. Only one DNA strand acts as template;
  4. RNA nucleotides attracted to exposed bases;
  5. (Attraction) according to base pairing rule;
  6. RNA polymerase joins (RNA) nucleotides together;
  7. Pre-mRNA spliced to remove introns.
87
Q

Describe the structure of proteins. (5 marks)

A
  1. Polymer of amino acids;
  2. Joined by peptide bonds;
  3. Formed by condensation;
  4. Primary structure is order of amino acids;
  5. Secondary structure is folding of polypeptide chain due to hydrogen bonding;
  6. Tertiary structure is 3-D folding due to hydrogen bonding and ionic /
    disulfide bonds;
  7. Quaternary structure is two or more polypeptide chains.
88
Q

Describe how proteins are digested in the human gut.

4 marks

A
  1. Hydrolysis of peptide bonds;
  2. Endopeptidases break polypeptides into smaller peptide chains;
  3. Exopeptidases remove terminal amino acids;
  4. Dipeptidases hydrolyse / break down dipeptides into amino acids.
89
Q

Mitochondrial disease (MD) often causes muscle weakness (lines 1–3). Use your
knowledge of respiration and muscle contraction to suggest explanations for this
effect of MD. (3 marks)

A
  1. Reduction in ATP production by aerobic respiration;
  2. Less force generated because fewer actin and myosin interactions in
    muscle;
  3. Fatigue caused by lactate from anaerobic respiration.
90
Q

Species richness and an index of diversity can be used to measure biodiversity within a
community.
(a) What is the difference between these two measures of biodiversity? (1 mark)

A

Species richness measures only number of (different) species / does not measure
number of individuals.

91
Q

The ecologists captured insects from a number of sites on the island. Suggest how
they decided where to take their samples.

A
  1. Draw grid over (map of) area;

2. Select squares / coordinates at random.

92
Q

What two measurements are needed to calculate an index of diversity? (2 marks)

A
  1. Number of (individuals of) each species;
    Accept: ‘population’ for ‘number’
    Total number of individuals / number of species;
    Accept: ‘species richness’
    MP2 allows for other types of diversity index
93
Q

The Amazonian forest today contains a very high diversity of bird species.
• Over the last 2 000 000 years, long periods of dry climate caused this forest to
separate into a number of smaller forests.
• Different plant communities developed in each of these smaller forests.
• Each time the climate became wetter again, the smaller forests grew in size and
merged to reform the Amazonian forest.
(a) Use the information provided to explain how a very high diversity of bird species
has developed in the Amazonian forest.
(5 marks)

A
  1. No interbreeding / gene pools are separate / geographic(al) isolation;
  2. Mutation;
  3. Different selection pressures / different foods / niches / habitats;
    Neutral: different climates
  4. Adapted organisms survive and breed / differential reproductive success;
  5. Change / increase in allele frequency / frequencies;
94
Q

Aspartic acid and proline are both amino acids. Describe how two amino acids differ
from one another. You may use a diagram to help your description. (1)

A
  1. Have different ‘R’ group
95
Q

Deletion of the sixth base (G) in the sequence shown in the diagram above would
change the nature of the protein produced but substitution of the same base would
not. Use the information in the table and your own knowledge to explain why. (3)

A
  1. Substitution would result in CCA / CCC / CCU;
  2. (All) code for same amino acid / proline;
  3. Deletion would cause frame shift / chang
96
Q

Messenger RNA (mRNA) is used during translation to form polypeptides.
Describe how mRNA is produced in the nucleus of a cell.
(6 marks)

A
  1. Helicase;
  2. Breaks hydrogen bonds;
  3. Only one DNA strand acts as template;
  4. RNA nucleotides attracted to exposed bases;
  5. (Attraction) according to base pairing rule;
  6. RNA polymerase joins (RNA) nucleotides together;
  7. Pre-mRNA spliced to remove introns.
97
Q

Describe the structure of proteins. (5 marks)

A
  1. Polymer of amino acids;
  2. Joined by peptide bonds;
  3. Formed by condensation;
  4. Primary structure is order of amino acids;
  5. Secondary structure is folding of polypeptide chain due to hydrogen bonding;
  6. Tertiary structure is 3-D folding due to hydrogen bonding and ionic /
    disulfide bonds;
  7. Quaternary structure is two or more polypeptide chains.
98
Q

Species richness and an index of diversity can be used to measure biodiversity within a
community.
(a) What is the difference between these two measures of biodiversity? (1 mark)

A

Species richness measures only number of (different) species / does not measure
number of individuals.

