AQA A Level Biology Paper 1 Revision Exam: Key Concepts and Practice Questions, Exams of Advanced Education

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AQA A LEVEL BIOLOGY PAPER 1

REVISION EXAM

Water removed from the reactants joining two molecules together forming a chemical bond Condensation The addition of water to the reactants to break a chemical bond between 2 molecules Hydrolysis

1. Add Benedict's reagent. 2. Heat the solution in a water bath for 5 minutes at 95 degrees Celsius. 3. Change from blue to brick red as CuO formed Test for Reducing Sugars (3) Smaller units from which larger molecules are made Monomer
2. Add 2cm³ of food sample then add 2cm³ of dilute HCl and heat.
3. Add 2cm³ of NaHCO3 then do test for reducing sugars. Non-Reducing Sugars (2) Add drops of iodine to starch solution. Colour change to blue-black Test for Starch (1)
4. Mix Test solution with ethanol.
5. Shake for 1 minute then add water.
6. Cloudy white emulsion Test for Lipids (3)
7. Obtain equal volumes of test solution and NaOH then add a few drops of biuret solution (dilute copper (II) sulphate solution).
8. Colour change to mauve/purple Test for Proteins (2)
9. Very high resolution.
10. Needs thin and dead specimen.

3. Artefacts can occur (remnant left on object during prep, such as air bubbles)
4. Uses magnets to focus on specimen
5. Uses electrons fired at sample.
6. Is not in colour Transmission Electron Microscope (5)
7. Inhibitor is similar in shape to substrate so it impermanently binds to the active site.
8. Prevents ESC from forming, slowing rate Competitive inhibition (2)
9. Molecule will bind to allosteric site.
10. Binding causes a change in active site.
11. Permanently preventing further ESC. Non-competitive inhibition (3)
12. DNA helicase breaks the hydrogen bonds between the base pairs
13. 2 single strands formed as the double helix "unzips".
14. Free DNA nucleotides in the nucleoplasm bond to the complementary bases on the strand.
15. DNA polymerase forms phosphodiester bonds between adjacent DNA nucleotides via condensation reaction with the hydrolysis of ATP, forming the phosphate backbone DNA Replication: Semiconservative (4)
16. ATP stores or releases only a small amount of energy at a time, so no energy is wasted as heat.
17. Small and soluble so easily transported
18. Easily broken down, so energy is released instantaneously
19. Can be quickly re-made
20. Can make other molecules more reactive via phosphorylation
21. ATP can't pass out of cell, so the cell always has an immediate supply of energy. Describe 6 properties of ATP that make it a good energy source. (6) Prevents the cell from drying out. Allows bacteria to stick to each other Slime capsule (2)

Used for attachment of a cell to a surface Fimbria Involved in bacterial conjugation Pilli Invagination of cell membrane. Site of cell respiration (prokaryotes) Mesosome The ability to distinguish two points apart Resolution

1. Lower resolution than TEM
2. 3D image
3. Does not require thin samples Scanning Electron Microscope The mass of organelles at the bottom of the test tube after centrifugation. Pellet Cold. Low temperature slows enzyme activity, minimising self digestion by reducing metabolic rate. Isotonic. Salt and sugar concentration kept the same, minimising organelle size change due to osmosis. Buffered. Minimum changes in pH, so prevents enzymes in organelles denaturing. Solution Required for cell fractionation (6 Marks)
4. Homogenisation. Breaking up cells by blending the sample to create a homogenate.
5. Filtering. Filtering the large, unwanted sil, producing the filtrate.
6. Ultracentrifugation. Spin in a centrifuge so components separate out by weight. Heavier near the bottom of the tube.
7. Supernatant is removed and spun again at higher speed. Separation of Organelles From The Cells (4) The solution not including the pellet at the bottom of the test tube after centrifugation. Supernatant

1. Cell wall forms, dividing the two genetically identical daughter cells.
2. Same circular DNA. Binary Fission 3 Nuclei, Chloroplasts, Mitochondria, Lysosomes, Endoplasmic Reticulum, Ribosomes Order of Organelles (Pass me a taco chief) Mitosis acronym Cell grows and carrys out its normal function Interphase Cells grow to normal size. Organelles replicate and genes are expressed to make proteins needed. Interphase G DNA and histones replicated. Interphase S Spindle fibres are made Interphase G
3. DNA winds up making chromosones from chromatin.
4. Centrioles appear at opposite poles of the cell. 3. Nucleolus disappears Prophase (3)
5. Nuclear envelope disappears.
6. Chromosomes align along the equator of the cell.
7. Spindle fibres connect centrioles to chromosomes Metaphase (3)
8. Spindle fibres contract pulling daughter chromosomes to opposite poles of the cell. Anaphase (1)

