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CIE A-Level Biology 9700 Paper 2 Practice Paper (AS structured questions) – 60 marks, 1 hour 15 minutes, with answers and mark scheme
TutorZone original practice paper – CIE A-Level Biology 9700 Paper 2 (AS structured questions)
6 questions, 30 parts – 60 marks – 1 hour 15 minutes – calculator allowed
Paper comparison
| Paper | Time | Marks | Role |
|---|---|---|---|
| Paper 2 (AS structured questions) | 1 hour 15 minutes | 60 | this paper |
| Paper 1 (multiple choice) | 1 hour | 40 | AS multiple choice |
| Paper 4 (A Level structured questions) | 2 hours | 100 | A Level paper |
| Paper 5 (planning, analysis and evaluation) | 1 hour 15 minutes | 30 | practical paper |
Data and formulae
| Quantity | Value |
|---|---|
| Magnification | magnification = image size / actual size |
| Length conversion | 1 mm = 1000 micrometres; 1 micrometre = 1000 nm |
| Percentage change in mass | (final mass – initial mass) / initial mass x 100 |
| Cardiac output | cardiac output = stroke volume x heart rate |
| Mitotic index | mitotic index = number of dividing cells / total number of cells |
| Mean relative molecular mass of one amino acid | 110 |
Questions
Total: 40 marks · 2 h · calculator allowed
Q 1 · Enzymes: catalase kinetics, active sites and denaturation [10]
(a) In an investigation 2.0 g of potato tissue (a source of catalase) was added to 20 cm3 of hydrogen peroxide solution at five different concentrations at a constant 25 degrees C. The volume of oxygen collected in 60 seconds was 12 cm3 at 0.10 mol dm-3, 24 cm3 at 0.20, 36 cm3 at 0.30, 42 cm3 at 0.40 and 44 cm3 at 0.50. State the independent variable. [1]
(b) Use the data to calculate the rate of oxygen production, in cm3 s-1, at 0.10 and at 0.20 mol dm-3. [2]
(c) Describe the trend in the rate as the concentration of hydrogen peroxide increases, using the data. [2]
(d) Explain why the rate does not increase much above 0.40 mol dm-3. [2]
(e) Predict and explain the effect on the rate of oxygen production if the potato tissue is boiled for five minutes before the investigation. [3]
Q 2 · Cell structure and microscopy: magnification, actual size and organelle function [10]
(a) A mitochondrion measures 48 mm in an electron micrograph; its actual length is 0.75 micrometres. Calculate the magnification and state the formula you used. [2]
(b) On another micrograph taken at x5,000, a chloroplast measures 30 mm across. Calculate its actual diameter in micrometres. [2]
(c) With a light microscope at x400 the field of view is 0.20 mm across and 40 cells of average size lie along that diameter. Calculate the mean diameter of one cell in micrometres. [2]
(d) Explain why an electron microscope is needed to see the internal structure of a mitochondrion clearly. [2]
(e) State one function of a mitochondrion and explain how its structure helps it carry out that function. [2]
Q 3 · Cell membranes and transport: osmosis in potato cylinders, water potential and plasmolysis [10]
(a) Five potato cylinders of initial mass 2.00 g were placed in 0.0, 0.2, 0.4, 0.6 and 0.8 mol dm-3 sucrose solution for 40 minutes. Their masses afterwards were 2.36, 2.20, 2.02, 1.86 and 1.70 g. Explain why the cylinders are blotted dry before weighing. [1]
(b) Calculate the percentage change in mass for the cylinder in 0.80 mol dm-3 sucrose. [2]
(c) Describe and explain the trend in the change in mass as sucrose concentration increases, using the data. [2]
(d) Use the data to estimate the sucrose concentration at which the potato tissue has the same water potential as the solution, and explain your reasoning. [2]
(e) Explain, in terms of the cell, why the mass falls in 0.80 mol dm-3 sucrose and name the phenomenon. [3]
Q 4 · The cell cycle and mitosis: root tip squash, mitotic index and stage durations [10]
(a) A student made a squash of an onion root tip and counted 200 cells: 150 in interphase, 18 in prophase, 12 in metaphase, 8 in anaphase and 12 in telophase; the cell cycle lasts 24 hours. Explain why a root tip is used rather than a mature leaf. [2]
(b) Calculate the mitotic index of the root tip cells and express it as a percentage. [2]
(c) State one feature of metaphase that can be seen under the microscope. [2]
