Is this experiment to check enzyme reaction rate according to substrate concentration correct?

Is this experiment to check enzyme reaction rate according to substrate concentration correct?

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We just had an experiment to check enzyme reaction rate according to it's substrate concentration.
In the experiment, we used a varying amount of substrate and the same amount of enzyme:

(1.5mm enz, 4.0mm water, 0.5mm subs)
(1.5mm enz, 3.5mm water, 1.0mm subs)
(1.5mm enz, 2.5mm water, 2.0mm subs)
(1.5mm enz, 1.5mm water, 3.0mm subs)
(1.5mm enz, 0mm water, 4.5mm subs)

We check the rate by looking at the optical density after 3 minutes.

Upon staring at this, I couldn't help but wonder: We don't only change the substrate concentration! We also change it's amount!

While 3.0/4.5 is a higher concentration than 0.5/4.5, it's also a higher amount resulting in longer reaction time for sure. If we check after 3 minutes, there's a chance that the 0.5mm subs would already be gone.

If we really wished to look at the rate according to concentration, shouldn't we have only changed the amount of water? It also changes the concentration of the enzyme but there is there any way to isolate either of them?

I think there are a few important answers in your question

  • The concentration of enzyme is the same in each reaction chamber. The volume of each chamber is 6ml, and you are using 1.5 ml of enzyme in a final volume of 6 ml, so the dilution factor (the same in all chambers) is 1 in 4. So if stock enzyme concentration is 4 micromolar (as an example), the concentration of enzyme is 1 micromolar in each reaction chamber.
  • The concentration of substrate varies in each reaction chamber. You MUST take the dilution factor into account when calculating the final substrate concentration. You do no tell me what the stock substrate concentration is, but if it is 1mM as an example, then in reaction chamber 1 you are taking 0.5 ml of a 1mM solution in a final volume of 6 ml, which is a 1 in 12 dilution. The substrate concentration in the reaction chamber is therefore 1/12 mM (about 83.33 micromolar). In a similar way, the concentration in the final reaction chamber (4.5 ml in a final volume of 6 ml, dilution factor of 1.33) is about 750 micromolar.

  • You are right, measuring the reaction velocity by taking a single time point after 3 minutes is dangerous! You must of course measure the initial rate if you want to analyze results according to the Michalis-Menten equation. The assumption you are making is that the rate is linear over a period of 3. minutes for all substrate concentrations. If this is not so, you will need to redesign your experiment to take a number of time points and take the tangent to the curve. I think you must be interested in determining the Michaelis constant.

  • In the language of enzyme kinetics you are doing a discontinuous assay with a single time point after 3 minutes as a measure of the initial rate at a series of substrate concentrations. A better approach if available is to use a continuous assay, for example by continuously monitoring the change of absorbance of NADH with time at 340nm. But of course continuous assay not always available.

  • I am surprised that you use water to dilute. Where is the buffer? And how control pH in your chamber? Maybe the enzyme and substrate stock solutions are made in buffer, is this the explanation?

  • You should also check that the velocity versus enzyme concentration is linear at all substrate concentrations.

Brent Cornell

When designing an experiment to test the effect of factors affecting enzyme activity, the three key decisions to be made are:

  • Which factor to investigate (i.e. the independent variable)
  • Which enzyme / substrate reaction to use
  • How to measure the enzyme activity (i.e. the dependent variable)

Choosing the Independent Variable

The main factors which will affect the activity of an enzyme on a given substrate are:

  • Temperature (use water baths to minimise fluctuations)
  • pH (acidic or alkaline solutions)
  • Substrate concentration (choose range to avoid saturation)
  • Presence of inhibitor (type of inhibitor will be enzyme-specific)

Selecting an Enzyme and Substrate

Selection will depend on availability within the school, however certain enzymes can be extracted from common food sources

Examples of common enzyme-catalysed reactions include:

Measuring Enzyme Activity

The method of data collection will depend on the reaction occurring – typically most reactions are measured according to:

It has been shown experimentally that if the amount of the enzyme is kept constant and the substrate concentration is then gradually increased, the reaction velocity will increase until it reaches a maximum. After this point, increases in substrate concentration will not increase the velocity (delta A/delta T). This is represented graphically in Figure 8.

It is theorized that when this maximum velocity had been reached, all of the available enzyme has been converted to ES, the enzyme substrate complex. This point on the graph is designated Vmax. Using this maximum velocity and equation (7), Michaelis developed a set of mathematical expressions to calculate enzyme activity in terms of reaction speed from measurable laboratory data.

