# Enzyme Kinetics

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Hi!

I am having a few problems calculating some required enzyme kinetic values. I've investigated the activity over 3 minutes of alcohol dehydrogenase at varying [S]. I used a spectrophotometer for this. From each of those graphs i've calculated the gradient (A/min) and converted this using the beer-lambert law and the reaction volume to give me velocity in "moles of product formed/min". I used this to then plot a lineweaver burke plot to get Vmax and Km values.

One of the questions asks me to calculate the Kcat value. This is where i'm stuck. I have the following data at hand:

Vmax = 2.442 x 10E-2 μmoles/min

Amount of enzyme used each time = 0.05mL

Concentration of supplied enzyme = 4μg/mL

Molecular Weight of Enzyme = 150,000 Mr

The enzyme however exists as a homotetramer, and has 4 identical active sites. I know I need to take that into consideration but i'm generally confused. Could someone talk me through step by step please? Thanks!

- Hidden

I am having a few problems calculating some required enzyme kinetic values. I've investigated the activity over 3 minutes of alcohol dehydrogenase at varying [S]. I used a spectrophotometer for this. From each of those graphs i've calculated the gradient (A/min) and converted this using the beer-lambert law and the reaction volume to give me velocity in "moles of product formed/min". I used this to then plot a lineweaver burke plot to get Vmax and Km values.

One of the questions asks me to calculate the Kcat value. This is where i'm stuck. I have the following data at hand:

Vmax = 2.442 x 10E-2 μmoles/min

Amount of enzyme used each time = 0.05mL

Concentration of supplied enzyme = 4μg/mL

Molecular Weight of Enzyme = 150,000 Mr

The enzyme however exists as a homotetramer, and has 4 identical active sites. I know I need to take that into consideration but i'm generally confused. Could someone talk me through step by step please? Thanks!

- Hidden

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#2

The equation for Kcat is

Vmax = Kcat x [Et] where [Et] is the total concentration of enzyme. Rearrange this equation to find Kcat.

Vmax = Kcat x [Et] where [Et] is the total concentration of enzyme. Rearrange this equation to find Kcat.

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(Original post by

The equation for Kcat is

Vmax = Kcat x [Et] where [Et] is the total concentration of enzyme. Rearrange this equation to find Kcat.

**Eloades11**)The equation for Kcat is

Vmax = Kcat x [Et] where [Et] is the total concentration of enzyme. Rearrange this equation to find Kcat.

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#4

(Original post by

I am unsure how to calculate Et though given that i'm dealing with a homotetramer

**hidden088**)I am unsure how to calculate Et though given that i'm dealing with a homotetramer

Ignore me if this is completely wrong.

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#5

**hidden088**)

I am unsure how to calculate Et though given that i'm dealing with a homotetramer

You're investigating the properties of this specific enzyme, because it's a homotetramer it doesn't affect the calculations in any way, unless you're trying to compare these results with another enzyme (which would be less accurate).

(Original post by

If they are related in a linear way, wouldn't you just multiply [Et] by 4, as the original equation (probably) assumes that you only have 1 active site.

Ignore me if this is completely wrong.

**thegodofgod**)If they are related in a linear way, wouldn't you just multiply [Et] by 4, as the original equation (probably) assumes that you only have 1 active site.

Ignore me if this is completely wrong.

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#6

(Original post by

Not necessarily, a substrate will have different affinities/successful collisions for an enzyme with 4 active sites than an enzyme with only 1 active site, hence comparing the properties wouldn't be accurate.

**Eloades11**)Not necessarily, a substrate will have different affinities/successful collisions for an enzyme with 4 active sites than an enzyme with only 1 active site, hence comparing the properties wouldn't be accurate.

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(Original post by

The exact same way as you would with any other enzyme. You have to calculate the Vmax and Km normally, and because the substrate can bind to any of the four active sites doesn't really change anything. Unless you're in a situation where you are trying to investigate the effects of one active (of which this enzyme may exist elsewhere, but we have no access to it) then even so, the chances of the substrates having the exact same affinity for both enzymes is low.

You're investigating the properties of this specific enzyme, because it's a homotetramer it doesn't affect the calculations in any way, unless you're trying to compare these results with another enzyme (which would be less accurate).

Not necessarily, a substrate will have different affinities/successful collisions for an enzyme with 4 active sites than an enzyme with only 1 active site, hence comparing the properties wouldn't be accurate.

**Eloades11**)The exact same way as you would with any other enzyme. You have to calculate the Vmax and Km normally, and because the substrate can bind to any of the four active sites doesn't really change anything. Unless you're in a situation where you are trying to investigate the effects of one active (of which this enzyme may exist elsewhere, but we have no access to it) then even so, the chances of the substrates having the exact same affinity for both enzymes is low.