(d) Use the data to calculate how long anaphase lasts, in minutes, in a 24-hour cell cycle. [2]
(e) Explain why cells should not be chosen only because they are darkly stained or large when counting. [2]
Q 5 · Gas exchange and transport: cardiac output, the response to exercise and the Bohr shift [10]
(a) An athlete has a stroke volume of 70 cm3 and a heart rate of 75 beats per minute at rest, and 110 cm3 and 150 beats per minute during vigorous exercise. Define cardiac output. [1]
(b) Calculate the cardiac output at rest in both cm3 per minute and dm3 per minute. [2]
(c) Calculate the cardiac output during exercise and the percentage increase compared with rest. [2]
(d) Explain how exercising muscle obtains more oxygen, referring to the shift of the oxygen dissociation curve. [3]
(e) Explain why both the rate and the depth of breathing increase during exercise. [2]
Q 6 · Nucleic acids and protein synthesis: base composition, codons and translation [10]
(a) In a sample of double-stranded DNA, adenine makes up 28 per cent of all bases. Calculate the percentage of thymine, guanine and cytosine, and state the basis of your answer. [2]
(b) Explain how the two polynucleotide chains of DNA are held together and why the pairing rule matters. [2]
(c) The coding region of a gene contains 1215 nucleotides. Calculate the number of codons and the number of amino acids in the polypeptide produced. [2]
(d) State two roles of the ribosome in translation. [2]
(e) If one nucleotide is replaced by a different purine (a substitution mutation), explain the possible effects on the structure of the protein. [2]
Answers and mark scheme
| Q | Part | Answer | Mark points | marks |
|---|---|---|---|---|
| 1 | (a) | The concentration of the hydrogen peroxide solution (mol dm-3). | concentration of hydrogen peroxide, 1 mark | 1 |
| 1 | (b) | rate = volume divided by time: 0.10 gives 12 / 60 = 0.20 cm3 s-1 and 0.20 gives 24 / 60 = 0.40 cm3 s-1, which is a 100 per cent increase. | 0.10: 12 / 60 = 0.20 cm3 s-1, 1 mark 0.20: 24 / 60 = 0.40 cm3 s-1, 1 mark |
2 |
| 1 | (c) | Between 0.10 and 0.40 mol dm-3 the rate rises from 0.20 to 0.70 cm3 s-1, roughly in proportion to concentration; above 0.40 the rate levels off, reaching only 0.733 cm3 s-1 at 0.50 mol dm-3. | rise between 0.10 and 0.40 with two quoted values, 1 mark plateau above 0.40 with a quoted value, 1 mark |
2 |
| 1 | (d) | All the active sites of the enzyme molecules are occupied, so enzyme concentration is the limiting factor; adding more substrate provides no extra active sites, so the rate of enzyme-substrate complex formation cannot increase further. | all active sites occupied, 1 mark enzyme / active site availability is the limiting factor, 1 mark |
2 |
| 1 | (e) | The rate falls to almost zero. Boiling denatures the enzyme: the extra kinetic energy breaks the hydrogen bonds and other forces holding the tertiary structure, so the active site changes shape, substrate molecules can no longer form enzyme-substrate complexes and no catalysis occurs. | predicts a rate of about zero, 1 mark denaturation / tertiary structure disrupted, 1 mark active site shape changed so no enzyme-substrate complexes form, 1 mark |
3 |
| 2 | (a) | magnification = image size divided by actual size; 48 mm = 48000 micrometres, so the magnification is 48000 / 0.75 = x64,000. | formula image size divided by actual size, 1 mark x64,000, 1 mark |
2 |
| 2 | (b) | actual size = image size divided by magnification = 30 mm / 5,000 = 0.006 mm, which is 6.0 micrometres. | 30 / 5,000 = 0.006 mm, 1 mark conversion to 6.0 micrometres, 1 mark |
2 |
| 2 | (c) | mean diameter = 0.20 mm / 40 = 0.0050 mm = 5.0 micrometres. | 0.20 / 40 = 0.0050 mm, 1 mark 5.0 micrometres, 1 mark |
2 |
| 2 | (d) | Electrons have a much shorter wavelength than visible light, so the resolving power is much higher and closer structures can be distinguished; the folded inner membrane (cristae) therefore becomes visible. | shorter wavelength / higher resolution, 1 mark cristae or inner membrane detail resolved, 1 mark |
2 |
