AQA Biology Mark Scheme Questions

Describe one way in which the structure of the DNA of a gene may be changed as a result of a mutation

- deletion/ substitution/ addition

- of nucleotide / base

Example of mutagens

- high energy ionised particles/ x-rays/ UV

- tobacco tar

- mustard gas - chemical mutagen

- high energy radiation - gamma rays

Explain how exposure to a mutagenic agent may result in an inactive enzyme being produced by a cell

- change in sequence of nucleotides

- changes sequence of amino acids

different tertiary structure

- inactive enzyme if shape of active site is changed

- ESC'S don't form

Explain how the mutation of a gene can result in an organism lacking a particular enzyme

- mutation changes sequence of nucleotide bases and therefore the amino acid sequence.

- frame shift

- incorrect codons

- incorrect amino acids brought to ribosome

- tertiary structure changes

- non-functional protein

Effect of deletion of a base?

- frame shift

- changes sequence of amino acids

- substitution alters one codon

- no effect if new triplet codes for same amino acid= degenerate nature of genetic code

Describe a plasmid

- circular dna

- contains only a few genes

- separate from main bacterial DNA

Explain the function of a vector

- transfers foreign dna/gene into bacteria/ host cell

Use your knowledge of monoclonal antibodies to suggest how this antibody stops the growth of a tumour.

- antibody has specific tertiary structure

- complementary shape to receptor protein

- mutation in intron

Name two structures in a eukaryotic cell that cannot be identified using an optical microscope

- mitochrondrian

- ribosome

Describe how the RER is involved in the production of enzymes.

- ribosomes

- to synthesise proteins

Describe how the Golgi apparatus is involved in the secretion of enzymes.

- modifies proteins

purpose of: No air-sealing grease was applied to either surface of the leaf.

- means stomata open which allows normal c02 uptake

purpose: The lower surface of the leaf was covered in air-sealing grease that

prevents gas exchange.

- grease stops c02 uptake through stomata

purpose: Both the lower surface and the upper surface of the leaf were covered in air-sealing grease that prevents gas exchange.

- stops all co2 uptake

- shows sealing is effective

The stomata close when the light is turned off.

Explain the advantage of this to the plant.

- because water is lost through the stomata

- closure prevents/ reduces water loss and decreases the wp gradient

- maintains the water content of cells

Suggest how this uptake of carbon dioxide continues even when the lower surface of the leaf is sealed

- CO2 uptake through upper surface of the leaf

Describe how you would test a piece of food for the presence of lipid.

- dissolve in alchohol, then add water

- white emulsion

Name the organelles you might find in an animal cell

1)Plasma cell surface membrane

2)Rough/smooth endoplasmic reticulum

3)nucleus (nucleolus)

4)lysosome

5)ribosome

6)Golgi apparatus

7)cytoplasm

8)mitochondrion

Name the organelles you might find in a plant cell

1)Plasma cell surface membrane

2)cellulose wall

3)chloroplast

4)rough/smooth endoplasmic reticulum

5)mitochondrion

6)Golgi apparatus

7)vacuole

8)cytoplasm

9)nucleus

10) ribosome

what is a pathogen?

a disease causing organism

Explain what is meant by a polymer. [1]

Molecule) made up of many identical/similar molecules/monomers/ subunits;

Describe how two-way chromatography is carried out. [2]

Run chromatogram then turn through 90/right angle; With a different solvent;

How is fish gills adapted for gaseous exchange?

1 Large surface area provided by lamellae/filaments;

2 Increases diffusion/makes diffusion efficient;

3 Thin epithelium/distance between water and blood;

4 Water and blood flow in opposite directions/countercurrent;

5 (Point 4) maintains concentration gradient (along gill)

/equilibrium not reached;

6 As water always next to blood with lower concentration

of oxygen;

7 Circulation replaces blood saturated with oxygen;

8 Ventilation replaces water (as oxygen removed);

The cell-surface membrane can be seen with a transmission electron microscope but not with an optical microscope.

Explain why.

Electron microscope has higher resolution (than optical microscope);

No organelles are visible in the cytoplasm of this red blood cell. Suggest why.

Cytoplasm of red blood cell filled with haemoglobin;

Before the cell was examined using the electron microscope, it was stained. This stain caused parts of the structure of the cell-surface membrane to appear as two dark lines.

