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Enzymes and Metabolism · Paper 2 — Free Biology SL Practice Questions

1MasterySAQ-SEnzyme kinetics and inhibition5 marksPaper 2~8 min
A researcher investigates an enzyme-catalysed reaction. The graph below shows reaction rate against substrate concentration under two conditions: no inhibitor (control) and with an inhibitor present. Graph description: Michaelis–Menten curves. Control curve: hyperbolic, Vmax=100 μmol min−1V_{max} = 100\,\mu\text{mol}\,\text{min}^{-1}, Km=2 mMK_m = 2\,\text{mM}. Inhibitor curve: hyperbolic, Vmax=60 μmol min−1V_{max} = 60\,\mu\text{mol}\,\text{min}^{-1}, Km=2 mMK_m = 2\,\text{mM}.
(a)
State the effect of the inhibitor on VmaxV_{max}. [1 mark]
(b)
State the effect of the inhibitor on KmK_m. [1 mark]
(c)
Using your answers to (a) and (b), identify the type of inhibition and explain how the inhibitor affects enzyme function. [2 marks]
(d)
A second inhibitor is tested. This inhibitor increases KmK_m from 2 mM2\,\text{mM} to 8 mM8\,\text{mM} but leaves VmaxV_{max} unchanged. Deduce what would happen to the reaction rate if substrate concentration were raised to a very high value, and explain how this differs from the behaviour of the first inhibitor. [1 mark]
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Solutions

2MasterySAQ-SEnzyme kinetics and inhibition6 marksPaper 2~9 min
The enzyme lactate dehydrogenase (LDH) catalyses the reduction of pyruvate to lactate in human muscle cells during anaerobic respiration. A student investigates the effect of an inhibitor on the rate of this reaction. The student measures the initial reaction rate at various substrate (pyruvate) concentrations, both without the inhibitor and with a fixed concentration of the inhibitor. The results are shown in the graph below.
(a)
State the type of inhibition shown by the inhibitor. [1 mark]
(b)
The Michaelis-Menten constant (KmK_m) for LDH without the inhibitor is 0.8 mmol dm−30.8\,\text{mmol dm}^{-3}. Using the graph, estimate the KmK_m value when the inhibitor is present. [2 marks]
(c)
Calculate the factor by which KmK_m has changed when the inhibitor is present. [1 mark]
(d)
Explain how the type of inhibition identified in (a) affects the apparent KmK_m without changing Vmax⁡V_{\max}. [2 marks]
diagram

Solutions

3ChallengeSAQ-LEnzyme kinetics and inhibition6 marksPaper 2~9 min
A student investigates the effect of an inhibitor on the enzyme catalase, which breaks down hydrogen peroxide (H2O2\text{H}_2\text{O}_2) into water and oxygen. The initial rate of oxygen production is measured at different H2O2\text{H}_2\text{O}_2 concentrations. The results are shown in the table below. [H2O2][\text{H}_2\text{O}_2] / mM — Rate without inhibitor / mL O2 min−1\text{mL O}_2\,\text{min}^{-1} — Rate with inhibitor / mL O2 min−1\text{mL O}_2\,\text{min}^{-1} 5 — 1.2 — 0.6 10 — 2.0 — 1.0 20 — 3.8 — 1.5 40 — 4.2 — 1.9 80 — 4.5 — 2.1 The Michaelis–Menten equation is: v=Vmax⁡[S]Km+[S]v = \dfrac{V_{\max}[\text{S}]}{K_m + [\text{S}]}
(a)
The student plots a graph of rate (vv) against [H2O2][\text{H}_2\text{O}_2] for both conditions. (i) State the value of Vmax⁡V_{\max} for catalase in the absence of the inhibitor, as indicated by the data. [1]
(ii) Determine whether the inhibitor is competitive or non-competitive, using evidence from the data table. [2 marks]
(b)
Explain the molecular mechanism by which a competitive inhibitor reduces the rate of an enzyme-catalysed reaction. [2 marks]
(c)
Evaluate the biological significance of enzyme inhibition in the regulation of a named metabolic pathway, referring to the type of inhibition identified in (a)(ii). [2 marks]
diagram

Solutions

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