Cambridge A Level Biology 9700 — 2018 Oct/Nov Paper 1 · Variant 3
9700/13/O/N/18 · 40 questions · 40 marks · ≈45 min
The question paper and its mark scheme, free to read here and free to download. This is Cambridge’s own paper, exactly as it was sat.
Question paper20 pages




















Mark scheme3 pages
Answers below. Sit the paper first if you are practising.



Questions as text
Q1 · Which statement explains why it is necessary to use an electron microscope to see the…
1 Which statement explains why it is necessary to use an electron microscope to see the cristae of a mitochondrion? A The magnification of the electron microscope is greater than that of the light microscope. B The membranes of the cristae are separated by a distance greater than 200 nm. C The resolution of a student microscope using daylight is too low. D The wavelength of an electron beam is longer than the wavelength of light.
Mark scheme: C
Q2 · A prokaryotic cell which is 0.25 µm in diameter, is magnified 50 000 times on an electron…
2 A prokaryotic cell which is 0.25 µm in diameter, is magnified 50 000 times on an electron micrograph. How big will its diameter be in the electron micrograph? A 1.25 × 10–1 mm B 1.25 × 100 mm C 1.25 × 101 mm D 1.25 × 102 mm
Mark scheme: C
Q3 · When making measurements in experiments, which methods could have parallax errors?
3 When making measurements in experiments, which methods could have parallax errors? 1 using a calibrated eyepiece graticule to measure length 2 using a measuring cylinder to measure volume 3 using a ruler to measure length of a shoot A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
Mark scheme: D
Q4 · The diagram is a drawing from an electron micrograph of a typical animal cell
4 The diagram is a drawing from an electron micrograph of a typical animal cell. X What is the function of the membrane system labelled X? A lipid synthesis only B protein synthesis and transport C protein synthesis only D protein transport only
Mark scheme: B
More questions on Cells as the basic units of living organisms
Q5 · The diagram is a drawing from an electron micrograph of a typical animal cell
5 The diagram is a drawing from an electron micrograph of a typical animal cell. 1 2 3 Where would nucleic acid be found? A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
Mark scheme: A
More questions on Cells as the basic units of living organisms
Q6 · The diagram shows some similarities between chloroplasts, mitochondria and typical…
6 The diagram shows some similarities between chloroplasts, mitochondria and typical prokaryotes. 1 2 3 4 5 Which row is correct? 1 2 3 4 5 A chloroplasts circular DNA mitochondria linear DNA prokaryotes B mitochondria 70S ribosomes chloroplasts linear DNA prokaryotes C mitochondria linear DNA chloroplasts 70S ribosomes prokaryotes D prokaryotes 70S ribosomes mitochondria 70S ribosomes chloroplasts
Mark scheme: D
More questions on Cells as the basic units of living organisms
Q7 · A student carried out tests on the same volume of four different solutions to investigate…
7 A student carried out tests on the same volume of four different solutions to investigate the presence of protein, starch and reducing sugar in each. The results are shown in the table. solution Benedict’s solution biuret reagent iodine solution 1 blue purple yellow 2 orange pale purple blue-black 3 orange purple yellow 4 red pale blue yellow Which conclusion can be drawn from these results? A Solution 1 has a lower protein and lower reducing sugar content than solution 2. B Solution 2 has less starch compared to solutions 1, 3 and 4. C Solution 3 has the most protein and the least starch. D Solution 4 has a high reducing sugar content and no starch.
Mark scheme: D
Q8 · Which words correctly complete the description of amylose?
8 Which words correctly complete the description of amylose? The ««1«« amylose is a ««2«« made up of ««3«« of the ««4«« glucose. 1 2 3 4 A macromolecule monomer polymers disaccharide B macromolecule monomer polymers monosaccharide C polysaccharide polymer monomers disaccharide D polysaccharide polymer monomers monosaccharide
Mark scheme: D
Q9 · Which statements about glycosidic bonds are correct?
