How to calculate the turnover number?

I understand that Kcat (turnover number) = Vmax/total enzyme concentration. However the formula I have been given is Kcat = specific activity/molecular mass of enzyme.

What is the relationship between these two different formulas - why do they give the same answer?

Definition of Kcat that I am using: specific activity * molecular mass of the enzyme Specific activity = units of enzyme/ mass of protein in mg

• Are you sure your formula refers to the molecular weight of the enzyme, and not the total mass of enzyme in the reaction? It would be helpful to post a copy of the question you are working on. – J-- Nov 17 '18 at 19:12
• @J-- I've uploaded the formulae – Anya Nov 17 '18 at 22:18
• Please do not upload text as images. It is impossible to index and it discriminates against people with sight defects. – David Nov 18 '18 at 10:47
• Yes, and especially don't post images of sideways text. – canadianer Nov 19 '18 at 23:12
• Do not post images as questions. There cannot be indexed by the search and pose problems for persons with sight problems. Please correct this to avoid closure and subsequent deletion of the question. – Chris Nov 20 '18 at 5:53

Let's do some unit analysis:

$$\pu t$$ = time

$$\pu U$$ = units = $$\mathrm{mol_{product}/t}$$

$$\pu s=$$ specific activity $$\mathrm{= U/g_{protein}}$$

Therefore:

$$\mathrm{s = mol_{product} / (g_{protien} \times{t})}$$

$$\mathrm{k_{cat}= s \times MW_{protien}}$$

And we know that:

$$\mathrm{MW_{protein} = g_{protein} / mol_{protein}}$$

Substituting definitions of specific activity and MWprotein:

$$\mathrm{k_{cat}= mol_{product} \times g_{protein} / (g_{protien} \times t \times mol_{protein})}$$

$$\mathrm{k_{cat} = mol_{product} / (mol_{protein} \times t)}$$

Cancel out the molar terms:

$$\mathrm{k_{cat} = t^{-1}}$$

This yields the correct unit for kcat ($$t^{-1}$$). You can think of this as the number of product molecules formed per enzyme molecule, per unit time.