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Form 6 Biology Practice Paper: Human Physiology, Immunity and Biotechnology
Form 6 Biology – 12 questions – 100 marks – 75 minutes
Form 6 Biology paper: 12 questions, 100 marks, 75 minutes on cardiac output, ventilation, kidney filtration, immunity, PCR, plasmids and allele frequency, with full worked answers.
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Paper details Form 6 Biology practice paper with 12 structured questions on physiology, immunity and gene technology; full answers and two printable PDFs.
Section A: Human Physiology and Homeostasis (Q1-Q4)
Answer all questions; show the expression, substitution and unit for every calculation.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A student at rest has a heart rate of 75 beats per minute and a stroke volume of 70 mL. (a) Calculate the cardiac output in L/min. (b) During exercise the heart rate rises to 150 beats per minute and the stroke volume to 110 mL. Calculate how many times the cardiac output increases. (c) Explain the physiological importance of the rise in cardiac output for skeletal muscle. |
(a) Cardiac output = stroke volume times heart rate = 70 mL times 75 = 5250 mL/min = 5.25 L/min. (b) 110 times 150 = 16 500 mL/min = 16.5 L/min; 16.5 divided by 5.25 = 3.14 times. (c) More oxygenated blood reaches skeletal muscle each minute, so oxygen and glucose are delivered faster and carbon dioxide and lactate are removed faster, supporting a higher rate of aerobic respiration. Marking notes: formula and unit (3 marks), ratio (3 marks), physiological meaning (3 marks). |
| 2 | A student has a tidal volume of 500 mL, a breathing rate of 14 per minute and a dead space of about 150 mL. (a) Calculate the pulmonary ventilation rate in L/min. (b) Calculate the alveolar ventilation rate in L/min. (c) Explain why slow deep breathing gives better alveolar ventilation than fast shallow breathing. |
(a) 500 times 14 = 7000 mL/min = 7.0 L/min. (b) (500 – 150) times 14 = 4900 mL/min = 4.9 L/min. (c) Each breath leaves 150 mL in the dead space; deep breaths bring in a larger fresh volume, so a greater fraction of each breath reaches the alveoli and gas exchange is more efficient. Marking notes: pulmonary ventilation (3 marks), alveolar ventilation (3 marks), explanation (2 marks). |
| 3 | Two cube models A and B have sides of 2 cm and 1 cm. (a) Calculate the surface area to volume ratio of each cube, in per cm. (b) State how many times the ratio changes when the side length is halved. (c) Use this to explain why large multicellular organisms need a circulatory system and a gas exchange system. |
(a) A: surface area = 6 times 2 times 2 = 24 cm2, volume = 8 cm3, ratio = 3.0 per cm; B: surface area = 6 cm2, volume = 1 cm3, ratio = 6.0 per cm. (b) The ratio rises from 3.0 to 6.0, that is 2.0 times. (c) Volume grows with the cube of length while surface area grows with the square, so diffusion alone cannot supply inner cells; a circulatory system transports substances and a gas exchange system adds diffusion surface. Marking notes: both calculations (4 marks), factor (2 marks), diffusion limit (3 marks). |
| 4 | An adult has a glomerular filtration rate of 125 mL/min. (a) Calculate the volume of filtrate produced per day in litres. (b) If 1.5 L of urine is produced per day, calculate the percentage of water reabsorbed. (c) Explain the importance of reabsorption for water balance. |
(a) 125 times 60 times 24 = 180 000 mL = 180 L per day. (b) Reabsorbed = 180 – 1.5 = 178.5 L; 178.5 divided by 180 times 100 = 99.2 percent. (c) Reabsorption, controlled by ADH, keeps plasma water potential and blood volume stable, preventing dehydration or water overload. Marking notes: daily filtrate (3 marks), percentage (3 marks), homeostasis (2 marks). |
Section B: Immunity and Defence (Q5-Q8)
Data: haemoglobin binds 1.34 mL of oxygen per gram; in Hardy-Weinberg equilibrium p + q = 1.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | A person has 15 g of haemoglobin per dL of blood; each gram binds 1.34 mL of oxygen; total blood volume is 5 L. (a) Calculate the oxygen carried per dL of blood. (b) Calculate the total oxygen carried by the whole blood, in mL. (c) Explain why carbon monoxide poisoning causes tissue hypoxia even when arterial oxygen partial pressure is normal. |
(a) 15 times 1.34 = 20.1 mL per dL. (b) 5 L = 50 dL; 20.1 times 50 = 1005 mL. (c) Carbon monoxide binds haemoglobin far more strongly than oxygen and shifts the dissociation curve left, so tissues cannot unload oxygen even though plasma oxygen partial pressure is normal. Marking notes: per dL (3 marks), total with unit conversion (3 marks), competitive binding (2 marks). |
| 6 | After a first infection the antibody titre is 1:10. The antibody concentration then doubles every 12 hours for 4 days. (a) Calculate the antibody titre after 4 days. (b) Describe the roles of plasma cells and memory cells in this process. (c) Explain why a second infection by the same pathogen raises the titre faster and higher. |
(a) 4 days = 96 hours; 96 divided by 12 = 8 doublings; titre = 10 times 2 to the power 8 = 2560, that is 1:2560. (b) Plasma cells secrete large amounts of antibody; memory cells do not secrete antibody but survive long term and recognise the same antigen. (c) Memory cells divide and differentiate into plasma cells rapidly on re-exposure, so the secondary response has a short lag, a higher peak and lasts longer. Marking notes: doublings and titre (4 marks), cell roles (2 marks), secondary response (2 marks). |
