IB Biology HL Theme C: Interaction and interdependence questions

Exam-style multiple choice and written questions with full markschemes. Check multiple choice answers straight away, or type, photograph or draw a written answer and get it marked.

Type

Multiple choice Written

Difficulty

Easy Medium Hard

Question 1

Paper 2Hard16 marks

The reaction S → P occurs in a cell and is catalysed by enzyme E. To investigate the reaction, equal volumes containing the same initial amount of S and no P were held at the same temperature and pH, either with E or without E. Table 1 shows the measured concentration of P at different times. Values are reported to the nearest 0.1 mmol dm⁻³.

Time / min	[P] with E / mmol dm⁻³	[P] without E / mmol dm⁻³
0	0.0	0.0
1	2.8	0.1
2	5.1	0.2
4	6.9	0.4
8	7.8	0.8
16	8.0	1.7
32	7.9	3.2
64	8.0	5.5
128	8.0	7.8

Figure 1 shows schematic energy profiles for the conversion of S to P.

(a)

Calculate the mean rate of formation of P during the first 2.0 min, with E and without E. Include units.

[3]

(b)

Describe the change in [P] in each mixture during the first 8 min.

[2]

(c)

Using Figure 1, calculate how much E lowers the activation energy.

[1]

(d)

Explain how E causes the faster formation of P and state one benefit of this increase in reaction rate for cells.

[4]

(e)

A cell process Q requires a supply of P at an average rate of at least 1.2 mmol dm⁻³ min⁻¹ during the first 2.0 min. Using your results from (a), determine whether the reaction with E and the reaction without E meet this requirement.

[2]

(f)

A student concludes that “E increases the total amount of P that can be produced from a fixed starting amount of S, not just the rate at which P appears”. Evaluate this conclusion using the data at 2 min and 128 min, and any other relevant data in Table 1.

[4]

Question 2

Paper 2Easy6 marks

During aerobic respiration, electrons pass along the electron transport chain in the inner mitochondrial membrane. Answer the following questions about the proton gradient and ATP synthesis.

(a)

State how energy released as electrons pass along the electron transport chain is used, and state the direction in which protons are moved across the inner mitochondrial membrane.

[2]

(b)

State the difference in proton concentration across the inner mitochondrial membrane that results.

[1]

(c)

State the direction in which protons move through ATP synthase.

[1]

(d)

Explain how the movement of protons through ATP synthase results in ATP synthesis.

[2]

Question 3

Multiple choiceEasy1 mark

Equal amounts of GP were added to four reaction mixtures. ATP and reduced NADP were added as indicated; all other conditions and the time interval were identical. The amount of TP formed after 1 min was measured.

Mixture	ATP added	Reduced NADP added	TP formed / μmol
1	Yes	No	0.0
2	No	Yes	0.0
3	Yes	Yes	2.4
4	No	No	0.0

Which combination was present in the mixture that formed TP?

Show answer and explanation

Answer: C. Only mixture 3 formed TP, and it contained both ATP and reduced NADP. This matches the conversion of GP into TP, which uses ATP and reduced NADP.

A: assumes ATP alone is sufficient; mixture 1 formed no TP. B: assumes reduced NADP alone is sufficient; mixture 2 formed no TP. D: selects the mixture without either compound, which formed no TP.

Remember: GP conversion to TP requires both ATP and reduced NADP.

Question 4

Paper 2Easy5 marks

During a simplified accounting of the Calvin cycle in a leaf, six molecules of triose phosphate are produced. One molecule leaves as net product; the remainder is used to regenerate RuBP.

(a)

Determine the number and fraction of the triose phosphate molecules used to regenerate RuBP.

[2]

(b)

State the number of RuBP molecules regenerated from these five triose phosphate molecules.

[1]

(c)

Explain the role of ATP in RuBP regeneration and why regenerated RuBP is needed for the Calvin cycle to continue.

[2]

Question 5

Multiple choiceHard1 mark

During a controlled observation, blood CO₂ is raised briefly and then allowed to fall. The measurements are:

Stage	Blood CO₂ (relative units)	Blood pH	Ventilation rate (dm³ min⁻¹)
Baseline	1.0	7.40	6
After CO₂ rises	1.3	7.32	12
Recovery	1.1	7.38	7

Which statements are consistent with the data and feedback control of ventilation?

I. The initial CO₂ increase can lower blood pH by increasing blood acidity.
II. The fall in pH causes brainstem chemoreceptors to signal the diaphragm and intercostal muscles to reduce ventilation.
III. Increased ventilation can remove CO₂ from the blood, raising pH and reducing chemoreceptor stimulation so ventilation falls.

Show answer and explanation

Answer: B. Increased blood CO₂ can lower pH. Brainstem chemoreceptors respond to the pH change by signalling the diaphragm and intercostal muscles to increase ventilation; greater CO₂ removal then helps pH recover and reduces the stimulus.

A: stops at the initial pH change and omits the feedback described in III.
C: reverses the response in II and omits the supported CO₂–pH relationship in I.
D: includes II, which incorrectly says ventilation decreases when pH falls.
Remember: increased ventilation removes CO₂, helping blood pH rise.

Question 6

Multiple choiceEasy1 mark

A hospital restricts antibiotics to cases in which they are needed. Some pathogenic bacterial strains are already resistant to several antibiotics. How can this careful use slow the increase of multiresistant strains?

Show answer and explanation

Answer: D. Antibiotic use selects for resistant bacteria because susceptible bacteria are more likely to be killed. With less unnecessary antibiotic use, the selective advantage of multiresistant bacteria is reduced, slowing their increase in frequency.

A: Treats antibiotics as directly causing resistance mutations rather than selecting bacteria that already have resistance.
B: Confuses reduced selection with blocked inheritance; resistant bacteria can still pass resistance to offspring.
C: Assumes resistance disappears without antibiotics; reduced use does not itself remove resistance.
Remember: antibiotics select among bacterial variants; they do not direct bacteria to become resistant.

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Time / min	[P] with E / mmol dm⁻³	[P] without E / mmol dm⁻³
0	0.0	0.0
1	2.8	0.1
2	5.1	0.2
4	6.9	0.4
8	7.8	0.8
16	8.0	1.7
32	7.9	3.2
64	8.0	5.5
128	8.0	7.8

Time / min	[P] with E / mmol dm⁻³	[P] without E / mmol dm⁻³
0	0.0	0.0
1	2.8	0.1
2	5.1	0.2
4	6.9	0.4
8	7.8	0.8
16	8.0	1.7
32	7.9	3.2
64	8.0	5.5
128	8.0	7.8

Time / min	[P] with E / mmol dm⁻³	[P] without E / mmol dm⁻³
0	0.0	0.0
1	2.8	0.1
2	5.1	0.2
4	6.9	0.4
8	7.8	0.8
16	8.0	1.7
32	7.9	3.2
64	8.0	5.5
128	8.0	7.8