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What happens to Km when enzyme concentration is *very* high?

What happens to Km when enzyme concentration is *very* high?


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Let Km be an empirical measurement of a certain enzyme with concentration [E]. Theoretically, this value is constant and shouldn't vary when [E] goes up or down. Now let [E']=10*Km. Under this concentration of enzyme, it's clear that if [S]=Km, V0 cannot be 1/2*Vmax (as there's only enough substrate to saturate 1/10-th of the enzyme molecules). What's happening here, and how does this relate to the derivation of the Michaelis-Menten equation?


Michaelis-Menten is often challenging for students because of the importance of the conditions that must exist in order for the model to hold. Walking through the derivation and the relevance of the assumptions can be helpful. Berg's Biochemistry is available via NCBI, and the section on Michaelis-Menten kinetics does a good job here, illustrating the following points:

Assumption 1: The rates of all the relevant reactions are linear.

This assumption is what allows us to write down the equations the derivation starts with. We look at each separate reaction, and say that the rate of that reaction is equal to some rate constant times the concentration of each thing on the left side. From the equation $E + S ightleftharpoons ES ightarrow E + P$, we get the following

  1. $E + S ightarrow ES$

    • $Rate = k_1[E][S]$
  2. $ES ightarrow E + S$

    • $Rate = k_{-1}[ES]$
  3. $ES ightarrow E + P$

    • $Rate = k_2[ES]$

This assumption (the rate is linear) only matches the experimental evidence early on in the reaction, while most of the substrate is free $[S]$ (rather than bound up in $ES$). Remember this. It will come into play when we evaluate assumption 3.

Assumption 2: The $ES ightarrow E + P$ reaction is irreversible

In order for us to combine the above rate equations to derive the Michaelis-Menten equation, the only source of ES should be from $E + S ightarrow ES$. We can't have any measurable $E + P ightarrow ES$. Experimentally, for enzymes that follow Michaelis Menten kinetics, this is true when $[E]$ and $[P]$ are small relative to [ES]. So here we have our first requirement that $[E]$ be small. Biochemical reactions tend to not involve very high changes in free energy. The in situ free energy of ATP hydrolysis is the high end for a reaction in the cell, and that releases about $50kJ mol^{-1}$. We're not talking about combustion here. So in fact, these reactions are often reversible at appreciable levels of $[E]$ and $[P]$

Assumption 3: ES formation and breakdown is at steady state

This assumption is often described in a confusing way, i.e. the entire reaction ($E + S ightarrow E + P$) is at steady state. Lets drill down to what we're actually talking about. In order to derive the MM equation. We need the rate of formation of $ES$ to equal the rate of breakdown of $ES$, that is, for $frac{d[ES]}{dt} = 0$. That is steady state. Let's say that in words: the change in $[ES]$ needs to be zero. We only achieve steady state at an early point in the reaction (remember, we need most of the substrate to be free $[S]$ for assumption 1 to hold), if $[S]$ is substantially greater than $[E]$. $E$ needs to be saturated with $S$ while $S$ is still mostly in the free form. This is the key combination of assumptions that the situation you described violates.

Once we meet the three assumptions above, we can write down the following equality:

$k_1[E][S] = (k_{-1} + k_2)[ES]$

This equation says that the rate of formation of $ES$ (the rate from equation 1 at the top) is equal to the combined rate of the breakdown of $ES$, either into $E + S$ or $E + P$ (equations 2 and 3 at the top). Once we have this equality, we can derive the Michaelis-Menten equation: $V_o = V_{max} frac{[S]}{[S] + K_m}$, where $K_m = frac{k_{-1} + k_2}{k_1}$. Again, see Biochemistry to walk through this.


The derivation of the Michaelis-Menten equation makes a number of assumptions. The exact assumptions depend a bit on how you actually do the derivation, but most modern formulations depend on the steady state approximation. (For example, see here.)

In your situation, where the concentration of enzyme greatly exceeds the concentration of substrate, that assumption is violated. You'll never reach a steady-state. As such, applying the Michaelis-Menten equation to this particular situation gives inaccurate results. You certainly can model the kinetics of the reaction in these conditions, but you need to use other equations and approaches.

