Bio....2Q - maryam2009

[ArchivalUser]

Consider a reaction that can be catalized by one of two enzymes,A and B,with the following kinetics

.....................Km(M).................Vmax(mmol/min)
A...................5*10^-6...............20
B...................5*10^-4...............30

At a concentration of 5*10^-4 M substrate,the velocity of the reaction catlized by enzyme A will be:

A.10
B.15
C.20
D.25
E.30

please explain your answer.Thank you

[ArchivalUser]

ans, C, I think v max doesnot change with sub con

[ArchivalUser]

v=v max[s]/Km+[s]
c is ans

[ArchivalUser]

Agree with CC........

[ArchivalUser]

the answer is C

As dr_eurekha mentioned

....the Vmax depends on Number of the enzymes and absolutely unrelated to Substrate concentrations....at the concentration 5*10^-4 ,enzyme A works at its Vmax,which is 20 mmol/min.

Thank you

[ArchivalUser]

And you

[ArchivalUser]

Hello,

Because I am a little confused, could you please help me understand the question? Does it ask about the velocity of the reaction or the Vmax of the reaction?

Thank you so much,

Elpida

[ArchivalUser]

Vmax is the maximum velocity or rate at which the enzyme catalyzed a reaction...and it depends only on Number of the enzymes not substate concentrations,so increased or decreased [substreaes] will not change the amount of Vmax
....so in this case at the new concentration (5*10^-4) ,enzyme A works at its Vmax,which is 20 mmol/min.

.

[ArchivalUser]

Vmax is the maximum velocity or rate at which the enzyme catalyzed a reaction...and it depends only on Number of the enzymes not *substrate concentrations,so increased or decreased* [substrates] will not change the amount of Vmax
....so in this case at the new concentration (5*10^-4) ,enzyme A works at its Vmax,which is 20 mmol/min.

[ArchivalUser]

Thank you so much for your explanation. Is the substrate concentration power to -4 or power to 4?

Thank you again, I appreciate your help