The problem can be solved with the help of Michaelis Menten kinetics.
The equation can be written as
[tex] V_{o}= \frac{V_{max}[S]}{K_{m}+ [S]} [/tex]
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,
[tex] V_{o}= \frac{V_{max}\times 3[S]}{K_{m}+ 3[S]} [/tex]
We have been given that Km = [S]
Let us write [S] in place of Km in our equation. We get,
[tex] V_{o}= \frac{V_{max}\times 3[S]}{[S]+ 3[S]} [/tex]
On adding the values on the denominator we get,
[tex] V_{o}= \frac{V_{max}\times 3[S]}{4[S]} [/tex]
We can cancel out [S] .
[tex] V_{o}= \frac{V_{max}\times 3}{4} [/tex]
[tex] V_{o}= \frac{3}{4}V_{max} [/tex]
[tex] V_{o}= 0.75 V_{max} [/tex]
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