S5 Biology Practice Paper (Set 1): Genetics And DNA Replication, Heart And Circulation, Gas Exchange And Ventilation, Osmosis And Water Potential, Energy Flow And Population Estimates, Experimental Design (with Answers) - TutorZone

S5 Biology Practice Paper (Set 1): Genetics and DNA Replication, Heart and Circulation, Gas Exchange and Ventilation, Osmosis and Water Potential, Energy Flow and Population Estimates, Experimental Design (with answers)

2026-11-15

S5 Biology

S5 Biology paper (Set 1) with answers: genetics and DNA replication, cardiac output, ventilation and gas exchange, osmosis and water potential, energy flow.

Paper details　S5 Biology practice paper (Set 1) with full answers: Mendelian ratios, DNA base pairing and semi-conservative replication, codons, cardiac output, ventilation, osmosis and water potential, energy pyramids, Lincoln index and experimental design.

Section A: Genetics and DNA Replication

Study each item and answer the questions.

#	Question	Answer and marking notes
1	Two heterozygous tall pea plants (Tt x Tt) are crossed, where T is dominant and t is recessive. State the genotypic ratio and the phenotypic ratio of the offspring. If 400 offspring are produced, how many are expected to be short?	Answer: Genotypes 1 TT : 2 Tt : 1 tt; phenotypes 3 tall : 1 short. Short plants = 400 x 1/4 = 100 (tall = 300).
2	In a double-stranded DNA molecule, adenine (A) makes up 30% of all bases. Find the percentage of thymine (T), guanine (G) and cytosine (C), and state the base-pairing rule.	Answer: T = 30%; G + C = 100 – 30 – 30 = 40%, so G = C = 20%. Pairing rule: A pairs with T and G pairs with C.
3	A DNA molecule with 1000 base pairs undergoes three rounds of semi-conservative replication. Find the total number of DNA molecules produced and the total number of base pairs in all products, and state how many original strands each daughter molecule keeps.	Answer: Molecules = 2^3 = 8; total base pairs = 8 x 1000 = 8000; each daughter molecule keeps 1 original strand (semi-conservative).
4	An mRNA is translated into a polypeptide of 60 amino acids, not counting the stop codon. Find the minimum number of bases needed in the mRNA and the number of tRNA molecules involved, assuming each tRNA is used once.	Answer: Bases = 60 x 3 = 180; tRNA molecules = 60.

Section B: Heart and Circulation

Study each item and answer the questions.

#	Question	Answer and marking notes
5	An athlete at rest has a heart rate of 75 beats per minute and a stroke volume of 70 cm3. Calculate the cardiac output in L/min and write down the formula used.	Answer: Cardiac output = 75 x 70 = 5250 cm3/min = 5.25 L/min. Formula: cardiac output = heart rate x stroke volume.
6	During vigorous exercise the same athlete reaches a heart rate of 150 beats per minute and a stroke volume of 110 cm3. Find the cardiac output in L/min and how many times larger it is than at rest (2 decimal places).	Answer: 150 x 110 = 16500 cm3/min = 16.5 L/min; 16.5 / 5.25 = 3.14 times.
7	Compare the thickness of the muscular wall of the left and right ventricles, and explain how this difference suits the functions of the pulmonary and systemic circulations.	Answer: The left ventricular wall is much thicker. The systemic circuit is long and has high resistance, so a higher pressure is needed; the pulmonary circuit is short, so a lower pressure from the right ventricle suffices and excessive pressure could damage the alveolar capillaries.
8	Blood flows at about 30 cm/s in the aorta and about 0.03 cm/s in the capillaries. How many times slower is the flow in the capillaries, and why is this change useful for the exchange of gases and nutrients?	Answer: 30 / 0.03 = 1000 times. Slower flow means blood stays longer in the capillaries, giving more time for exchange of gases and nutrients with tissue fluid (the huge total cross-sectional area of capillaries lowers the velocity).

Section C: Gas Exchange and Ventilation

Study each item and answer the questions.

#	Question	Answer and marking notes
9	A person has a tidal volume of 500 cm3 and a breathing rate of 12 breaths per minute. Find the pulmonary ventilation rate in L/min.	Answer: 500 x 12 = 6000 cm3/min = 6.0 L/min.
10	Inhaled air contains 21% oxygen and exhaled air contains 16%. The person breathes 500 cm3 per breath at 12 breaths per minute. Find the oxygen consumption per minute in cm3.	Answer: Oxygen per breath = 500 x (21% – 16%) = 25 cm3; per minute = 25 x 12 = 300 cm3.
11	The alveolar wall and capillary wall together are about 2 um thick in a healthy person but about 6 um in a long-term smoker. Assuming other factors are unchanged, what fraction of the normal diffusion rate remains, and why?	Answer: About one third. The diffusion rate is inversely proportional to thickness; tripling the thickness cuts the rate to one third, so gas exchange is much less efficient.
12	Each 100 cm3 of blood contains 15 g of haemoglobin, and each gram of haemoglobin can bind 1.34 cm3 of oxygen. Find the maximum oxygen carried by 100 cm3 of blood in cm3, and state two factors that affect how much oxygen haemoglobin carries.	Answer: 15 x 1.34 = 20.1 cm3. Factors include the partial pressure of oxygen and temperature (also acceptable: blood pH, carbon dioxide concentration).

