Vi= VmaxS/Km+S[1+I/Ki]
Degree of inhibition= 1+(I/Ki)
Vmax= 270nmoles/litre/min
Km = 0.002M
Ki= 0.00015M
1) I= 0.002, S= 0.002
Vi= 270×0.002/0.002+0.002[1+0.002/0.00015]
=0.54/0.002+(0.002×14.33)
=0.54/[0.002+0.028]
=0.54/0.03
Vi=18 nmoles/litre/min
Degree of inhibition alpha= 1+(I/Ki)
= 1+(0.002/0.00015)
= 14.33
2) I= 0.00002, S= 0.0004
Vi= 270× 0.0004/0.002+0.0004[1+0.00002/0.00015]
= 0.108/0.002+(0.0004×1.13)
= 0.108/(0.002+0.00045)
= 0.108/0.0024
Vi= 45 nmoles/litre/min
Degree of inhibition alpha= 1+(0.00002/0.00015)
= 1.13
3) I= 0.00005, S= 0.0075
Vi= (270×0.0075)/0.002+0.0075[1+(0.00005/0.00015)]
= 2.025/0.002+0.0075(0.0002/0.00015)
= 2.025/0.002+0.0075×1.33
= 2.025/0.002+0.01
= 2.025/0.012
Vi= 168.75 nmoles/litre/min
Degree of inhibition alpha= 1+(0.00005/0.00015)
= 0.0002/0.00015
= 1.33
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