The Michaelis constant Km is defined as the substrate concentration at 1/2 the maximum velocity. This is shown in Figure 8. Using this constant and the fact that Km can also be defined as:

K +1 , K -1 and K +2 being the rate constants from equation (7). Michaelis developed the following

Michaelis constants have been determined for many of the commonly used enzymes. The size of Km tells us several things about a particular enzyme.

  • A small Km indicates that the enzyme requires only a small amount of substrate to become saturated. Hence, the maximum velocity is reached at relatively low substrate concentrations.
  • A large Km indicates the need for high substrate concentrations to achieve maximum reaction velocity.
  • The substrate with the lowest Km upon which the enzyme acts as a catalyst is frequently assumed to be enzyme's natural substrate, though this is not true for all enzymes.

MCAT Biochemistry Quiz

Question 1
Which of the following statements about enzyme kinetics is FALSE?
A. An increase in the substrate concentration (at constant enzyme concentration) leads to proportional increases in the rate of the reaction.
B. Most enzymes operating in the human body work best at a temperature of 37°C.
C. An enzyme–substrate complex can either form a product or dissociate back into the enzyme and substrate.
D. Maximal activity of many human enzymes occurs around pH 7.4.

A: Most enzymes in the human body operate at maximal activity around a temperature of 37°C and a pH of 7.4, which is the pH of most body fluids. In addition, as characterized by the Michaelis–Menten equation, enzymes form an enzyme–substrate complex, which can either dissociate back into the enzyme and substrate or proceed to form a product. So far, we can eliminate choices (B), (C), and (D), so let’s check choice (A). An increase in the substrate concentration, while maintaining a constant enzyme concentration, leads to a proportional increase in the rate of the reaction only initially. However, once most of the active sites are occupied, the reaction rate levels off, regardless of further increases in substrate concentration. At high concentrations of substrate, the reaction rate approaches its maximal velocity and is no longer changed by further increases in substrate concentration.

Question 2
Which two polysaccharides share all of their glycosidic linkage types in common?
A. Cellulose and amylopectin
B. Amylose and glycogen
C. Amylose and cellulose
D. Glycogen and amylopectin

D: Glycogen and amylopectin are the only polysaccharide forms that demonstrate branching structure, making them most similar in terms of linkage. Both glycogen and amylopectin use α-1,4 and α-1,6 linkages. Cellulose uses β-1,4 linkages and amylose does not contain α-1,6 linkages.

Question 3
Which of the following is correct about fat-soluble vitamins?
I. Vitamin E is important for calcium regulation.
II. Vitamin D protects against cancer because it is a biological antioxidant.
III. Vitamin K is necessary for the posttranslational introduction of calcium-binding sites.
IV. Vitamin A is metabolized to retinal, which is important for sight.
A. III only
B. I and II only
C. III and IV only
D. II, III, and IV only

C: Vitamin A is metabolized to retinal, which is important for sight. Vitamin D is metabolized to calcitriol, which is important for calcium regulation. Vitamin E is made up of tocopherols, which are biological antioxidants. Vitamin K is necessary for the introduction of calcium binding sites, such as during the posttranslational modification of prothrombin.

Question 4
Topoisomerases are enzymes involved in:
A. DNA replication and transcription.
B. posttranscriptional processing.
C. RNA synthesis and translation.
D. posttranslational processing.

A: Topoisomerases, such as prokaryotic DNA gyrase, are involved in DNA replication and mRNA synthesis (transcription). DNA gyrase is a type of topoisomerase that enhances the action of helicase enzymes by the introduction of negative supercoils into the DNA molecule. These negative supercoils facilitate DNA replication by keeping the strands separated and untangled.

Question 5
Which of the following is true of diffusion and osmosis?
A. Diffusion and osmosis rely on the electrochemical gradient of only the compound of interest.
B. Diffusion and osmosis rely on the electrochemical gradient of all compounds in a cell.
C. Diffusion and osmosis will proceed in the same direction if there is only one solute.
D. Diffusion and osmosis cannot occur simultaneously.

A: The movement of any solute or water by diffusion or osmosis is dependent only on the concentration gradient of that molecule and on membrane permeability.