That's not to say that the Km doesn't apply in those situations. The Km of an enzyme is a property of the enzyme that applies even in the high-enzyme regime - it's just that the "conventional" interpretation of Km ("the concentration of substrate at which reaction rate is half maximal") doesn't apply. That statement is only true in the particular conditions which the Michaelis-Menten derivation holds.


The effect of high concentrations of enzyme on the Michaelis-Menten equation has been considered by M. F. Chaplin, "The effect of enzyme concentration on the Michaelis-Menten equation" published in the January 1981 issue of Trends in Biochemical Sciences, page VI, under the heading 'Textbook Error' (but, as far as I can determine, this article is not indexed in Pubmed or the like).

The author draws two important conclusions:

  1. The assumption that the total enzyme concentration, $E_o$, must be less than the initial substrate concentration ($S_o$) is unnecessarily restrictive. The Michaelis-Menten equation holds, to a very good approximation, when $E_o$ is greater than $S_o$ but is less than the Michaelis constant (!)
  2. At high enzyme concentrations, a modified equation is proposed $$ v_i = frac{V_{max}[S_o]}{K_m + [S_o] + [E_o]}$$

The derivation of this equation is given in the cited reference.

Edit

  • A Dropbox link to the Chaplin reference. It does not seem to be available from anywhere, including the TIBS site (and it does not appear to be indexed in PubMed or the like)
  • A user asked (question now deleted) if $K_m$ is in any way dependent on E${_o}$

Setting

$$ K^{'}_m = K_m + E_o$$

it is seen that the substrate concentration at half-V$_{max}$ is $K^{'}_m$

that is the 'true' Michaelis constant will indeed depend on $E_o$ 'just a little bit' as the User suggested under the less restrictive assumptions considered by M.F. Chaplin (where $K_m$ is defined algebraically above).

  • Perhaps the User would consider un-deleting this question? IMO an interesting one.

Enzymes

7.2.3 Specific Activity

Enzyme concentrations are often given in terms of “units” rather than in mole or mass concentration. We rarely know the exact mass of the enzyme in a sample, since it is generally prepared via isolation of the enzyme from microorganisms, or animal or plant tissues, and often contains a great deal of noncatalytic protein, the amount of which may vary from sample to sample. Hence a different approach must be adopted, and enzyme concentration is reported in units of specific activity. A “unit” is defined as the amount of enzyme (eg, microgram) that gives a certain amount of catalytic activity under specified conditions (eg, producing 1.0 micromole of product per minute in a solution containing a substrate concentration sufficiently high to be in the “saturation” region, as shown in Fig. 7.9 where [ES] and [E] are relatively invariant).

Thus different suppliers of enzymes may have preparations with different units of activity, and care must be taken in analyzing kinetic data. Thus a purified enzyme preparation will have a higher specific activity than a crude preparation often a protein is considered pure when a constant specific activity is reached during the purification steps.

The activity is given by the amount of product formed or substrate consumed in the reaction mixture, under the conditions specified (temperature, pH, buffer type, substrate and enzyme concentrations, etc.). If the molecular weight of the enzyme is known, the specific activity can also be defined as


On a plot of Velocity versus Substrate Concentration ( V vs. [S]), the maximum velocity (known as Vmax) is the value on the Y axis that the curve asymptotically approaches. It should be noted that the value of V max depends on the amount of enzyme used in a reaction. Double the amount of enzyme, double the Vmax . If one wanted to compare the velocities of two different enzymes, it would be necessary to use the same amounts of enzyme in the different reactions they catalyze. It is desirable to have a measure of velocity that is independent of enzyme concentration. For this, we define the value Kcat , also known as the turnover number. Mathematically,

To determine Kcat, one must obviously know the Vmax at a particular concentration of enzyme, but the beauty of the term is that it is a measure of velocity independent of enzyme concentration, thanks to the term in the denominator. Kcat is thus a constant for an enzyme under given conditions. The units of K cat are ( ext

Figure 4.7.2: Common turnover numbers

Another parameter of an enzyme that is useful is known as KM , the Michaelis constant. What it measures, in simple terms, is the affinity an enzyme has for its substrate. Affinities of enzymes for substrates vary considerably, so knowing KM helps us to understand how well an enzyme is suited to the substrate being used. Measurement of KM depends on the measurement of Vmax. On a V vs. [S] plot, KM is determined as the x value that give Vmax/2. A common mistake students make in describing V max is saying that KM = Vmax/2. This is, of course not true. KM is a substrate concentration and is the amount of substrate it takes for an enzyme to reach Vmax/2. On the other hand Vmax/2 is a velocity and is nothing more than that. The value of KM is inversely related to the affinity of the enzyme for its substrate. High values of KM correspond to low enzyme affinity for substrate (it takes more substrate to get to Vmax ). Low KM values for an enzyme correspond to high affinity for substrate.


