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Enzymatic error rate

Enzymatic error rate


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I am aware that each enzyme generate a certain amount of misproducts. This is well documented, for example, for the DNA polymerase.

I am interested in enzyme involved in biochemical processes, so for example methytransferase, oxidoreductase, carboxylase and so on. I have been looking for any reference that could report the error rate or the amount of misproduct for any of these enzymes, but I was not successful.

Does anyone know a reference reporting the error rate, or any other information about the misproduct generate by these enzymes?

EDIT

I will try to clarify a bit better my question. Let's assume that we are considering an enzyme that add a methyl group to a molecule, we can choose propionic acid for example. In theory the correct product would derive from the addition of a methyl to the C3. For the biochemistry study we know that the activation energy is decreased by an enzyme for example by orienting the molecules in the right way to react with the second molecule. Therefore, if the enzyme in question will not be able to orient the molecule in the right way there is a certain probability that there will be a mis-product. For example the methyl group could be added to the C2 and not the C3. Now. the probability for this process is quite low, but it cannot be excluded from happening. My question is, if there is anything known about this type of error rates in the different type of enzymes.

I hope this make things a bit more clear

EDIT 2

Lets consider the imagine below that was taken from here

In B you can see the interaction of the active site of the enzyme with a substrate (for the example it does not matter if it is the real substrate or an inhibitor). I would say that among al possible interaction that this active site can make with substrates there are some much more likely than others. An hypothetical chart could be created considering the free energy of the complex enzyme-substrate, higher the free energy lower the stability. Therefore among all possible enzyme-substrate combination we will have just few possibilities that have a low free energy. In the first place we will find the "correct" substrate, then we will find eventual competitors and other molecules that have similar chemical features as the substrate. Even though with low probability these interaction, enzyme-competitors, will happen. This might led to the formation of "unwanted product" that I refereed to in my question as "misproduct".

I am interested in reading some paper that are considering thie phenomenon and also quantify its rate. I am also curious if this phenomenon can happen with different rates in different enzymes.


First it is important to define what actually an error is. What you call misproduct is actually a right product for a different substrate. The enzyme has a low specificity; but you won't say hexokinase is error prone because it has wide substrate specificity. The question is - is specificity critical?

Coming to the example of DNA polymerase, you should note that DNA polymerase only adds a nucleotide to the DNA strand. Here the template DNA strand acts like a co-enzyme and it is the template DNA that provides specificity to the reaction. In this case the specificity is really critical and there are mechanisms that help in ensuring that.

Other enzymes also make errors and the probability of making errors depends on different parameters. Take the example of restriction enzymes that are supposed to cut at a specific site. When this specificity is lost we say that the enzyme is showing a "star activity" i.e. non-specific cleavage. This depends on different parameters such as buffer composition, enzyme concentration etc.

In general specificity is provided because of high affinity of binding towards one molecule compared to the other. These binding reactions are second order - a high affinity substrate will bind to the enzyme even at a low concentration, however if you increase the concentration of the low affinity substrate or the enzyme, then at some point the rate of binding may be significant enough to create a "wrong" product.

Sometimes there are cases where the enzyme can bind to a wrong substrate but cannot create a product. This is the typical case of a competitive inhibitor. It is also possible that a certain molecule is good at binding but the conversion rate is slower.

To conclude, these kinds of "errors" happen with many enzymes but it is very critical in case of DNA polymerization. You can get information on star activity of restriction enzymes (just google it).


Rubisco is an enzyme that is involved in the biochemical process of carbon fixation in photosynthesis, but also has an economically important error (photorespiration) which incorporates the wrong molecule (oxygen) into RuBP, therefore wasting the energy captured by the plant and producing toxic downstream products.

Rubisco's error rate in carbon fixation ranges from approximately 1:10 to 1:80.


Chris is right here, polymerases are generating sequences, the enzymes your asking about do a single type of reaction, so you would be assessing its activity not its fidelity. A good resource I've found for looking at qc data for enzymes like these are just the product inserts from the manufacturer. NEB for example generally seems to provide sufficient qc data in my opinion. For example:

https://www.neb.com/~/media/Catalog/All-Products/B2D1E3A4F480489EAD7A1CD43E164DA0/Datacards%20or%20Manuals/M0233Datasheet-Lot0031209-1.pdf


Since chemical reactions are governed by quantum physics, there is a whole lot of products that can come out of mixing together several multi-atom molecules. And all of those will have non-zero probability of occurrence. But this probability is extremely low at normal temperatures, so you don't see those products.

Enzymes lower activation energy and so increase chance of reaction, but only for certain substances and products (spatial position of reagents can influence outcome). In case of DNA replication, when wrong nucleotide is inserted, it happens because enzyme/template nucleotide complex has not enough specificity. All 4 nucleotides have rather high chance of participating in the reaction (but errors are still on order of $10^{-6}$ and lower).


DNA polymerase

A DNA polymerase is a member of a family of enzymes that catalyze the synthesis of DNA molecules from nucleoside triphosphates, the molecular precursors of DNA. These enzymes are essential for DNA replication and usually work in groups to create two identical DNA duplexes from a single original DNA duplex. During this process, DNA polymerase "reads" the existing DNA strands to create two new strands that match the existing ones. [1] [2] [3] [4] [5] [6] These enzymes catalyze the chemical reaction

DNA polymerase adds nucleotides to the three prime (3')-end of a DNA strand, one nucleotide at a time. Every time a cell divides, DNA polymerases are required to duplicate the cell's DNA, so that a copy of the original DNA molecule can be passed to each daughter cell. In this way, genetic information is passed down from generation to generation.

