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### Topic: Absorbance-time graph and enzyme activity  (Read 32792 times)

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##### Absorbance-time graph and enzyme activity
« on: June 11, 2013, 12:47:58 PM »
Hi I have some questions about substrate concentration and enzyme activity. In my lab, we were task to find the absorbance of 0.3mM, 1.5mM, 2.1mM, 2.7mM and 3.0mM of a substrate concentration at a constant concentration of enzyme. So after plotting them and drawing the best fit line for each of them, I went to get the gradient which would give me my rate of change of absorbance for the various substrates. I realized it gave me an increasing rate of change until 3.0mM where the rate stayed the same.

So after checking out the reasoning behind this, I came across this link http://www.chemguide.co.uk/organicprops/aminoacids/enzymes2.html explaining that it stopped increasing as there was just not enough enzyme to catalyze the reaction.

However, I was thinking about this say I have a 10mM (call this A) and a 5mM (call this B) substrate concentration in 1 L of water. And say at my fixed enzyme concentration can only catalyze 2 mmoles per second. So initially for both A and B would produce the same amount of products per unit time giving the same absorbance reading. But after 2 seconds B would only have 1 mmole of substrate to react. While for A there is still plenty of substrate to continue getting catalyzed. So in this case wouldn't A start to give a higher absorbance reading as now the concentration of the products is greater? Like this http://postimg.org/image/4c7zxpbiv/

So shouldn't the concentration affect the enzyme activity still? And I don't quite get how my absorbance-time graph was linear  if there was not enough substrate say in the A B example i have another solution with 1mM of substrate. Then wouldn't it just plateau almost immediately as the enzyme would catalyze it in 1 second (cos the enzyme can catalyze up to 2 mmoles/s so when I have a 1mM in 1 L of water, then it would be consumed in 1 second only?)?

I think I have a misconception about this somewhere but I'm not too sure where I went wrong. Thanks for the help

#### Yggdrasil

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##### Re: Absorbance-time graph and enzyme activity
« Reply #1 on: June 11, 2013, 07:26:22 PM »
We generally study enzyme kinetics under two different conditions, which allow us to study different aspects of the enzyme's activity.  The first regime under which enzymes are studied is the steady state regime (aka Michaelis Menten kinetics).  Under these conditions, you have many more substrate molecules than enzyme molecules, so each enzyme will undergo multiple turnover events (i.e. it will interact with many substrate molecules).  This is the case for the experiments you are performing.

The other regime where we study enzyme kinetics is the pre-steady state regime.  Under these conditions, you have many more enzyme molecules than substrate molecules, so each enzyme will catalyze at most one reaction.  This is like the case you are describing in your hypothetical.

Usually when doing enzyme kinetics measurements, you take steps to ensure that you stay in either the steady state or pre-steady state regime.  For measuring the Michaelis-Menten kinetics of an enzyme, you would ensure that the concentration of substrate is high enough that it does not get significantly depleted during the reaction (for example, consider 10mM vs 5mM for an enzyme preparation whose Vmax is 0.2mmol per second).  Under these conditions you expect to see the concentration of product increase linearly with time.  If the substrate gets depleted during the reaction, you see the slope of the product vs time graph gradually decrease with time (i.e. the curve is concave down).  Here, you would estimate the initial reaction velocity (the initial slope before the graph begins to curve) and use that value for the Michaelis-Menten plot.

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##### Re: Absorbance-time graph and enzyme activity
« Reply #2 on: June 12, 2013, 12:23:16 AM »
We generally study enzyme kinetics under two different conditions, which allow us to study different aspects of the enzyme's activity.  The first regime under which enzymes are studied is the steady state regime (aka Michaelis Menten kinetics).  Under these conditions, you have many more substrate molecules than enzyme molecules, so each enzyme will undergo multiple turnover events (i.e. it will interact with many substrate molecules).  This is the case for the experiments you are performing.

The other regime where we study enzyme kinetics is the pre-steady state regime.  Under these conditions, you have many more enzyme molecules than substrate molecules, so each enzyme will catalyze at most one reaction.  This is like the case you are describing in your hypothetical.

Usually when doing enzyme kinetics measurements, you take steps to ensure that you stay in either the steady state or pre-steady state regime.  For measuring the Michaelis-Menten kinetics of an enzyme, you would ensure that the concentration of substrate is high enough that it does not get significantly depleted during the reaction (for example, consider 10mM vs 5mM for an enzyme preparation whose Vmax is 0.2mmol per second).  Under these conditions you expect to see the concentration of product increase linearly with time.  If the substrate gets depleted during the reaction, you see the slope of the product vs time graph gradually decrease with time (i.e. the curve is concave down).  Here, you would estimate the initial reaction velocity (the initial slope before the graph begins to curve) and use that value for the Michaelis-Menten plot.