Hydrogen Ion Concentration (pH)

Because most enzymes are proteins, they are sensitive to changes in the hydrogen ion concentration or pH. Enzymes may be denatured by extreme levels of hydrogen ions (whether high or low) any change in pH, even a small one, alters the degree of ionization of an enzyme&rsquos acidic and basic side groups and the substrate components as well. Ionizable side groups located in the active site must have a certain charge for the enzyme to bind its substrate. Neutralization of even one of these charges alters an enzyme&rsquos catalytic activity.

An enzyme exhibits maximum activity over the narrow pH range in which a molecule exists in its properly charged form. The median value of this pH range is called the optimum pH of the enzyme (part (b) of Figure (PageIndex<2>)). With the notable exception of gastric juice (the fluids secreted in the stomach), most body fluids have pH values between 6 and 8. Not surprisingly, most enzymes exhibit optimal activity in this pH range. However, a few enzymes have optimum pH values outside this range. For example, the optimum pH for pepsin, an enzyme that is active in the stomach, is 2.0.

Michaelis &ndash Menten Equation

Leonor Michaelis and Maud Menten postulated that the enzyme first combines reversibly with its substrate to form an enzyme-substrate complex in a relatively fast reversible step:


In the next step, this ES complex is breaks down in to the free enzyme and the reaction product P:


Since the second step is the rate limiting step, the rate of overall reaction must be proportional to the concentration of the ES that reacts in the second step. The relationship between substrate concentration, [S] and Initial velocity of enzyme, V0 (Fig. 1) has the same general shape for most enzymes (it approaches a rectangular hyperbola). This can be expressed algebraically by the Michaelis-Menten equation. Based on their basic hypothesis that the rate limiting step in enzymatic reactions is the breakdown of the ES complex to free enzyme and product, Michaelis and Menten derived an equation which is


The necessary terms in this reaction are [S], V0, Vmax, and Km (Michaelis constant),. All these terms can be measured experimentally.

Lineweaver &ndash Burke plot

In 1934, Lineweaver and Burke made a simple mathematical alteration in the process by plotting a double inverse of substrate concentration and reaction rate.


For enzymes obeying the Michaelis-Menten relationship, the &ldquodouble reciprocal&rdquo of the V0 versus [S] from the first graph,(fig1) yields a straight line (Fig. 2). The slope of this straight line is KM /Vmax, which has an intercept of 1/Vmax on the 1/V0 axis, and an intercept of -1/KM on the 1/[S] axis. The double-reciprocal presentation, also called a Lineweaver-Burk plot. The main advantage of Lineweaver-Burk plot is to determine the Vmax more accurately, which can only be approximated from a simple graph of V0 versus [S] (Fig 1).

Fig2: Lineweaver-Burk plot.

Adapted from David L. Nelson, Michael M. Cox , Lehninger principles of biochemistry, 4th edition.

Deriving the Michaelis-Menten Equation

Memorize this derivation as soon as your encounter it in your text, and you will be able to read the remainder of the chapter with far greater understanding. For other suggestions on how to make your study of biochemistry easier, see Learning Strategies .

A simple model of enzyme action:

We would like to know how to recognize an enzyme that behaves according to this model. One way is to look at the enzyme's kinetic behavior -- at how substrate concentration affects its rate. So we want to know what rate law such an enzyme would obey. If a newly discovered enzyme obeys the rate law derived from this model, then it's reasonable to assume that the enzyme reacts according to this model. It's not proof that the model is correct, but at least it tells us that kinetics does not rule it out.

Let's derive a rate law from this model.

For this model, let V 0 be the initial velocity of the reaction. Then

The maximum velocity V max occurs when the enzyme is saturated -- that is, when

all enzyme molecules are tied up with S, or

So V max = k cat [E] total . (3)

We want to express V 0 in terms of measurable quantities, like [S] and [E] total , so we can see how to test the mechanism by experiments in kinetics. So we must replace [ES] in (2) with measurables.

During the initial phase of the reaction, as long as the reaction velocity remains constant, the reaction is in a steady state , with ES being formed and consumed at the same rate. During this phase, the rate of formation of [ES] equals its rate of consumption. According to model (1),

Rate of formation of ES = k 1 [E][S].

Rate of consumption of ES = k -1 [ES] + k cat [ES].