Before replication can take place, an enzyme called helicase unwinds the DNA molecule from its tightly woven form, in the process breaking the hydrogen bonds between the nucleotide bases. This opens up or "unzips" the double-stranded DNA to give two single strands of DNA that can be used as templates for replication in the above reaction.


#74 Identifying sources of error

It is very important to understand the difference between experimental errors and 'mistakes'. A mistake is something that you do incorrectly, such as misreading the scale on a thermometer, or taking a reading at the wrong time, or not emptying a graduated pipette fully. Do not refer to these types of mistake when you are asked to comment on experimental errors.


You've already seen, on post # 70 , that every measuring instrument has its own built-in degree of uncertainty in the values you read from it. You may remember that, in general, the size of the error is half the value of the smallest division on the scale.

Errors can also occur if there were uncontrolled variables affecting your results. For example, if you were doing an investigation into the effect of leaf area on the rate of transpiration, and the temperature in the laboratory increased while you were doing your experiment, then you can't be sure that all the differences in rate of transpiration were entirely due to differences in leaf area.

Systematic and random errors

Systematic errors are ones that are the same throughout your investigation, such as intrinsic errors in the measuring instruments you were using.

Random errors are ones that can differ throughout your investigation. For example, you might be doing an osmosis investigation using potato strips taken from different parts of a potato, where perhaps the cells in some parts had a higher water potential than in others. Or perhaps the temperature in the room was fluctuating up and down.

Spotting the important sources of error

You should be able to distinguish between significant errors and insignificant ones. For example, a change in room temperature could have a significant effect on the rate of transpiration (Investigation 4) but it would not have any effect at all on the number of stomata on the upper and lower surface of a leaf (investigation 3).

Another thing to consider is how well a variable has been controlled. If you were doing an enzyme investigation using a water bath to control temperature, then you should try to be realistic in estimating how much the temperature might have varied by. If you were using a high-quality, electronically controlled water bath, then it probably did not vary much, but if you were using a beaker and Bunsen burner then it is likely that temperature variations could indeed be significant.

Tips

During your course:

• Every time you do an investigation, work out and write down the uncertainty in all the types of measurement that you make.
• Every time you do an investigation, think carefully about any errors that may be die to lack of control of variables - which ones might genuinely be significant!

Inthe exam:

• If you are asked about an investigation that seems familiar. It is tempting just to try to recall what the main errors were in the investigation that you did before. This is not a good idea, because the investigation in the exam may not be quite the same. Always think about the actual investigation in the examination question, and think through what the significant sources of error are.

Suggesting improvements

You may be asked to suggest how the investigation you have just done, or an investigation that has been described, could be improved. Your improvements should be aimed at getting more valid or reliable results to the question that the investigation was trying to answer - do not suggest improvements that would mean you would now be trying to answer a different question. For example, if you were doing an investigation to investigate the effect of leaf area on the rate of transpiration, don't suggest doing something to find out the effect of the wind speed on the rate of transpiration.

The improvements you suggest could include controlling certain variables that were not controlled, or controlling them more effectively. For example, you may suggest that the investigation could be improved by controlling temperature. To earn a mark, you must also say how you would control it, for example by placing sets of test-tubes in a thermostatically controlled water bath.

You could also suggest using better methods of measurement. For example, you might suggest using a colorimeter to measure depth of colour, rather than using your eyes and a colour scale.

It is almost always a good idea to do several repeats in your investigation and then calculate a mean of your results. For example, if you are measuring the effect of light intensity on the rate of transpiration, then you could take three sets of readings for the volume of water taken up by your leafy shoot in one minute at a particular light intensity. The mean of these results is more likely to give you the true value of the rate of transpiration than anyone individual result.

Tips

During your course:

• If time allows, try to do at least two (and possibly three) repeats when you do an investigation.
• As you do an investigation, be thinking all the time about how reliable or accurate your measurements and readings are. Think about what you would like to be able to do to improve their reliability or accuracy.

In the exam:

• Be very precise in suggesting how you could improve the investigation - for example, don't just say you would control a particular variable, but say how you would control it.


RESULTS

  1. The enzyme catalase kickstarts the catabolic reaction of the substrate Hydrogen Peroxide breaking down into water and oxygen.
  1. Factors tested for their effect on this reaction are surface area, temperature, pH, and the reuse of materials. Surface area made the reaction quicker, temperature increased the reaction until it started to denature the enzymes, when the pH became more acidic it denatured many of the enzymes and made the reaction much slower, and when the pH became more basic it denatured few of the enzymes and only slightly decreased the reaction speed.
  2. A possible explanation for the difference between the reaction rate of the liver and potato is the potato contains less of the enzyme catalase.
  3. H2O2 can be broken down by catalysts other than those found in living systems as demonstrated in procedure A.
  4. While it is hard to be certain if the experiment would change if the liver of a dog was used instead, if we assume that the optimum temperature of the liver that was used was 37 degrees, then the optimum temperature would have been 3 degrees higher.
  5. As it can be determined from the lab results, catalase is reusable as in part C the reused liver still had a reaction that was almost at the same level as that of the unused liver. Also, a catalyst by definition is neither a reactant or product meaning it’s not used up in the reaction.
  6. The graph shows the decrease in energy needed to start a reaction once an enzyme is added to the reaction.

The Scientific Method

As scientists, biologists apply the scientific method. Science is not simply a list of facts, but is an approach to understanding the world around us. It is use of the scientific method that differentiates science from other fields of study that attempt to improve our understanding of the world.

