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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number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.