What is Km in an enzyme?

In Enzymology, KM is referred to the Michaelis Constant . This constant has a simple operational definition, considering the Michaelis-Menten equation (in honor to Leonor Michaelis and Maud Menten , two of the principal "parents" of enzymology): v o = Vmax [S] / KM + [S], where v o is the initial velocity of the reaction, Vmax is the maximal velocity, [S] is the substrate concentration, and KM the Michaelis Constant.

The Michaelis Constant is explained as: at the substrate concentration where [S] = KM, yields v o (the initial velocity of the reaction) = Vmax / 2 so that KM is the substrate concentration at which the reaction velocity is half-maximal . Therefore, if an enzyme has a small value of KM, it achieves maximal catalytic eficiency at low substrate concentration. The magnitud of KM vareis widely with the identity of the enzyme and the nature of the substrate. Furthermore, is a function of temperature and pH.

The Michaelis constant can be expressed as: KM = k-1 / k1 + k2 / k1 = Ks + k2 / k1 (where, k-1 is the enzyme-substrate complex disociation constant into enzyme and substrate, k1 is the enzyme-substrate complex asociation constant, and k2 is the enzyme-substrate complex disociation constant to form product and the enzyme). Since Ks is the dissociation constant of the Michaelis complex, as Ks decreases, the enzyme's affinity for substrate increases. KM is therefore also a measure of the affinity of the enzyme for its substrate providing k2 / k1 is small compared with Ks, that is, k2 Wiki User


Effect of pH on Enzymes

For every enzyme, there is an optimum pH value, at which the specific enzyme functions most actively. Any change in this pH significantly affects the enzyme activity and/or the rate of reaction. To know more about the relation between pH and enzymes, and/or the effect of pH on enzymes, go through this write-up.

For every enzyme, there is an optimum pH value, at which the specific enzyme functions most actively. Any change in this pH significantly affects the enzyme activity and/or the rate of reaction. To know more about the relation between pH and enzymes, and/or the effect of pH on enzymes, go through this write-up.

Enzymes are proteinaceous catalysts, which speed up the rate of a biochemical reaction. They reduce the activation energy that is essential for starting any type of chemical reaction. With a low energy requirement for activation, the reaction takes place faster. The overall performance of an enzyme depends on various factors, such as temperature, pH, cofactors, activators, and inhibitors. You might have a fair idea regarding the effect of pH on enzymes. But why and how does pH and temperature affect enzymes?

Why pH Affects Enzyme Activity?

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The rate of a chemical reaction and/or the enzyme activity is greatly influenced by the structure of the enzyme. Or in other words, a change in the structure of the enzyme affects the rate of reaction. When pH of a particular medium changes, it leads to an alteration in the shape of the enzyme. Not only enzymes, the pH level may also affect the charge and shape of the substrate.

Within a narrow pH range, changes in the structural shapes of the enzymes and substrates may be reversible. But for a significant change in pH levels, the enzyme and the substrate may undergo denaturation. In such cases, they cannot identify each other. Consequently, there will be no reaction. This is why pH is a determining factor of enzyme activity.

How Does pH Affect Enzymes?

Each and every enzyme is characterized by an optimum pH. At this specific pH level, a particular enzyme catalyzes the reaction at the fastest rate than at any other pH level. For example, the enzyme pepsin (a protease enzyme) is most active at an acidic pH, whereas the enzyme trypsin (another protease enzyme) performs best at a slightly alkaline pH. Thus, the optimum pH of an enzyme is different from that of another enzyme.

When we study pH, it is clearly defined as the measurement for the acidic or alkaline nature of a solution. To be more precise, pH indicates the concentration of dissolved hydrogen ions (H + ) in the particular solution. An increase or decrease in the pH changes the ion concentration in the solution.