The scientific method is a systematic approach to problem solving. Although some argue that there is not one single scientific method, but a variety of methods each of these approaches, whether explicit or not, tend to incorporate a few fundamental steps: observing, questioning, hypothesizing, predicting, testing, and interpreting results of the test. Sometimes the distinction between these steps is not always clear. This is particularly the case with hypotheses and predictions. But for our purposes, we will differentiate each of these steps in our applications of the scientific method.

You are already familiar with the steps of the scientific method from previous lab experiences. You will need to use your scientific method knowledge in today’s lab in creating hypotheses for each experiment, devising a protocol to test your hypothesis, and analyzing the results. Within the experimentation process it will be important to identify the independent variable, the dependent variable, and standardized variables for each experiment.


The Kinetics of Enzyme Action (With Diagram)

In most instances, the association of the enzyme with the substrate is so fleeting that the complex is ex­tremely difficult to detect.

Yet, as early as 1913, L. Michaelis and M. L. Menten postulated the existence of this transient complex.

On the basis of their obser­vations with the enzyme invertase, which catalyzes the hydrolysis of sucrose to glucose and fructose, they proposed that enzyme-catalyzed reactions were char­acterized by a sequence of phases that involves:

(1) The formation of a complex (ES) between the enzyme (E) and the substrate (S)

(2) The modification of the sub­strate to form the product (P) or products, which briefly remain associated with the enzyme (EP) and

(3) The release of the product or products from the en­zyme that is,

In this equation, it is assumed that the combination of enzyme and substrate is reversible is the rate con­stant for the formation of ES (the dimensions of the rate constant are seconds- 1 ), and k2 is the rate con­stant for the dissociation of ES. After the ES complex is formed, S is converted to P with the rate constant k3. As long as the concentration of P remains negligi­ble (as it would be at the outset of catalysis) or if P is in some manner quickly removed from the system, then it is not necessary to consider the negligible reverse flux from E + P to ES.

Figure 8-4 depicts the relationship that exists be­tween substrate concentration and the rate at which reaction products appear for enzyme-catalyzed reac­tions of this type. The curve describes the initial rate of product formation at a fixed enzyme concentration when the substrate concentration is varied on succes­sive trials.

At low concentrations of substrate, the ini­tial velocity of the reaction (i.e., v0) is directly propor­tional to the substrate concentration (i.e., follows first-order kinetics). However, as the substrate con­centration is increased, the reaction velocity levels off, approaching a maximum value.

At this high substrate concentration, the reaction velocity is limited by the amount of time required to convert bound substrate (i.e., ES) to product (P) and free enzyme (E). The curve now follows zero-order kinetics. The curve of Figure 8-4 is called a Michaelis-Menten curve, and for an idealized enzyme-catalyzed reaction is a rectan­gular hyperbola.

Michaelis and Menten are also credited with the first mathematical study of the relationship between substrate concentration and reaction rates. They in­troduced two particularly useful mathematical expres­sions that for any enzyme relate [S] to V (velocity) and permit quick comparisons of various enzyme- catalyzed reactions the two expressions are now called the Michaelis-Menten constant and the Michaelis-Menten equation. They are derived as fol­lows.

[S] = the concentration of free substrate, S (We may assume that the amount of available sub­strate is so great that the amount combined with the enzyme may be ignored in compari­son. Hence, total substrate concentration and the free substrate concentration are the same)

[E]T =the total concentration of enzyme, E

[E] = the concentration of free enzyme (that not complexed with substrate)

[ES] = the concentration of enzyme-substrate com­plex (Because the total amount of enzyme present is assumed to be very small in com­parison with the total amount of substrate, a significant proportion of the total enzyme may be involved in the ES complex. Hence, separate terms for the total and free enzyme concentrations are warranted)

[P] = the concentration of product, P.

The total enzyme concentration is given by:

The ratio of constants given in equation 8-22 may be set equal to a new constant, KM, which is the Michaelis-Menten constant. Thus, the Michaelis- Menten constant is a constant that relates the steady- state concentrations of enzyme, enzyme-substrate complex, and substrate.

Each enzyme-catalyzed reaction reveals a charac­teristic KM value, and this value is a measure of the tendency of the enzyme and the substrate to combine with each other. In this sense, the KM value is an index of the affinity of the enzyme for its particular sub­strate.

Some representative KM values are given in Table 8-2. It is to be stressed that the greater the af­finity of an enzyme for its substrate, the lower the KM value. This is because the KM value is numerically equal to the substrate concentration at which half of the enzyme molecules are associated with substrate (i.e., in the ES form).

The relationships derived above are based on reac­tions in which a single substrate molecule is bound to the enzyme. However, they also apply in situations where more than one substrate is bound to the en­zyme, as long as the concentrations of all but one sub­strate species are held constant or are not rate- limiting (i.e., present in large excess).

For example, even the pioneering studies of Michaelis and Menten involved a bimolecular enzyme-catalyzed reaction. They employed the enzyme invertase, which forms a complex with one molecule of sucrose and one mole­cule of water and releases glucose and fructose as products. However, in this reaction (as in nearly all hy­dropses), the concentration of water remains virtu­ally unaltered during the course of the reaction.

Ku values are seldom determined using the rela­tionship given in equation 8-22, because the enzyme- substrate concentration [ES] cannot easily be mea­sured. Some additional mathematical manipulations may be carried out to convert equation 8-22 into a more useful form.

As seen in Figure 8-4, the initial reaction velocity increases with substrate concentration until the en­zyme concentration becomes limiting, and at that time, the initial reaction velocity approaches a maxi­mum value (i.e., Vmax). This occurs because nearly all the enzyme is maintained in the ES form. Thus, [ES] will equal [E]T, and k3[E]T corresponds to Vmax.