Suggest an explanation for the appearance of the cell-surface membrane as two dark lines

Membrane has phospholipid bilayer;

Stain binds to phosphate/glycerol;

. On inside and outside of membrane;

Describe how substances move across cell-surface membranes by facilitated

diffusion.

Carrier/channel protein;

2. (Protein) specific/complementary to substance;

3. Substance moves down concentration gradient;

suggest and explain one advanatge of asexual reproduction and sexual reproduction

asexual = fewer stages so quicker, only one parent involved so can colonise new environment

sexual= increases genetic diversity so increases chances of survival

The hypha supporting the spore-carrying sporangium is vertical. Suggest one advantage of the hypha being vertical.

spores spread/ dispersed further

A scientist wanted to calculate the mean volume of Rhizopus spores. Describe how she could use Figure 5 do this.

You may assume the spores are perfectly spherical.

Measurediameteroflargenumberofspores;

2. Dividemeasuredvaluesby700(tofindtrue diameter);

3. Referencetousingvolumeofsphere;

The scientists decided to collect animals from the soil samples for 40 minutes. Suggest how the scientists decided that 40 minutes was an appropriate time.

1. Repeat soil sorting for different times and record number of species collected;

2. Find optimum time / time beyond which further sorting does not lead to increase in animal species found;

The scientists calculated a correlation coefficient, R, from their data. They found R = 0.50, with P <0.0001

Explain the meaning of the result of their calculations.

Shows a positive correlation;

2. There probability of getting this correlation by chance is less than 0.0001;

3. Correlation is highly significant;

Suggest the role of the mRNA base triplets UGA, UAG and UAA.

Stop translation;

2. Result in detachment of polypeptide chain from ribosome;

give the formula for calculating the percentage saturation of haemoglobin with oxygen

oxygenated haemoglobin × 100

maximum saturation

Suggest one advantage of this change in the affinity of haemoglobin for oxygen.

Ensures rapid/more intake of oxygen in lungs/release of oxygen in tissues;

The scientists dried the plant samples in an oven at 100 °C.

Give two reasons why they used this temperature.

Evaporates all water

2. (But) does not burn (organic compounds);

Contrast the structure of a bacterial cell and the structure of a human cell.

1.Bacterial cell is much smaller than a human

cell;

2. Bacterial cell has a cell wall but human cell does not;

3. Bacterial cell lacks a nucleus but human cell has a nucleus;

4. Bacterial cell lacks membrane-bound organelles but human cell has membrane- bound organelles;

5. Bacterial ribosomes smaller than human ribosomes / bacteria have 70S ribosomes whereas humans have 80S ribosomes;

6. Bacterial DNA is circular but human DNA is linear;

7. Bacterial DNA is 'naked' whereas human DNA is bound to histones/proteins;

Figure 1 shows glycogen granules present in skeletal muscle.

Explain their role in skeletal muscle.

1. As a store of glucose OR

To be hydrolysed to glucose;

2. For respiration/to provide ATP;

During vigorous exercise, the pH of skeletal muscle tissue falls. This fall in pH leads to a reduction in the ability of calcium ions to stimulate muscle contraction.

Suggest how.

Low pH changes shape of calcium ion receptors

2. Fewer calcium ions bind to tropomyosin

3. Fewer tropomyosin molecules move away;

4. Fewer binding sites on actin revealed;

5. Fewer cross-bridges can form

OR

Fewer myosin heads can bind;

A teacher studying these data with her students told her class that no definite

conclusions could be drawn when comparing the mean values in Figure 3.

Suggest why the teacher said this.

1. No error bars/SD;

2. To show if overlap occurs so difference (in means) is not significant/due to chance

Give one way in which benign tumours differ from malignant tumours.

Cells of benign tumours cannot spread to other parts of the body/metastasise

OR

Cells of benign tumours cannot invade neighbouring tissues;

Give two features of a climax community.

1. Same species present (over long time) / stable community (over long time);

2. Abiotic factors (more or less) constant (over time);

3. Populations stable (around carrying capacity);

A student investigated the monohybrid inheritance of eye shape in fruit flies. Two fruit flies with bar (narrow) eyes were crossed. Of the offspring, 1538 had bar eyes and 462 had round (normal) eyes.

09.1

[2 marks]

Using suitable symbols, give the genotypes of the parents. Explain your answer.

