Question

Enzymes. When k2 is very small compared to k–1 the KM becomes a Kd (or in some textbooks this is called a KS ) for the ES complex. A Kd is really an equilibrium constant (Keq ) for a dissociation reaction in which the products are the dissociated form of the reactants. Write the reaction for which this Kd equals the Keq for your reaction. Make a generalization about the relationship between KM and enzyme-substrate binding affinity.

When k2 is very small compared to k1 and k–1, what process is rate determining?

Answer 1

A small Km indicates that the enzyme requires only a small amount of substrate to become saturated.

Hence, the maximum velocity is reached at relatively low substrate concentrations

. A large Km indicates the need for high substrate concentrations to achieve maximum reaction velocity.

enzymes break down and bond molecules (substrates) they fit into eachother like a lock and key one enzyme will only fit togeather with one type of substrate and so there are many differnt enzymes that have different active zones for so substrates can find an enzyme that it fits in to.

The KM does not vary with enzyme concentration .

Km is the concentration of substrate at which the enzyme will be running at "half speed".

If you doubled the amount of enzyme, sure the Vmax is going to increase..

1/2 Vmax will increase too, obviously.

But Km, the amount of substrate at which half of the enzymes are working and half of the enzymes are bored and txting on their iphones, will remain the same.

These problems are typically set up such that there is an overwhelming amount of substrate, relatively little enzyme, and the empty enzyme starts processing the substrate by randomly bumping into it and sticking to it.

So doubling the amount of enzyme simply doubles the number of workers who still randomly bump into the enormous amounts of substrate at half of their capacity.