The Michaelis-Menten equation models the hyperbolic relationship between [S] and the initial reaction rate V? for an enzyme-catalyzed, single-substrate reaction E + S ? ES ? E + P. The model can be more readily understood when comparing three conditions: [S] << K?, [S] = K?, and [S] >> K?. Match each statement with the condition that it describes. Note that "rate" refers to initial velocity V? where steady state conditions are assumed. [E?????] refers to the total enzyme concentration and [E????] refers to the concentration of free enzyme. [S] << K? [S] = K? [S] >> K? Not true for any of these conditions Almost all active sites are empty. The rate is directly proportional to [S]. Half of the active sites are filled with S. [ES] is much higher than [E????]. This condition rarely occurs for most in vivo enzymes. Increasing [E?????] will lower K?. Answer Bank
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In this condition, [ES] is much higher than [Efree] and almost all active sites are empty. ** Show more…
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The Michaelis-Menten equation models the hyperbolic relationship between [S] and the initial reaction rate V0 for an enzyme-catalyzed, single-substrate reaction E + S ⇌ ES → E + P. The model can be more readily understood when comparing three conditions: [S] << Km, [S] = Km, and [S] >> Km. Match each statement with the condition that it describes. Note that "rate" refers to initial velocity V0 where steady state conditions are assumed. [Etotal] refers to the total enzyme concentration and [Efree] refers to the concentration of free enzyme. [S] << Km [S] = Km [S] >> Km Not true for any of these conditions The rate is directly proportional to [S]. Almost all active sites are empty. Half of the active sites are filled with S. This condition rarely occurs for most in vivo enzymes. Increasing [Etotal] will increase Km. [ES] is much higher than [Efree]. Answer Bank
Shaiju T.
The Michaelis-Menten equation models the hyperbolic relationship between [S] and the initial reaction rate V0 for an enzyme-catalyzed, single-substrate reaction E + S ⇌ ES ⟶ E + P. The model can be more readily understood when comparing three conditions: [S] << Km, [S] = Km, and [S] >> Km. Match each statement with the condition that it describes. Note that "rate" refers to initial velocity V0 where steady state conditions are assumed. [Etotal] refers to the total enzyme concentration and [Efree] refers to the concentration of free enzyme. [S] << Km [S] = Km [S] >> Km Not true for any of these conditions Almost all active sites are empty. The rate is directly proportional to [S]. The rate is half of the maximum rate. [ES] is much higher than [Efree]. This condition rarely occurs for most in vivo enzymes. Increasing [Etotal] will increase Km. Answer Bank
Adi S.
The rate equation for an enzyme subject to competitive inhibition is $$V_{0}=\frac{V_{\max }[\mathrm{S}]}{\alpha K_{\mathrm{m}}+[\mathrm{S}]}$$,Beginning with a new definition of total enzyme as $$\left[\mathbf{E}_{t}\right]=[\mathbf{E}]+[\mathbf{E S}]+[\mathbf{E I}]$$ and the definitions of $a$ and $K_{1}$ provided in the text, derive the rate equation above. Use the derivation of the Michaelis-Menten equation as a guide.
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