9 Which statements about glycosidic bonds are correct? 1 They occur by condensation reactions between the OH groups of two molecules. 2 They occur only between glucose molecules. 3 The bonds can only be formed between carbon 1 and carbon 4 or carbon 1 and carbon 6 of adjacent molecules. 4 Hydrolysis of the bonds releases energy. A 1, 2 and 3 B 1 and 4 C 2 and 4 D 3 and 4
Mark scheme: B
Q10 · What is the general formula for glycogen?
10 What is the general formula for glycogen? A (C5H10O5)n B (C5H10O6)n C (C6H10O5)n D (C6H12O6)n
Mark scheme: C
Q11 · Which statements are correct?
11 Which statements are correct? 1 In a triglyceride molecule the bonds between the fatty acids and the glycerol molecule are called ester bonds. 2 Triglycerides are formed by a condensation reaction between the hydroxyl group of a fatty acid molecule and one of the carboxyl groups of the glycerol molecule. 3 Triglycerides are insoluble in water because the fatty acid carbon chain is non-polar. A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
Mark scheme: C
Q12 · The table shows the diameter of some atoms when they form bonds
12 The table shows the diameter of some atoms when they form bonds. single bond double bond atom / nm / nm H 0.060 – O 0.132 0.110 N 0.140 0.120 C 0.154 0.134 The approximate length of the amino acid shown was estimated using the figures in the table. R O H N C C O H H H 0.7 nm What would be the approximate length of a dipeptide formed using this amino acid? A 0.9 nm B 1.2 nm C 1.4 nm D 1.7 nm
Mark scheme: B
Q13 · Which statement about the typical modes of action of a competitive inhibitor and a…
13 Which statement about the typical modes of action of a competitive inhibitor and a non-competitive inhibitor is correct? A Competitive inhibitors can bind to alternative (allosteric) sites of an enzyme, non-competitive inhibitors have an irreversible effect on the enzyme. B Competitive inhibitors have exactly the same shape as the substrate, non-competitive inhibitors can have any shape. C Competitive inhibitors may be used to regulate enzyme activity, non-competitive inhibitors have no functions in enzyme regulation. D Competitive inhibitors will not alter Vmax, non-competitive inhibitors will reduce Vmax.
Mark scheme: D
Q14 · Some animals produce antimicrobial proteins which protect them from pathogens
14 Some animals produce antimicrobial proteins which protect them from pathogens. These proteins could be used to kill human pathogens, however when used as a medicine they are broken down by protein-digesting enzymes. Replacing one of the amino acids found in the protein with one that had been synthesised in the laboratory resulted in a modified protein that was not broken down. What could explain why this modified protein was not broken down by the protein digesting enzymes? 1 The modified protein has a different tertiary structure to the original protein. 2 The modified protein is not complementary in shape to the enzyme’s active site. 3 The modified protein is unable to induce a fit with the protein digesting enzyme. A 1, 2 and 3 B 1 and 2 only C 2 and 3 only D 2 only
Mark scheme: A
Q15 · An investigation was carried out on the effect of temperature on an enzyme-catalysed…
15 An investigation was carried out on the effect of temperature on an enzyme-catalysed reaction. The enzyme and its substrate were initially placed into separate test-tubes and raised to the temperature required. They were then mixed and placed into four tubes A, B, C and D. These tubes were incubated for the time and at the temperature stated. The mass of the product formed was then measured. In which tube was the rate of reaction highest? incubation incubation mass of time / s temperature / °C product / µg A 30 25 2.5 B 30 45 5.0 C 600 25 32.0 D 600 45 10.0
Mark scheme: B
Q16 · In an experiment the proteins in the cell surface membrane of a living cell were labelled…
16 In an experiment the proteins in the cell surface membrane of a living cell were labelled with a fluorescent dye. A small area of the membrane was observed using a microscope. A laser was then used to permanently destroy the fluorescence of the dye molecules in this field of view. The fluorescence in this field of view was later seen to return. What does this experiment show? A Proteins are found only on the outer surface of cell surface membranes. B Proteins in the outer layer of a bilayer do not penetrate into the inner layer. C Proteins move freely in the phospholipids of a bilayer. D The cell surface membrane of the cell is a bilayer.
