The below graph is the result of an experiment containing 0.10 M enzyme and vari

Question

The below graph is the result of an experiment containing 0.10 M enzyme and various substrate concentrations. The activity of the enzyme was measured with each substrate concentration and plotted as Vo vs [S] 4. 75 50 25 2 4 6 8 10 12 14 [S] (mM) What is Vmax for this enzyme with this substrate? Circle the answer. 0.80 M/min a. 10 M/min 0.050 mM/min 30 M/min 50 mM/min b. What is the Km of this enzyme for this substrate? Circle the answer 10p31 1.5 mM 0.50 M 1.5 M 1.0 mM What is the kcat (turnover number) for this enzyme with this substrate? Cirele the answer 5.0 x 10/s c. 5.0 x 10 /min 1.0/min 1.0 x 10 /s 5.0 x 10 /min

Explanation / Answer

The answers are to the best of my knowledge:

A)The answer should be 0.050mM/min as the max value of V is 50uM/min as seen in the above graph which converts to this answer when units are changed.

B) The answer to this should be 1.5mM as km is the value of substrate corresponding to Vmax/2 which is 25.

C) I am not sure about this one but its 5.0x102/min according to me from what i can gather from the internet. sorry in case this is wrong.

Feel free to leave a comment down below for any further query.

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