These ions alter the structure of the enzymes and at times the substrate, either due to formation of additional bonds or breakage of already existing bonds. Ultimately, the chemical makeup of the enzyme and substrate are changed. Also, the active site of the enzyme is changed, after which the substrate can no longer identify the enzyme. For more information on enzymes, you can refer to enzyme substrate complex.

Consider a case when the reaction is adjusted at a pH level different from the optimum value. Over here, the rate of reaction or the activity of enzymes will not be the same as the previous one. At times, you will notice that there is no reaction at all. This occurs when there are changes in the structure of the active site and the substrate. Hence, for the chemical reaction to take place, you need to adjust the pH of the solution in such a way that it is suitable for both, the enzyme and the substrate. This way, the effect of pH on enzyme activity can be studied practically.

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Vmax, Km, Kcat, Ki help.

I'm struggling to understand what the concept of these terms are especially when math is involved.

Things like Km(app) = 3.2x10 -5 M --> what does this mean to a biochemist? I know Km is the [S] at 1/2 Vmax but what does the value indicate?

Can someone give me an overall explanation of what these terms actually mean? Thanks.

Km is a measure of how easy the substrate associates with the enzyme. A high Km means that you need more substrate to reach a certain reaction rate, while a low Km means the opposite.

Kcat, or k2 or turnover number (they all mean the same thing) is a measure of how many substrates one (1) enzyme can convert into a product per second. Vmax is simply Kcat times the enzyme concentration.

Ki is like Km, but for an inhibitor. It measures the affinity the inhibitor has for the enzyme and if Ki is low, that means the affinity is high (you need a lower concentration to reach a certain inhibition), and the opposite for a high Ki.

edit: Km(app) is the Km when an inhibitor is involved. For example, if you have a reversible competetitive inhibitor, the Vmax will stay the same but Km will increase, giving you an apparant Km. This is because you need more substrate to compete with the inhibtor, but if you get enough substrate you will still be able to reach the same Vmax. However, if you have an irreversible inhibitor, Vmax will decrease, but Km will stay the same.


What happens to Km when enzyme concentration is *very* high? - Biology

Figure 1. Enzymes lower the activation energy of the reaction but do not change the free energy of the reaction.

A substance that helps a chemical reaction to occur is called a catalyst, and the molecules that catalyze biochemical reactions are called enzymes. Most enzymes are proteins and perform the critical task of lowering the activation energies of chemical reactions inside the cell. Most of the reactions critical to a living cell happen too slowly at normal temperatures to be of any use to the cell. Without enzymes to speed up these reactions, life could not persist. Enzymes do this by binding to the reactant molecules and holding them in such a way as to make the chemical bond-breaking and -forming processes take place more easily. It is important to remember that enzymes do not change whether a reaction is exergonic (spontaneous) or endergonic. This is because they do not change the free energy of the reactants or products. They only reduce the activation energy required for the reaction to go forward (Figure 1). In addition, an enzyme itself is unchanged by the reaction it catalyzes. Once one reaction has been catalyzed, the enzyme is able to participate in other reactions.

The chemical reactants to which an enzyme binds are called the enzyme’s substrates. There may be one or more substrates, depending on the particular chemical reaction. In some reactions, a single reactant substrate is broken down into multiple products. In others, two substrates may come together to create one larger molecule. Two reactants might also enter a reaction and both become modified, but they leave the reaction as two products. The location within the enzyme where the substrate binds is called the enzyme’s active site. The active site is where the “action” happens. Since enzymes are proteins, there is a unique combination of amino acid side chains within the active site. Each side chain is characterized by different properties. They can be large or small, weakly acidic or basic, hydrophilic or hydrophobic, positively or negatively charged, or neutral. The unique combination of side chains creates a very specific chemical environment within the active site. This specific environment is suited to bind to one specific chemical substrate (or substrates).

Active sites are subject to influences of the local environment. Increasing the environmental temperature generally increases reaction rates, enzyme-catalyzed or otherwise. However, temperatures outside of an optimal range reduce the rate at which an enzyme catalyzes a reaction. Hot temperatures will eventually cause enzymes to denature, an irreversible change in the three-dimensional shape and therefore the function of the enzyme. Enzymes are also suited to function best within a certain pH and salt concentration range, and, as with temperature, extreme pH, and salt concentrations can cause enzymes to denature.

