IB Biology HL C1.1 Enzymes and metabolism 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.

Question 1

Multiple choiceMedium1 mark

A purified enzyme forms a complex with each of four substrates. Shape-change scores compare each molecule’s shape in the complex with its unbound shape; 0 means no change, and a positive score means a change.

SubstrateEnzyme shape-change scoreSubstrate shape-change score
P21
Q20
R01
S00

Which substrate’s results are consistent with induced-fit binding?

Show answer and explanation

Answer: A. Induced-fit binding involves a change in shape of both the enzyme and the substrate. For P, both shape-change scores are positive, so both molecules change shape during complex formation.
B: uses the enzyme’s positive score but overlooks that the substrate’s score is zero.
C: uses the substrate’s positive score but overlooks that the enzyme’s score is zero.
D: shows no shape change in either molecule and therefore is not induced-fit binding.
Remember: induced-fit binding involves shape changes in both interacting molecules.

Question 2

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
00.00.0
12.80.1
25.10.2
46.90.4
87.80.8
168.01.7
327.93.2
648.05.5
1288.07.8

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

Relative energy / kJ mol−1−20020406080SPwithout Ewith EProgress of reaction
(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]

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