Ohh I think I get it. Can I try examining the trend of an increasing absorbance with concentration up till a certain point?

The enzyme ability to catalyze the substrate is dependent on the concentration of the substrate. So when the substrate conc is low, less can be catalyzed. So as we increase the substrate conc, so does the amount of products (which are colored we used our substrate as ONPG so it would undergo hydrolysis to form yellow ONP) formed. However, the enzyme has a limited ability to catalyze the substrate so once the concentration is too high, only a few molecules can be reacted per unit time. Hence further increasing the concentration would not yield a significant increase in products formed.

Does this explanation make sense to you? Thanks so much Yggdrasil!

#### Babcock_Hall

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##### Re: Absorbance-time graph and enzyme activity
« Reply #3 on: June 12, 2013, 10:57:24 AM »
There are two very separate issues here, and I think it makes sense to disentangle them first.  Unless I am told otherwise, if I am given some data of rate versus substrate concentration, I will assume:
(1) We are working under initial velocity conditions, meaning that perhaps 1-3% of the substrate has been consumed (converted into product).  Therefore depletion of substrate has not occurred to any significant extent and the rate will not go down.
(2) We are working under steady-state conditions.  For many situations, steady state is reached in less than a second.  Steady-state just means that the concentration of the complex between the enzyme and substrate, [E•S], is not changing  much in time.

If both of these conditions are met, then the graph of rate versus concentration of substrate will be a rectangular hyperbola, meaning linear at low concentration of substrate and gradually leveling off.  When substrate concentration is very high, all of the enzyme active sites are occupied (the enzyme is "saturated"), and higher concentrations of substrate can no longer affect the rate.
« Last Edit: June 12, 2013, 11:35:30 AM by Babcock_Hall »

#### Yggdrasil

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##### Re: Absorbance-time graph and enzyme activity
« Reply #4 on: June 12, 2013, 11:01:31 AM »
Ohh I think I get it. Can I try examining the trend of an increasing absorbance with concentration up till a certain point?
Yes, you want to be looking only at the initial rate of reaction, not the final concentration of product.

Quote
The enzyme ability to catalyze the substrate is dependent on the concentration of the substrate. So when the substrate conc is low, less can be catalyzed. So as we increase the substrate conc, so does the amount of products (which are colored we used our substrate as ONPG so it would undergo hydrolysis to form yellow ONP) formed. However, the enzyme has a limited ability to catalyze the substrate so once the concentration is too high, only a few molecules can be reacted per unit time. Hence further increasing the concentration would not yield a significant increase in products formed.

Does this explanation make sense to you? Thanks so much Yggdrasil!

Yes, at low substrate concentrations, only a few enzyme molecules have substrate bound, and the reaction will proceed slowly.  Here, the reaction rate is basically limited by the time it takes for an enzyme molecule to diffuse around the solution and find a substrate molecule to react with (i.e. the binding time).  As you increase the substrate concentration, it becomes easier for enzymes to find substrate molecules and time an enzyme spends waiting for a substrate to bind decreases, increasing the overall reaction rate.  However, at a certain point, the intrinsic rate of catalysis becomes the rate limiting step in the process, and decreasing the binding time by increasing the substrate concentration will no longer increase the reaction rate.  At these high substrate concentration, there's essentially no waiting time between dissociation of the product and binding of a new substrate molecule, and the rate of reaction reflects the maximum rate at which the enzymes can work.

In other words, at low concentrations of substrate the reaction rate depends on the rate at which substrates bind to the enzyme (which of course is dependent on the concentration of substrate).  At high substrate concentration, the reaction rate depends on the intrinsic rate of catalysis by the enzyme (e.g. how fast the enzyme can break or form the chemical bonds), which does not depend on the concentration of substrate.

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##### Re: Absorbance-time graph and enzyme activity
« Reply #5 on: June 12, 2013, 12:45:34 PM »
Ohh I think I get it. Can I try examining the trend of an increasing absorbance with concentration up till a certain point?
Yes, you want to be looking only at the initial rate of reaction, not the final concentration of product.

