Question

The Michaelis‑Menten equation models the hyperbolic relationship between [S] and the initial reaction rate V 0 V0 for an enzyme‑catalyzed, single‑substrate reaction E + S − ⇀ ↽ − ES ⟶ E + P E+S↽−−⇀ES⟶E+P . The model can be more readily understood when comparing three conditions:

[ S ] < [ S ]>>Km
[ S ] = K m

Match each statement with the condition that it describes. Note that "rate" refers to initial velocity V 0 where steady state conditions are assumed. [ E total ] refers to the total enzyme concentration and [Efree] refers to the concentration of free enzyme.

1. [ S ] < 2. [ S ]>>Km
3. [ S ] = K m
4. Not true for any of these conditions.

A. [Efree] is about equal to [Etotal]
B. Half of the active sites are filled with S.
C. [ES] is much lower than [Efree]
D. Almost all active sites will be filled.
E. Increasing [Etotal] will lower Km.
F. Reaction rate is independent of [S]

Answer 1

Answer:

The Michaelis‑Menten equation is given as

v₀ = Kcat X [E₀] X [S] / (Km + [S])

where,

Kcat is the experimental rate constant of the reaction; [s] is the substrate concentration and

Km is the Michaelis‑Menten constant.

Explanation:

See attached image for a detailed explanation

Answer 2

The Michaelis-Menten equation is presented as follows

$V_0 = \dfrac{K_{cat} \times [E_0] \times [S] }{\left(K_m+[S] \right)}$

$K_{cat}$ = The experimental reaction rate

[S] = The concentration of the substrate

$K_m$ = The Michaelis-Menten constant

1. [S] << $K_m$

When the concentration of the substrate concentration is very low, we get;

$V_0 = \dfrac{K_{cat} \times [E_0] \times [S] }{\left(K_m \right)}$

From the above equation, we have;

The rate, V₀, is directly proportional to the [S]

Where;

$[E_{total}]$ = $[E_0]$ = $[E_{free}]$ + [ES]

Given that there is a low substrate is low,  [ES] is also low, therefore;

$[E_{total}]$ ≈ $[E_{free}]$

Almost all the reaction sites are empty

Therefore, we have;

$[E_{free}]$ is about equal to $[E_{total}]$

[ES] is much lower than $[E_{free}]$

2. When [S] = $K_m$, we get;

$V_0 = \dfrac{K_{cat} \times [E_{total}] \times [S] }{\left([S]+[S] \right)} = \dfrac{1}{2} \times K_{cat} \times [E_{total}]$

The rate of the reaction is 50% of the maximum rate, therefore, the substrate is bound to half of the enzyme's active site, therefore;

$[E_{free}]$ = [ES]

Half of the reaction sites are filled with S

Therefore, we have;

Half of the active sites are filled with S

3. [S] >> $K_m$, therefore; ($K_m$ + [S]) ≈ [S], which gives;

$V_0 = \dfrac{K_{cat} \times [E_0] \times [S] }{\left([S] \right)} = K_{cat} \times [E_0]$

Almost all the active sits are filled

The reaction rate no longer depends on [S], and increasing the concentration of the substrate will not increase the rate

The reaction rarely occurs for most vivo enzymes

Almost all the active sites will be filled

The reaction rate is independent of [S]

4. Not true for any of these conditions

Increasing $[E_{total}]$  will lower $K_m$

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