Enzyme kinetics: Quick recap E + S ? ES ? E + P k1, k-1, k2 Vary substrate concentration... • What happens at zero [S]? • What happens at maximum [S]? • What could you do to improve kcat? • What could you do to improve Km? • What could you do to improve Vmax? • What could you do to change apparent Km? • What are some assumptions of Michaelis-Menten? Initial velocity, V0 (M/min) Substrate concentration, [S] (mM) Vmax 1/2 Vmax Km Michaelis-Menten plot V0 = (Vmax [S]) / (Km + [S])
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The Michaelis-Menten model is a common framework used to describe the rate of enzymatic reactions. It involves the formation of an enzyme-substrate complex (ES) and its conversion to product (P). Show more…
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An enzyme that follows simple Michaelis-Menten kinetics has an initial reaction velocity of 10 μmol·min⁻¹ when the substrate concentration is five times greater than the KM. What is the Vmax of this enzyme? Vmax = 1.1 μmol·min⁻¹
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For an enzyme that displays Michaelis-Menten kinetics, what is the reaction velocity, $\left.V \text { (as a percentage of } V_{\max }\right)$, observed at the following values? (a) $[\mathrm{S}]=K_{\mathrm{M}}$ (b) $[\mathrm{S}]=0.5 K_{\mathrm{M}}$ (c) $[\mathrm{S}]=0.1 K_{\mathrm{M}}$ (d) $[\mathrm{S}]=2 K_{\mathrm{M}}$ $(\mathrm{e})[\mathrm{S}]=10 K_{\mathrm{Y}}$
The rate of a simple enzyme reaction is given by the standard Michaelis-Menten equation: \[\text { rate }=V_{\max }[\mathrm{S}] /\left(K_{\mathrm{M}}+[\mathrm{S}]\right)\] If the $V_{\max }$ of an enzyme is 100 \mumole/sec and the $K_{M}$ is $1 \mathrm{mM},$ at what substrate concentration is the rate $50 \mu \mathrm{mole} / \mathrm{sec} ?$ Plot a graph of rate versus substrate $(\mathrm{S})$ concentration for $[\mathrm{S}]=0$ to $10 \mathrm{mM}$. Convert this to a plot of 1/rate versus $1 /[\mathrm{S}] .$ Why is the latter plot a straight line?
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