CIE A Level Biology 9700 Paper 4 Practice Paper – 100 Marks, 2 Hours

Practice paper · CIE A Level Biology 9700

Paper 4 structured (A Level)

Questions

Q 1 · Photosynthesis and limiting factors

(a) State where the light-dependent stage takes place in a chloroplast. [2]

(b) When light intensity rises from 20 to 40 units the oxygen rate rises from 0.15 to 0.3 cm³ per minute. Explain why. [4]

(c) At 60 units, calculate the total oxygen released in 120 minutes (cm³). [3]

(d) Above 60 units the curve becomes horizontal. Explain. [4]

(e) Write the balanced equation for photosynthesis and explain the sharp fall in rate above 40 °C. [6]

(f) Design an experiment using an aquatic plant to measure the effect of light intensity, and state two controlled variables. [6]

Q 2 · Genetics, mutation and gene expression

(a) Define a gene mutation. [3]

(b) The HBB coding region has 450 bases; in a patient one base changes from A to T. Calculate this as a percentage of the coding region and state the effect on the protein. [5]

(c) How many amino acids does this coding region code for? Show your working. [4]

(d) Some substitutions do not change the protein. Explain why. [3]

(e) Two pea plants (RrYy x RrYy) produce 800 offspring. Using 9:3:3:1, calculate the expected number of each phenotype. [5]

(f) Explain how two processes during meiosis increase genetic variation in offspring. [5]

Q 3 · Gas exchange and transport

(a) Fish gills use a counter-current arrangement. Explain how this raises oxygen uptake. [4]

(b) Stroke volume is 70 cm³ and heart rate is 72 per minute. Calculate cardiac output in dm3 min-1. [4]

(c) During exercise stroke volume rises to 110 cm³ and heart rate to 150. Calculate cardiac output and how many times the resting value it is. [5]

(d) Tidal volume is 500 cm³, dead space 150 cm³ and breathing rate 12 per minute. Calculate alveolar ventilation as a percentage of total ventilation. [4]

(e) Explain the benefit of the Bohr shift to exercising muscle. [4]

(f) A patient haemoglobin falls from 15 to 8 g/dL. Each gram of haemoglobin binds 1.34 cm³ of oxygen. Calculate the percentage fall in oxygen content per 100 cm³ of blood. [4]

Q 4 · Energy flow and ecosystems

(a) Grass holds 50,000 kJ, rabbits 5,000 kJ and foxes 400 kJ. Calculate the transfer efficiency between each level. [3]

(b) An ecosystem has GPP of 2400 kJ m-2 yr-1 and respiration of 1560 kJ m-2 yr-1. Calculate NPP and its percentage of GPP. [4]

(c) Explain why food chains rarely have more than five trophic levels. [5]

(d) Explain the difference between nitrification and nitrogen fixation in the nitrogen cycle, giving one organism for each. [5]

(e) Explain why peat bogs and forests act as carbon sinks. [4]

(f) Explain how intensive farming reduces farmland biodiversity and suggest one mitigation. [4]

Answers

Q 1 · Photosynthesis and limiting factors

(a) On the thylakoid membranes. [2]

(b) Light intensity is the limiting factor; more photosystems are excited, water is photolysed faster and more ATP and reduced NADP reach the Calvin cycle. [4]

(c) 0.45 cm³ per minute x 120 = 54.0 cm³. [3]

(d) Light intensity is no longer limiting; the limiting factor becomes carbon dioxide concentration or temperature, and the photosystems are saturated. [4]

(e) 6CO2 + 6H2O -> C6H12O6 + 6O2 (light, chlorophyll); heat denatures enzymes, the active site shape changes and both light reactions and Calvin cycle enzymes fail. [6]

(f) Illuminate Elodea with an LED at set distances/intensities, collect oxygen for 30 minutes with a capillary tube or bubble count, hold 25 °C in a water bath and repeat three times for a mean; control CO2 (same hydrogencarbonate solution) and shoot length. [6]

