Hello I need help with these qutestion that has many parts ! The following data

Question

Hello I need help with these qutestion that has many parts !

The following data were obtained from a single substrate enzyme that obey Michaleis- Menten kinetics:

Without inhbitors (X ) Vmax (micromole/min) = 50 and Km (micromole/Liter) = 10

With 100 micromole X inhbitiors Vmax (micromole/ min) = 50 and Km = 25 micromoles/liters.

If 10 mircrograms of enzyme were uesed in the acivity assay, which had total voulme of 2 ml and Molecular weight = 50,000 daltons (g/mol)

( a) what kind of inhbitor is X ?

(b)calculate the dissociation constant for X?

(c) what the catalytic rate constant ?

(d) what is the initial velocity of enzyme-catalyzed reaction at substate concerntration of 20 micromole in the absence of any inhibitor and in the presences of 100 micromole inhbitor X?

Explanation / Answer

(a) Here Vmax is same both in presence of inhibitor and in absence of inhibitor. However the value of Km increases. Hence this is an example of competitive inhibitor.

(c) Kcat = Vmax / [E]

Vmax = 50 microM / min

where [E] = concentration of enzyme

mass of enzyme = 10 microgram

Moles of enzyme = mass / molecular mass = 10 microgram / 50000 g/mol = 2x10-4 micromol

volume, V = 2 mL = 2x10-3 L

Hence [E] =  2x10-4 micromol / 2x10-3 L = 0.1 microM

Kcat = Vmax / [E] = 50 microMmin / 0.1 microM = 500 min-1 (answer)

(d) In absence of inhibitor: [S] = 20 microM

V max = 50 microM/min

Km = 10 microM/L

V = Vmax x [S] / (Km + [S]) =  (50 microM/min)x 20 microM / (10 microM/L + 20 microM)

= 33.33 microM / min (answer)

In presence of inhibitor: [S] = 20 microM

V max = 50 microM/min

Km = 25 microM/L

V = Vmax x [S] / (Km + [S]) =  (50 microM/min)x 20 microM / (25 microM/L + 20 microM)

= 22.22 microM / min (answer)

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