- parents must be Bb

- because both parents have bar eyes, but have some offspring with round eyes, so parents must be carriers of recessive allele for round eyes;

Suggest two reasons why observed ratios are often not the same as expected

Fertilisation is random OR

Fusion of gametes is random;

2. Small/not large population/sample;

3. Selection advantage/disadvantage/lethal alleles;

Suggest and explain two reasons why a poor supply of phosphate ions results in poor growth of plants

(Required to) make ATP/glucose phosphate, so less respiration/less energy for growth;

2. (Required to) make nucleotides, so less DNA/mRNA/tRNA for cell division/production of protein (for growth);

3. (Required to) make RuBP/NADP, so less CO2 fixed/reduced into sugar;

4. (Required to) make phospholipids for membranes;

Explain how you could determine the total amount of carbon dioxide secreted at 30 °C during the period of recording.

Determine the area under the curve;

Suggest an explanation for the effect of temperature on the rate of carbon dioxide release.

1. Enzymes/metabolism faster;

2. Higher rate of respiration and carbon dioxide production/release;

3. Spiracles open more often/remain open to excrete/get rid of carbon dioxide/get more oxygen;

Each type of labelled antibody binds specifically to one of the proteins. Explain why

1. Each protein has a different tertiary structure;

2. (Each) antibody has a specific antigen/binding/variable region/site;

3. So, (each antibody) forms different antigen- antibody complex

OR

(each antibody) only binds to complementary (protein);

The scientists used an optical microscope to measure the number of capillaries in thin sections cut from samples of heart muscle.

Describe the method they would have used to find the mean number of capillaries per mm2

Measure diameter of field of view and calculate area;

2. Using micrometer slide and eyepiece graticule;

3. Count number of capillaries in large number of fields of view and calculate mean;

4. Select fields of view randomly

The scientists gave their results as ratios. Explain the advantage of giving the results of this investigation as a ratio.

allows comparison

How is O2 able to bind to haemoglobins?

Fe2+ in haem group of quarternary structure binds to oxygen molecules - each haemoglobin can transport 4 molecules

Why do haemoglobins of different species have different affinities towards oxygen?

Each species produce haemoglobins with different amino acid sequences, resulting in varying affinities

Heat from respiration helps mammals to maintain a constant body temperature.

Use this information to explain the relationship between the surface area to volume ratio of mammals and the oxygen dissociation curves of their haemoglobins.

Smaller mammal has greater surface area to volume ratio;

Smaller mammal / larger SA:Vol ratio more heat lost (per unit body mass);

Smaller mammal / larger SA:Vol ratio has greater rate of respiration / metabolism;

Oxygen required for respiration so (haemoglobin) releases more oxygen / oxygen released more readily / haemoglobin has lower affinity;

Haemoglobin is a protein. Explain why a mature red blood cell cannot make haemoglobin.

No nucleus/ DNA;

(Nucleus) codes for protein/ can't make RNA;

OR No ribosomes / rough endoplasmic reticulum;

Protein is made/ synthesised/ translated (on ribosomes);

OR No mitochondria;

(Mitochondria) supply energy/ ATP for making proteins;

Explain the advantage to people living at high altitude

(Haemoglobin) has lower affinity for oxygen / more oxygen released;

In / to the cells / tissues;

Explain how glycogen's structure relates to its function (4 marks).

1. Insoluble so does not draw water into cells (by osmosis);

2. (Insoluble so) does not diffuse out of cells;

3. Compact so a lot can be stored in a small space:

4. (More) highly branched (than starch) so more rapidly broken down to form monomers of glucose;

Describe how the structure of a protein depends on the amino acids it contains (5 marks).

1. Structure is determined by (relative) position of amino acid/R group/interactions;

(accept for 'interactions', hydrogen bonds / disulfide bridges / ionic bonds / hydrophobic hydrophilic interactions)

2. Primary structure is sequence/order of amino acids;

3. Secondary structure formed by hydrogen bonding (between amino acids);

(Accept alpha helix/β-pleated sheet for 'secondary structure')

4. Tertiary structure formed by interactions (between R groups);

5. Creates active site in enzymes

OR

Creates complementary/specific shapes in antibodies/carrier proteins/receptor (molecules);

6. Quaternary structure contains >1 polypeptide chain

OR

Quaternary structure formed by interactions/bonds between polypeptides;

(accept prosthetic (group))

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

1. Lowers activation energy;

2. Induced fit causes active site (of enzyme) to change shape;

3. (So) enzyme-substrate complex causes bonds to form/break;

(accept: description, of induced fit)

(accept: enzyme-substrate complex causes stress/strain on bonds)

Describe how a non-competitive inhibitor can reduce the rate of an enzyme-controlled reaction (3 marks).