By substituting Vmax for k3[E]T in equation 8-27, we obtain

Equation 8-28 is the Michaelis-Menten equation. From this equation, it may be seen that when the sub­strate concentration is numerically equal to the KM value of the enzyme, then the reaction velocity is equal to one-half the maximum value. For example, if [S] and Km are both equal to 3, then equation 8-27 simpli­fies to

Consequently, the Michaelis-Menten constant for an enzyme may be determined from the substrate con­centration at which the reaction velocity proceeds at one-half its maximum value (Fig. 8-5).

From the above, we may conclude that the maxi­mum velocity of an enzyme-catalyzed reaction de­pends on the affinity between the enzyme and its sub­strate (i.e., the KM value). The higher the KM value is, the lower the affinity of the enzyme for the substrate and the smaller the KM value, the greater the affinity.

In many instances, the same substrate may enter ei­ther of several different enzyme-catalyzed reactions occurring in cells. Which of the alternative reactions predominates in the cell depends in part on the KM val­ues of the respective enzymes and the concentration of available substrate.

At very low substrate concen­trations, the specific reaction catalyzed by the enzyme with the lowest KM will dominate, whereas at higher substrate concentrations, the reaction catalyzed by the enzyme having the greatest KM value can domi­nate (if enough of the enzyme is present).

Therefore, which of several different metabolic pathways is actu­ally followed by the initial substrate in the course of a series of enzyme-catalyzed reactions may be regu­lated by controlling the amount of available substrate. The relationship between re­action velocity and substrate concentration for two en­zymes that act on the same substrate is depicted in Figure 8-6.

To obtain the KM value of an enzyme experimen­tally, it is necessary to determine v0 for a series of sub­strate concentrations. In practice the evaluation of KM from a plot similar to that in Figure 8-6 is difficult, because the precise value of Vmax cannot be deter­mined. This is because the curve is a rectangular hy­perbola that approaches Vmax asymptotically. In 1934, H. Lineweaver and D. Burk introduced a different form of the Michaelis-Menten equation (8-28) that simplifies the determination of KM from experimen­tally obtained values of [S] and v0. In this form, the ac­quired data are used to construct a straight line rather than a hyperbola. By taking the inverse of equation 8- 28 we obtain

The terms are now arranged in the form of the gen­eral equation of a straight line, y = ax + b, in which a is the slope of the line and b is the intercept on the y axis. Thus, for equation 8-31, 1/v0 serves as the y axis and 1/[S] as the x axis. Consequently, the y intercept will be 1/vmax, the slope will be KM/Vmax, and the x inter­cept will be -1/KM. The graph showing this relation­ship (known as a double reciprocal plot) is presented in Figure 8-7.


TEXTBOOK TREATMENTS

In deriving the Michaelis-Menten equation, all textbooks stress sufficiently that s0 must be larger than e0 to satisfy the approximation that the substrate concentration, s, always equals s0. In regard to the inhibition kinetics, however, this same prerequisite with respect to the inhibitor concentrations seems to be tacitly postulated in deriving the rate equation, in which the concentration of inhibitor, i, is expressed as its initial concentration, i0. This treatment ignores the fact that some of the inhibitor is present as enzyme·inhibitor complex. For example, i0, not i, is used to determine a value for the inhibition constant Ki using graphical approaches such as a Dixon plot (1/ν versus i), and the student should be made aware that i0 = i is an approximation. In transient kinetics, which is taught to students in advanced courses, the concentrations of enzyme may not always be low enough to satisfy the approximation that i = i0. In the enzymology textbooks that we have read, no attention is paid to the discrepancy between i0 and i when the equation for a Dixon plot is derived.


Error Rate Comparison during Polymerase Chain Reaction by DNA Polymerase

As larger-scale cloning projects become more prevalent, there is an increasing need for comparisons among high fidelity DNA polymerases used for PCR amplification. All polymerases marketed for PCR applications are tested for fidelity properties (i.e., error rate determination) by vendors, and numerous literature reports have addressed PCR enzyme fidelity. Nonetheless, it is often difficult to make direct comparisons among different enzymes due to numerous methodological and analytical differences from study to study. We have measured the error rates for 6 DNA polymerases commonly used in PCR applications, including 3 polymerases typically used for cloning applications requiring high fidelity. Error rate measurement values reported here were obtained by direct sequencing of cloned PCR products. The strategy employed here allows interrogation of error rate across a very large DNA sequence space, since 94 unique DNA targets were used as templates for PCR cloning. The six enzymes included in the study, Taq polymerase, AccuPrime-Taq High Fidelity, KOD Hot Start, cloned Pfu polymerase, Phusion Hot Start, and Pwo polymerase, we find the lowest error rates with Pfu, Phusion, and Pwo polymerases. Error rates are comparable for these 3 enzymes and are >10x lower than the error rate observed with Taq polymerase. Mutation spectra are reported, with the 3 high fidelity enzymes displaying broadly similar types of mutations. For these enzymes, transition mutations predominate, with little bias observed for type of transition.

Figures

Representative agarose gel electrophoresis images…

Representative agarose gel electrophoresis images of products of PCR amplification of 24 unique…


Kinetics of Enzymatic Reactions | Biochemistry

A catalyst — whether chemical or biological — acts by increasing the velocity of reactions. We will now examine the kinetics of enzymatic reactions. The velocity of an enzymatic reaction is generally fol­lowed up by measuring the quantity of substrate transformed per unit time one can also record the quantity of the product (or one of the products) formed by the reaction.

Samples of the reaction mixture are therefore collected at dif­ferent times and titrations made by chemical, chromatographic, colorimetric or manometric methods depending on each case.