1. Spindle fibres disperse.
2. Nuclear envelope forms.
3. Chromatids uncoil to chromatin Telophase (3)
4. Cytoplasm constricts separating the cells into two.
5. Membrane forms creating two new genetically identical daughter cells. Cytokinesis (2) number of cells in mitosis/total number of cells Mitotic index
6. Circular DNA replicates and both copies attach to the cell membrane.
7. Plasmids also replicate. Binary fission 1 (2) Cell membrane grows between the two DNA molecules and pinches them inwards dividing the cell into 2. Binary Fission 2 (1)
8. Attach to host cell via attachment protein and inject nucleic acid into the cell.
9. This gives the instructions to construct the virus Virus Replication 1 (2)
10. The virus is then assembled and leaves the cell, taking the phospholipid bilayer with it.
11. This creates holes in the cell and kills it. Virus Replication 2 (2) Fluid: All the components can move around. Mosaic: Many different components all fit together Fluid mosaic Structure [Fluid][Mosaic] (2) The passive transport of large molecules such as amino acids and sugars, but they require integral proteins to pass through. Facilitated Diffusion The passive movement of particles from a high concentration to a region of low concentration, down a concentration gradient.

Diffusion

1. Add antibody that is specific to antigen. After leaving them to bind, wash the surface to remove unattached antibodies.
2. Add a second antibody with an enzyme attached.
3. Second antibody binds with first antibody
4. Add colourless substrate of enzyme, which the enzyme will act upon to give a coloured product.
5. The intensity of the colour is relative to the amount of antigen present. ELISA Test (Enzyme linked Immunosorbent Assay) (5) Small, non-polar molecules (excluding water) freely diffuses in and out of cells through gaps between phospholipids. Simple Lipid Diffusion The passive movement of water molecules across a partially permeable membrane from a high water potential to a low water potential, down a concentration gradient. (Distilled water = 0Ψ) Osmosis Equal concentration solution to the cell Isotonic Solution of higher concentration to cell. Hypertonic Solution of a lower concentration to cell. Hypotonic The movement of substances across a cell membrane through a carrier protein against a concentration gradient, from a low to high concentration with the expenditure of ATP. Active transport
6. The simultaneous transport of two different substances through one carrier protein.
7. One substance (usually an ion) moves down its concentration (passive) whilst the other substance is transported against its concentration gradient (active transport)

Co-transport (2) A molecule, usually a glycoprotein or protein that stimulates an immune response. Antigen A protein molecule that can bind specifically to an antigen Antibody

1. Ingestion (phagocytosis).
2. Antigen Presentation.
3. Clonal Expansion.
4. Differentiation (T-cells and cell mediated immunity).
5. Differentiation (B-cells and antibody mediated immunity) Stages of response to a non-self antigen found in body fluid Vaccination of a significant portion of a population providing a measure of protection for the individuals who have not developed immunity/have not been vaccinated. Herd immunity Antibodies produced from cloned plasma B lymphocytes Monoclonal Antibodies
6. Hydrogen bonds between DNA base pairs are broken
7. One DNA strand acts as a template
8. Free RNA nucleotides align by complementary base pairing (uracil is used instead of thymine)
9. RNA polymerase forms phosphodiester bonds between adjacent RNA nucleotides joining them together.
10. Pre-mRNA is spliced, removing introns then joining the exons.
11. mRNA moves out of nucleus via nuclear pore Polypeptide synthesis (In nucleus) (6) [Transcription] Adaptation. How it works. Example. Repeat twice for 6 marks 6 marker layout for Xerophytic plants
12. mRNA attaches to ribosome
13. tRNA molecule with complementary anticodon and desired amino acid

moves to the ribosome and binds to the codon.