99
Q

The ecologists captured insects from a number of sites on the island. Suggest how
they decided where to take their samples.

A
  1. Draw grid over (map of) area;

2. Select squares / coordinates at random.

100
Q

What two measurements are needed to calculate an index of diversity? (2 marks)

A
  1. Number of (individuals of) each species;
    Accept: ‘population’ for ‘number’
    Total number of individuals / number of species;
    Accept: ‘species richness’
    MP2 allows for other types of diversity index
101
Q

The Amazonian forest today contains a very high diversity of bird species.
• Over the last 2 000 000 years, long periods of dry climate caused this forest to
separate into a number of smaller forests.
• Different plant communities developed in each of these smaller forests.
• Each time the climate became wetter again, the smaller forests grew in size and
merged to reform the Amazonian forest.
(a) Use the information provided to explain how a very high diversity of bird species
has developed in the Amazonian forest.
(5 marks)

A
  1. No interbreeding / gene pools are separate / geographic(al) isolation;
  2. Mutation;
  3. Different selection pressures / different foods / niches / habitats;
    Neutral: different climates
  4. Adapted organisms survive and breed / differential reproductive success;
  5. Change / increase in allele frequency / frequencies;
102
Q

Aspartic acid and proline are both amino acids. Describe how two amino acids differ
from one another. You may use a diagram to help your description. (1)

A
  1. Have different ‘R’ group
103
Q

Deletion of the sixth base (G) in the sequence shown in the diagram above would
change the nature of the protein produced but substitution of the same base would
not. Use the information in the table and your own knowledge to explain why. (3)

A
  1. Substitution would result in CCA / CCC / CCU;
  2. (All) code for same amino acid / proline;
  3. Deletion would cause frame shift / chang
104
Q

Messenger RNA (mRNA) is used during translation to form polypeptides.
Describe how mRNA is produced in the nucleus of a cell.
(6 marks)

A
  1. Helicase;
  2. Breaks hydrogen bonds;
  3. Only one DNA strand acts as template;
  4. RNA nucleotides attracted to exposed bases;
  5. (Attraction) according to base pairing rule;
  6. RNA polymerase joins (RNA) nucleotides together;
  7. Pre-mRNA spliced to remove introns.
105
Q

Describe how proteins are digested in the human gut.

4 marks

A
  1. Hydrolysis of peptide bonds;
  2. Endopeptidases break polypeptides into smaller peptide chains;
  3. Exopeptidases remove terminal amino acids;
  4. Dipeptidases hydrolyse / break down dipeptides into amino acids.
106
Q

Suggest how the change in the anticodon of a tRNA leads to MD (lines 10–13).
(3 marks)

A
  1. Change to tRNA leads to wrong amino acid being incorporated into
    protein;
  2. Tertiary structure (of protein) changed;
  3. Protein required for oxidative phosphorylation / the Krebs cycle, so less /
    no ATP made.
107
Q

If someone has MD, the concentration of lactate in their blood after exercise is
usually much higher than normal (lines 15–17). Suggest why.

