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Enzyme Kinetics & Steady State

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For a MM reaction
k1=5x10^7
k-1=2X10^4
k2=4x10^2

i calculated Km and Ks to be equal at about 4 x10^-4M

Does substrate binding achieve equilibrium or the steady state?

Could you clarify when you have ph=pkA how you get whether it is -1/2 or +1/2 or in the other case +1/3 or -1/3. How do you tell whether it is + or -

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ENZYME KINETICS -- EQUILIBRIUM VS. STEADY STATE CONDITIONS

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for a Michaelis-Menten reaction:

k1=5x10^7
k-1=2x10^4
k2=4x10^2

Km = (k-1 + k2)/k1 = 4x10^-4

Km = Ks + (k2/k1), therefore:

Ks = Km - (k2/k1) = 4x10^-4

Therefore, your calculations are correct. Km and Ks are about equal at 4x10^-4 M.

Question: Does substrate binding achieve equilibrium or the steady state?

Response: The assumption made in the equilibrium derivation is that k-1 >> k2. In other words, the ES complex can be assumed to dissociate back to E + S almost exclusively, and not to E + P. That means that E, S and ES are assumed to be in equilibrium. That's why we base our calculations on Ks, ...

Solution Summary

484-word solution includes calculations for a Michaelis-Menten reaction and a conceptual explanation of the pH=pkA equation.

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