1. Attaches to the enzyme at a site other than the active site;

(accept 'attaches to allosteric/inhibitor site')

2. Changes (shape of) the active site

OR

Changes tertiary structure (of enzyme);

3. (So active site and substrate) no longer complementary so less/no substrate can fit/bind;

(accept 'no longer complementary so less/no enzyme- substrate complexes form')

(accept abbreviations of enzyme-substrate complex.)

Describe how a competitive inhibitor can reduce the rate of an enzyme-controlled reaction (3 marks).

1. Attaches to the active site;

2. Changes (shape of) the active site

OR

Changes tertiary structure (of enzyme);

3. (So active site and substrate) no longer complementary so less/no substrate can fit/bind;

(accept 'no longer complementary so less/no enzyme- substrate complexes form')

(accept abbreviations of enzyme-substrate complex.)

Describe the structure of DNA (5 marks).

1. Polymer of nucleotides;

(accept 'Polynucleotide')

2. Each nucleotide formed from deoxyribose, a phosphate (group) and an organic/nitrogenous base;

3. Phosphodiester bonds (between nucleotides);

4. Double helix/2 strands held by hydrogen bonds;

5. (Hydrogen bonds/pairing) between adenine, thymine and cytosine, guanine;

Describe the process of semi-conservative replication (5 marks).

1. DNA helicase breaks hydrogen bonds between complementary bases (on the DNA strand);

2. Each exposed polynucleotide strand then acts like a template;

3. Free nucleotides that have been activated by ATP bind to their complementary bases;

4. DNA polymerase joins the nucleotides together by making phosphodiester bonds;

5. (Semi-conservative because) the new DNA molecules contain half of the original DNA and half of the new DNA;

Describe how an ATP molecule is formed from its component molecules (4 marks).

1. and 2. Accept for 2 marks correct names of three components adenine, ribose/pentose, three phosphates;;

(accept for 1 mark, correct name of two components)

(accept for 1 mark, ADP and phosphate/Pi)

(ignore adenosine)

(accept suitably labelled diagram)

3. Condensation (reaction);

(ignore phosphodiester)

4. ATP synthase;

(reject ATPase)

ATP is formed via a condensation reaction between adenine and three phosphates, catalysed by ATP synthase

Explain five properties that make water important for organisms (5 marks).

1. A metabolite in condensation/hydrolysis/ photosynthesis/respiration;

2. A solvent so (metabolic) reactions can occur

OR

A solvent so allowing transport of substances;

3. High (specific) heat capacity so buffers changes in temperature;

(for 'buffer' accept 'resist')

4. Large latent heat of vaporisation so provides a cooling effect (through evaporation);

(reject latent heat of evaporation)

5. Cohesion (between water molecules) so supports columns of water (in plants);

(for 'columns of water' accept 'transpiration stream')

(do not credit 'transpiration' alone but accept description of 'stream')

(for 'columns of water' accept 'cohesion-tension (theory)')

6. Cohesion (between water molecules) so produces surface tension supporting (small) organisms;

(for cohesion accept hydrogen bonding)

Describe the roles of iron ions, sodium ions, and phosphate ions in cells (5 marks).

(must have MP1 for 5 max)

(3 max for sodium and 3 max for phosphate)

IRON IONS:

1. Haemoglobin binds/associates with oxygen

OR

Haemoglobin transports/loads oxygen;

(ignore reference to 2+ or 3+ in Fe2+ or Fe3+)

SODIUM IONS:

2. Co-transport of glucose/amino acids (into cells);

3. (Because) sodium moved out by active transport/Na - K pump;

4. Creates a sodium concentration/diffusion gradient;

5. Affects osmosis/water potential;

PHOSPHATE IONS:

6. Affects osmosis/water potential;

(accept 5. OR 6. - not both)

7. Joins nucleotides/in phosphodiester bond/in backbone of

DNA/RNA/in nucleotides;

8. Used in/to produce ATP;

(reject 'energy produced')

9. Phosphorylates other compounds (usually) making them more reactive;

10. Hydrophilic/water soluble part of phospholipid bilayer/membrane;

Describe how a sample of chloroplasts could be isolated from leaves (4 marks).