A
  1. Mitochondria / aerobic respiration not producing much / any ATP;
  2. (With MD) increased use of ATP supplied by increase in anaerobic
    respiration;
  3. More lactate produced and leaves muscle by (facilitated) diffusion
108
Q

Suggest how the production of a protein with one amino acid missing may lead
to a genetic disorder such as Ellis-van Creveld syndrome.
(2)

A
  1. Change in tertiary structure / active site;
    Neutral: change in 3D shape / structure
  2. (So) faulty / non-functional protein / enzyme;
109
Q

A mutation can lead to the production of a non-functional enzyme. Explain how (6 marks)

A
  1. Change / mutation in base / nucleotide sequence (of DNA / gene);
  2. Change in amino acid sequence / primary structure (of enzyme);
  3. Change in hydrogen / ionic / disulfide bonds;
  4. Change in the tertiary structure / shape;
  5. Change in active site;
  6. Substrate not complementary / cannot bind (to enzyme / active site) / no
    enzyme-substrate complexes form.
    Accept: no E S complexes form
110
Q

Some populations of animals that have never been hunted show very low levels of
genetic diversity.
Other than hunting, suggest two reasons why populations might show very low
levels of genetic diversity. (2)

A
  1. Population might have been very small / genetic bottleneck;
  2. Population might have started with small number of individuals / by one
    pregnant female / founder effect;
  3. Inbreeding.
111
Q

what is meant by genetic diversity?

A

Number of different alleles of each gene.

112
Q

Give three ways in which courtship behaviour increases the probability of successful
mating. (3 marks)

A
  1. Recognise / identify / attract same species;
    Ignore: references to letting them produce fertile offspring
  2. Stimulates / synchronises mating / production / release of gametes;
  3. Recognition / attraction of mate / opposite sex;
    Accept finding a mate
    Accept: gender
  4. Indication of (sexual) maturity / fertility / receptivity / readiness to mate;
  5. Formation of a pair bond / bond between two organisms (to have / raise young).
113
Q

What is meant by a hierarchy?

A
  1. Groups within groups

2. No overlap (between groups);

114
Q

Explain the role of independent segregation in meiosis. (2)

A
  1. (To provide) genetic variation;
    Genetic variation must be directly stated and not implied
  2. (Allows) different combinations of maternal and paternal
    chromosomes / alleles;
115
Q

What is a species?

A
  1. Group of similar organisms
  2. Reproduce / produce offspring;
  3. That are fertile;

‘Produce fertile offspring’ = 2 marks

116
Q

What is meant by phylogenetic group

A

(Grouped according to) evolutionary links / history / relationships /
common ancestry;

117
Q

Some cancer cells have a receptor protein in their cell-surface membrane that binds
to a hormone called growth factor. This stimulates the cancer cells to divide.
Scientists have produced a monoclonal antibody that stops this stimulation.
Use your knowledge of monoclonal antibodies to suggest how this antibody stops
the growth of a tumour. (3)

A
  1. Antibody has specific tertiary structure / binding site / variable region;
  2. Complementary (shape / fit) to receptor protein / GF / binds to receptor protein / to GF;
  3. Prevents GF binding (to receptor).
118
Q

A mutation of a tumour suppressor gene can result in the formation of a tumour.
Explain how.

A
  1. (Tumour suppressor) gene inactivated / not able to control / slow down
    cell division;
    Ignore: references to growth
  2. Rate of cell division too fast / out of control
119
Q

Give the 2 types of molecule from which a ribosome is made (2)

A

1) protein

2) RNA

120
Q

Describe the role of a ribosome in the production of a polypeptide. Do not include transcription in your answer (3)

A
  1. mRNA binds to ribosome
  2. allows tRNA with anticodons to bind
  3. catalyses formation of polypeptide bond between amino acids
121
Q

In a eukaryotic cell, the base sequence of the mRNA might be different from the sequence of the pre-mRNA. Explain why (2)

A
  1. introns in pre-mRNA

2. splicing

122
Q

Describe the structure of glycogen (2)

A
  1. polysaccharide of alpha glucose

2. joined by glycosidic bonds

123
Q

Suggest how glycogen acts as a source of energy (2)

A
  1. hydrolysed to glucose

2. glucose is used in respiration

124
Q

Suggest and explain two ways the cell0surface membranes of the cells lining the uterus may be adapted to allow rapid transport of nutrients (2)

A
  1. membrane folded so increased surface area
  2. larger number of protein channels for facilitated diffusion
  3. large number of protein carriers for active transport
  4. large number of protein for co-transport
125
Q