3. Amino acids join by peptide bonds with the use of ATP
4. tRNA released after amino acid joined to polypeptide
5. The ribosome moves along the mRNA to form the polypeptide Polypeptide synthesis (In ribosomes) (5) [Translation]
6. Substrate binds to active site
7. ESC forms
8. Active site changes shape slightly distorting hydrogen bonds in the substrate.
9. Reduces activation energy Describe the Induced fit model of enzyme action and how an enzyme acts as a catalyst. (3 marks) A change in the nucleotide base sequence. Mutations (Normal = BEAST) Substitution in a codon turns it into a stop codon Substitution (Nonsense) Causing a different amino acid to be coded for Missense (FEAST) Results in a codon coding for the same amino acid creating degenerate code. Silent Nucleotide is gained from a DNA strand, creating a right frameshift. Addition (BREAST) A nucleotide is lost so frameshift to the left. Deletion (BEST) When one or more bases are repeated so shift to the right. Duplication (BEEAST) A group of bases become separated from the DNA, then join back but inverted. Inversion (BEATS) Breaks the end peptide bond

Exopeptidase Breaks the middle peptide bonds Endopeptidase Tidal volume x Ventilation rate pulmonary ventilation =

1. Proteins remain in the plasma
2. This reduces the water potential
3. Water moves to blood by osmosis
4. Returns to blood via lymphatic system Tissue Fluid is formed from blood at the arteriolar end of a capillary bed. Explain how water from tissue fluid is returned to the circulatory system. (4)
5. Co-transport
6. Hydrolysis of ATP
7. Na+ and H+ bind to carrier protein
8. Protein changes shape and moves Na+ and H+ across membrane. 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. NHE actively transports one sodium ion into the cell in exchange for one proton (hydrogen ion) out of the cell. Use your knowledge of transport across cell membranes to suggest how NHE3 does this. (4)
9. High salt results in low water potential in tissue fluid
10. Less reabsorbed back into capillaries at venule end by osmosis 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)
11. To digest protein
12. So they can absorb amino acids for growth OR destroy toxins. Some proteases are secreted as extracellular enzymes by bacteria. Suggest one advantage to a bacterium of secreting an extracellular protease in its natural environment. Explain your answer. (2)

1. In metaphase II, homologous chromosomes are randomly aligned along the equator.
2. They are randomly pulled to different poles of the cell Variation in Meiosis [Independent assortment of homologous chromosomes] (2)
3. External intercostals contract
4. Diaphragm contracts
5. Pressure decreases in ribcage as thorax volume increases.
6. Air moves from high to low pressure (Forced Expiration is just the opposite of this) Inspiration (4)
7. In meiosis I/II, chromatids of each pair become twisted and form bivalents.
8. Chiasmata form causing tension in parts of chromatin, thus equal parts if chromatids break off and rejoin, forming recombinants chromatids with a new allele combination. Variation in Meiosis (Crossing over) (2)
9. The rate of diffusion is directly proportional to the membrane surface area
10. Concentration gradient is inversely proportional to the membrane thickness. Fick's Law (2) Domain, kingdom, phylum, class, order, family, genus, species Taxonomy order (King phillip came over for good spaghetti) The number of different species in a community Species richness The relationship between the number of species in a community and the number of individuals in each species Index of diversity

The number of different alleles in each gene. Genetic Diversity A group of similar organisms that can breed and produce fertile offspring. Species Anatomical, physiological, behavioural Types of adaptations (3) Alleles towards the mean more likely to survive. Environment conditions are stable Human birth weight Stabilising Selection Transport of sap (a very concentrated solution of dissolved sucrose and amino acids) in the phloem vessels. Translocation stroke volume x heart rate Cardiac output

1. Phagosome fuses with lysosome;
2. Virus destroyed by lysozymes
3. Antigen (from virus) are displayed on the cell membrane Describe how phagocytosis of a virus leads to presentation of its antigens. (2)
4. Transpiration.
5. Water leaves the leaf creating low pressure.
6. Water is sucked up from a high pressure (roots) to a low pressure.
7. Column of water does not break because of high tensile strength.
8. Strong lignin walls preventing xylem vessels from collapsing. Xylem movement (5)
9. Sucrose actively transported to phloem vessels decreasing water potential.
10. Ions actively transported out of xylem vessel, increasing water potential.
11. Water diffuses from xylem to phloem.
12. Causes hydrostatic pressure so sap in phloem pushed downwards (mass

flow).