1. Break open cells/tissue and filter

OR

Grind/blend cells/tissue/leaves and filter;

(accept homogenise and filter)

2. In cold, same water potential/concentration, pH controlled solution;

(accept for 'same water potential/ concentration', isotonic)

(accept for 'pH controlled', buffered)

3. Centrifuge/spin and remove nuclei/cell debris;

4. (Centrifuge/spin) at high(er) speed, chloroplasts settle out;

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.

Describe three properties of this solution and explain how each property prevented damage to the organelles (3 marks).

1. (Ice) cold to prevent/reduce enzyme activity;

2. Buffered to prevent denaturing of enzyme/protein;

(accept description of buffer)

(accept: prevent change of tertiary structure)

3. Same water potential/ Ψ to prevent lysis/bursting (of organelle);

(accept: isotonic for same water potential)

(reject: references to turgor or plasmolysis or crenation)

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

1. TEM uses electrons and optical light:

2. TEM allows a greater resolution;

3. (So with TEM) smaller organelles / named cell structure can be observed

OR

greater detail in organelles / named cell structure can be

observed;

('clearer' is not equivalent to 'detail')

4. TEM view only dead / dehydrated specimens and optical (can) view live specimens;

(accept 'Only optical can view live specimens')

5. TEM does not show colour and optical (can);

(accept 'Only optical can show colour')

6. TEM requires thinner specimens;

7. TEM requires a more complex/time consuming preparation;

(accept 'TEM requires a more difficult preparation')

8. TEM focuses using magnets and optical uses (glass) lenses;

(ignore references to artefacts)

Describe binary fission in bacteria (3 marks).

1. Replication of (circular) DNA;

(accept nucleoid)

(reject chromosome)

(reject mitosis)

2. Replication of plasmids;

3. Division of cytoplasm (to produce daughter cells);

(ignore genetically identical)

Describe the appearance and behaviour of chromosomes during mitosis (5 marks).

(During prophase)

1. Chromosomes coil / condense / shorten / thicken / become visible;

2. (Chromosomes) appear as (two sister) chromatids joined at the centromere;

(During metaphase)

3. Chromosomes line up on the equator / centre of the cell;

4. (Chromosomes) attached to spindle fibres;

5. By their centromere;

(During anaphase)

6. The centromere splits / divides;

7. (Sister) chromatids / chromosomes are pulled to opposite poles / ends of the cell / separate;

(During telophase)

8. Chromatids / chromosomesuncoil / unwind / become longer / thinner.

Name and describe five ways substances can move across the

cell-surface membrane into a cell (5 marks).

1. (Simple) diffusion of small/non-polar molecules down a concentration gradient;

2. Facilitated diffusion down a concentration gradient via protein carrier/channel;

3. Osmosis of water down a water potential gradient;

4. Active transport against a concentration gradient via protein carrier using ATP;

5. Co-transport of 2 different substances using a carrier protein;

(for any answer accept a correct example)

(for 'carrier protein' accept symport OR cotransport protein)

The movement of substances across cell membranes is affected by membrane structure. Describe how (5 marks).

1. Phospholipid (bilayer) allows movement/diffusion of non- polar/lipid-soluble substances;

2. Phospholipid (bilayer) prevents movement/diffusion of polar/ charged/lipid-insoluble substances

OR

(Membrane) proteins allow polar/charged substances to cross the membrane/bilayer;

3. Carrier proteins allow active transport;

4. Channel/carrier proteins allow facilitated diffusion/co-transport;

5. Shape/charge of channel / carrier determines which substances move;

6. Number of channels/carriers determines how much movement;

7. Membrane surface area determines how much diffusion/movement;

Describe how HIV is replicated (4 marks).

1. Attachment proteins attach to receptors on helper T cell/lymphocyte;

2. Nucleic acid/RNA enters cell;

3. Reverse transcriptase converts RNA to DNA;

4. Viral protein/capsid/enzymes produced;

5. Virus (particles) assembled and released (from cell);

Describe how a phagocyte destroys a pathogen present in the blood (3 marks).