Sodium ions from salt (sodium chloride) are absorbed by cells lining the gut. Some of these cells have membranes with a carrier protein called NHE3. NHE3 actively transports one sodium ion into the cell in exchange for one promo (hydrogen ion) out of the cell. Use your knowledge of transport across cell membranes to suggest how NHE3 does this (3)

A
  1. co-transport
  2. uses hydrolysis of ATP
  3. sodium ion and proton bind to the protein
  4. protein changes shape
126
Q

High absorption of salt from the diet can result in a higher than normal concentration of salt in the blood plasma entering capillaries. This can lead to a build-up of tissue fluid. Explain how (2)

A
  1. higher salt results in lower water potential of tissue fluid
  2. so less water returns to capillary by osmosis at venue end
127
Q

Describe how bacteria divide (2)

A
  1. binary fission
  2. replication of circular DNA
  3. division of cytoplasm to produce 2 daughter cells
  4. each with single copy of circular DNA
128
Q

Suggest one advantage to a bacterium of secreting an extracellular protease in its natural environment (2)

A
  1. to digest protein

2. so they can absorb amino acids for protein synthesis

129
Q

Mammals have some cells that produce extracellular proteases. They also have cells with membrane-bound dipeptidases. Describe the action of these membrane-bound dipeptidases and explain their importance (2)

A
  1. hydrolyse peptide bonds to release amino acids

2. amino acids can cross cell membrane

130
Q

Suggest why human ATP synthase is not inhibited and bacterial synthase is inhibited (1)

A
  1. human ATP synthase has a different shape active site to bacterial ATP synthase
131
Q

Give 3 environmental variables that should be controlled when growing the plants before treatment with the different sprays (2)

A
  1. carbon dioxide concentration
  2. light intensity
  3. temperature
132
Q

The scientists incubated the flasks containing the leaf discs at 26C and gently shook the flasks. Suggest one reason why the scientists ensured the temperature remained constant and one reason why the leaf discs were shaken (2)

A
  1. temperature - so rate of diffusion remains constant

2. shaking - so all surfaces of the leaf discs are exposed

133
Q

Describe how phagocytosis of a virus leads to presentation of its antigens (3)

A
  1. phagosome fuses with lysosome
  2. virus destroyed by lysozymes
  3. peptides from virus are displayed on the cell membrane
134
Q

Describe how presentation of a virus antigen leads to secretion of an antibody against this virus antigen (3)

A
  1. helper T cells bind to the antigen
  2. this stimulates a specific B cell
  3. B cell clones
135
Q

State 3 comparisons of genetic diversity that the scientists used in order to generate classifications based on other comparisons of genetic characteristics (3)

A
  1. comparison of DNA sequence
  2. base sequence of mRNA
  3. amino acid sequence
136
Q

Explain 3 ways in which an insects tracheal system is adapted for efficient gas exchange (3)

A
  1. tracheoles have thin walls so short diffusion pathway
  2. large number of tracheoles so large surface area
  3. tracheae provide tubes full of air so fast diffusion
137
Q

Suggest ways to improve a scientific drawing (2)

A
  1. dont shade
  2. no sketching, only single lines
  3. dont cross labelling lines
  4. add measurement or scale
  5. label more feature
138
Q

Contrast how an optical microscope and a transmission electron microscope work and contrast the limitations of their use when studying cells (6)

A
  1. TEM uses electrons and optical use light
  2. TEM allows a greater resolution
  3. with TEM smaller organelles can be observed
  4. TEM view only dead/dehydrated specimens and optical can view live specimens
  5. TEM does not show colour and optical can
  6. TEM requires thinner specimens
  7. TEM requires a more complex/time consuming preparation
  8. TEM focuses using magnets and optical uses glass lenses
139
Q

The nucleus and chloroplast of a plant cell both contain DNA.
Give THREE ways in which the DNA in a chloroplast is different from the DNA in the nucleus. (3)

