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Relative concentration of enzyme vs reaction product

Relative concentration of enzyme vs reaction product


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I was reading a paper in which a recombinant protein (His-6 tagged) is expressed in E Coli (BL21 DE3). The yield of the enzyme isolated from the culture is reported as 10-30 mg per L of bacterial culture. (After purification by Ni-affinity Chromatography)

First, is this a reasonable concentration for an enzyme? I thought it was rather high.

Second, when they measure the concentration of the product of the enzymatic reaction that was found to be merely 1-3 mg/Litre of bacterial culture.

Is this typical or unusual for the product concentration to be an order of magnitude smaller than the enzyme concentration? I assumed that since enzymes were in catalytic amounts they ought to be at lower concentrations?

PS. They use SDS-PAGE followed by the Bradford method for determining the protein titres. How accurate is the Bradford method?


Remember that the concentrations you report are in mass units rather than mole units. If the protein has a mass of 40,000 Da, reasonable for many proteins, then 20 mg/liter means $0.5$ x $10^{-6}$ mol/liter. A typical small-mass product of an enzyme reaction might have a mass about 200, so its apparently lower mass concentration of 2 mg/liter is $10$ x $10^{-6}$ mol/liter, 20 times higher than the enzyme concentration in terms of numbers of molecules per liter. The mole or molecule ratios are what you should think about in terms of catalytic function.

Furthermore, from what you provide we don't know whether the enzyme was produced under circumstances that allowed it to express its activity; there may have been limitations on substrate availability, presence of inhibitors that would be removed in later purification, and so forth. In this scenario, evidently to obtain enzyme for downstream use, the emphasis is typically on getting large amounts of the specific enzyme regardless of whether some reaction product is formed. I don't find the enzyme concentration particularly high for something that the investigators are trying to over-express as in this case.

The Bradford reaction is a mainstay of standard biochemistry, as it's simple and straightforward. Different proteins can bind the reagent to somewhat different extents, so there is a second-order issue of whether the protein used for the standard curve and that being measured are similar enough in that respect. In practice, what's important is to provide enough information in the methods that are described so that someone could repeat the work to determine the relation between the Bradford measurement and some other more accurate method for this particular protein.


Relative concentration of enzyme vs reaction product - Biology

Concentration Effects

Effects of Substrate Concentration

We examined reaction rate at different substrate concentrations. Our standard tubes contained 1 mL of hydrogen peroxide, and different groups ran reactions with more or less substrate to compare to that. Because each group tried different volumes of substrate, we got extremely variable results which proved difficult to interpret. In general, with much less substrate we got slower reactions and with more, we got faster reactions. What we expect to see is that the rate should level off once it reaches the maximum (when every peroxidase enzyme has a hydrogen peroxide molecule to interact with -- at that point, adding more hydrogen peroxide can't increase the reaction rate because all of the available enzymes are busy).

Effects of Enzyme Concentration

We also examined reaction rate when we increased the amount of turnip extract in our tubes. Since the turnip extract contained our peroxidase enzyme, this increased the enzyme concentration in our reactions. The graph above shows that as we increased the amount of enzyme in the tubes, our reaction rate increased. At some point, the maximum raction rate should be achieved and we should see the curve level off (see the substrate concentration curve above for an example), but based on this data, more than 3 mL of turnip extract is required to reach a 1:1 ratio of enzyme to substrate molecules.

Concentration Effects Summary

Enzyme reaction rates have set limits: they can't drop past zero (no reaction) and they can't speed up past 100%. When there is a 1:1 ratio of enzyme and substrate, the reaction is going as fast as it can. Add more enzyme, and there won't be substrate for it to react with. Add more substrate and there won't be enzyme free to work with it. Thus, when adding either substrate or enzyme, you should expect to see the see the reaction rate curve level off when the maximum is reached.