Quote
The enzyme ability to catalyze the substrate is dependent on the concentration of the substrate. So when the substrate conc is low, less can be catalyzed. So as we increase the substrate conc, so does the amount of products (which are colored we used our substrate as ONPG so it would undergo hydrolysis to form yellow ONP) formed. However, the enzyme has a limited ability to catalyze the substrate so once the concentration is too high, only a few molecules can be reacted per unit time. Hence further increasing the concentration would not yield a significant increase in products formed.

Does this explanation make sense to you? Thanks so much Yggdrasil!

Yes, at low substrate concentrations, only a few enzyme molecules have substrate bound, and the reaction will proceed slowly.  Here, the reaction rate is basically limited by the time it takes for an enzyme molecule to diffuse around the solution and find a substrate molecule to react with (i.e. the binding time).  As you increase the substrate concentration, it becomes easier for enzymes to find substrate molecules and time an enzyme spends waiting for a substrate to bind decreases, increasing the overall reaction rate.  However, at a certain point, the intrinsic rate of catalysis becomes the rate limiting step in the process, and decreasing the binding time by increasing the substrate concentration will no longer increase the reaction rate.  At these high substrate concentration, there's essentially no waiting time between dissociation of the product and binding of a new substrate molecule, and the rate of reaction reflects the maximum rate at which the enzymes can work.

In other words, at low concentrations of substrate the reaction rate depends on the rate at which substrates bind to the enzyme (which of course is dependent on the concentration of substrate).  At high substrate concentration, the reaction rate depends on the intrinsic rate of catalysis by the enzyme (e.g. how fast the enzyme can break or form the chemical bonds), which does not depend on the concentration of substrate.

Oh! This makes so much sense now in other explanations I found for catalysts a similar explanation is also given for why at high concentrations the rate of production of products remain the same. Does this mean they all use the initial velocity too?

But actually, the enzyme only provides an alternative reaction pathway with a lower activation energy. However, isn't it still possible for the substrate to react with the water without the enzyme lowering that reaction pathway? So by increasing the concentration shouldn't the initial velocity still increase as even though the number of substrate being catalyze is the same in both cases, won't there be more active collision between the ONPG and the water for the more concentrated solution? So shouldn't the rate of the more concentrated one still be slightly higher?

Thanks again Yggdrasil

#### Yggdrasil

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##### Re: Absorbance-time graph and enzyme activity
« Reply #6 on: June 12, 2013, 02:53:47 PM »
Oh! This makes so much sense now in other explanations I found for catalysts a similar explanation is also given for why at high concentrations the rate of production of products remain the same. Does this mean they all use the initial velocity too?

Yes, this refers only to the initial velocity.  The concentration of product will of course level off at different values for different initial concentrations of substrate.

Quote
But actually, the enzyme only provides an alternative reaction pathway with a lower activation energy. However, isn't it still possible for the substrate to react with the water without the enzyme lowering that reaction pathway? So by increasing the concentration shouldn't the initial velocity still increase as even though the number of substrate being catalyze is the same in both cases, won't there be more active collision between the ONPG and the water for the more concentrated solution? So shouldn't the rate of the more concentrated one still be slightly higher?

Yes, you are correct about this.  In general, enzyme catalyzed reactions occur much more quickly than uncatalyzed reactions, but it the uncatalyzed reaction occurs at a significant rate, then you will see this effect.

#### Babcock_Hall

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##### Re: Absorbance-time graph and enzyme activity
« Reply #7 on: June 12, 2013, 05:41:24 PM »
Often one runs a blank containing no enzyme.  When I have worked with ONPG, the absorbance of the blank was small.

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##### Re: Absorbance-time graph and enzyme activity
« Reply #8 on: June 13, 2013, 12:35:24 PM »
Often one runs a blank containing no enzyme.  When I have worked with ONPG, the absorbance of the blank was small.

Perhaps it's because I only started timing when all the enzyme was added into the cuvette. So before I even start t=0s, some enzyme has reacted. Causing the absorbance to be slightly greater than 0AU?

Thanks for the help everyone

#### Babcock_Hall

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##### Re: Absorbance-time graph and enzyme activity
« Reply #9 on: June 13, 2013, 01:00:13 PM »
If the ONPG has a little bit of free ortho-nitrophenol, then even at time zero there would be some absorbance.  A second explanation for a small, nonzero value of the absorbance at time zero would be hydrolysis of ONPG prior to the addition of enzyme.

The blank I mentioned is run identically to the unknowns (same reagents and same time), but it simply has no enzyme.