You're investigating the properties of this specific enzyme, because it's a homotetramer it doesn't affect the calculations in any way, unless you're trying to compare these results with another enzyme (which would be less accurate).

Not necessarily, a substrate will have different affinities/successful collisions for an enzyme with 4 active sites than an enzyme with only 1 active site, hence comparing the properties wouldn't be accurate.

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#8

(Original post by

It makes sense but I can not make sense of why i've been asked this question. I've been given the molecular weight, and it's stressed that it's a homotetramer (with identical monomers), and asks me to work out the turnover number for each active site.

**hidden088**)It makes sense but I can not make sense of why i've been asked this question. I've been given the molecular weight, and it's stressed that it's a homotetramer (with identical monomers), and asks me to work out the turnover number for each active site.

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(Original post by

Ok, you didn't mention you had to calculate it for each active site. In which case you simply divide Kcat 4. What value did you get for Kcat?

**Eloades11**)Ok, you didn't mention you had to calculate it for each active site. In which case you simply divide Kcat 4. What value did you get for Kcat?

I've done this:

My lineweaver burke plot gave me a Y-intercept of 4.10E+7 which gives me a Vmax of

**2.442E-8 moles of NADH formed/min.**

I used 0.05mL of Enzyme in every cuvette (4μg/mL stock solution) = 0.2μg = 2E-7 grams of enzyme were used per cuvette. It has a molecular weight of 150,000 so:

Moles = 2E-7g / 150,000

= 1.33 x 10E-12 moles of enzyme used

Kcat = Vmax / [E0]

Kcat = 2.442E-8 / 1.33 x 10E-12

Kcat = 18,300

As you said:

Kcat (per active site) = 18,300/4 = 4575/min

Seem correct?

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#10

(Original post by

My mistake! Sorry

I've done this:

My lineweaver burke plot gave me a Y-intercept of 4.10E+7 which gives me a Vmax of

I used 0.05mL of Enzyme in every cuvette (4μg/mL stock solution) = 0.2μg = 2E-7 grams of enzyme were used per cuvette. It has a molecular weight of 150,000 so:

Moles = 2E-7g / 150,000

= 1.33 x 10E-12 moles of enzyme used

Kcat = Vmax / [E0]

Kcat = 2.442E-8 / 1.33 x 10E-12

Kcat = 18,300

As you said:

Kcat (per active site) = 18,300/4 = 4575/min

Seem correct?

**hidden088**)My mistake! Sorry

I've done this:

My lineweaver burke plot gave me a Y-intercept of 4.10E+7 which gives me a Vmax of

**2.442E-8 moles of NADH formed/min.**I used 0.05mL of Enzyme in every cuvette (4μg/mL stock solution) = 0.2μg = 2E-7 grams of enzyme were used per cuvette. It has a molecular weight of 150,000 so:

Moles = 2E-7g / 150,000

= 1.33 x 10E-12 moles of enzyme used

Kcat = Vmax / [E0]

Kcat = 2.442E-8 / 1.33 x 10E-12

Kcat = 18,300

As you said:

Kcat (per active site) = 18,300/4 = 4575/min

Seem correct?

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(Original post by

That seems fine, don't forget your units though!

**Eloades11**)That seems fine, don't forget your units though!

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#12

Another (albeit weird, long-winded and kind of pointless :P ) way you could look at it which may help you to get your head around is:

1.33 x 10E-12 moles of enzyme used - how many molecules is this? Use avogardo's number

1.33 x 10E-12 moles x 6.022E+23 = 8E+11 molecules (you can think of this as the number of homotetramers present)

Thus the number of active sites you have is 3.2E+12.

Do the same with your vMax value - how many molecules of NADH is that formed?

2.442E-8 moles of NADH x 6.022E+23 = 1.47E+16 molecules of NADH

Now just look at the ratio between molecules of NADH formed and the molecules of active sites:

1.47E+16 molecules of NADH / 3.2E+12 active sites = 4593.75 molecules/min turnover

Maybe that will help you bit!

1.33 x 10E-12 moles of enzyme used - how many molecules is this? Use avogardo's number

1.33 x 10E-12 moles x 6.022E+23 = 8E+11 molecules (you can think of this as the number of homotetramers present)

Thus the number of active sites you have is 3.2E+12.

Do the same with your vMax value - how many molecules of NADH is that formed?