5. Sucrose is unloaded in the roots via active transport or diffusion.
6. Ions pumped into xylem via active tranport, decreasing water potential.
7. Osmosis from phloem to xylem
8. Water and ions made up by cohesion tension. Mass flow Hypothesis (8)
9. The base sequence of DNA
10. The base sequence of mRNA
11. The amino acid sequence of proteins State three comparisons of genetic diversity that the scientists used in order to generate classification Y. (3)
12. A reactant in hydrolysis, photosynthesis
13. High heat capacity so buffers changes in temperature.
14. Large latent heat of vaporisation so provides a cooling effect through evaporation.
15. Cohesion between water molecules so supports columns of water in plants.
16. Cohesion between water molecules so produces surface tension supporting small organisms Explain five properties that make water important for organisms. (5)
17. Tracheoles have thin walls so short diffusion distance to cells
18. Large number of tracheoles so short diffusion distance to cells;
19. Large number of tracheoles so large surface area (for gas exchange)
20. Body can be moved (by muscles) to move air so maintains diffusion/concentration gradient for oxygen/carbon dioxide

The adult damselfly uses a tracheal system for gas exchange. Explain three ways in which an insect's tracheal system is adapted for efficient gas exchange. (3) Glucose is high concentration in the ileum, low in the epithelial, but we want to increase the rate of transport. Sodium-potassium pump actively transports in a K+ to transport out Na+ creating a concentration gradient. Na+ and glucose pass through a co transporter in the phospholipid bilayer of the epithelial cell. Co-Transport (Not mark scheme specific, only here as a guidance) Mitosis:

1. Division of body cells
2. results in 2 diploid daughter cells
3. genetically identical
4. used for growth, repair, and cell replacement Meiosis:
5. germline cells divide into sex cells
6. 4 haploid daughter cells
7. genetic variation due to:
8. crossing over
9. random orientation
10. used for sexual reproduction Similarities
11. Start as diploid cells
12. Same basic stages prophase, metaphase, anaphase, telophase Compare and contrast mitosis and meiosis [Differences Mitosis(4), Meiosis(6)] [Similarities (2)]
13. A condensation reaction joins monomers by forming a chemical bond and releases water.
14. A hydrolysis reaction breaks a chemical bond between monomers by adding water.
15. Example: A-glucose —> Starch, glycogen. Alpha 1,4 Glycosidic bonds formed
16. Example: B-glucose —> cellulose beta 1,4 glycosidic bonds

Describe the chemical reactions involved in the conversion of polymers to monomers and monomers to polymers. Give two named examples of polymers and their associated monomers. (5)

1. Micelles include bile salts and fatty acids.
2. Makes fatty acids more soluble in water
3. Brings fatty acids towards the lining
4. Fatty acids absorbed by diffusion Describe the roles of micelles in the absorption of fats into the cells lining the ileum (3).
5. Hydrolysis
6. Of glycosidic bonds
7. Starch to maltose by amylase
8. Maltose to glucose by maltase
9. Membrane-bound (disaccharidase/maltase) (Add context to where such digestion happens in relation to digestive system) Describe the complete digestion of starch by a mammal. (4)
10. Nuclear envelope and pores
11. Chromatin
12. Nucleolus
13. Holds genetic material to code for polypeptides
14. DNA replication occurs here
15. Production of mRNA/tRNA
16. Production of ribosomes Describe the structure and the function of the nucleus. (4) [Structure (3)] [Function(4)]
17. Both polysaccharides
18. Both contain glycosidic bonds (between monomers); 3 Both contain carbon, hydrogen and oxygen
19. Starch has a-glucose and cellulose has B- glucose;
20. Starch molecule is helical whereas coiled and cellulose

molecule is straight;

6. Starch molecule is branched and cellulose is unbranched Compare and contrast the structure of starch and the structure of cellulose. (6)
7. Sucrose actively transported/co transported with H+ into phloem (cell)
8. By companion cells;
9. Lowers water potential in phloem and water enters from xylem by osmosis
10. Produces higher hydrostatic pressure
11. Mass flow to respiring cells
12. Unloaded from phloem by active transport Describe the transport of carbohydrate in plants. (5)
13. Muscle contracts
14. Lumen constricts Explain how an arteriolar can reduce the blood flow into capillaries. (2) 1 and 2. Adenine, ribose and 3 phosphates
15. Condensation reaction 4 Via ATP synthase Describe how an ATP molecule is formed from its component molecules. (4)
16. Increasing Pi concentration, more ESCs are formed.
17. At or above 40 mmol dm-3 all active sites occupied. Explain the change in ATP concentration with increasing inorganic phosphate concentration. (2)
18. Larger organisms have a smaller surface area:volume ratio
19. Overcomes long diffusion pathway Explain the advantage for larger animals of having a specialised system that facilitates oxygen uptake. (2)