For many years, scientists thought that enzyme-substrate binding took place in a simple “lock and key” fashion. This model asserted that the enzyme and substrate fit together perfectly in one instantaneous step. However, current research supports a model called induced fit (Figure 2). The induced-fit model expands on the lock-and-key model by describing a more dynamic binding between enzyme and substrate. As the enzyme and substrate come together, their interaction causes a mild shift in the enzyme’s structure that forms an ideal binding arrangement between enzyme and substrate.

When an enzyme binds its substrate, an enzyme-substrate complex is formed. This complex lowers the activation energy of the reaction and promotes its rapid progression in one of multiple possible ways. On a basic level, enzymes promote chemical reactions that involve more than one substrate by bringing the substrates together in an optimal orientation for reaction. Another way in which enzymes promote the reaction of their substrates is by creating an optimal environment within the active site for the reaction to occur.

Figure 2. The induced-fit model is an adjustment to the lock-and-key model and explains how enzymes and substrates undergo dynamic modifications during the transition state to increase the affinity of the substrate for the active site.

Careers in Action: Pharmaceutical Drug Developer

Figure 3. Have you ever wondered how pharmaceutical drugs are developed? (credit: Deborah Austin)

Enzymes are key components of metabolic pathways. Understanding how enzymes work and how they can be regulated are key principles behind the development of many of the pharmaceutical drugs on the market today. Biologists working in this field collaborate with other scientists to design drugs.

Consider statins for example—statins is the name given to one class of drugs that can reduce cholesterol levels. These compounds are inhibitors of the enzyme HMG-CoA reductase, which is the enzyme that synthesizes cholesterol from lipids in the body. By inhibiting this enzyme, the level of cholesterol synthesized in the body can be reduced. Similarly, acetaminophen, popularly marketed under the brand name Tylenol, is an inhibitor of the enzyme cyclooxygenase. While it is used to provide relief from fever and inflammation (pain), its mechanism of action is still not completely understood.

How are drugs discovered? One of the biggest challenges in drug discovery is identifying a drug target. A drug target is a molecule that is literally the target of the drug. In the case of statins, HMG-CoA reductase is the drug target. Drug targets are identified through painstaking research in the laboratory. Identifying the target alone is not enough scientists also need to know how the target acts inside the cell and which reactions go awry in the case of disease. Once the target and the pathway are identified, then the actual process of drug design begins. In this stage, chemists and biologists work together to design and synthesize molecules that can block or activate a particular reaction. However, this is only the beginning: If and when a drug prototype is successful in performing its function, then it is subjected to many tests from in vitro experiments to clinical trials before it can get approval from the U.S. Food and Drug Administration to be on the market.

Many enzymes do not work optimally, or even at all, unless bound to other specific non-protein helper molecules. They may bond either temporarily through ionic or hydrogen bonds, or permanently through stronger covalent bonds. Binding to these molecules promotes optimal shape and function of their respective enzymes. Two examples of these types of helper molecules are cofactors and coenzymes. Cofactors are inorganic ions such as ions of iron and magnesium. Coenzymes are organic helper molecules, those with a basic atomic structure made up of carbon and hydrogen. Like enzymes, these molecules participate in reactions without being changed themselves and are ultimately recycled and reused. Vitamins are the source of coenzymes. Some vitamins are the precursors of coenzymes and others act directly as coenzymes. Vitamin C is a direct coenzyme for multiple enzymes that take part in building the important connective tissue, collagen. Therefore, enzyme function is, in part, regulated by the abundance of various cofactors and coenzymes, which may be supplied by an organism’s diet or, in some cases, produced by the organism.


A general rule of thumb for most chemical reactions is that a temperature rise of 10°C approximately doubles the reaction rate. To some extent, this rule holds for all enzymatic reactions. After a certain point, however, an increase in temperature causes a decrease in the reaction rate, due to denaturation of the protein structure and disruption of the active site (part (a) of Figure 18.14 "Temperature and pH versus Concentration"). For many proteins, denaturation occurs between 45°C and 55°C. Furthermore, even though an enzyme may appear to have a maximum reaction rate between 40°C and 50°C, most biochemical reactions are carried out at lower temperatures because enzymes are not stable at these higher temperatures and will denature after a few minutes.

