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The velocity of a reaction is 40 µM/min at a substrate concentration of 1.5 mM. The...

The velocity of a reaction is 40 µM/min at a substrate concentration of 1.5 mM. The enzyme is known to reach maximal velocity at 220 µM/min. What would the velocity be at a substrate concentration of 5 mM?

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Homework Answers

Answer #1

We know that,

V = (Vmax × [S]) ÷ (KM + [S])

Where,

V = Velocity of reaction at particular concentration of substrate, [S]

Vmax = maximal velocity = 220 µM/min

[S] = Substrate concentration

Km = Michaelis constant (a measure of enzyme/substrate affinity)

For solving this, first finding the Km value when substrate concentration is 1.5 mM

V1 = 40 µM/min

[S1] = 1.5 mM

Vmax = 220 µM/min

V = (Vmax × [S]) ÷ (Km + [S])

40 = (220 × 1.5) ÷ ( Km + 1.5)

40Km + 60 = 330

40Km = 330 - 60

Km = 270/40 = 6.75 mM

Now, putting Km in the second reaction where concentration of substrate is 5 mM.

[S] = 5 mM, we have to find V = ?

V = (220 × 5) ÷ (6.75 + 5)

V = 1100/ 11.75

V = 93.62

Thus velocity at substrate concentration 5 mM is 93.63 µM/min.

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