Question

An enzyme with a Km of 1.2x10-4 M was assayed at an initial sustrate concentration of 0.02 M. By 30 seconds, 2.7 μmoles/liter of product had been produced. How much product will be present at (a) 3 min and (b) 5.3 minutes? (c) What percent of the original substrate is used up at (c) 3 min and (d) 5.3 minutes?

Answer 1

Ater 3 minute

product =
`1.629* 10^(-5)`Molar

susbtrate = 0.081%

after 5.3 minutes

product =
`2.879* 10^(-5) M`

susbtrate = 0.01435

Step-by-step explanation:

Given data:

`Km = 1.2* 10^(-4) M`

Sustrate concentration = 0.02 M

Amount of product
`2.7\mu moles/ltr`

Rate = dP/dT = product formed per time

`= (2.7 * 10^(-6) moles)/(30) = 9 * 10^{-8`

we know rate
`= (v_(max)[S])/(Km + S)`

hence
`9* 10^(-8) = (v_(max)(0.02))/((1.2* 10^(-4) + 0.02))`

`v_(max) = 9.054* 10^(-8)`

now after 3 min = 180 sec

product formed
`dP = rate * dT = 9.054* 10^(-8)* 180 = 1.629* 10^(-5)`Molar,

% substrate utilised
`= ((1.629* 10^(-5)))/(0.02) * 100 = 0.081`%

after 3.5 minutes = 318

`dP = 9.054x10^(-8)* 318 = 2.879* 10^(-5) M`

%
`of susbtarte\ utilised =(2.879* 10^(-5))/(0.02) * 100 = 0.0143` %