2.442E-8 moles of NADH x 6.022E+23 = 1.47E+16 molecules of NADH

Now just look at the ratio between molecules of NADH formed and the molecules of active sites:

1.47E+16 molecules of NADH / 3.2E+12 active sites = 4593.75 molecules/min turnover

Maybe that will help you bit!

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(Original post by

Another (albeit weird, long-winded and kind of pointless :P ) way you could look at it which may help you to get your head around is:

1.33 x 10E-12 moles of enzyme used - how many molecules is this? Use avogardo's number

1.33 x 10E-12 moles x 6.022E+23 = 8E+11 molecules (you can think of this as the number of homotetramers present)

Thus the number of active sites you have is 3.2E+12.

Do the same with your vMax value - how many molecules of NADH is that formed?

2.442E-8 moles of NADH x 6.022E+23 = 1.47E+16 molecules of NADH

Now just look at the ratio between molecules of NADH formed and the molecules of active sites:

1.47E+16 molecules of NADH / 3.2E+12 active sites = 4593.75 molecules/min turnover

Maybe that will help you bit!

**cptbigt**)Another (albeit weird, long-winded and kind of pointless :P ) way you could look at it which may help you to get your head around is:

1.33 x 10E-12 moles of enzyme used - how many molecules is this? Use avogardo's number

1.33 x 10E-12 moles x 6.022E+23 = 8E+11 molecules (you can think of this as the number of homotetramers present)

Thus the number of active sites you have is 3.2E+12.

Do the same with your vMax value - how many molecules of NADH is that formed?

2.442E-8 moles of NADH x 6.022E+23 = 1.47E+16 molecules of NADH

Now just look at the ratio between molecules of NADH formed and the molecules of active sites:

1.47E+16 molecules of NADH / 3.2E+12 active sites = 4593.75 molecules/min turnover

Maybe that will help you bit!

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I also have to plot a Eadie-Hofstee plot (v

However I am unsure how to go about this. My velocity data (v) is currently in moles/min, and my [S] data is in mM, i'm not sure if I need to convert any units to plot this graph. Also what units would I get out? The Y-intercept value of Vmax will come out in moles/min but what about the gradient?

*vs*v/[S]) where the Y-intercept = Vmax, and the gradient = -Km.However I am unsure how to go about this. My velocity data (v) is currently in moles/min, and my [S] data is in mM, i'm not sure if I need to convert any units to plot this graph. Also what units would I get out? The Y-intercept value of Vmax will come out in moles/min but what about the gradient?

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#15

(Original post by

I also have to plot a Eadie-Hofstee plot (v

However I am unsure how to go about this. My velocity data (v) is currently in moles/min, and my [S] data is in mM, i'm not sure if I need to convert any units to plot this graph. Also what units would I get out? The Y-intercept value of Vmax will come out in moles/min but what about the gradient?

**hidden088**)I also have to plot a Eadie-Hofstee plot (v

*vs*v/[S]) where the Y-intercept = Vmax, and the gradient = -Km.However I am unsure how to go about this. My velocity data (v) is currently in moles/min, and my [S] data is in mM, i'm not sure if I need to convert any units to plot this graph. Also what units would I get out? The Y-intercept value of Vmax will come out in moles/min but what about the gradient?

^{-1}data by 1000, so you have mmol min

^{-1}, and mmol on the other side as well

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(Original post by

It would be easier to work with if you have the same units, so multiply the mol min

**thegodofgod**)It would be easier to work with if you have the same units, so multiply the mol min

^{-1}data by 1000, so you have mmol min^{-1}, and mmol on the other side as well
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#17

(Original post by

Okay I now have mmol/min on the Y-axis and v/[S] (in mM) on the X-axis, math is quite a weak-spot for me however and i'm not sure what the units the gradient would be in

**hidden088**)Okay I now have mmol/min on the Y-axis and v/[S] (in mM) on the X-axis, math is quite a weak-spot for me however and i'm not sure what the units the gradient would be in

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#18

**hidden088**)

Okay I now have mmol/min on the Y-axis and v/[S] (in mM) on the X-axis, math is quite a weak-spot for me however and i'm not sure what the units the gradient would be in

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#19

how do you work out Vmax and Km from the Michaelis-Menton graph and the Lineweaver-Burke equation? i need to use these on my my results but havn't got a clue of what to do :/

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#20

(Original post by

how do you work out Vmax and Km from the Michaelis-Menton graph and the Lineweaver-Burke equation? i need to use these on my my results but havn't got a clue of what to do :/

**Perky perks**)how do you work out Vmax and Km from the Michaelis-Menton graph and the Lineweaver-Burke equation? i need to use these on my my results but havn't got a clue of what to do :/

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