1. Engulfs;

(accept endocytosis)

2. Forming vesicle/phagosome and fuses with lysosome;

3. Enzymes digest/hydrolyse;

(accept lysozymes for 'enzymes')

Explain how HIV affects the production of antibodies when AIDS develops in a person (3 marks).

1. Less/no antibody produced;

2. (Because HIV) destroys helper T cells;

(accept 'reduces number' for 'destroys')

3. (So) few/no B cells activated / stimulated

OR

(So) few/no B cells undergo mitosis/differentiate/form plasma cells;

Describe how B lymphocytes would respond to vaccination (3 marks).

1. B cell (antibody) binds to (viral) specific/complementary receptor/antigen;

(accept B cell forms antigen-antibody complex)

2. B cell clones

OR

B cell divides by mitosis;

3. Plasma cells release/produce (monoclonal) antibodies (against the virus);

4. (B/plasma cells produce/develop) memory cells;

(accept B cell undergoes clonal selection/expansion)

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

1. (First) antibody binds/attaches /complementary (in shape) to antigen;

2. (Second) antibody with enzyme attached is added;

3. (Second) antibody attaches to antigen;

(accept (second) antibody attaches to (first) antibody (indirect ELISA test))

4. (Substrate/solution added) and colour changes;

(only award if enzyme mentioned)

During vaccination, each animal is initially injected with a small volume of venom. Two weeks later, it is injected with a larger volume of venom.

Use your knowledge of the humoral immune response to explain this vaccination programme (3 marks).

1. B cells specific to the venom reproduce by mitosis;

2. (B cells produce) plasma cells and memory cells;

3. The second dose produces antibodies (in secondary immune response) in higher concentration and quickly

OR

The first dose must be small so the animal is not killed;

Describe how vaccination can lead to protection against viruses (6 marks).

1. Antigen / bacterium binds to surface protein / surface receptor on a (specific / single) B cell.

2. (Activated) B cell divides by mitosis / produces clone;

3. (Division) stimulated by cytokines / by T cells;

4. B cells / plasma cells release antibodies;

5. (Some) B cells become memory cells;

6. Memory cells produce plasma / antibodies faster;

Describe the difference between active and passive immunity (5 marks).

1. Active involves production of antibody by plasma cells / memory cells;

2. Active involves memory cells, passive does not;

3. Passive involves antibody introduced into body from outside/named source;

4. Active long term, because antibody produced in response to antigen;

5. Passive short term, because antibody (given) is broken down;

6. Active (can) take time to develop / work, passive fast acting;

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

1. Water and blood flow in opposite directions;

2. Maintains diffusion/concentration gradient of oxygen;

(accept oxygen concentration always higher (in water))

3. (Diffusion) along length of lamellae/filament/gill/capillary;

Describe the gross structure of the human gas exchange system and how we breathe in and out (6 marks).

1. Named structures - trachea, bronchi, bronchioles, alveoli;

2. Above structures named in correct order

OR

Above structures labelled in correct positions on a diagram;

3. Breathing in - diaphragm contracts and external intercostal muscles contract;

4. (Causes) volume increase and pressure decrease in thoracic cavity (to below atmospheric, resulting in air moving in);

5. Breathing out - Diaphragm relaxes and internal intercostal muscles contract;

6. (Causes) volume decrease and pressure increase in thoracic cavity (to above atmospheric, resulting in air moving out);

Explain three ways in which an insect's tracheal system is adapted for efficient gas exchange (3 marks).

1. Tracheoles have thin walls so short diffusion distance to cells;

2. Highly branched / large number of tracheoles so short diffusion distance to cells;

3. Highly branched / large number of tracheoles so large surface area (for gas exchange);

4. Tracheae provide tubes full of air so fast diffusion (into insect tissues);

5. Fluid in the end of the tracheoles that moves out (into tissues) during exercise so faster diffusion through the air to the gas exchange surface;

OR

Fluid in the end of the tracheoles that moves out (into tissues) during exercise so larger surface area (for gas exchange);

6. Body can be moved (by muscles) to move air so maintains diffusion / concentration gradient for oxygen / carbon dioxide;

Describe the processes involved in the absorption and transport of digested lipid molecules from the ileum into lymph vessels (5 marks).