| 2 | (e) | It is the site of aerobic respiration and ATP production; the inner membrane is folded into cristae, which increases the surface area for the enzymes of the respiratory chain and ATP synthase, so ATP is produced faster. | ATP production / aerobic respiration, 1 mark cristae increase surface area for enzymes, 1 mark |
2 |
| 3 | (a) | Drops of solution on the surface would add extra mass, so the change in mass would be inaccurate. | surface water would add mass, 1 mark | 1 |
| 3 | (b) | (1.70 – 2.00) / 2.00 x 100 = -15.0 per cent, a loss of 15.0 per cent. | correct difference over 2.00 g, 1 mark -15.0 per cent, 1 mark |
2 |
| 3 | (c) | At low concentrations the mass increases (+18.0 per cent at 0.0 and +10.0 per cent at 0.2) because the solution has a higher water potential than the cells, so water enters by osmosis; at 0.6 and 0.8 mol dm-3 the mass falls (-7.0 and -15.0 per cent) as water leaves the cells. | gain at low concentration explained by osmosis, 1 mark loss at high concentration explained, 1 mark |
2 |
| 3 | (d) | About 0.425 mol dm-3. The percentage change is positive at 0.40 (+1.0) and negative at 0.60 (-7.0), so there is no net water movement where it crosses zero; interpolating gives 0.40 + 0.20 x (1.0 / 8.0) = 0.425 mol dm-3. | identifies the zero point between 0.40 and 0.60, 1 mark interpolates to 0.425 mol dm-3, 1 mark |
2 |
| 3 | (e) | The solution has a lower water potential than the cell sap, so water leaves the cell by osmosis through the partially permeable membrane; the vacuole shrinks and the protoplast pulls away from the cell wall. This is plasmolysis, so the mass of the tissue falls. | compares water potentials, 1 mark water leaves by osmosis through a partially permeable membrane, 1 mark names plasmolysis with vacuole shrinking, 1 mark |
3 |
| 4 | (a) | The root tip is meristematic tissue where cells are actively dividing, so cells at every stage of mitosis can be found; mature leaf cells are differentiated and no longer divide, so almost all would be in interphase. | meristem / actively dividing cells, 1 mark leaf cells differentiated and not dividing, 1 mark |
2 |
| 4 | (b) | mitotic index = cells in mitosis divided by total cells = 50 / 200 = 0.25, which is 25 per cent. | 50 over 200, 1 mark 0.25, 1 mark |
2 |
| 4 | (c) | The chromosomes, each of two sister chromatids, are lined up along the equator of the cell, and they are so condensed that individual chromosomes can be seen. | chromosomes at the equator, 1 mark chromosomes condensed and individually visible, 1 mark |
2 |
| 4 | (d) | anaphase is 8 / 200 = 0.040 of the cycle; 24 hours is 1440 minutes, so 0.0400 x 1440 = 57.6 minutes, about 58 minutes. | ratio and 1440 minutes, 1 mark 57.6 minutes, 1 mark |
2 |
| 4 | (e) | Choosing dark or large cells is subjective and would systematically over-represent dividing cells, whose condensed chromosomes stain more strongly, so the mitotic index would be too high; cells should be selected at random or systematically across the field so the sample is representative. | systematic bias towards dividing cells, 1 mark random or systematic sampling, 1 mark |
2 |
| 5 | (a) | The volume of blood pumped out by one ventricle each minute, measured in cm3 per minute or dm3 per minute. | volume per minute with a unit, 1 mark | 1 |
| 5 | (b) | cardiac output = stroke volume x heart rate = 70 x 75 = 5,250 cm3 per minute, which is 5.25 dm3 per minute. | 5,250 cm3 per minute, 1 mark = 5.25 dm3 per minute, 1 mark |
2 |
| 5 | (c) | 110 x 150 = 16,500 cm3 per minute (16.5 dm3 per minute); the increase is (16,500 – 5,250) / 5,250 x 100 = 214 per cent, about 3.14 times the resting value. | 16,500 cm3 per minute, 1 mark 214 per cent, 1 mark |
2 |
| 5 | (d) | Respiration in muscle releases more carbon dioxide, so blood pH falls; the affinity of haemoglobin for oxygen decreases, so the dissociation curve shifts to the right (the Bohr shift) and more oxygen is unloaded at the same partial pressure of oxygen. | more carbon dioxide / lower pH, 1 mark lower affinity, curve shifts right, 1 mark more oxygen unloaded at the same partial pressure, 1 mark |
3 |
| 5 | (e) | Rising carbon dioxide concentrations stimulate the respiratory centre in the medulla and the chemoreceptors, increasing ventilation; this maintains the concentration gradients of oxygen and carbon dioxide between alveoli and blood so that diffusion can supply the extra oxygen and remove the extra carbon dioxide. | carbon dioxide stimulates the respiratory centre, 1 mark maintains diffusion gradients, 1 mark |