So in the steady state, k -1 [ES] + k cat [ES] = k 1 [E][S]. (4)

Remember that we are trying to solve for [ES] in terms of measurables, so that we can replace it in (2). First, collect the kinetic constants in (4):

To simplify (5), first group the kinetic constants by defining them as K m :

and then express [E] in terms of [ES] and [E]total:

Substitute (6) and (7) into (5):

Solve (8) for [ES]: First multiply both sides by [ES]:

Then collect terms containing [ES] on the left:

Factor [ES] from the left-hand terms:

and finally, divide both sides by (K m + [S]):

Substitute (9) into (2): V 0 = k cat [E] total [S]/(K m + [S]) (10)

Recalling (3), substitute V max into (10) for k cat [E] total :

V 0 = V max [S]/(K m + [S]) (11)

This equation expresses the initial rate of reaction in terms of a measurable quantity, the initial substrate concentration. The two kinetic parameters, V max and K m , will be different for every enzyme-substrate pair.

Equation (11), the Michaelis-Menten equation, describes the kinetic behavior of an enzyme that acts according to the simple model (1). Equation (11) is of the form

y = ax/(b + x) (does this look familiar?)

This is the equation of a rectangular hyperbola, just like the saturation equation for the binding of dioxygen to myoglobin.

Equation (11) means that, for an enzyme acting according to the simple model (1), a plot of V 0 versus [S] will be a rectangular hyperbola. When enzymes exhibit this kinetic behavior, unless we find other evidence to the contrary, we assume that they act according to model (1), and call them Michaelis-Menten enzymes.

QUIZ for USM Students

Quiz at first class on enzyme kinetics: Derive equation (11) from model (1).

Is this experiment to check enzyme reaction rate according to substrate concentration correct? - Biology


Hydrogen peroxide (H2O2) is a common by-product of metabolic reactions. In high concentration it is toxic therefore, its accumulation in cells would be harmful. Most tissues, however, contain the enzyme catalase, which catalyzes the breakdown of peroxide to water and oxygen as follows: The reaction is extremely rapid. The action of the enzyme can be demonstrated easily by the evolution of oxygen in the form of gas bubbles when an extract of a tissue containing the enzyme is added to a dilute solution of hydrogen peroxide. We will use homogenized (ground-up) chicken or beef liver as a source of catalase. Catalase

  • Catalase - general information Classrooms of 21st century
  • Catalase - An Extraordinary Enzyme
  • Heat denature enzyme
  • Structure of Catalase image and graph
  • Links -


    Amylase References:
    • Amylase - Independent study by Amy Caruana (Student at Kean Univ)
    • Amylase - results from U. North Dakota
    • Enzymes amylase, temp, pH, substrate concentration
    • Benedicts test for reducing sugar - image of result
    • Amylase How does it work?
    • Amylase - starch hydrolysis
    • Salivary amylase - pH
    Pepsin is a protease that begins digestion of proteins, breaking them into peptides and amino acids. Pepsinogen, is secreted by gastric glands of the stomach into the stomach. There, in the acid environment of the stomach, pepsinogen is converted into pepsin.

    Although both pepsin and trypsin are proteases, they require quite different conditions of acidity and alkalinity for their action.

      Pepsin References:
      • Pepsin -
      • Pepsin - molecule of the month
      • pH optimum -
      • Pepsin - mentions pH optimum
      Trypsin is a protease secreted into the small intestine by the pancreas. As pepsin, trypsin digests proteins into peptides and amino acids and is made and secreted in an inactive form, trypsinogen.

      Although both pepsin and trypsin are proteases, they require quite different conditions of acidity and alkalinity for their action.

        Trypsin References:
        • Trypsin Encyclopedia description
        • Illustration search for trypsin within this large file
        • Images Google search engine

        An Analogy

        Suppose you are interested in purchasing a Pizza store and wish to investigate how productive the store is without the present owner knowing because, you fear the owner will raise the price. So, instead of going into the store and watching what happens and asking to examine the books that record expenses and profits, you decide to watch the store from outside.

        You observe how often trucks arrive with pizza dough, pizza toppings (cheese, pepporoni, etc.), and other supplies. You also observe how often workers leave the store to deliver pizzas to customers.

        In this analogy, the pizza supplies are the reactants and the boxed pizzas that are delivered to customers are the end products. The workers within the store that shape the dough, add the toppings and place the pizzas in ovens and finally in boxes are the equivalent of the enzymes.

        Although we don't actually see the workers doing their job, we can infer that if the store is using large quantities of reactants (dough and toppings) and / or making large numbers of end products (pizzas) that the workers (enzymes) must be very active.

