Also find the Vmax and Km when no inhibitor is present, with A inhibitor and wit

Question

Also find the Vmax and Km when no inhibitor is present, with A inhibitor and with B inhibitor.

50. The table presents the rates of reaction at specific substrate concentrations for an enzyme that displays classical Michaelis-Menten kinetics. Two sets of inhibitor data are also included. Determine the Km and Vmax for the uninhibited enzyme. [S] (mM) 1.3 2.6 6.5 13.0 26.0 Without inhibitor 2.50 4.00 6.30 7.60 9.00 With inhibitor A 1.17 2.10 4.00 5.70 7.20 With inhibitor B 0.62 1.42 2.65 3.12 3.58 51. Determine the type of inhibition exhibited by each inhibitor in Question 50. What kind of affinity would these inhibitors have for ES vs E? Also find the Vmax and Km when no inhibitor is present, with A inhibitor and with B inhibitor.

Explanation / Answer

1) No inhibitor is present

we know that

Vo = Vmax S / ( Km + S )

Vo2 / Vo1 = S2 ( Km + S1 ) / S1 ( Km + S2 )

consider Vo= 4 and 9

9/4 = 26 ( Km + 2.6 ) / 2.6 ( Km + 26 )

solving we get

Km = 4.1935 M

consider at Vo = 9

9 = Vmax x 26 / ( 26 + 4.1935 )

Vmax = 10.45 uM/s

so Km = 4.1935 M and Vmax = 10.45 uM/s

2) WIth inhibitor A

Vo2 / Vo1 = S2 ( Km + S1 ) / S1 ( Km + S2 )

consider Vo= 2.1 and 7.2

7.2/2.1= 26 ( Km + 2.6 ) / 2.6 ( Km + 26 )

solving we get

Km = 9.60 M

consider at Vo = 7.2

7.2 = Vmax x 26 / ( 26 + 9.60 )

Vmax = 9.86 uM/s

so Km = 9.60 M and Vmax = 9.86 uM/s

3) with inhibitor B

Vo2 / Vo1 = S2 ( Km + S1 ) / S1 ( Km + S2 )

consider Vo= 1.42 and 3.58

3.58/ 1.42 = 26 ( Km + 2.6 ) / 2.6 ( Km + 26 )

solving we get

Km = 5.288 M

consider at Vo = 3.58

3.58 = Vmax x 26 / ( 26 + 5.288 )

Vmax = 4.31 uM/s

so Km = 5.288 M and Vmax = 4.31 uM/s

4)

A is competitive inhibitor as Km is increased and Vmax almost remains the same .

B is non competitive inhibitor as Km is almost same and Vmax is decreased .

A binds to E but not to ES

B has same affinity for both E and ES

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