A
  1. DNA shorter in chloroplast
  2. Fewer genes in chloroplast
  3. DNA in chloroplast circular not linear
  4. DNA in chloroplast not associated with proteins/histones
  5. Introns absent in the DNA in chloroplast but present in nuclear DNA
140
Q

Give differences between the structure of DNA nucleotide and the structure of RNA nucleotide

A
  1. DNA has thymine base. RNA doesn’t. RNA has Uracil instead

2. Deoxyribose in DNA and ribose in RNA

141
Q

Not all mutations in the nucleotide sequence of a gene cause a change in the structure of a polypeptide.
Give two reasons why (2)

A
  1. Triplets code for the same amino acid
  2. Occurs in introns/non-coding sequence

Accept: DNA/code/triplets are degenerate

142
Q

What is a monoclonal antibody

A

Antibodies with the same tertiary structure
OR
Antibody produced from identical/clones plasma cells/B cells/B lymphocytes

143
Q

Give one example of using monoclonal antibodies in a medical treatment (1)

A
  1. Targets/binds/carries drug/medicine to specific cells/antigens/receptors
    ORRR
    Block antigens/receptors on cells
144
Q

Describe the role of antibodies in producing a positive result in an ELISA test (4)

A
  1. First antibody attaches/binds to complimentary antigen
  2. Second Antibody with enzyme attached is added
  3. Second antibody attaches to antigen
  4. Substrate/solution is added and colour changes
145
Q

Suggest a method other than using a colorimeter that the student could use to measure the quantity of a reducing sugar in a solution. (2)

A
  1. Filter and dry (the precipitate)

2. Find the mass/weight

146
Q

Use of a colorimeter in this investigation would improve the repeatability of the students result. Give one reason why (1)

A
  1. Quantative (colour change is subjective)

2. Standardises the method

147
Q

Explain why it is more useful to calculate an index of diversity than to record species of richness (2)

A
  1. Index of diversity also measures abundance/number of each species
  2. It’s is useful because there may be many/few of some species
148
Q

Suggest how the scientist measure the rate of flow in the river

A 
1. Movement of floating object over known distance and time 
OR 
time to fill container of known volume 
OR 
use of a data logging device
149
Q

Name an organelle found in both a chloroplast and a prokaryotic cell (1)

A

Ribosome

150
Q

A biologist separated cell components to investigate organelle activity. She prepared a suspension of the organelles in a solution that prevented damage to the organelles. (3)

A
  1. Ice cold to prevent/reduce enzyme activity
  2. Buffered to prevent denaturing the enzyme/protein
  3. Same water potential/isotonic to prevent bursting of organelle
151
Q

Describe and explain the advantage of the counter current principle in gas exchange across a fish gill (3)

A
  1. Water and blood flow in opposite directions
  2. Maintains diffusion/concentration gradient of oxygen
    OR
    oxygen concentration always higher in water
  3. Diffusion along length of lamellae/filament/gill/capillary
152
Q

Explain how the active site of an enzyme causes a high rate of reaction (3)

A
  1. Lowers activation energy
  2. Induced fit causes active site to change shape
  3. So e-z complexes causes bonds to form/break
153
Q

Describe a biochemical test to confirm the presence of protein in a solution (2)

A
  1. Add biuret reagent

2. Positive= purple/lilac/violet/mauve

154
Q

A dipeptide consists of two amino acids joined by a peptide bond. Dipeptides may differ in the type of amino acids they contain.
Describe two other ways in which all dipeptides are similar and one way in which they might differ. (3)

A

Sim:

  1. Amine group at the end
  2. Carboxyl group at the end
  3. Two R groups
  4. All contain C,H,N,O

Diff:
1.variable/diff R groups

155
Q

Suggest why preventing formation of spindle fibres stopped the cell cycle (2)

A 
1. Chromosome/centromeres cannot attach to spindle
OR 
chromosomes cannot line up 
2. So no metaphase 
OR 
3. Chromatids cannot separate 
4. No anaphase
156
Q

Function of ATP hydrolase

A

1) Catalyses the hydrolysis of ATP into ADP+Pi
2) to release energy needed for the transport of substance by carrier proteins, process is active transport so requires energy