Continuous analysis techniques may be used sometimes, particularly spectrophotometric methods (for ex­ample, one can record the variations of optical density at a given wavelength if the reaction is amenable to it, which is especially the case with NAD or NADP dehydrogenases) or titrimetric methods (for instance, the pH variations occur­ring during the reaction can be balanced by continuous addition of a base or acid, addition which may be automatically controlled by a pH meter).

Enzymatic activity may be expressed in several ways:

1. The enzymatic unit is the quantity of enzyme which catalyses the trans­formation of 1 micromole of substrate in one minute, at 25° in optimal condi­tions of pH and substrate concentration (but in certain cases one millimole, or one milligram of substrate transformed per minute, are also used).

2. The ratio of the number of enzymatic units to the amount (generally to one mg) of protein, gives the specific activity (which, obviously, increases as enzyme purification progresses).

3. The molar activity is the number of moles of substrate transformed (or product formed) per mole of enzyme per minute for this mode of expression one must have a highly purified enzyme of known molecular weight.

I. Initial Velocity of the Reaction:

If the quantity of substrate transformed is plotted against time, it is observed (Fig. 2-2) that the first portion of the curve is a straight line with constant slope the curve then bends, i.e. velocity decreases and finally becomes zero when all the substrate which was to be transformed (as a function of the equilibrium of the reaction) has disappeared.

It is therefore advantageous to measure the velocity at the beginning of the reaction, when only a small quantity of the substrate has been transformed and when the amounts of reaction products are sufficiently small for the reverse reaction to be negligible (in practice a large excess of substrate is added) the slope is then maximum and the initial velocity of the reaction (v0) can be determined.

The reaction velocity (v) at a given time (t) is d[P]/dt (where P is the quantity of product formed in the reaction). It can be determined by measuring the angle of the tangent to the curve at time t. The tangent at the origin gives v0, the initial velocity of the reaction.

Some factors can influence the velocity of an enzymatic reaction.

II. Effect of Enzyme Concentration on Kinetics:

If several experiments are carried out with increasing quantities of enzymes, it is observed that after a given time (t1), the quantity of substrate transformed is larger when more enzyme is present, provided one remains in the straight line portion of the curve (i.e. provided initial velocities are measured in each experiment) on the contrary, at time t2, the proportionality between velocity and enzyme concentration ceases (see fig. 2-3 A).

The variation of initial velocity can also be plotted against enzyme con­centration (after carrying out several experiments with increasing enzyme concentrations): it is observed that within certain limits, initial velocity is proportional to enzyme concentration (see fig 2-3 B) theoretically, for high enzyme concentrations, the velocity curve should bend and velocity should become constant, but in general this phenomenon is not observed because of the limitation imposed by the solubility of protein macromolecules (a 0.01 M solution of a protein of molecular weight 100 000 would require a solution of 1000 g/liter) therefore, in the curve of figure 2-3 B, the plateau is not reached although — the enzyme is a molecule which reacts with the substrate and one would expect that it would influence reaction velocity in the same manner as the substrate (see fig. 2-4).

The curve of figure 2-3 B shows that reaction velocity is nil when enzyme concentration is nil but it was said earlier that enzymes catalyze reactions which could proceed even in their absence and sometimes with appreciable velocities.

While studying velocity as a function of enzyme concentration, it is usual to carry out a “control” or “blank” test without the enzyme the values obtained at different times in this blank test are subtracted from the values observed under various enzyme concentrations. The curve obtained with these corrected values represents exclusively the result of the enzymatic action, (and v = 0 for [E] = 0 is therefore justified).

The proportionality between reaction velocity and enzyme concentration has important practical applications because it enables the estimation of the relative concentrations of a given enzyme in cell homogenates without the necessity of purifying the enzyme.

One has, on the one hand, v = Const. X [E], and on the other hand v = [P]/t, therefore:

If the same time is used to study the reaction, t is a constant and the product t X Const., i.e., the product of two constants will give another constant (which may be denoted by C). In these conditions enzyme concentration is therefore directly proportional to the quantity of product formed ([E] = [P]/C).

Let us consider for example, a homogenate which can yield x mg of P in a given time if another homogenate can yield 2x mg of P in the same time, we conclude that it contains twice the quantity of enzyme catalysing the formation of P.

Such measurements of relative enzyme concentrations are carried out sys­tematically (on blood for example) in clinical laboratories, because certain variations enable the diagnosis of a pathological condition.

In research laboratories, these determinations often represent the initial tests carried out while studying the regulation of the biosynthesis of an enzyme. It must be noted however, that the increase of active enzyme concentration can be due to either an increased synthesis of enzyme or an activation of the pre-existing enzyme molecules.

III. Effect of Substrate Concentration on Kinetics:

If enzyme concentration is maintained constant while substrate concentra­tion [S] is varied, a rapid increase of velocity is observed first but if [S] continues to increase, the curve bends and for higher values of [S], velocity ceases to increase and tends asymptotically to a maximum value (Vmax), as can be seen in figure 2-4.

1. Enzyme-Substrate Complex:

For a better understanding of the changes taking place when substrate and enzyme concentrations are varied, one must study in more details the reaction catalyzed by an enzyme.

When this reaction is written (admitting of course, that there can be more than one substrate and more than one product), we get the false impression that the enzyme will modify the velocity of this reaction by its mere presence in the medium.

But in reality it par­ticipates in the reaction by forming transitorily a specific enzyme-substrate complex. The existence of such complexes was suggested early this century, but it is only about thirty years later that their reality could be proved, first in the case of the peroxidase of horseradish (whose porphyrin prosthetic group has a characteristic absorption spectrum which is modified during the forma­tion of the complex with the substrate), then for other enzymes, by various methods, using physicochemical techniques like the study of visible or ultraviolet absorption spectra, fluorescence, nuclear magnetic resonance etc. or by equilibrium dialysis.

