Question

6. An enzyme with a km of 0.06mmol/L hydrolyzed a substrate of a concentration 0.03 mmol/L. The initial velocity ws 0.0015 mmol/L.min!. Calculate the substrate concentration which gives an initial velocity of 0.003 mmol/L. min-1 7. Urease hydrolyzed urea at [s]=0.03mmol/L with a km of 0.06 mmol/L. The initial velocity observed was 1.5X10-3 mmol/L.min Calculate the initial velocity of the enzyme reaction when using [s]=0.12mmol/L

Answer 1

1. Given,

S1=0.003, Km=0.06, Vo1=0.0015, Vo2=0.003, S2=?

Answer 1: We know that,

V0= Vmax [S1] / Km + [S1]

0.0015= Vmax * 0.03 / 0.06 + 0.03

0.015= Vmax * 0.03 / 0.09

0.015 * 0.09= Vmax * 0.03

0.00135 = Vmax * 0.03

Vmax = 0.00135 / 0.03

Vmax =0.045

0.03 = 0.045 * S2 / 0.06 + S2

S2= 0.12 mmol/L

2. Given,

S1= 0.03, Km= 0.06, V_o1= 1.5 * 10^-3, S2= 0.12 mmol/L, V₀₂=?

Answer 2:- We know that, V₀₁= V_max [S1] / Km + [S1]

1.5 * 10^-3 = V_max * 0.03 / 0.06 + 0.03

1.5 * 10^-3 = V_max * 0.03 / 0.09

0.000135 = V_max * 0.03

V_max = 0.000135 / 0.03

V_max = 0.0045

V₀₂ = 0.0045 * 0.12 / 0.06 + 0.12

V₀₂ = 0.00054 / 0.18

V₀₂ = 3 * 10^-3 mmol/L.min^-1