Section D: Osmosis and Water Potential

Study each item and answer the questions.

#	Question	Answer and marking notes
13	Four potato strips, each of initial mass 4.0 g, are placed in sucrose solutions of different concentrations for 30 minutes. The strip in 0.30 mol/dm3 sucrose becomes 3.4 g. Find the percentage change in mass and decide whether the solution is hypotonic, isotonic or hypertonic to the potato cells.	Answer: (3.4 – 4.0) / 4.0 x 100% = -15.0%. The loss in mass shows water left the cells, so the solution is hypertonic.
14	The strip placed in distilled water becomes 4.6 g. Find the percentage change in mass and explain why the cell does not burst.	Answer: (4.6 – 4.0) / 4.0 x 100% = +15.0%. Water enters the cell by osmosis, and the cell wall provides mechanical support that stops the cell over-expanding and bursting.
15	Results: +9.0% at 0.10 mol/dm3, +2.0% at 0.20 mol/dm3 and -5.0% at 0.30 mol/dm3. Use linear interpolation to estimate the concentration of the isotonic solution in mol/dm3 (2 decimal places).	Answer: The zero-change concentration lies between 0.20 and 0.30: 0.20 + 0.10 x 2.0 / (2.0 + 5.0) = 0.229, i.e. about 0.23 mol/dm3.
16	Explain why the potato strip becomes soft in 0.40 mol/dm3 sucrose and name the phenomenon.	Answer: The external solution is hypertonic, so water leaves the vacuole and cytoplasm by osmosis; the protoplast shrinks away from the cell wall (loss of turgor), so the strip softens. This is plasmolysis.

Section E: Energy Flow and Population Estimates

Study each item and answer the questions.

#	Question	Answer and marking notes
17	Producers in an ecosystem fix 20 000 kJ of energy, and the transfer efficiency is 10% per trophic level. Find the energy available to the primary, secondary and tertiary consumers in kJ.	Answer: Primary consumers 20 000 x 10% = 2000 kJ; secondary 200 kJ; tertiary 20 kJ.
18	A Lincoln index estimate is made: 120 snails are caught, marked and released; a second sample of 200 snails contains 15 marked individuals. Estimate the population size.	Answer: Population = (120 x 200) / 15 = 1600 snails.
19	Producers absorb 1 000 000 kJ of photosynthetically active radiation and the net primary production is 12 000 kJ. Find the photosynthetic efficiency in % (1 decimal place).	Answer: 12 000 / 1 000 000 x 100% = 1.2%.
20	Atmospheric carbon dioxide rose from 280 ppm to 420 ppm. Find the percentage increase, and give one human activity responsible and one measure that could slow the rise.	Answer: (420 – 280) / 280 x 100% = 50%. Human activity: burning fossil fuels. Measure: switch to renewable energy (also acceptable: planting trees).

Section F: Experimental Design and Data Processing

Study each item and answer the questions.

#	Question	Answer and marking notes
21	A student measures the number of gas bubbles produced per minute in a catalase experiment. The three results are 32, 34 and 33. Find the mean.	Answer: (32 + 34 + 33) / 3 = 33.0 bubbles.
22	If the theoretical value is 32 bubbles and the experimental mean is 33, find the percentage error in % (1 decimal place).	Answer: (33 – 32) / 32 x 100% = 3.125%, which is 3.1% to 1 decimal place.
23	A microscope has a x10 eyepiece and a x40 objective. Find the total magnification; if the image is 20 mm long, find the actual length of the specimen in um.	Answer: Total magnification = 10 x 40 = 400; actual length = 20 mm / 400 = 0.05 mm = 50 um.
24	A student investigates the effect of temperature on yeast fermentation, using the number of bubbles per minute as the indicator. State the independent variable, the dependent variable and two controlled variables that must be kept constant.	Answer: Independent variable: temperature. Dependent variable: number of bubbles per minute. Controlled variables (any two): concentration and volume of the yeast suspension and glucose solution, reaction time, type and mass of yeast used.

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