1. Micelles contain bile salts and fatty acids/monoglycerides;

2. Make fatty acids/monoglycerides (more) soluble (in water)

OR

Bring/release/carry fatty acids/monoglycerides to cell/lining (of the iluem)

OR

Maintain high(er) concentration of fatty acids/monoglycerides to cell/lining (of the ileum);

(accept lipid/fat for fatty acid/ monoglyceride)

4. Fatty acids/monoglycerides absorbed by diffusion;

(reject if absorbed by facilitated diffusion)

(ignore if micelles themselves are being absorbed)

5. Triglycerides (re)formed (in cells);

(accept chylomicrons form)

6. Vesicles move to cell membrane;

(accept exocytosis for 'vesicles move')

Describe the role of enzymes in the digestion of proteins in a mammal (4 marks).

1. (Reference to) hydrolysis of peptide bonds;

2. Endopeptidase act in the middle of protein/polypeptide

OR

Endopeptidase produces short(er) polypeptides/ increase number of ends;

3. Exopeptidases act at end of protein/polypeptide

OR

Exopeptidase produces dipeptides/amino acids;

4. Dipeptidase acts on dipeptide/between two amino acids OR

Dipeptidase produces (single) amino acids;

Describe the process of the cardiac cycle (6 marks).

1. Diastole - blood flows into the atria;

2. AV valves open and blood flows into ventricles;

3. Pressure within ventricles is lower than that of the arteries so SL valves close ('dub' sound);

4. Atrial systole - atria contract, forcing any remaining blood into ventricles;

5. Ventricular systole - ventricles contract;

6. Pressure increases so AV valves close ('lub' sound);

7. (Once pressure exceeds that of the arteries) blood flows into arteries;

Explain how water from tissue fluid is returned to the circulatory system (4 marks).

1. (Plasma) proteins remain;

2. (Creates) water potential gradient

OR

Reduces water potential (of blood);

3. Water moves (to blood) by osmosis;

4. Returns (to blood) by lymphatic system;

Describe the cohesion-tension theory of water transport in the xylem (5 marks).

1. Water lost from leaf because of transpiration / evaporation of water (molecules) / diffusion from mesophyll / leaf cells;

OR

Transpiration / evaporation / diffusion of water (molecules) through stomata / from leaves;

2. Lowers water potential of mesophyll / leaf cells;

3. Water pulled up xylem (creating tension);

4. Water molecules cohere / 'stick' together by hydrogen bonds;

5. (forming continuous) water column;

6. Adhesion of water (molecules) to walls of xylem;

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

1. In source / leaf sugars actively transported into phloem;

osmosis;

2. By companion cells;

3. Lowers water potential of sieve cell / tube and water enters by osmosis;

4. Increase in pressure causes mass movement (towards sink /

root);

5. Sugars used / converted in root for respiration for storage.

Describe how a gene is a code for the production of a polypeptide (3 marks).

1. (Because) base/nucleotide sequence;

2. (In) triplet(s);

3. (Determines) order/sequence of amino acid sequence/primary structure (in polypeptide);

Because base sequences in triplets determines the order of the amino acid sequence in polypeptides

Compare and contrast the DNA in a eukaryotic cell with the DNA in the prokaryotic cell (5 marks).

Eukaryote v prokaryote

COMPARISONS:

1. Nucleotide structure is identical;

2. Nucleotides joined by phosphodiester bond;

OR

Deoxyribose joined to phosphate (in sugar, phosphate backbone);

3. DNA in mitochondria / chloroplasts same / similar (structure) to DNA in prokaryotes;

CONTRASTS:

4. (Associated with) histones/proteins v no histones/proteins;

5. Linear v circular;

6. No plasmids v plasmids;

7. Introns v no introns;

8. Long(er) v short(er);

Describe how mRNA is formed by transcription in eukaryotes (5 marks).

1. Hydrogen bonds (between DNA bases) break;

2. (Only) one DNA strand acts as a template;

3. (Free) RNA nucleotides align by complementary base pairing;

4. (In RNA) Uracil base pairs with adenine (on DNA)

OR

(In RNA) Uracil is used in place of thymine;

5. RNA polymerase joins (adjacent RNA) nucleotides;

6. (By) phosphodiester bonds (between adjacent nucleotides);

7. Pre-mRNA is spliced (to form mRNA)

OR

Introns are removed (to form mRNA);

Describe how a polypeptide is formed by translation of mRNA (6 marks).