157
Q

2 features in cell specialised for absorption

A 
Villi increases surface area for enzymes to work on
Thin walls (one cell thick) so short diffusion pathway
158
Q

How do amino acids join to form a polypeptide

A

Peptide bonds are formed between amino acid molecules in a condensation reaction

159
Q

Describe the role of micelles in the absorption of fats into the cells lining in the ilium (3)

A

Micelles increase the surface area for lipase to act on, which means faster hydrolysis action by lipase. Micelles are water soluble vesicle and so deliver fatty acids and glycerol to the epithelial cells of the ileum for absorption

160
Q

Explain the reason for increase in pressure and in rate of blood flow in the aorta

A

Atria and ventricle relaxed blood enters at low pressure through veins pulmonary vein and vena cava into the aorta
Blood pressure begins to increase as blood flows into the aorta causing AV valves to become opened allowing for the blood to enter the ventricles

161
Q

Two variable to keep constant

A

Water temp

Conc of acid

162
Q

Why higher conc could cause a tumour

A

Cells go through DNA rep earlier. More DNA rep. Rep is uncontrollable results in mass of cells

163
Q

Describe the process of binary fission

A

1) circular DNA and plasmids replicate. Main DNA loop is only replicated once but plasmids can be replicated many times.
2) cells gets bigger and DNA loops move to opposite poles
3) cytoplasm divides and new cell wall begins to form
4) two daughter cells produced each with one copy of circular DNA and variable number of copies of plasmids

164
Q

Describe and explain the effects of increasing carbon dioxide concentration on the dissociation of oxyhaemaglobin

A

1) increases/more oxygen dissociation/unloading
OR
decreases haemaglobins affinity for oxygen

2) by decreasing blood Ph/ increasing acidity

165
Q

Name the two scientists who proposed models of the chemical structure of DNA and DNA replication

A

Watson and crick

166
Q

Role of single stranded DNA fragments

Role of DNA nucleotide

A

Role of DNA fragments:

1) template
2) determines order of nucleotide/bases

Role of nucleotide:
1) forms complementary pairs/DNA strand

167
Q

Describe and explain the role of antibodies in stimulating phagocytosis

A

1) bind to antigens/ are markers
2) cause clumping/agglutination
OR
Attract phagocytes

168
Q

Explain why the coloured water moved up the stalk. (Investigation of movement of water in leaf stalks)

A

1) water evaporates/is transpired from leaves
2) water potential gradient creates tension/pull up of water
3) hydrogen bonds maintain column

169
Q

Describe and explain three ways in which the scientist would ensure he used aseptic techniques to move each cube of agar onto a new agar plate

A

1) wash hands to kill bacteria/wear gloves
2) burning Bunsen burner close by to create an upward current of air
3) disinfect workplace to prevent contamination
4) flame the instruments to prevent contamination
5) slightly lift the lid to prevent entry of microbes

170
Q

Give two differences between mitosis and meiosis

A

1) mitosis= 1 division
Meiosis = 2 divisions
2) mitosis produces genetically identical daughter cells, daughter cells are genetically different in meiosis
3) two cells produced in mitosis, 4 cells produced in meiosis
4) diploid to diploid OR haploid to haploid in mitosis, diploid to haploid in meiosis
5) separation of homologous chromosomes only in meiosis
6) crossing over and independent segregation in meiosis

171
Q

Describe the role of iron ions, sodium ions and phosphate ions in cells

A

Iron:
1) haemaglobin bonds with oxygen
Transport oxygen

Sodium:

2) co transport of glucose/amino acid into cells
3) because sodium moved out by active transport/Na-K pump
4) creates sodium concentration/diffusion gradient
5) affects osmosis/water potential

Phosphate:

6) affects osmosis/water potential
7) joins nucleotide (sugar phosphate backbone)
8) used in/to produce ATP
9) phosphorylates other compounds usually making them more reactive
10) hydrophilic

AQA A Level Biology (yr 12 content) (17 decks)

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