# The Michaelis-Menten Equation

The enzyme (E) and substrate (S) react to form the enzyme-substrate complex (ES).

**E + S <=> ES <=> P + E**

**k _{1}**: the forward rate of E + S to ES

**k**: the forward rate of ES to P + E

_{2}**k**: the reverse rate

_{-1}**k**: the reverse rate

_{-2}During the initial reaction where [P] is small (usually the first 60 seconds), k_{-2} will be negligible. During this time [S] will be essentially constant.

__Formation of ES__: V_{1} = k_{1}[E][S]

__Degradation of ES__: V_{2} = k_{-1}[ES] + k_{2}[ES] = [ES] (k_{-1 }+ k_{2})

**Steady state** occurs when V_{1} = V_{2}.

Setting the above equations equal yields the following:

**K _{m}** = [E][S]/[ES] = (k

_{-1 }+ k

_{2})/ k

_{1}

**K _{m}** is the

**Michaelis-Menten Constant**

__Observations:
__

If K

_{m}is

**high**=> [ES] is low =>

**low**substrate affinity

If K

_{m}is

**low**=> [ES] is large =>

**high**substrate affinity

These correlations are only true if k_{2 }<< other rate constants. That is, when k_{2 }is rate-limiting: k_{-1}/k_{1 } ≈ K_{m}. i.e., when the conversion of ES into P + E is the rate-limiting step.

[E] and [ES] are very difficult to determine experimentally. Instead, use the total enzyme concentration **E**_{T} defined as **[E _{T}] = [E] + [ES]**

Substitute this into the equation of **K _{m}** and rearrange:

K_{m} = ([Et]/[ES] – 1)[S]

Next, eliminate [ES]. Recall that **V _{0 } = k_{2}[ES]**. When there’s a high concentration of substrate, the reaction achieves its maximum velocity at which point all the enzyme is associated to an ES complex. The total concentration of enzyme

**E**then becomes the same as the concentration of ES.

_{T }Thus, an alternate expression for Vmax is **V _{max } = k_{2}[E_{T}] **: all enzyme is in ES complex.

Substituting these relations in to the K_{m} equation above yields:

K_{m} = (V_{max}/V_{0}-1)[S] => **V _{0} = V_{max}[S]/(K_{m} + [S])**

This is the** Michaelis-Menten Equation. **It’s graph is a **hyperbolic** curve, as seen before experimentally

What predictions are now possible?

- When [S] is low => V
_{0}≈ (V_{max * }[S])/K_{m}. This means Vis proportional to [S]. This is not good for a cell. The reaction becomes very sensitive to changes in concentration of substrate. However, the Enzyme will never reach its maximum velocity, and thus will not work at its full potential._{0} - When [S] is high => V
_{0}≈ V_{max}. This means V_{0}is proportional to V_{max }, and thus the cell can not fine tune the activity of the enzyme because the E is always working at its maximum capacity, which is a wasteful degradation of substrate even when the product of the reaction is not needed. - When
**[S] = V**=> V_{max}_{0}≈ V_{max}/2. This is**optimal**for the cell because the E can both be sensitive to changes in [S], but it can also reach its maximum velocity.

This latter observation sets up an alternate definition of K_{m}: **the [S] needed to reach half of maximum velocity.**

Note: as will be discussed in a future lecture, Km does not apply to **allosteric** enzymes. Those have a different property, which will be explained.

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