Figure 18.14 Temperature and pH versus Concentration

(a) This graph depicts the effect of temperature on the rate of a reaction that is catalyzed by a fixed amount of enzyme. (b) This graph depicts the effect of pH on the rate of a reaction that is catalyzed by a fixed amount of enzyme.

At 0°C and 100°C, the rate of enzyme-catalyzed reactions is nearly zero. This fact has several practical applications. We sterilize objects by placing them in boiling water, which denatures the enzymes of any bacteria that may be in or on them. We preserve our food by refrigerating or freezing it, which slows enzyme activity. When animals go into hibernation in winter, their body temperature drops, decreasing the rates of their metabolic processes to levels that can be maintained by the amount of energy stored in the fat reserves in the animals’ tissues.


An Overview of Existing Individual Unit Operations

Solmaz Aslanzadeh , . Mohammad J. Taherzadeh , in Biorefineries , 2014

1.3.2 Factors Affecting the Enzymatic Process

Substrate concentration affects the primary rate and yield of enzymatic hydrolysis. High substrate concentrations can result in substrate inhibition, which significantly lowers the hydrolysis rate [ 40 ]. Problems relating to reduction in heat and mass transfer efficiency, rheological problems, and increased inhibitor concentration have been reported in association with high solid-loading [ 89 ]. Released during the hydrolysis, sugars, primarily consisting of cellobiose and glucose, inhibit the cellulase activity cellobiose at a concentration of 6 g/l can reduce the cellulase activity by 60% [ 84 ]. Glucose also decreases the cellulase activity, but it has a smaller effect than cellobiose does a glucose concentration of 3 g/l can cause a 75% reduction of β-glucosidase activity [ 88 ].

Lignocellulosic materials possess two different types of surface area, internal and external. The internal surface area relies on the capillary structure of cellulosic fibers, whereas the external area depends on the size and shape of the particles. Generally, dry cellulosic fibers are small in size, about 15-40 μm hence, they have a substantial external specific surface area, namely, 0.6-1.6 m 2 /g. The internal surface area of dried cellulosic fibers is smaller compared to the external surface area [ 84 ] because of pores collapse due to drying, and this is an important reason not to dry lignocellulosic materials after pretreatment. The interaction between the enzymes and the accessible surface area of the lignocellulosic material can be a limiting factor in the enzymatic process [ 90 ].


What Effects of Concentrations of a Substrate on Enzyme Lab Answers

An enzyme is described as a biological catalyst that speeds up the rate of a chemical reaction. In order for an enzyme to perform its given job, it needs what is known as a substrate to bind to the active site of the enzyme so that the enzyme can speed up the reaction of the substrate. In this given experiment, it was to be tested what impact the concentration level of the substrate will have on the reaction rate.

The enzyme catalase, found in potato juice, was used for the catalyst along with a substrate known as hydrogen peroxide (H2O2). The job of catalase in this experiment was to accelerate the breakdown of hydrogen peroxide into water and oxygen gas. After looking and recording how different concentrations of a substrate affect the enzyme’s activity, the results were put into a chart and graph to show our observations.

The given chart and graph show the different concentration levels of the substrate that were used, along with the time it took for the reaction to occur.

An enzyme is supposed to speed up the reaction, but our observations show that the concentration of the substrate also had an effect on how fast the reaction could occur. When there were zero concentrates of hydrogen peroxide, meaning the solvent was only water, there was no reaction with the catalase. As the concentration of hydrogen peroxide increases, the rate of the reaction began to increase.

This occurred because adding amounts of hydrogen peroxide gave the catalase the opportunity to break down the hydrogen peroxide, whereas when there was no H2O2 present, there was nothing for the catalase to react with. The catalase needs its specific substrate to bind with in order for a reaction to occur.

The shape of the chart itself shows the decrease in needed reaction time, proving our analysis to be correct. If we had continued to test out different concentrations of our substrate, the reaction time would have continued to decrease until the enzyme hit what is known as the saturation level, which means all the enzymes are working and the reaction cannot continue to increase because there is no location at that time for the substrates to bind to. For this experiment, the saturation point was not reached but could have if the experiment had continued.

From our observations, it could be concluded that the higher the concentration of substrate, Hydrogen Peroxide there was in the reaction, the less time was needed for the reaction to occur.


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Comments:

  1. Denis

    What an interesting answer

  2. Abdul-Tawwab

    I want you to say that you are not right.



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