To understand the relationship between velocity and substrate concentration for an enzyme with a Hill coefficient of 2, we first rewrite the Hill equation for n = 2 states as follows,

If we let [S] = s₁, where s₁is an arbitrary concentration, we can find V as,

Now we want to know what happens to V if the substrate concentration is halved. Mathematically, we can express this as s₂ = s₁/2. To understand this, we use the Hill equation as,

We now want to see if there is a simple way that v₂ relates to v₁ because this will tell us how halving the substrate concentration has changed the velocity (e.g. v₂ = 2v₁). To clear the compound fraction in v₂, we multiply the numerator and denominator by 4 to get,

Now compare this expression to v₁,

Notice that v₂ ≠ 1/4v₁ because the constant 4 is not being multiplied by the entire denominator, i.e. because. In fact, there is no constant c such that v₂ = cv₁.

Convince yourself this is true by setting v₂ = cv₁ and try to solve for c. You will not be able to find c as a constant. Therefore, halving the substrate concentration does no alter the velocity in a simple linear fashion.

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