Mark scheme: C
Q17 · In plants adapted to cold conditions, their cell surface membranes change as the weather…
17 In plants adapted to cold conditions, their cell surface membranes change as the weather gets colder, allowing the plants to carry out exocytosis. Which change occurs in their cell surface membranes? A a decrease in the ratio of proteins to saturated phospholipids B a decrease in the ratio of unsaturated phospholipids to saturated phospholipids C an increase in the ratio of proteins to unsaturated phospholipids D an increase in the ratio of unsaturated phospholipids to saturated phospholipids
Mark scheme: D
Q18 · Which statement describes endocytosis?
18 Which statement describes endocytosis? A movement across a membrane against the concentration gradient and requiring energy B movement across a membrane down the concentration gradient using a carrier molecule C movement across a membrane into a cell using a vesicle and requiring energy D movement across a membrane using a vesicle and requiring no energy
Mark scheme: C
Q19 · An indicator is colourless in acid and pink in alkali
19 An indicator is colourless in acid and pink in alkali. In an experiment a petri dish of agar was prepared using an acidic solution of this indicator. A disc of agar 1 cm in diameter was removed from the centre to create a well. A white card showing circular marker lines 1 cm apart was placed underneath the petri dish. 1 cm3 alkali solution was put into the well in the agar and a stop-watch was started. A circular disc of pink colour appeared and spread through the agar. It reached the first marker line in a short time but took longer to reach the second marker line and a very long time to reach the third marker line. What explains these observations? A facilitated diffusion of alkali solution B facilitated diffusion of the indicator C simple diffusion of alkali solution D simple diffusion of the indicator
Mark scheme: C
Q20 · How many copies of each DNA molecule will be found in a cell at each stage of the mitotic…
20 How many copies of each DNA molecule will be found in a cell at each stage of the mitotic cell cycle? G2 of metaphase cytokinesis interphase A 1 1 1 B 1 2 2 C 2 1 1 D 2 2 2
Mark scheme: D
More questions on Replication and division of nuclei and cells
Q21 · Centromere, chromatids and telomeres are parts of a chromosome in a eukaryotic cell
21 Centromere, chromatids and telomeres are parts of a chromosome in a eukaryotic cell. Which row correctly describes these structures? centromere chromatid telomere A binds the two sister contains histone proteins protects the tips of each chromatids of a and a molecule of chromatid from shortening at chromosome together double-stranded DNA each mitotic division B site of attachment of contains histone proteins prevents the genes nearest spindle microtubules and a molecule of the ends of each chromatid single-stranded DNA from being lost C site of production of contains a molecule of contains DNA with many spindle microtubules double-stranded DNA with no repeats of short base associated histone proteins sequences D splits at metaphase contains a molecule of ensures that when DNA is of mitosis single-stranded DNA with no replicated, the ends of the associated histone proteins molecule are included
Mark scheme: A
Q22 · Which bases found in nucleic acids are purines?