1. Blood and water flow in opposite directions
2. Concentration gradient maintained along length of lamella Explain how the counter-current principle allows efficient oxygen uptake in the fish gas exchange system. (2)
3. In fish, blood leaving (V) has more oxygen than water leaving (E)
4. But in humans, blood leaving (V) has less oxygen than air leaving (E)
5. So fish remove a greater proportion from the oxygen they take in. A student studied Figure 3 and concluded that the fish gas exchange system is more efficient than the human gas exchange system. Use Figure 3 to justify this conclusion. (3)
6. Disinfect surfaces and instruments
7. Disinfect hands Describe two precautions the student should take when clearing away after the dissection. (2) Feature: DNA Difference: DNA is circular in chloroplast but nuclear DNA is linear Give one feature of the chloroplast that allows protein to be synthesised inside the chloroplast and describe one difference between this feature in the chloroplast and similar features in the rest of the cell. (2) [Feature][Difference]
8. One glycerol and three fatty acids
9. Condensation reactions and removal of three molecules of water
10. Ester bonds formed Describe how a triglyceride molecule is formed (3)
11. Spin supernatant at very high speeds
12. CENP-W separates depending on molecular mass Explain how ultracentrifugation separates CENP-W from other molecules. (2) Sucrose

What is the only non-reducing dissacharide sugar? A small repeating unit from which larger polymers are made. What is a monomer? (1)

1. Filter and dry the precipitate
2. Find mass Suggest another method other than colorimetry, that a student could use to measure the quantity of reducing sugar in a solution. (2)
3. Insoluble in water so doesn't affect water potential
4. Branched alpha helix so makes molecule compact
5. Branched for large SA for fast hydrolysis. Starch is a carbohydrate often stored in plant cells. Describe and explain 3 features of starch that make it good storage molecule.
6. Polysaccharide of a-glucose
7. Branched structure Describe the structure of glycogen. (2) Maltose - Glucose + Glucose Lactase - Glucose + Galactose Sucrose - Glucose + Fructose Maltose, Lactose, Sucrose [Disaccharide - Monosaccharide + Monosaccharide]
8. Lowers activation energy
9. Induced fit causes active site of enzyme to change shape
10. So ESC causes bonds to break/form. Explain how the active site of an enzyme causes a high rate of reaction. (3)
11. Condensation reaction
12. Between NH2 and COOH Describe how a peptide bond is formed between two amino acids to form a dipeptide (2)
13. Condensation reaction
14. Between phosphate and deoxyribose
15. Catalysed by DNA polymerase

Describe how a phosphodiester bond is formed between two nucleotides within a DNA molecule. (3)

1. Joins adjacent DNA nucleotides
2. Catalyses condensation reactions and formation of phosphodiester bonds between adjacent DNA nucleotides Describe the role of DNA polymerase in the semi-conservative replication of DNA. (2) Watson and Crick Name the two scientists who proposed models of the chemical structure of DNA and of DNA replication. (1)
3. DNA has antiparallel strands
4. Shape of the nucleotides are aligned differently
5. Enzymes have active sites with specific shapes
6. Only substrates with complementary shape can bind with the active site of DNA polymerase. Use figure 2 and your knowledge of enzyme action to explain why the arrows point in opposite directions. (4)
7. Na+ ions lower the water potential of blood
8. Water would move into the blood by osmosis from tissue fluid
9. Increasing the blood volume A high concentration of sodium in the blood can affect blood volume and cause hypertension. Use your knowledge of water potential to suggest how high sodium concentrations in the medicines taken could affect blood volume. (3) Iron ions
10. Haemoglobin binds/associates with oxygen Sodium ions
11. Co-transport of glucose into cells
12. Because sodium moved out by active transport via Na - K pump Phosphate ions