        Types of assay

        All enzyme assays measure either the consumption of substrate or production of product over time. A large number of different methods of measuring the concentrations of substrates and products exist and many enzymes can be assayed in several different ways. Biochemists usually study enzyme-catalysed reactions using four types of experiments: [2]

        (1) Initial rate experiments. When an enzyme is mixed with a large excess of the substrate, the enzyme-substrate intermediate builds up in a fast initial transient. Then the reaction achieves a steady-state kinetics in which enzyme substrate intermediates remains approximately constant over time and the reaction rate changes relatively slowly. Rates are measured for a short period after the attainment of the quasi-steady state, typically by monitoring the accumulation of product with time. Because the measurements are carried out for a very short period and because of the large excess of substrate, the approximation free substrate is approximately equal to the initial substrate can be made. The initial rate experiment is the simplest to perform and analyze, being relatively free from complications such as back-reaction and enzyme degradation. It is therefore by far the most commonly used type of experiment in enzyme kinetics.

        (2) Progress curve experiments. In these experiments, the kinetic parameters are determined from expressions for the species concentrations as a function of time. The concentration of the substrate or product is recorded in time after the initial fast transient and for a sufficiently long period to allow the reaction to approach equilibrium. We note in passing that, while they are less common now, progress curve experiments were widely used in the early period of enzyme kinetics.

        (3) Transient kinetics experiments. In these experiments, reaction behaviour is tracked during the initial fast transient as the intermediate reaches the steady-state kinetics period. These experiments are more difficult to perform than either of the above two classes because they require rapid mixing and observation techniques.

        (4) Relaxation experiments. In these experiments, an equilibrium mixture of enzyme, substrate and product is perturbed, for instance by a temperature, pressure or pH jump, and the return to equilibrium is monitored. The analysis of these experiments requires consideration of the fully reversible reaction. Moreover, relaxation experiments are relatively insensitive to mechanistic details and are thus not typically used for mechanism identification, although they can be under appropriate conditions.

        Enzyme assays can be split into two groups according to their sampling method: continuous assays, where the assay gives a continuous reading of activity, and discontinuous assays, where samples are taken, the reaction stopped and then the concentration of substrates/products determined.

        Catalase and Hydrogen Peroxide Experiment

        How do living cells interact with the environment around them? All living things possess catalysts, or substances within them that speed up chemical reactions and processes. Enzymes are molecules that enable the chemical reactions that occur in all living things on earth. In this catalase and hydrogen peroxide experiment, we will discover how enzymes act as catalysts by causing chemical reactions to occur more quickly within living things. Using a potato and hydrogen peroxide, we can observe how enzymes like catalase work to perform decomposition, or the breaking down, of other substances. Catalase works to speed up the decomposition of hydrogen peroxide into oxygen and water. We will also test how this process is affected by changes in the temperature of the potato. Is the process faster or slower when compared to the control experiment conducted at room temperature?


        What happens when a potato is combined with hydrogen peroxide?



        1. Divide the potato into three roughly equal sections.
        2. Keep one section raw and at room temperature.
        3. Place another section in the freezer for at least 30 minutes.
        4. Boil the last section for at least 5 minutes.
        5. Chop and mash a small sample (about a tablespoon) of the room temperature potato and place into beaker or cup.
        6. Pour enough hydrogen peroxide into the cup so that potato is submerged and observe.
        7. Repeat steps 5 & 6 with the boiled and frozen potato sections.

        Observations & Results

        Watch each of the potato/hydrogen peroxide mixtures and record what happens. The bubbling reaction you see is the metabolic process of decomposition, described earlier. This reaction is caused by catalase, an enzyme within the potato. You are observing catalase breaking hydrogen peroxide into oxygen and water. Which potato sample decomposed the most hydrogen peroxide? Which one reacted the least?

        You should have noticed that the boiled potato produced little to no bubbles. This is because the heat degraded the catalase enzyme, making it incapable of processing the hydrogen peroxide. The frozen potato should have produced fewer bubbles than the room temperature sample because the cold temperature slowed the catalase enzyme&rsquos ability to decompose the hydrogen peroxide. The room temperature potato produced the most bubbles because catalase works best at a room temperature.


        Catalase acts as the catalyzing enzyme in the decomposition of hydrogen peroxide. Nearly all living things possess catalase, including us! This enzyme, like many others, aids in the decomposition of one substance into another. Catalase decomposes, or breaks down, hydrogen peroxide into water and oxygen.

        Want to take a closer look? Go further in this experiment by looking at a very small sample of potato combined with hydrogen peroxide under a microscope!

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        Watch the video: Biochemistry I Michaelis Menten Problem 2 (November 2022).