Chemistry: 1[s] = km for a simple enzymatic reaction. when substrate concentration is tripled, the rate becomes vmax.

It s either amino acids or glucose...

1[s] = km for a simple enzymatic reaction. when

paradisetiyae6903

13.03.2022

The problem can be solved with the help of Michaelis Menten kinetics.

The equation can be written as

$V_{o}= \frac{V_{max}[S]}{K_{m}+ [S]}$

Where Vo = Initial velocity of the enzyme. This is also known as initial rate of the reaction.

Vmax = Maximum velocity

[S] = Substrate concentration

Km = Michaelis constant for the given enzyme.

When the substrate concentration is tripled, [S] becomes 3[S]

Let us plug this in Michaelis equation.

The new equation that we get is,

$V_{o}= \frac{V_{max}\times 3[S]}{K_{m}+ 3[S]}$

We have been given that Km = [S]

Let us write [S] in place of Km in our equation. We get,

$V_{o}= \frac{V_{max}\times 3[S]}{[S]+ 3[S]}$

On adding the values on the denominator we get,

$V_{o}= \frac{V_{max}\times 3[S]}{4[S]}$

We can cancel out [S] .

$V_{o}= \frac{V_{max}\times 3}{4}$

$V_{o}= \frac{3}{4}V_{max}$

$V_{o}= 0.75 V_{max}$

From the above equation, we can see that initial rate of the reaction (V₀) becomes 0.75 times Vmax

Therefore, when substrate concentration is tripled, the rate becomes 0.75 Vmax

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