C1.1 Enzymes and metabolism

The inhibition cannot be overcome by increasing substrate concentration, which is the critical clue in this question.

When increasing substrate concentration does not restore enzyme activity, the inhibitor cannot be competing with the substrate for the active site. This rules out competitive inhibition.

A non-competitive inhibitor binds to an allosteric site — a region separate from the active site. This causes a conformational change in the enzyme's shape, distorting the active site so the substrate cannot bind effectively, regardless of how much substrate is present.

This scenario describes feedback inhibition — a common regulatory mechanism where the end-product of a metabolic pathway inhibits an earlier enzyme (typically the first committed step) to prevent overproduction. This is a form of non-competitive inhibition.

The correct answer is: "The final product binds to an allosteric site on the first enzyme, causing a conformational change that reduces catalytic activity."

We need to identify the inhibition mechanism where doubling substrate concentration fails to restore enzyme activity.

"The final product is a competitive inhibitor... overcome by adding more substrate" is eliminated — the question explicitly states doubling substrate does NOT restore activity, which is the defining feature that rules out competitive inhibition.

"The final product denatures the first enzyme by breaking its peptide bonds" is eliminated — denaturation is caused by extreme heat or pH, not by a metabolic product. Also, denaturation breaks the 3D structure, not specifically peptide bonds.

"The final product acts as a coenzyme that increases activation energy" is eliminated — coenzymes assist enzyme function and do not increase activation energy; this contradicts basic enzyme chemistry.

The remaining option — binding to an allosteric site causing a conformational change — correctly describes non-competitive inhibition, which cannot be overcome by additional substrate.

The rate increases from $[H_2{O}_2]$ of 5 to 40 mmol/dm³, but between 40 and 100 mmol/dm³, the rate is essentially constant at approximately 10.1 cm³/min — this is the plateau ($V_{max}$).

When the plateau is reached, every enzyme active site is occupied by substrate at all times. Adding more $H_2{O}_2$ cannot increase the rate because there are no free active sites available — the enzyme is operating at maximum capacity ($V_{max}$).

Denaturation would cause the rate to fall at higher concentrations. Here the rate simply levels off, which is characteristic of saturation rather than denaturation.

The correct answer is that all enzyme active sites are fully occupied, which is the mechanistic explanation for the $V_{max}$ plateau.

We need to explain why the reaction rate plateaus at high $[H_2{O}_2]$ rather than continuing to increase.

"Catalase has been denatured at high substrate concentrations" is eliminated — substrate molecules do not denature enzymes. Denaturation is caused by extreme temperature, pH, or chemical agents, not by high concentrations of a normal substrate.

"The reaction has reached chemical equilibrium" is eliminated — enzyme-catalysed reactions can be driven by continuous substrate addition. The plateau here reflects a kinetic limit of the enzyme, not thermodynamic equilibrium. Also, the oxygen continues to be produced (just at a constant rate).

"Increasing substrate concentration always proportionally increases the reaction rate" is eliminated — the data clearly contradicts this; the rate does not increase between 80 and 100 mmol/dm³.

The correct answer is enzyme active site saturation — all active sites are occupied, so no additional substrate can be processed, giving the characteristic $V_{max}$ plateau.