Question

1) For the Vmax obtained in question 26, calculate the turnover number assuming that 1 x 10-4 mol of enzyme were used.

Vmax was found to be 0.681 mM min -1

answer is 20.43 min -1

2) You do an enzyme kinetic experiment and calculate a Vmax of 100 μmol of product per minute. If each assay used 0.1 mL of an enzyme solution that had a concentration of 0.2 mg/mL, what would be the turnover number if the enzyme had a molecular weight of 128,000 g/mol?

Answer: The number of moles of enzyme is 1.56 x 10^-10. The turnover number is 10,700 sec -1.

@Frostbite

Answer 1

Hey again @Summersnow8

for 1):
Within the MM-framework the maximum reaction rate is assumed to happen when all enzyme is bound as enzyme-substrat complex. \(V_{max}\) is therefore define as the following:
\[\Large V_{\max}=k_{cat} \times [E]_0\]
Now I can however see we got a unit problem. In order for the units to cancel you need to have a volume of enzyme concentration.

for 2): you simply need to units to cancel. The issue is I show you then I give the answer away but think about the equation I wrote in 1) and think about how all units cancel to give you the answer. Write if this is insufficient help, and I try help some more! :)

Answer 2

@Frostbite
1) Converting Vmax units:
\[0.681 \frac{ mM }{ \min } \times \frac{ 1 mole }{ 1000 mM } = 0.000681 \frac{ mole }{ \min }\]

\[V _{\max} = \kappa _{cat} \times \left[ E \right]_{0}\]
\[0.000681 \frac{ mole }{ \min } = \kappa _{cat} \times 0.0001 moles \]
\[\kappa _{cat} = \frac{ 0.000681 \frac{ moles }{ \min} }{0.0001 moles }\]
\[\kappa _{cat} = 6.81 \min ^{-1}\]

I don't know how to get the answer to be 20.43 min -1

Answer 3

2) Data:

\[V _{\max} = 100 μmol/min \]
\[V _{\max} = 100 μmol / min \times \frac{ 1 mol }{ 1000000 μmol }\]
\[V _{\max} = 1 \times 10^{-4} mol/min\]
\[V _{\max}= 1 \times 10^{-4} \frac{ mol }{ \min} \times \frac{ 1 \min }{ 60 \sec } = 1.666\times10^{-6} \frac{ mol }{ \sec }\]


unit conversions:
\[\left[ E \right]_{0}= 0.1 mL \times 0.2 \frac{ mg }{ mL } \times \frac{ 1 g }{ 1000 mg} \times \frac{ mol }{ 128,000 g }\]
\[\left[ E \right]_{0} = 1.5625 \times10^{-10} moles\]

\[V _{\max} = k_{cat} \times \left[ E \right]_{0}\]
\[k _{cat} = \frac{ V _{\max} }{ \left[ E \right]_{0}}\]
\[k _{cat} = \frac{ 1.666\times10^{-6} \frac{ moles }{ \sec } }{ 1.5625\times10^{-10} moles } = 10679 \sec ^{-1} = 10700 \sec ^{-1}\]

Answer 4

in 1) what is your volume of enzyme?

Answer 5

1x10^-4 moles?

Answer 6

V max = 0.681 mM / min

Answer 7

@Frostbite, I don't see a volume given for this particular question

Answer 8

I guess I am unsure where I am making the mistake because I was able to solve question 2, but not question 1 :/

Answer 9

The problem lies in the very first line:
\[0.681 \frac{ mM }{ \min } \times \frac{ 1 mole }{ 1000 mM } = 0.000681 \frac{ mole }{ \min }\]
When you want to found the amount of mole/min we need a volume for this to be true.

Answer 10

Can you see what I mean? :)

Answer 11

@Frostbite, yes I do.... I am suppose to use the Vmax from question 26. This is 26, but I am confused

Answer 12

I am wondering which "volume" to use and if I use a volume how would I cancel them out to get an answer that is min -1

Answer 13

Can you help me with question 2? @caozeyuan