Question

enzyme‑catalyzed, single‑substrate reaction E + S − ⇀ ↽ − ES ⟶ E + P . The model can be more readily understood when comparing three conditions: [ S ] << K m , [ S ] = K m , and [ 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 [ E free ] refers to the concentration of free enzyme.

Answer 1

Complete Question

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Answer:

[S]<<KM             |   [S]=KM                  |  [S]>>KM                     | Not true

____________  |   Half of the active  | Reaction rate is         | Increasing

$[E_{free}]$ is about   |    sites are filled of  |    independent of      |  $[E_{Total}]$ will                                            

 equal to $[E_{total}]$. |                                 |   [S]                             | lower KM

_____________________________________________|____________

[ES] is much       |                                 | Almost all active

 lower than         |                                 | sites are filled

$[E_{free}]$                  |                                 |

Explanation:

Generally the combined enzyme$[ES]$ is mathematically represented as

                   $[ES] = \frac{[E_{total}][S]}{K_M + [S]}----(1)$

for Michaelis-Menten equation

Where $[S]$ is the substrate concentration and $K_M$ is the Michaelis constant

Considering the statement $[S] < < K_M$

  Looking at the equation $[S]$ is denominator so it can be ignored(it is far too small compared to $K_M$)  hence the above equation becomes

               $[ES] = \frac{[E_{total}][S]}{K_M}$

Since $[S]$ is less than $K_M$ it means that $\frac{[S]}{K_M} < < 1$

so it means that $[ES] < < [E_{total}]$

  What this means is that the  number of combined enzymes$[ES]$ i.e the number of occupied site is very small compared to the the total sites $[E_{total}]$  i.e the total enzymes concentration which means that the free sites [$E_{free}]$  i.e the concentration of free enzymes is almost equal to $[E_{total}]$

Considering the second statement

      $[S] = K_M$

So  this means that equation one would now become

           $[ES] = \frac{[E_{total}][S]}{2[S]} = \frac{[E_{total}]}{2}$

So this means that half of the active sites that is the total enzyme concentration are filled with S

Considering the Third Statement

      $[S] >>K_M$

In this case the $K_M$ in the denominator of equation 1 would be neglected and the equation becomes

       $[ES] = \frac{[E_{total}] [S]}{[S]} = [E_{total}]$

This means that almost all the sites are occupied with substrate

 The rate of this reaction is mathematically defined as

             $v =\frac{V_{max}[S]}{K_M [S]}$

Where v is the rate of the reaction(also know as the velocity of the reaction at a given time t) and $V_{max}$  is he maximum velocity of the reaction

In this case also the $K_M$ at the denominator would be neglected as a result of the statement hence the equation becomes

                $v = \frac{V_{max}[S]}{[S]} = V_{max}$

So it means that the reaction does not depend on the concentration of substrate [S]

For the final statement(Not True ) it would match with condition that states that increasing $[E_{total}]$ will lower $K_M$

This is because $K_M$ does not depend on enzyme concentration it is a property of a enzyme