Find the initial velocity for an enzymatic reaction when vmax = 6.5 x 10–5 mol•sec–1 , [s] = 3.0 x 10–3 m, and km = 4.5 x 10–3 m. a) not enough information is given to make this calculation b) 2.6 x 10–5 mol•sec–1 c) 1.4 x 10–2 mol•sec–1 d) 8.7 x 10–3 mol•sec–1 e) 3.9 x 10–5 mol•sec–1

correct option is the option B= 2.6 × 10^-5 mol.sec^-1

Explanation:

Check attached file/picture for the plot of graph of reaction velocity against substrate concentration.

Initial velocity can be defined as the velocity at the beginning of an enzyme-catalyzed reaction. It is usually denoted as V°.

In the equation of reaction below;

A + B----------> [AB] ------------>Product

<--------

Where the forward reaction is k1 that is the rection producing the intermediate [AB].

And, the backward reaction is k-1 that is the arrow pointing back to the Reactants.

K2 is the arrow from the intermediate [AB] to the products.

Note that k1 and k-1

represent rapid non-covalent dissociation of substrate from enzyme active site. Also, K2 is the rate constant for the formation of the products.

In order to calculate the initial velocity,V° we need to use the Michaelis- Menteen equation.

V° = Vmax ×(S)/ Km + (S)-------------------------------------------------------------(1).

From the question the parameters given are;

Vmax = 6.5 x 10–5 mol•sec–1 , [S] = 3.0 x 10–3 M, and KM = 4.5 x 10–3 M.

Therefore, slotting in the above parameters into the equation (1) above, we have;

V° = 6.5 x 10–5 mol•sec–1 × 3.0 x 10–3 M / 4.5 x 10–3 M + 3.0 x 10–3 M.

Initial velocity,V° = 2.6 x 10-5 moles per sec..

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