# Enzyme Kinetics

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#1
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
0
8 years ago
#2
The equation for Kcat is

Vmax = Kcat x [Et] where [Et] is the total concentration of enzyme. Rearrange this equation to find Kcat.
0
#3
The equation for Kcat is

Vmax = Kcat x [Et] where [Et] is the total concentration of enzyme. Rearrange this equation to find Kcat.
I am unsure how to calculate Et though given that i'm dealing with a homotetramer
0
8 years ago
#4
(Original post by hidden088)
I am unsure how to calculate Et though given that i'm dealing with a homotetramer
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. 1
8 years ago
#5
(Original post by hidden088)
I am unsure how to calculate Et though given that i'm dealing with a homotetramer
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).

(Original post by 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. 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.
1
8 years ago
#6
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.
Ahh - that makes sense, when you look at how oxygen molecules bind to haemoglobin and the oxygen-haemoglobin dissociation curve 0
#7
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.
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.
0
8 years ago
#8
(Original post by 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.
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?
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#9
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?
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? 0
8 years ago
#10
(Original post by 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? That seems fine, don't forget your units though! 0
#11
That seems fine, don't forget your units though! Thank you very much! It's still a bit confusing to me but hopefully i'll get it soon! 0
8 years ago
#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?

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

0
#13
(Original post by 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?

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

It makes sense now!! Thank you!!
0
#14
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?
0
8 years ago
#15
(Original post by 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?
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 0
#16
(Original post by 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 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
0
8 years ago
#17
(Original post by 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
My knowledge of maths has now reached the end-point  0
8 years ago
#18
(Original post by 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
If I understand this correctly, you've got 'v' on the Y-axis, and v/[S] on the X-axis, this would mean your gradient is: v/(v/[S]) = [S] 0
8 years ago
#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 :/
0
8 years ago
#20
(Original post by 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 :/
You can't calculate Km and Vmax from the Michaelis-Menten graph, but if you plot the reciprocals of the velocity against substrate concentration (Lineweaver-Burke plot) then the x intercept will be -1/Km, and the y intercept will be 1/Vmax.
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