The modes of interactions between the 2 partners of the complex are now determined very accurately by X-ray diffraction analysis of crystals of enzyme-substrate complexes.

The basic reaction describing the enzymatic catalysis is written:

This representation is valid for the initial period of the reaction when con­centration of P is sufficiently low for the reverse reaction to be neglected. The quantity of P formed will depend directly on the E —S complex concentration so that one is led to study the variations of E — S concentration as a function of the increase of [S]. Posing the problem in this manner, it is clear that in the beginning, if [S] increases, [E-S] will also increase and reaction velocity will therefore increase.

But if [S] further increases, it is obvious that [E – S] cannot continue to increase beyond a certain maximum value which depends on the quantity of enzyme available. [E] becomes the limiting factor of the formation of the E —S complex, and an increase of [S] ceases to affect v: the plateau is reached (see fig. 2-4).

2. Michaelis-Menten Equation:

These two authors carried out a satisfactory quantitative study of the varia­tions of the velocity of an enzymatic reaction as a function of substrate concentra­tion.

This study is based on the representation:

and on the fact that equilibrium between E, S and E — S is a rapid process compared to the reaction E — S → E + P (which therefore controls the velocity of the enzymatic reaction).

For the study of the velocity of a reaction, the Menten-Michaelis equation offers the advantage that it is not necessary to know either [E] or [E — S], concentrations which are often difficult to deter­mine (and which were even more so in 1913 at the time of these two authors).

As mentioned above, the velocity of the enzymatic reaction is proportional to the E —S complex concentration, therefore v = k3 [E—S]. As for the maximum velocity (V max), it is observed for a substrate concentration such that the entire enzyme is bound to the substrate the maximum value of [E — S] is equal to the total enzyme concentration [ET]: Vmax — k3 [ET].

We can write:

The equilibrium constant or Michaelis constant for the dissociation of the complex E — S is:

where [E] is the free enzyme concentration, equal to [ET] — [E — S], Substitut­ing [E] by this value, we can write:

This may be written in the form:

It may be noted that, having determined experimentally the variations of v as a function of [S], one can use the Michaelis-Menten equation to obtain the value of Vmax and calculate Km.

The knowledge of enzyme concentration is not necessary Km can therefore be determined on incompletely purified prepara­tions. If for a given substrate, an identical Km is found in two preparations, it may be assumed that they contain the same enzyme.

Taking v = Vmax/2, we have:

The Michaelis constant (or equilibrium constant of the dissociation of the E -S complex) is therefore equal to the substrate concentration for which velocity is half the maximum velocity (see fig. 2-4). In general, the Km values range between 10 -2 M and 10 -8 M. It must be noted that Km is a measure of the affinity of the enzyme for its substrate.

The stronger the E — S interaction, the larger the quantity of enzyme combined with its substrate in the form E — S, and the smaller the quantity of free enzyme therefore [E] will be small and [E — S] large consequently, Km will be small.

The affinity of an enzyme for a substrate is equal to 1/Km. If Km is large (a high substrate concentration is therefore required to obtain a velocity equal to Vmax/2), affinity is low if Km is small, affinity is high (a small substrate concentration is sufficient to have a velocity equal to Vmax/2).

Examining the Michaelis-Menten equation written in the form one may note that when [SJ is very large compared to Km, Km can be neglected, and v tends to Vmax. On the contrary, when [S] is very small compared to Km, v is equal to Vmax , which shows that v is proportional to [S].

This does confirm the experimental results represented by the curve of figure 2-4.

Enzymatic catalysis comprises two steps:

1) The formation of E—S which implies the recognition of S by the active site of the enzyme. This first step is characterized by the Km which, as mentioned earlier, is equal to

2) The chemical operation of the decomposition of E – S into E + P.

This second step is characterized by Vmax.

v can be plotted as a function of [S] on the basis of experimental results a branch of an equilateral hyperbola is obtained (figure 2-4). Vmax (at least an approximate value) is obtained for large values of [S].

To determine Km the point Vmax/2 is plotted on the vertical axis and from the corresponding point of the curve a perpendicular is drawn to the horizontal axis. But the value of Vmax being only approximate, the determinations of Vmax/2 and thence Km are rather inaccurate.

Furthermore, a large number of experimental points are needed to trace such a curve, particularly at the low substrate concentrations where precision is not very high. But there is another graphical representation which requires only a small number of experimental points obtained at high substrate concentrations where precision in the determination of initial velocity is much greater.

3. Lineweaver-Burk Representation:

The Michaelis-Menten equation may be written:

The latter form is an equation of the type y = ax + b (where 1/v and 1/[S] are the variables y and x, while Km/Vmax and 1/ Vmax are the constants a and b). Then, by plotting the values of 1/v (the reciprocals of velocities determined experimentally) as a function of 1/[S], we obtain a straight line which — as seen in figure 2-5 — intersects the vertical axis (1/v) at the point 1/Vmax (since 1/[S] = 0, we have 1/v = 1/Vmax).

This straight line intersects the horizontal axis (1/[S]) at the point – 1/Km if 1/v = 0, we have:

This graphical representation is more convenient for determining the Michaelis constant (Km) and the maximum velocity (Vmax) of the reaction for a given enzyme concentration. Moreover, as will be seen later, it permits the distinction between a competitive inhibitor and a non-competi­tive inhibitor.


