Question

Please submit answer in units of micromolars and show full work.

Sre E B NOTE: Conversions: I M = 103 mM= 10 UM 10' nM An enzyme-catalyzed reaction has a KM ofl mM and a Vmax of 5 nM s. What is the reaction velocity when the substrate concentration is 0.25 mM?

Answer 1

According to Michaelis and Menten Equation -

Vo = Vmax[S]/ Km +[S]

Given that Km = 1mM = 10³ uM

Vmax = 5 * 10^-3 uM s-1

[S] = 0.25 mM or 0.25*10³ uM

Vo = 5 * 10^-3 * 0.25 * 10³ / 1 * 10³ + 0.25*10³

= 5 *10^-3 * 0.25 * 10³ /1.25 *10³

= 5 *10^-3 * 0.25 * 10³ *10^-3 / 1.25

= 4 *10^-3 * 0.25 * 10³ * 10^-3

1 * 10^-3 * 10⁶ * 10^-6

= 1 * 10^-3 uM/s

So initial velocity is 1 * 10^-3 uM/s

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