Q 2 · Genetics, mutation and gene expression

(a) A change in the base sequence of DNA, such as substitution, insertion or deletion, which may change a codon and therefore the polypeptide. [3]

(b) 1 / 450 x 100 = 0.222 %; the codon changes from GAG to GTG so glutamic acid becomes valine; the beta chain charge changes and red cells sickle at low oxygen. [5]

(c) 450 / 3 = 150 codons; one is a stop codon which codes for no amino acid, so 150 – 1 = 149 amino acids. [4]

(d) The code is degenerate: several codons code for the same amino acid, so the substitution is synonymous and the primary structure is unchanged. [3]

(e) 9/16 x 800 = 450; 3/16 x 800 = 150; 3/16 x 800 = 150; 1/16 x 800 = 50; total 800. [5]

(f) (1) Crossing over in prophase I swaps segments between non-sister chromatids, producing recombinant chromosomes; (2) independent assortment in metaphase I gives 2^23 combinations in humans. Both make gametes genetically different. [5]

Q 3 · Gas exchange and transport

(a) Water and blood flow in opposite directions, so water pO2 stays above blood pO2 along the whole lamella; the diffusion gradient is maintained everywhere and blood leaves up to about 80 % saturated. [4]

(b) 70 x 72 = 5040 cm3 min-1; / 1000 = 5.04 dm3 min-1. [4]

(c) 110 x 150 = 16500 cm3 min-1 = 16.5 dm3 min-1; 16500 / 5040 = 3.27 times. [5]

(d) Total 500 x 12 = 6000 cm3 min-1; alveolar (500 – 150) x 12 = 4200 cm3 min-1; 4200 / 6000 = 70 %. [4]

(e) Muscle CO2 lowers blood pH and raises temperature, so haemoglobin affinity for oxygen falls, the curve shifts right and more oxygen is unloaded at a lower pO2. [4]

(f) 15 x 1.34 = 20.1 cm3/dL; 8 x 1.34 = 10.72 cm3/dL; (20.1 – 10.72) / 20.1 x 100 = 46.7 %. [4]

Q 4 · Energy flow and ecosystems

(a) Grass to rabbit: 5,000 / 50,000 x 100 = 10 %; rabbit to fox: 400 / 5,000 x 100 = 8 %. [3]

(b) NPP = 2400 – 1560 = 840 kJ m-2 yr-1; 840 / 2400 x 100 = 35 %. [4]

(c) Only about 10 % of energy passes to the next level, the rest is lost in respiration, excretion and uneaten parts; by level five less than 0.01 % remains, too little to support a viable population. [5]

(d) Fixation: Rhizobium reduces atmospheric N2 to ammonia/ammonium; nitrification: Nitrosomonas oxidises ammonium to nitrite and Nitrobacter nitrite to nitrate, the form plants absorb. [5]

(e) They fix more carbon by photosynthesis than they release by respiration and decomposition; bog acidity and low oxygen suppress decomposers so organic carbon accumulates. [4]

(f) Monoculture removes habitat and food variety; herbicides and insecticides kill non-target species; fertiliser runoff causes eutrophication. Mitigation: field margins, rotation, fewer pesticides. [4]

Marking scheme

Q 1 · Photosynthesis and limiting factors

(a) 2 marks

• thylakoid membranes
• holds photosystems I and II plus ATP synthase

(b) 4 marks

• light intensity is limiting
• faster photolysis gives more ATP / reduced NADP
• more GP reduced to TP
• rate rises until photosystems saturate

(c) 3 marks

• divide the 30-minute value by 30 (13.5 / 30 = 0.45)
• multiply by 120 to get 54
• unit cm³ stated

(d) 4 marks

• photosystems saturated
• limiting factor becomes CO2 or temperature
• Calvin cycle enzymes / CO2 fixation are the bottleneck
• rate stays at 0.45 cm3/min

(e) 6 marks

• equation balanced for C, H and O
• light and chlorophyll named
• denaturation changes active site shape
• no enzyme-substrate complexes form
• both light and dark stage enzymes affected
• fall is sharp, not gradual