22 Which bases found in nucleic acids are purines? A adenine and cytosine B guanine and adenine C thymine and cytosine D uracil and guanine
Mark scheme: B
More questions on Structure of nucleic acids and replication of DNA
Q23 · A length of double-stranded DNA contains 120 nucleotides and codes for a section of a…
23 A length of double-stranded DNA contains 120 nucleotides and codes for a section of a polypeptide. What is the maximum length of this section of a polypeptide? A 20 amino acids B 40 amino acids C 60 amino acids D 120 amino acids
Mark scheme: A
Q24 · Four students were asked to suggest a set of four labels to add to a plan diagram of a…
24 Four students were asked to suggest a set of four labels to add to a plan diagram of a transverse section of a dicotyledonous leaf. Which student suggested a correct set of labels? label 1 label 2 label 3 label 4 A lower phloem spongy palisade epidermis mesophyll mesophyll B phloem sieve upper spongy xylem vessel tubes epidermis mesophyll elements C spongy endodermis xylem vessel palisade mesophyll elements mesophyll D waxy cuticle xylem companion spongy cells mesophyll
Mark scheme: A
Q25 · Photomicrograph X shows a transverse section through a typical stem
25 Photomicrograph X shows a transverse section through a typical stem. Photomicrograph Y shows a longitudinal section through one type of cell shown in photomicrograph X. Which letter represents the cells from photomicrograph Y? photomicrograph X photomicrograph Y A B C D
Mark scheme: C
Q26 · By which process would water rise up through xylem vessels in a plant root when the shoot…
26 By which process would water rise up through xylem vessels in a plant root when the shoot has been removed? A capillarity, root pressure and transpiration pull B capillarity and root pressure only C capillarity and transpiration pull only D root pressure only
Mark scheme: B
Q27 · Which changes occur to the water potential and the volume of liquid in the phloem sieve…
27 Which changes occur to the water potential and the volume of liquid in the phloem sieve tube element when carbohydrate is moved into a sink? water potential volume of liquid of the sieve of the sieve tube element tube element A lowers decreases B lowers increases C raises decreases D raises increases
Mark scheme: C
Q28 · The graph shows the diameter of a tree trunk over time
28 The graph shows the diameter of a tree trunk over time. diameter of tree trunk 12:00 00:00 12:00 00:00 12:00 time of day Which statement explains these changes? A Cohesive tension forces increased during the day. B Columns of water break up during the day. C Mass flow of sucrose decreased during the day. D Root pressure increased during the day.
Mark scheme: A
Q29 · What is found in all blood vessels, lymph and tissue fluid?
29 What is found in all blood vessels, lymph and tissue fluid? 1 carbon dioxide 2 glucose 3 white blood cells 4 antibodies A 1, 2, 3 and 4 B 1, 2 and 3 only C 1, 3 and 4 only D 2 and 4 only
Mark scheme: A
Q30 · Which reactions will be taking place in blood that is passing through active tissues?
30 Which reactions will be taking place in blood that is passing through active tissues? 1 HbO8 → Hb + 4O2 2 HbO8 + H+ → HHb + 4O2 3 HCO3 – + H+ → H2CO3 4 H2O + CO2 → H2CO3 A 1, 2, 3 and 4 B 1, 2 and 4 only C 1, 3 and 4 only D 2 and 3 only
Mark scheme: B
Q31 · The graph shows the percentage saturation of haemoglobin with oxygen at a range of…
31 The graph shows the percentage saturation of haemoglobin with oxygen at a range of partial pressures of oxygen and at two partial pressures of carbon dioxide. One curve is at the partial pressure of carbon dioxide in the lungs and the other curve at the partial pressure of carbon dioxide in the liver. 100 L M percentage saturation of haemoglobin with oxygen J K 0 0 partial pressure of oxygen / kPa Which points on the curves correspond to the percentage saturation of haemoglobin at the lungs and at the liver? A L and J B L and K C M and J D M and K
Mark scheme: B
Q32 · A person moves from sea level to live at a high altitude
32 A person moves from sea level to live at a high altitude. After three months living at high altitude, which change would be found in this person when at rest? A increased volume of blood per heartbeat B increased breathing rate C increased heart rate D increased mass of red blood cells
Mark scheme: D
Q33 · The diagram shows three of the features found in tissues of the gas exchange system
33 The diagram shows three of the features found in tissues of the gas exchange system. ciliated goblet epithelium cells X cartilage Which structures of the gas exchange system could be represented at position X in the diagram? 1 bronchiole 2 trachea 3 bronchus A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
Mark scheme: D
Q34 · Chronic obstructive pulmonary disease (COPD) includes bronchitis and emphysema
34 Chronic obstructive pulmonary disease (COPD) includes bronchitis and emphysema. Which row shows the effects of bronchitis? T-helper cells bronchioles infection A destroyed inflamed absent B destroyed narrowed present C increased scarred present D increased stiffened absent
Mark scheme: C
Q35 · Which factors affect the distribution of the disease malaria?