Reviewmylife

This is an experiment to examine how the concentration of the substrate hydrogen peroxide affects the rate of reaction of the enzyme catalase.

Introduction

This is an A-level biology project. It helped me get an A grade for biology many years ago. The whole project is reproduced here for your reference.

Enzymes such as Catalase are protein molecules which are found in living cells. They are used to speed up specific reactions in the cells. They are all very specific as each enzyme just performs one particular reaction.

Catalase is an enzyme found in food such as potato and liver. It is used for removing Hydrogen Peroxide from the cells. Hydrogen Peroxide is the poisonous by-product of metabolism. Catalase speeds up the decomposition of Hydrogen Peroxide into water and oxygen as shown in the equations below.

Formula:

It is able to speed up the decomposition of Hydrogen Peroxide because the shape of it's active site matches the shape of the Hydrogen Peroxide molecule. This type of reaction where a molecule is broken down into smaller pieces is called an anabolic reaction.

  1. Gas Syringe
  2. Metal Stand
  3. Yeast Catalase
  4. Hydrogen Peroxide
  5. Test Tubes
  6. Beakers
  7. Test Tube Rack
  8. Stop Watch
  9. Pipette
  10. Pipette Filler
  11. Tap Water

To test out how the concentration of hydrogen peroxide affects the rate of reaction first set up the apparatus below.

[Aparatus picture not reproduced]

1. Add 2cm3 of yeast to one test tube. Add 4cm3 of hydrogen peroxide solution at a concentration of 20% to the other test tube. Use a pipette to measure out the volumes. It is very important to accurately measure the amounts of Hydrogen Peroxide, Yeast and water to ensure a fair test.

2. Pour the hydrogen peroxide solution into the test tube containing the yeast and immediately put the gas syringe bung on the end of the test tube, at the same time start the stopwatch.

3. Bubbles should start to rise up the tube and the gas syringe will move outwards, as soon as the gas syringe passes the 30cm3 mark stop the stopwatch and note the elapsed time down to the nearest 1/10th of a second.

4. Repeat the experiment with hydrogen peroxide concentrations of 16%, 12%, 10%, 8%, 4% and 0%. The 0% concentration of hydrogen peroxide solution is done as a control solution to show that at 0% concentration no reaction occurs. The different concentrations of Hydrogen Peroxide are made by adding tap water to the 20% Hydrogen Peroxide in the correct amounts. The table below shows what amounts of Hydrogen Peroxide and water are needed to make the solutions.

Concentration Of Hydrogen Peroxide

Volume Of Hydrogen Peroxide (cm3)

20% 4 0 16% 3.2 0.8 12% 2.4 1.6 10% 2 2 8% 1.6 2.4 4% 0.8 3.2 0% 0 4

5. Repeat all the tests at least three times so that an average can be obtained. Repeating the experiments several times will help to produce better and more accurate results as any inaccuracies in one experiment should be compensated for by the other experiments. Note all the results in a table such as the one below.

Hydrogen Peroxide Concentration 0% 4% 8% 10% 12% 16% 20%
Time Taken (Test 1)
Time Taken (Test 2)
Time Taken (Test 3)
Average of the Tests
Rate

The rate can then be worked out by

This gives the rate in cm3 of oxygen produced per second, this is because I am timing how long it takes to produce 30cm3 of oxygen. From these results a graph can be plotted with concentration on the x-axis and time taken on the y-axis.

I am using yeast catalase as opposed to catalase from apples, potatoes or liver because it is easier to get the desired amount of yeast catalase by simply measuring it off. To obtain catalase from a substance such as potato would involve crushing it and with that method you would never be sure of the concentration of the catalase. If the catalase was used up then another potato would have to be crushed and this could produce catalase of a totally different concentration which would lead to inaccuracies in the experiment making this an unfair test.

To ensure this is a fair test all the variables except for the concentration of Hydrogen Peroxide must be kept the same for all the experiments. Variables that must not be altered include:-

Temperature, yeast concentration, type of yeast, batch of yeast, volume of yeast, volume of hydrogen peroxide, air pressure and humidity.

When measuring the volumes of Hydrogen Peroxide, Yeast and Water the measurement should be taken by looking at the scale at an angle of 90 degrees to it to avoid any parallax error.

I predict that as the substrate concentration increases, the rate of reaction will go up at a directly proportional rate until the solution becomes saturated with the substrate hydrogen peroxide. When this saturation point is reached, then adding extra substrate will make no difference.

The rate steadily increases when more substrate is added because more of the active sites of the enzyme are being used which results in more reactions so the required amount of oxygen is made more quickly. Once the amount of substrate molecules added exceeds the number of active sites available then the rate of reaction will no longer go up. This is because the maximum number of reactions are being done at once so any extra substrate molecules have to wait until some of the active sites become available.

I carried out the above experiment and these results were obtained.

Hydrogen Peroxide Concentration 0% 4% 8% 10% 12% 16% 20%
Time Taken (Test 1) 47.3 18.4 17.3 14.5 10.6 9.7
Time Taken (Test 2) 43.3 19 16.7 14.9 11.2 10
Time Taken (Test 3) 52.2 17.2 18.5 11.2 8.6 7.8
Average of the Tests 47.6 18.2 17.5 13.5 10.1 9.2
Rate =30/Average (Cm3/second) 0 0.63 1.65 1.71 2.22 2.97 3.26

All the times are in seconds. The average results are all written down to one decimal place because although the stopwatch gives results to two decimal places it is impossible to get accurate times to two decimal places due to the fact that our reaction times are not fast enough to stop the stopwatch precisely. I then worked out the rates of the reactions with the equation

From these rates I was able to plot a graph of the rate of reaction against concentration of Hydrogen Peroxide.