1. (mRNA attaches) to ribosomes

OR

(mRNA attaches) to rough endoplasmic reticulum;

2. (tRNA) anticodons (bind to) complementary (mRNA) codons;

3. tRNA brings a specific amino acid;

4. Amino acids join by peptide bonds;

5. (Amino acids join together) with the use of ATP;

6. tRNA released (after amino acid joined to polypeptide);

7. The ribosome moves along the mRNA to form the polypeptide;

Define 'gene mutation' and explain how a gene mutation can have:

no effect on an individual

a positive effect on an individual (4 marks).

(Definition of gene mutation)

1. Change in the base/nucleotide (sequence of chromosomes/DNA);

For 4 marks at least one mark must be scored in each section of the answer.

2. Results in the formation of new allele;

(Has no effect because)

3. Genetic code is degenerate (so amino acid sequence may not change);

OR

Mutation is in an intron (so amino acid sequence may not change);

(accept description of 'degenerate', eg some amino acids have more than one triplet/codon)

4. Does change amino acid but no effect on tertiary structure;

5. (New allele) is recessive so does not influence phenotype;

(Has positive effect because)

6. Results in change in polypeptide that positively changes the

properties (of the protein)

OR

Results in change in polypeptide that positively changes a named protein;

7. May result in increased reproductive success

OR

May result in increased survival (chances);

Describe the process of crossing over and explain how it increases genetic diversity (4 marks).

1. Homologous pairs of chromosomes associate / form a bivalent;

2. Chiasma(ta) form;

3. (Equal) lengths of (non-sister) chromatids / alleles are exchanged;

4. Producing new combinations of alleles;

Bivalent: a pair of homologous chromosomes

Chaisma(ta): X-shaped points where two non-sister chromatids of homologous chromosomes are physically connected

REMEMBER🙆🏽‍♀️’Chaii’

Describe the process of photosynthesis (6 marks).

1. (LDR) in thylakoid membrane;

2. (Photolysis) - water converted to oxygen, 2e- and 2H+;

3. (Photoionisation) - electrons are excited and leave chlorophyll;

4. energy (from photoionisation) used in chemiosmosis;

5. (chemiosmosis) - protons pass through ATP synthase to form ATP

6. protons combine with the co-enzyme NADP to form NADPH;

7. (LIR) in stroma;

8. (Calvin cycle) - CO2 + RuBP combine with the enzyme Rubisco to make 2x GP;

9. GP reduced by ATP and NADPH to 2x triose phosphate;

Describe the process of respiration (6 marks).

1. (Glycolysis) phosphorylation (and splitting) of glucose using ATP;

2. Oxidation of triose phosphate to pyruvate;

3. Net gain of 2x ATP and NAD reduced;

4. (Link reaction) pyruvate is oxidised to form acetate and CO2 (NAD is reduced);

5. Acetate combines with co-enzyme A to form acetyl CoA;

6. (Krebs cycle) acetyl CoA combines with a 4-carbon molecule to form citrate;

7. Citrate loses CO2 and H2 to form a 4-carbon molecule that can combine with acetyl CoA;

8. (Oxidative phosphorylation) NADH and FADH reduce carrier proteins by donating their electrons;

9. Electrons travel along the ETC releasing energy;

10. This energy is used to pump protons through ATP synthase channel to produce ATP;

11. Oxygen is the final electron acceptor in the chain;

Explain why converting pyruvate to lactate allows the continued production of ATP by anaerobic respiration (2 marks).

1. Regenerates/produces NAD

OR

oxidises reduced NAD;

2. (So) glycolysis continues;

Describe the advantage of the Bohr effect during intense exercise (2 marks).

1. Increases dissociation of oxygen;

(Accept unloading/release/reduced affinity for dissociation)

2. For aerobic respiration at the tissues/muscles/cells

OR

Anaerobic respiration delayed at the tissues/muscles/cells

OR

Less lactate at the tissues/muscles/cells;

Describe the nitrogen cycle (6 marks).

1. (Ammonification) saprobionts feed on proteins / amino acids / urea;

2. And release ammonia into the soil;

3. (Nitrification) oxidation of ammonia to nitrite ions;

4. oxidation of nitrite ions to nitrate ions;

5. (Denitrification) anaerobic denitrifying bacteria convert soil nitrates into gaseous nitrogen;

6. (Nitrogen fixation) nitrogen-fixing bacteria convert gaseous nitrogen into nitrogen-containing compounds;