December 03, 2020, 06:04:37 AM
Forum Rules: Read This Before Posting


Topic: Calculating the activity of an enzyme at Vmax  (Read 288 times)

0 Members and 1 Guest are viewing this topic.

Offline yogurtspoon

  • Regular Member
  • ***
  • Posts: 13
  • Mole Snacks: +0/-0
Calculating the activity of an enzyme at Vmax
« on: October 02, 2020, 05:07:32 PM »
Here's the problem:

A student who performed the experiments in this lab exercise* found the Vmax of alkaline phosphatase to be 0.073 umole substrate hydrolyzed per minute. The concentration of alkaline phosphatase in the stock solution used in the experiment was 0.7 mg protein/mL. Calculate the activity of the enzyme at Vmax as umole substrate hydrolyzed per minute per mg of enzyme protein.

*We basically did a Pierce 660 assay to measure the rate of a enzyme-catalyzed reaction with & without the presence of an inhibitor. This problem refers to the procedure without the inhibitor. I think all the info you'd need to know to answer is that the total volume of each cuvette (containing the reaction mix) was 3mL. Each cuvette contained 0.1mL alkaline phosphatase. The amount of water and substrate varied between cuvettes but iirc we shouldn't need those numbers for this calculation.

Here's I've done so far:
C1V1=C2V2
(0.7 mg/mL)(0.1 mL) = (C2)(3 mL)
C2 = .0233 mg/mL

At first I thought I could just divide 0.073 umol/min by that number... but then I realized that the units would be wrong. It'd give me umol*mL/min*mg, and the problem wants the answer in umol/min*mg.

I'm not sure what volume I would have to multiply  C2 by though, because doesn't that calculation already account for the volume of the cuvette & the stock soln?

I feel like I'm misunderstanding or overlooking something simple. Any help would be appreciated.

Offline Babcock_Hall

  • Chemist
  • Sr. Member
  • *
  • Posts: 4406
  • Mole Snacks: +269/-18
Re: Calculating the activity of an enzyme at Vmax
« Reply #1 on: October 03, 2020, 10:02:42 AM »
How many milligrams of alkaline phosphatase are in the assay?

Offline yogurtspoon

  • Regular Member
  • ***
  • Posts: 13
  • Mole Snacks: +0/-0
Re: Calculating the activity of an enzyme at Vmax
« Reply #2 on: October 03, 2020, 03:57:18 PM »
That would just be (0.7 mg/mL)(0.1 mL) = 0.21 mg. But doesn't the C1V1 equation already account for that?

Or would I just do 0.073 umol/min divided by 0.21mg? That would get the right units, but in that case Idk if I get why I could just ignore the 3mL cuvette. Would it be because 0.21mg refers to the amount of protein instead of the concentration? So the volume of the solution it's in doesn't really matter?

Offline Babcock_Hall

  • Chemist
  • Sr. Member
  • *
  • Posts: 4406
  • Mole Snacks: +269/-18
Re: Calculating the activity of an enzyme at Vmax
« Reply #3 on: October 04, 2020, 09:53:07 AM »
I don't believe that it is appropriate to use a dilution formula in this case.  I think that the reason behind the definition of specific activity might lie in the desire to nullify differences in the volumes of different assays, presumably by different workers.

Sponsored Links