[Oishee]

Hello!

My Vmax value came out as 1004 micromol/min/mL. The enzyme's concentration used was 1 microgram/mL. The molecular weight of the enzyme was 50kDa. We had three problems to solve:

1) Convert Vmax into moles substrate converted per minute per mg of enzyme.

2) Convert Vmax into moles substrate converted per minute per enzyme molecule.

3) When you convert Vmax in this way, what value are you obtaining?
a. kcat
b. turnover number: the number of molecules catalysed per second per enzyme molecule
c. another way of expressing Vmax
d. Km

Now, this is what I've done so far:

1) 1.004 micromol/min/mL = 1.004x10^3 micromol/min/mg = 1004 micromol/min/mg

Do we convert to mol/min/mg then?

2) Number of mol of enzyme in 1 mg = 1/(50x10^3x10^3) = 2x10^-8
because molecular weight is 50x10^3 g/mol, which means 1/(50x10^3) mol/g. So, in 1 mg, times by 10^3.

Therefore, number of molecules of enzyme in 1 mg = 2x10^-8 x 6.02x10^23 = 1.204x10^16

So, Vmax in this format = (1004x10^-6)/(1.204x10^16) = 8.34x10^-20 mol/min/molecule

Am I correct?

3) I don't think its either a or b, and it can't be d. So is this really just another way of stating Vmax?

Please *delete me*

Warm regards,
Oishee

[Babcock_Hall]

Can you define the turnover number for us?

[Oishee]

Can you define the turnover number for us?

Wikipedia says:
en.wikipedia.org/wiki/Turnover_number

In enzymology, turnover number (also termed kcat) is defined as the maximum number of molecules of substrate that an enzyme can convert to product per catalytic site per unit of time (a turnover rate).

[Babcock_Hall]

The definition of turnover number should make clear the condition that the enzyme is saturated with substrate.  Perhaps that is the implication of the word "maximum" in the Wikipedia definition (so the definition is fine).  The mathematical definition of k_cat is V_max/[E_tot].  That should take care of question 3.  With respect to question 1, as you say converting µmoles to moles should do it.

[Oishee]

The definition of turnover number should make clear the condition that the enzyme is saturated with substrate.  Perhaps that is the implication of the word "maximum" in the Wikipedia definition (so the definition is fine).  The mathematical definition of k_cat is V_max/[E_tot].  That should take care of question 3.  With respect to question 1, as you say converting µmoles to moles should do it.

Wait, so… whatever we've done here IS in fact "just another way of expressing Vmax"!?

And the bizarre thing is that prior to these calculations, the paper says, "Convert from moles to micromoles."

Like, we had to work out Vmax in these units from change in absorbance/min, which I've done to get 1.004 micromol/min/mL. Yet he clearly states to convert to micromoles, but now he's asking for moles again?!

>_<

So in other words, what I've done is correct? But why did we need to do this? What does this show?

;___;
Sorry for being dumb.

[Oishee]

Hello!

My Vmax value came out as 1004 micromol/min/mL. The enzyme's concentration used was 1 microgram/mL. The molecular weight of the enzyme was 50kDa. We had three problems to solve:

1) Convert Vmax into moles substrate converted per minute per mg of enzyme.

2) Convert Vmax into moles substrate converted per minute per enzyme molecule.

3) When you convert Vmax in this way, what value are you obtaining?
a. kcat
b. turnover number: the number of molecules catalysed per second per enzyme molecule
c. another way of expressing Vmax
d. Km

Now, this is what I've done so far:

1) 1.004 micromol/min/mL = 1.004x10^3 micromol/min/mg = 1004 micromol/min/mg

Do we convert to mol/min/mg then?

2) Number of mol of enzyme in 1 mg = 1/(50x10^3x10^3) = 2x10^-8
because molecular weight is 50x10^3 g/mol, which means 1/(50x10^3) mol/g. So, in 1 mg, times by 10^3.

Therefore, number of molecules of enzyme in 1 mg = 2x10^-8 x 6.02x10^23 = 1.204x10^16

So, Vmax in this format = (1004x10^-6)/(1.204x10^16) = 8.34x10^-20 mol/min/molecule

Am I correct?

3) I don't think its either a or b, and it can't be d. So is this really just another way of stating Vmax?

Please *delete me*

Warm regards,
Oishee

Sorry, I mean 1.004 micromol/min/mL for the original Vmax value.

[Babcock_Hall]

I think in order to calculate k_cat, you would need to have the rate and the enzyme amount match in units. meaning that if the rate were in µmols per minute and the amount of enzyme were in µmols, then you would have it.  By the way, I usually try to express my rates in µmol per minute because it allows one to calculate activity and specific activity more easily.

[Oishee]

I think in order to calculate k_cat, you would need to have the rate and the enzyme amount match in units. meaning that if the rate were in µmols per minute and the amount of enzyme were in µmols, then you would have it.  By the way, I usually try to express my rates in µmol per minute because it allows one to calculate activity and specific activity more easily.

You're so ambiguous >_<

[Babcock_Hall]

I don't think that the calculation you have done is a calculation of k_cat.

[Oishee]

I don't think that the calculation you have done is a calculation of k_cat.

Ah, yes. That was the best blunt answer ever
I'm sorry for being silly.

So I'm presuming the numerical answers are okay otherwise? I still don't get why we had to calculate Vmax this way.