Question

In: Chemistry

3. The kinetics of an enzyme are studied in the absence and presence of an inhibitor...

3. The kinetics of an enzyme are studied in the absence and presence of an inhibitor (A). The intial rate is given as a function of substrat concentration in the table.

	V0	V0
[S] (mmol/L)	no Inhibitor	Inhibitor A
1.25	1.72	0.98
1.67	2.04	1.17
2.5	2.63	1.47
5.00	3.33	1.96
10.00	4.17	2.38

(a) What kind oof competitor is inhibition is involved (competitive, noncompetitive, uncompetitive)? [Please Provide Graph]

(b) Determine Vmax and Km in the absence and presence of the inhibitor.

Expert Solution

The best way to find this is to calculate the reciprocal values and make a graph from these ( 1/S , 1/V0)

The graph goes like this

This graph describes the behavior of a noncompetitive inhibition, if this is true then the values of Km must be the same between the two graphs and the Vmax of the inhibitor must be lower than the no inhibition one.

Let´s get the values of slope and intercept

remember the equation y = mx + b, m is the slope and b the intercept

for the no inhibition one

I choose to points ( x1 = 0.8 y1 = 0.58 , x2 = 0.1 y2 = 0.239)

m = y2 - y1 / (x2-x1)

the value of b is gotten from the straight line equation

y = mx + b

b = y - mx

Choose the values of x2 and y2

b = 0.239 - 0.488*0.1 = 0.19

b = 1 / Vmax

Vmax for no inhibition = 1 / 0.19 = 5.2353

Remember that the value of slope m = Km / Vmax

so if we have a value of 0.488 of m then

Km = m * Vmax = 5.2353 * 0.488 = 2.55

Now if we repeat these calculation for the inhibitor A we get

m = 0.857

b = 0.334

V max= 2.99

Km = 2.56

As you can see

Vmax no inhibitor is greater than Vmax inhibitor A

Km is very similar 2.55 vs 2.56

So the inhibition is non competitive

Finally here´s a graph of 1/S vs 1/V0, values has been extrapolated

queen_honey_blossom answered 2 years ago

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