35 Which factors affect the distribution of the disease malaria? 1 draining swamps 2 vaccination 3 antibiotics 4 migration of people due to conflict 5 climate A 1, 2 and 4 B 1, 4 and 5 C 2, 3 and 5 D 3, 4 and 5
Mark scheme: B
Q36 · Which diseases cannot be treated with antibiotics?
36 Which diseases cannot be treated with antibiotics? 1 cholera 2 HIV/AIDS 3 measles A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only
Mark scheme: D
Q37 · A West African country introduced a measles vaccination during a measles epidemic
37 A West African country introduced a measles vaccination during a measles epidemic. Later, it was realised that vaccinated children were more likely to survive childhood than unvaccinated children, even when there were no measles epidemics. The vaccine had given the children some protection against other pathogenic infections. Which statement could account for this extra protection? A B-lymphocytes produced memory cells which gave the children passive immunity to these infections. B Memory cells produced plasma cells which secreted anti-measles antibodies that bound to antigens that closely resembled measles antigens. C Memory cells produced plasma cells which secreted anti-measles antibodies that bound to any antigen. D T-lymphocytes produced memory cells which gave the children natural immunity against these other infections.
Mark scheme: B
Q38 · Which graph shows the changes in antibody concentration following a second exposure to…
38 Which graph shows the changes in antibody concentration following a second exposure to the same infection? 1000 antibody 100 A concentration / arbitrary units 10 1 0 10 20 30 40 50 60 70 80 time / days first exposure second exposure 1000 antibody 100 B concentration / arbitrary units 10 1 0 10 20 30 40 50 60 70 80 time / days first exposure second exposure 1000 antibody 100 C concentration / arbitrary units 10 1 0 10 20 30 40 50 60 70 80 time / days first exposure second exposure 1000 antibody 100 D concentration / arbitrary units 10 1 0 10 20 30 40 50 60 70 80 time / days first exposure second exposure
Mark scheme: A
Q39 · Where are antigens found?
39 Where are antigens found? on surface in blood of pathogen plasma A Jv Jv key B v x ¥ = antigens found Cc x v X = antigens not found D x x
Mark scheme: B
Q40 · Whooping cough is a highly infectious disease of the gas exchange system, caused by the…
40 Whooping cough is a highly infectious disease of the gas exchange system, caused by the bacterium Bordetella pertussis. Which method provides protection to infants against whooping cough and reduces the chance of developing this disease later in life? A a short course of more than one type of antibiotic B a six month course of one type of antibiotic C injections of antibody specific to Bordetella pertussis D injections of dead Bordetella pertussis bacteria
Mark scheme: D
What was in this paper
The subtopics covered by these 40 questions, and how many questions each got. Open one in a new tab to see every Cambridge question on it.
4Antibodies and vaccination3Cells as the basic units of living organisms3Transport mechanisms3Transport of oxygen and carbon dioxide3Factors that affect enzyme action2Fluid mosaic membranes2Movement into and out of cells2Structure of transport tissues2The gas exchange system2The microscope in cell studies2Antibiotics1Chromosome behaviour in mitosis1Infectious diseases1Mode of action of enzymes1Nucleic acids and protein synthesis1Protein synthesis1Proteins1Replication and division of nuclei and cells1Structure of nucleic acids and replication of DNA1Testing for biological molecules1The circulatory system1The immune system1What you needed in this session
Cambridge’s own grade thresholds for 2018 Oct/Nov, Paper 1 · Variant 3. A higher threshold means an easier paper — the bar moves with how the cohort did.