(f) 6 marks

• independent variable: light intensity
• dependent variable: oxygen volume in 30 min
• water bath at 25 °C
• same hydrogencarbonate concentration
• same species and shoot length
• three repeats, calculate a mean rate

Q 2 · Genetics, mutation and gene expression

(a) 3 marks

• change in DNA base sequence
• names a type (substitution / insertion / deletion)
• may change a codon and the amino acid sequence

(b) 5 marks

• 1 / 450 x 100
• answer 0.222 % (about 0.22 %)
• GAG -> GTG
• Glu -> Val
• beta chain charge changes, sickling at low pO2

(c) 4 marks

• 450 / 3 = 150 codons
• stop codon codes for no amino acid
• 150 – 1 = 149
• answer in amino acids

(d) 3 marks

• genetic code is degenerate
• amino acid sequence unchanged
• tertiary structure and function unchanged

(e) 5 marks

• 9/16 x 800 = 450 (both dominant)
• 3/16 x 800 = 150 for each single dominant class
• 1/16 x 800 = 50 (both recessive)
• four values sum to 800
• ratio matched to phenotype

(f) 5 marks

• crossing over in prophase I between non-sister chromatids
• recombinant chromosomes form
• independent assortment at metaphase I
• 2^23 = about 8.4 million combinations in humans
• random fertilisation adds more variation

Q 3 · Gas exchange and transport

(a) 4 marks

• water and blood flow in opposite directions
• gradient maintained along the lamella
• no equilibrium at the exit (unlike parallel flow)
• blood leaves near saturation

(b) 4 marks

• 70 x 72 = 5,040
• divide by 1,000
• answer 5.04
• unit dm3 min-1

(c) 5 marks

• 110 x 150 = 16,500
• 16.5 dm3 min-1
• 16,500 / 5,040
• answer 3.27 (about 3.3 times)
• units correct

(d) 4 marks

• 500 x 12 = 6,000
• (500 – 150) x 12 = 4,200
• 4,200 / 6,000
• answer 70 %

(e) 4 marks

• dissolved CO2 lowers pH
• Hb affinity for O2 falls (curve shifts right)
• more O2 unloaded at the same pO2
• more O2 for muscle respiration

(f) 4 marks

• 15 x 1.34 = 20.1
• 8 x 1.34 = 10.72
• difference 9.38
• 9.38 / 20.1 x 100 = 46.7 %

Q 4 · Energy flow and ecosystems

(a) 3 marks

• 5,000 / 50,000 x 100 = 10 %
• 400 / 5,000 x 100 = 8 %
• both far below 100 % due to respiration, excretion, uneaten parts

(b) 4 marks

• NPP = GPP – respiration
• 2,400 – 1,560 = 840
• 840 / 2,400 x 100
• answer 35 % with unit kJ m-2 yr-1

(c) 5 marks

• about 10 % transfer per level
• losses via respiration, excretion, uneaten parts
• available energy falls exponentially (0.1^4 = 0.0001)
• top consumers need large ranges
• too little energy for further levels

(d) 5 marks

• fixation: N2 -> NH3 / NH4+ by Rhizobium
• nitrification step 1: NH4+ -> NO2- by Nitrosomonas
• step 2: NO2- -> NO3- by Nitrobacter
• nitrate is the form plants absorb
• both are bacterial, aerobic processes

(e) 4 marks

• photosynthesis exceeds respiration and decomposition
• acid, anoxic bog suppresses decomposers
• organic matter accumulates as peat
• forest biomass stores much carbon

(f) 4 marks

• monoculture homogenises habitat
• pesticides kill non-target species
• fertiliser runoff causes eutrophication
• one concrete mitigation: margins or rotation

A TutorZone practice paper; questions and data are original, not publisher or board material.

Format follows the CIE 9700 Biology syllabus Paper 4: 2 h, 100 marks, structured questions.