When the concentration of Hydrogen Peroxide is increased, the rate of reaction increases at a directly proportional rate until the concentration of Hydrogen Peroxide reaches about 16%. If you double the concentration of Hydrogen Peroxide then the rate of reaction doubles as well. When the concentration is doubled from 8-16% the rate goes up from 1.65-2.97 Cm3 Oxygen produced per second, which is an increase of 1.8 times. I would expect the rate to increase two times if the Hydrogen Peroxide concentration is increased two times because there are twice as many substrate molecules which can join onto the enzymes active sites. The reason that the number is less than two times could be put down to the fact that at 16% the Enzyme's active sites may already be close to being saturated with Hydrogen Peroxide. There may also be some experimental error which causes the inaccuracies.

After 16% the increase in the rate of reaction slows down. This is shown by the gradient of the graph going down. At this point virtually all the active sites are occupied so the active sites are said to be saturated with Hydrogen Peroxide. Increasing the Hydrogen Peroxide Concentration after the point of saturation has been reached will not cause the rate of reaction to go up any more. All the active sites are being used so any extra Hydrogen Peroxide molecules will have to wait until an active site becomes available.

The theoretical maximum rate of reaction is when all the sites are being used but in reality this theoretical maximum is never reached due to the fact that not all the active sites are being used all the time. The substrate molecules need time to join onto the enzyme and to leave it so the maximum rate achieved is always slightly below the theoretical maximum. The time taken to fit into and leave the active site is the limiting factor in the rate of reaction.

The diagram below shows what happens.

To help make this experiment more accurate, I repeated it three times and then used the average of all the results to plot a graph with a line of best fit. I tried to keep all the variables except for the concentration of Hydrogen Peroxide the same for all the experiments. However, in reality it is impossible to keep all the variables precisely the same. For example:

a) There is a slight delay between pouring the Hydrogen Peroxide into the yeast, putting the bung on and starting the stopwatch. This will slightly affect all the results but as I carried out all the three steps in the same way for all the experiments it should not make any difference to the overall result.

b) It is also impossible to precisely measure out the amounts of Hydrogen Peroxide, Yeast and Water each time. As the scale on the pipettes shows the volume to the nearest mm3 the volume of the solutions that I used should be correct to the nearest mm3. The volume of gas in the test tube to start with is slightly affected by the amount which the bung is pushed down each time, if the bung is pushed down further then the volume in the tube will be less so the 30cm3 of gas is reached faster.

c) Due to the fairly slow speed of our reactions it is only possible to measure the time of the reaction to the nearest 0.1 second even though the stopwatch shows the measurements to the nearest 0.01 second.

The plotted results on the graph produce a straight line of best fit to begin with which then goes into a curve of steadily decreasing gradient. The only anomalies are the results at 8% and 10%. The result at 8% is slightly above the line of best fit and the 10% result is slightly below it. This is probably due to an experimental error involving one of the factors mentioned above.

This experiment could be improved in a number of ways. It could be repeated more times to help get rid of any anomalies. A better overall result would be obtained by repeating the experiment more times because any errors in one experiment should be compensated for by the other experiments.

Using more concentrations of Hydrogen Peroxide would have produced a better looking graph and I would have liked to use concentrations higher than 20% to extend the graph so that the maximum possible rate of reaction could be reached.

The problem of the delay between pouring in the Hydrogen Peroxide, bunging the test tube and starting the stopwatch could have been limited by getting another person to start the stopwatch when the hydrogen peroxide was poured into the tube.

Disclaimer

This is a real A-level school project and as such is intended for educational or research purposes only. Extracts of this project must not be included in any projects that you submit for marking. Doing this could lead to being disqualified from all the subjects that you are taking. You have been warned. If you want more help with doing your biology practicals then have a look at 'Advanced Level Practical Work for Biology' by Sally Morgan. If you want more detailed biology information then I'd recommend the book 'Advanced Biology' by M. Kent.

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13 Responses to “The Effect Of Substrate Concentration On The Activity Of The Enzyme Catalase”

Would it be possible to have the same/a similar title for my coursework? this is very helpful. Would be great to use this is a guideline. Obviously not to copy.

Hi H.H, yes you can use this title. It isn’t my title – it was actually a title set by the exam board during the year I did my A-Levels. Good luck with your coursework.

Am I allowed to use this for secondary data for my controlled assessment?

Hi Zoë, yes you can use the data as long as you reference this web page. Good luck!

Hi! I was wondering how i could reference this page in APA format. Since I don’t have an author’s name, I can barely reference this page properly…. Thanks!

The reaction that breaks down large molecules into smaller moleculs is catabolism not anabolic reaction.

and also, shouldnt your graph’s x axis title be ‘concentration of hydrogen peroxide’ instead of ‘concentration of yeast’?

No. The x axis is fine. Because he adds water to the yeast not to the peroxide to change the concentration. So the indepent variable is the yeast and the indepent variable should be the x-axis!

Its catabolic reaction not anabolic.

CATABOLIC=CUT
as my bio teacher says

As the concentration increased , the amount of oxygen decrease? Why you wrote that it is directly proportional?

Hi I was just wondering how to cite this page all I got is the page name there is barely anything to cite can you help me

Hi Skye, you could just cite the page URL, and page title. Thanks.

Seems to be useful article but on some points your not correct.
forexample you said

“This type of reaction where a molecule is broken down into smaller pieces is called an anabolic reaction”.

What I know is that this type of reaction should be catabolism and not anabolism as you said.
anabolism and catabolism are two types of metabolism where the anabolism being building reaction and the catabolism being breaking reaction