2 |
| 6 | (a) | A pairs with T, so thymine is also 28 per cent; the remaining 44 per cent is shared equally between G and C because they pair with each other, giving 22 per cent each. | T = 28 per cent, 1 mark G = C = 22 per cent, 1 mark |
2 |
| 6 | (b) | The bases are joined by hydrogen bonds: two between A and T, three between G and C, so G-C pairs are stronger; the fixed complementary pairing allows each strand to act as a template during replication, so the genetic information is copied accurately. | hydrogen bonds, two for A-T and three for G-C, 1 mark template for accurate replication, 1 mark |
2 |
| 6 | (c) | codons = 1215 / 3 = 405; one of these is a stop codon that codes for no amino acid, so the polypeptide has 405 – 1 = 404 amino acids. | 405 codons, 1 mark 404 amino acids after removing the stop codon, 1 mark |
2 |
| 6 | (d) | It is the site where mRNA and tRNA bind and it moves along the mRNA reading the codons; it catalyses the formation of peptide bonds between adjacent amino acids, building the polypeptide chain. | binding site and movement along mRNA, 1 mark catalyses peptide bond formation, 1 mark |
2 |
| 6 | (e) | The codon may change so that a different amino acid is inserted, altering the primary structure and therefore the hydrogen and ionic bonds that hold the tertiary structure, which can distort the active site and make the protein non-functional; however the genetic code is degenerate, so the new codon may still code for the same amino acid or the change may be in a non-critical region, leaving the protein unaffected. | amino acid may change and alter tertiary structure or active site, 1 mark degenerate code or non-critical position, may have no effect, 1 mark |
2 |
Topic breakdown
| Question | Topic | Marks |
|---|---|---|
| 1 | Enzymes: catalase kinetics, active sites and denaturation | 10 |
| 2 | Cell structure and microscopy: magnification, actual size and organelle function | 10 |
| 3 | Cell membranes and transport: osmosis in potato cylinders, water potential and plasmolysis | 10 |
| 4 | The cell cycle and mitosis: root tip squash, mitotic index and stage durations | 10 |
| 5 | Gas exchange and transport: cardiac output, the response to exercise and the Bohr shift | 10 |
| 6 | Nucleic acids and protein synthesis: base composition, codons and translation | 10 |
Common lost marks
- Giving a difference instead of a rate or a percentage because the time or the starting value is not used as the divisor.
- Mixing millimetres and micrometres in magnification calculations, which puts the answer out by a factor of one thousand.
- Saying only that water moves into or out of the cell without comparing water potentials, which loses the mechanism marks.
- Using the number of dividing cells as the denominator of the mitotic index, or forgetting to convert 24 hours into 1440 minutes.
Frequently asked
What does Paper 2 test?
Paper 2 is the AS structured paper: 1 hour 15 minutes, 60 marks and about six compulsory questions built around data, graphs and experimental contexts. This paper follows the public syllabus with enzyme kinetics, microscopy and cell structure, osmosis and water potential, mitosis, gas exchange and transport, and nucleic acids and protein synthesis.
How many significant figures should I give?
The data are usually given to two or three significant figures, so write answers to the same and always include the unit (cm3 s-1, micrometres, per cent, minutes, dm3 min-1). Do not round intermediate values; round only at the final step.
Where are marks most often lost?
Describing a result without explaining the mechanism (for example writing that water enters the cell without comparing water potentials), calculating a percentage change without dividing by the starting value, mixing units in magnification questions, and failing to quote the data provided.
This paper is written by TutorZone tutors from the public syllabus; all questions and data are original and not endorsed by CIE.
The answer booklet gives mark-scheme points for study use only; please credit the source when reusing.
