[mightysparks]

We did a lab on 'Effect of pH/temp/substrate concentration on enzyme activity' using alkaline phosphatase. I understand all of the calculations except for one bit in the sample calculation (in bold):

The Lab (for pH):
We had 10 tubes, 5 tests and 5 blanks. Each tube 1-5 had a different pH and 1.0mL of 7.5mM substrate (pNPP) and 1.0mL buffer.

We added 1.0mL alkaline phosphate to the tests. After incubating, we stopped the reaction by adding 1mL 2M NaOH to the tubes.

We added 1.0mL 2M NaOH to the blanks. Then added 1mL of alkaline phosphatase to the blanks.

Sample Calc
a=Ebc
.500 = 18300 x c x 1cm
c = .500/18300 M

c = .500/18300 x 10⁹ nM
ie .500/18300 x 10⁹ nmoles in 1000mL
? nmoles in 4mL

=.500/18300 x 10⁹ x 4/1000 nmoles produced in 10 min by 1 mL enzyme

=.500/18300 x 10⁹ x 4/1000 x 1/10 nmoles min^-1 mL^-1

=10.0nmoles pNP


Where the hell did 4mL come from?? But then in the sample given to us for the substrate concentration activity, we were given 3mL. I have no idea where these volumes are coming from.

[Borek]

As far as I can tell 1+1+1+1=4

[mightysparks]

Yeah I figured that, but I wasn't sure why that was 4mL and the substrate concentration one was 3mL.

The substrate concentration activity had a 'hint': To calculate the concentration of substrate for each tube, remember that the substrate concentration is the concentration of substrate in mmoles.L^-1 (mM) in the test mixture at the start of the enzyme incubation. This volume includes all the reagents except NaOH.

That activity also added a total of 4mL but in our prelab the example he gave us used 3mL in the calculation.

I probably should've put the lab for Activity 3 instead. In that one, we had 13 test tubes (1-7 were tests) and they each had 1.0mL of buffer, each had a differing amount of pNPP and water adding to 1mL (ie, test tube 2 had 0.1mL of pNPP and 0.9mL of water)

We then added the same amount of alkaline phosphatase and NaOH.

[mightysparks]

Urgh, nevermind I read the whole thing wrong. I get it now.