4. Joins nucleotides in the backbone of DNA
5. Used to produce ATP
6. Phosphorylates other compounds (glucose in glycolysis) making them more reactive Describe the roles of iron ions, sodium ions, and phosphate ions in cells. (6) [Iron 1][Sodium 2-3][Phosphate 4-6]
7. Lysosomes
8. Fuse with vesicle
9. Releases hydrolytic enzymes (lyzozymes) U. marinum cells ingest bacteria and digest them in the cytoplasm. Describe the role of one named organelle in digesting these bacteria.
10. Resolution too low
11. Because wavelength of light is too long Explain why it is not possible to determine the identity of the structures labelled X using an optical microscope. (2)
12. DNA in nucleus is code for protein
13. Ribosomes on rough endoplasmic reticulum produce protein
14. Mitochondria produce ATP for protein synthesis
15. Golgi apparatus package/modify protein
16. Golgi vessicles transport proteins and undergo exocytosis when they fuse with cell surface membrane Eukaryotic cells produce and release proteins. Outline the role of organelles in the production, transport and release of proteins from eukaryotic cells. Do not include details of transcription and translation in your answer. (5) 70s ribosomes Name an organelle found in both a chloroplast and a prokaryotic cell. (1)
17. TEM use electrons and optical use light
18. TEM allows a greater resolution

3. So with TEM smaller organelles can be observed
4. TEM view only dead specimens whereas optical can view live specimens
5. TEM does not show colour and optical does
6. TEM requires thinner specimens
7. TEM requires a more complex and more time consuming preparation
8. TEM focuses using magnets and optical uses glass lenses Contrast how an optical microscope and a transmission electron microscope work and contrast the limitations of their use when studying cells. (8)
9. Break open cells and produce homogenate;
10. Remove unbroken cells and larger debris; (Ultracentrifugation) Explain why the biologist used a blender and then filtered the mixture. (2) Cytoplasm of red blood cell is filled with haemoglobin. No organelles are visible in the cytoplasm of this red blood cell. Suggest why. (1)
11. Membrane has phospholipid bilayer
12. Stain binds to phosphate or glycerol
13. On inside and outside of membrane. 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. (3)
14. Add drop of water to glass slide
15. Obtain thin section of plant tissue and place on slide
16. Stain with iodine in potassium iodide.
17. Lower cover slip using mounted needle. Describe how you could make a temporary mount of a piece of plant tissue to observe the position of starch grains in the cells when using an optical light microscope. (4)
18. Examine over 20 cells
19. Repeat the count to ensure figures are correct

Describe and explain what the student should have done when counting cells to make sure that the mitotic index he obtained for this root tip was accurate. (2)

1. Two chromosomes that carry the same genes
2. But have different alleles What is a pair of homologous chromosomes? (2) Region where mitosis occurs Why should you take cells from the root tip? (1)
3. To make the sample thinner
4. To allow light through Why should you firmly squash the root tip using a mounting needle? (2)
5. Phospholipid bilayer allows diffusion of nonpolar, lipid-soluble substances
6. Carrier proteins allow active transport
7. Channel/carrier proteins allow facilitated diffusion/co-transport
8. Number of carrier/channel proteins determines how much movement
9. Shape of carrier proteins determines which substances are pumped in The movement of substances across cell membranes is affected by membrane structure. Describe how. (5)
10. There is a significant difference
11. As there is a less than 0.05 probability that the difference is due to chance P-value MS
12. Genetic material (DNA/RNA)
13. Enzyme
14. Matrix
15. Capsid
16. Attachment protein
17. Lipid envelope Label the Diagram (6)

1. Attachment proteins attach to receptors on helper T cell
2. RNA enters cell
3. Reverse transcriptase converts RNA to DNA which is inserted into the nucleus
4. DNA is transcribed into mRNA and this mRNA is translated to form viral protein, capsid and enzymes
5. Virus assembled and released from cell Describe how HIV is replicated. (5)
6. Tumour cells
7. Cells from transplants Give two types of cell, other than pathogens, that can stimulate an immune response. (2)
8. Less antibodies produced
9. Because HIV destroys helper T cells
10. So few B cells stimulated Explain how HIV affects the production of antibodies when AIDS develops in a person. (3)
11. The scientists could identify proteins that derive from the genetic code
12. They could then identify potential antigens to use in the vaccine Determining the genome of the viruses could allow scientists to develop a vaccine. Explain how. (2)
13. Bind to antigen
14. Antibodies cause agglutination Describe and explain the role of antibodies in stimulating phagocytosis. Do not include details about the process of phagocytosis. (2)
15. Bacterium binds to surface protein on a specific B cell.
16. B cell divides by mitosis and differentiates into plasma cells that produce specific antibodies.
17. Division stimulated by cytokines
18. Plasma cells release antibodies
19. Some B cells become memory cells
20. Memory cells produce antibodies faster

