Non-competitive inhibition causes substrate buildup upstream

Steps:

• Non-competitive inhibitors bind to an allosteric site on enzyme 2, reducing its activity without affecting substrate binding.
• This slows the conversion of substance X to substance Y in the pathway.
• Substance X continues to form from the reactant via enzyme 1 but accumulates due to reduced processing by enzyme 2.
• Downstream production of the end product decreases, but the pathway does not halt completely with a small inhibitor amount.

Why B is correct:

• Non-competitive inhibition lowers Vmax (maximum reaction rate) per Michaelis-Menten kinetics, causing substrate X to accumulate as enzyme 2 processes it less efficiently.

Why the others are wrong:

• A: Non-competitive inhibition involves reversible binding to an allosteric site, not denaturation of the enzyme.
• C: A small amount partially inhibits enzyme 2, so substance Y forms at a reduced rate, not zero.
• D: Enzyme 1 remains unaffected, so the initial reactant continues to be metabolized into substance X.

Final answer: B