| 7 | Trial data: 14 days after dose one the antibody titre is 1:320; 3 days after dose two it is already 1:5120. (a) Calculate how many times higher the titre is after the booster. (b) Explain why a booster produces a high antibody level so quickly. (c) Give one advantage and one limitation of a vaccine made from killed pathogens. |
(a) 5120 divided by 320 = 16 times. (b) Dose one built a memory cell pool; the booster presents antigen again, so memory cells proliferate and differentiate into plasma cells within days. (c) Advantage: the pathogen cannot replicate, so safety is high and it suits people with weak immunity. Limitation: immunogenicity is weaker, so repeated doses and adjuvants are usually needed. Marking notes: ratio (3 marks), memory cell mechanism (3 marks), advantage and limitation (3 marks). |
| 8 | A person produces 10 000 kJ of heat per day and loses it by radiation (6 000 kJ), evaporation (2 200 kJ) and conduction plus convection (1 800 kJ). (a) Calculate the percentage lost by each route. (b) Explain why the share lost by evaporation rises during strenuous exercise. (c) Explain why a prolonged fever of 41 degrees Celsius is life threatening. |
(a) Radiation 60 percent, evaporation 22 percent, conduction and convection 18 percent, totalling 100 percent. (b) Exercise raises heat production and body temperature; sweat glands secrete more sweat and its evaporation removes latent heat, so evaporation takes a larger share. Thermoregulation depends on this shift. (c) Enzymes denature at very high temperature, metabolic reactions lose control, and dehydration and nervous damage become likely. Marking notes: three percentages (4 marks), evaporation mechanism (2 marks), enzyme denaturation (2 marks). |
Section C: Biotechnology and Evolution (Q9-Q12)
Answer explanation questions in full sentences; state the unit for every percentage.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A polymerase chain reaction starts with 1 double stranded DNA molecule and runs for 30 cycles. (a) Calculate the number of DNA molecules at the end. (b) Explain why a heat stable Taq polymerase is used instead of human DNA polymerase. (c) State the role of the primers. |
(a) The number doubles each cycle, so 2 to the power 30 = 1,073,741,824 molecules. (b) The denaturation step heats the sample to about 95 degrees Celsius to separate the strands; Taq polymerase from thermophilic bacteria survives this and can be reused, while human polymerase would denature. (c) Primers bind complementary sequences at the two ends of the target, provide a free 3 prime end for the polymerase and define the region that is amplified. Marking notes: calculation (4 marks), heat stability (3 marks), primer role (2 marks). |
| 10 | A plasmid has 4 800 base pairs. A gene of 1 200 base pairs is inserted to form a recombinant plasmid. (a) Calculate the number of base pairs in the recombinant plasmid. (b) Calculate the percentage increase in size. (c) Explain why the plasmid and the gene must be cut with the same restriction enzyme. |
(a) 4 800 + 1 200 = 6000 base pairs. (b) 1 200 divided by 4 800 times 100 = 25.0 percent. (c) The same enzyme produces identical sticky ends, so the two fragments pair by complementary bases and DNA ligase can join them; different enzymes give incompatible ends. Marking notes: total length (3 marks), percentage (3 marks), sticky ends (2 marks). |
| 11 | Digestion of the recombinant plasmid gives three fragments of 6 000 bp, 2 000 bp and 500 bp, separated on a 1 percent agarose gel at 100 V for 30 minutes. (a) State which fragment travels furthest and explain why. (b) Calculate the length ratio of the longest to the shortest fragment. (c) Explain how the gel result confirms that the recombinant plasmid was built. |
(a) The 500 bp fragment travels furthest because the gel mesh retards long fragments more, so short fragments move further in gel electrophoresis. (b) 6 000 divided by 500 = 12 times. (c) A correct construct gives the expected three bands; a 4 800 bp band would mean no insert, and any other pattern means the insert is wrong or in the wrong orientation. Sizes are read against a DNA ladder. Marking notes: shortest fragment with reason (3 marks), ratio (3 marks), interpretation (3 marks). |
| 12 | A population of 1 000 individuals has 490 AA, 420 Aa and 90 aa genotypes. (a) Calculate the allele frequency of A and of a. (b) Use Hardy-Weinberg to calculate the expected number of each genotype and compare with the observed numbers. (c) Predict how the frequency of a changes if every aa individual dies before breeding. |
(a) Frequency of A = (490 times 2 + 420) divided by 2 000 = 1 400 divided by 2 000 = 0.70; frequency of a = 0.30. (b) Expected AA = 0.70 squared times 1 000 = 490; Aa = 2 times 0.70 times 0.30 times 1 000 = 420; aa = 0.30 squared times 1 000 = 90. Observed values match, so the allele frequency is in equilibrium. (c) Removing aa individuals removes a alleles, so the frequency of a falls and the frequency of A rises in the next generation. Marking notes: frequencies (3 marks), expected values and comparison (4 marks), direction (2 marks). |
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