In the UK, children are vaccinated against this disease. Describe how vaccination can lead to protection against bacterial meningitis. (6)

1. Active involves memory cells, passive does not
2. Active involves production of antibody by plasma cells and memory cells
3. Passive involves antibody introduced into body from outside such as antivenom
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, passive is fast acting. Describe the difference between active and passive immunity. (6)
7. Micelles contain bile salts and fatty acids
8. Make fatty acids more soluble in water
9. Fatty acids absorbed by simple lipid diffusion
10. Triglycerides reformed in cells
11. Vesicles move to cell membrane Describe the processes involved in the absorption and transport of digested lipid molecules from the ileum into lymph vessels. (5)
12. Structure is determined by position of amino acid and R group
13. Primary structure is sequence of amino acids
14. Secondary structure formed by hydrogen bonding between amino acids (alpha helix and B-pleated sheet)
15. Tertiary structure formed by interactions between R groups such as hydrogen bonds, disulphide bridges and ionic bonds
16. Quaternary structure contains >1 polypeptide chain and/or prosthetic. Describe how the structure of a protein depends on the amino acids it contains. (6)
17. Peptide bonds are hydrolysed
18. Endopeptidase act in the middle of polypeptide
19. Exopeptidases act at end of polypeptide
20. Dipeptidase acts on dipeptide producing single amino acids Describe the role of enzymes in the digestion of proteins in a mammal. (4)

1. Modifies triglycerides
2. Combines triglycerides with proteins
3. Packaged for release to undergo exocytosis Suggest how the Golgi apparatus is involved in the absorption of lipids. (3)
4. Thin so small diffusion pathway
5. Flat so large surface area:volume ratio Explain how two features of the body of the tubifex worm allow efficient gas exchange. (2)
6. More oxygen dissociation
7. By decreasing blood pH Describe and explain the effect of increasing carbon dioxide concentration on the dissociation of oxyhaemoglobin. (2)
8. Binding of first oxygen changes the quaternary structure of haemoglobin
9. Creates another binding site on Fe Binding of one molecule of oxygen to haemoglobin makes it easier for a second oxygen molecule to bind. Explain why. (2)
10. High rate of transpiration
11. Water lost through stomata causing high tension in xylem
12. Causes less water movement from xylem to phloem Phloem pressure is reduced during the hottest part of the day. Use information in the graph above along with your understanding of transpiration and mass flow to explain why. (3)
13. Water lost from leaf because of transpiration
14. Lowers water potential of leaf cells
15. Water pulled up xylem creating tension
16. Water molecules cohere together by hydrogen bonds
17. This forms a continuous water column
18. Adhesion of water molecules to walls of xylem Describe the cohesion-tension theory of water transport in the xylem. (6)
19. Because base nucleotide sequence is in triplets
20. Determines order of amino acid sequence in polypeptide

Describe how a gene is a code for the production of a polypeptide. Do not include information about transcription or translation in your answer. (2) Triplet base sequence that codes for a polypeptide Define exon. (1)

1. Triplets code for same amino acid (degenerate code)
2. Occurs in introns Not all mutations in the nucleotide sequence of a gene cause a change in the structure of a polypeptide. Give two reasons why. (2)
3. Free RNA nucleotides bind to complementary base pairs
4. Phosphodiester bonds form
5. By action of RNA polymerase Describe how mRNA is produced from an exposed template strand of DNA. Do not include DNA helicase or splicing in your answer. (2) The full range of different proteins that a cell is able to produce at a given time What is the proteome of a cell? (1)
6. RNA
7. Polypeptides What are ribosomes made of? (2) A factor that increases the rate of mutations Mutagenic agent
8. Same genus
9. Same common ancestor There are many different species of field mouse in Europe. Using a phylogenetic classification, all of these species have names that start with Apodemus. What information does this give about field mice? (2)
10. Index of diversity also measures population size of each species
11. So useful because may be many of some species and less of others. Explain why it is more useful to calculate an index of diversity than to record species richness. (2)