In order to study the effect of increasing the enzyme concentration upon the reaction rate, the substrate must be present in an excess amount i.e., the reaction must be independent of the substrate concentration. Any change in the amount of product formed over a specified period of time will be dependent upon the level of enzyme present. Graphically this can be represented as:

These reactions are said to be "zero order" because the rates are independent of substrate concentration, and are equal to some constant k. The formation of product proceeds at a rate which is linear with time. The addition of more substrate does not serve to increase the rate. In zero order kinetics, allowing the assay to run for double time results in double the amount of product.

Table I: Reaction Orders with Respect to Substrate Concentration
Order Rate Equation Comments
zero rate = k rate is independent of substrate concentration
first rate = k[S] rate is proportional to the first power of substrate concentration
second rate = k[S][S]=k[S] 2 rate is proportional to the square of the substrate concentration
second rate = k[S1][S2] rate is proportional to the first power of each of two reactants

The amount of enzyme present in a reaction is measured by the activity it catalyzes. The relationship between activity and concentration is affected by many factors such as temperature, pH, etc. An enzyme assay must be designed so that the observed activity is proportional to the amount of enzyme present in order that the enzyme concentration is the only limiting factor. It is satisfied only when the reaction is zero order.

In Figure 5, activity is directly proportional to concentration in the area AB, but not in BC. Enzyme activity is generally greatest when substrate concentration is unlimiting.

When the concentration of the product of an enzymatic reaction is plotted against time, a similar curve results, Figure 6.

Between A and B, the curve represents a zero order reaction that is, one in which the rate is constant with time. As substrate is used up, the enzyme's active sites are no longer saturated, substrate concentration becomes rate limiting, and the reaction becomes first order between B and C.

To measure enzyme activity ideally, the measurements must be made in that portion of the curve where the reaction is zero order. A reaction is most likely to be zero order initially since substrate concentration is then highest. To be certain that a reaction is zero order, multiple measurements of product (or substrate) concentration must be made.

Figure 7 illustrates three types of reactions which might be encountered in enzyme assays and shows the problems which might be enountered if only single measurements are made.

B is a straight line representing a zero order reaction which permits accurate determination of enzyme activity for part or all of the reaction time. A represents the type of reaction that was shown in Figure 6. This reaction is zero order initially and then slows, presumably due to substrate exhaustion or product inhibition. This type of reaction is sometimes referred to as a "leading" reaction. True "potential" activity is represented by the dotted line. Curve C represents a reaction with an initial "lag" phase. Again the dotted line represents the potentially measurable activity. Multiple determinations of product concentration enable each curve to be plotted and true activity determined. A single end point determination at E would lead to the false conclusion that all three samples had identical enzyme concentration.


Enzyme Kinetics Lab – The Relationship Between Enzyme and Substrate Concentrations and Rates of Reaction

Enzymes are catalysts which lower the activation of chemical reactions, thus making them happen more rapidly. In this experiment, different amount of enzyme and substrate were put in a test tube, then were observed using a spectrophotometer to see how fast the reacted to produce product. It was found that as the concentration of enzyme was increased, the speed of reaction increased. Likewise, it was found that as the concentration of substrate was increased, the speed of reaction increased. As there is more enzyme, it is able to react with more substrate at once, therefore increasing the rate of reaction. When there is more substrate, there will be just as much enzyme, also making the rate of the reaction increase. Thus, higher levels of enzyme or substrate mean there will be a higher turnover rate of product.

Introduction

A catalyst is a substance that lowers the activation energy of a chemical reaction (Zubay et al., 1995). In order words, this means that a catalyst makes chemical reactions happen faster than they normally do. Enzymes are catalysts that speed up reactions in living cells. For example, the enzyme carbonic anhydrase makes the chemical reaction CO2 + H2O —> H2CO3 happen 10 7 times faster than it normally does. In this example, the enzyme reacted with CO2. The substance which an enzyme catalyzes is called a substrate. Enzymes bind to substrates in order to speed the reaction in turning the substrate to a product. When there is a little amount of substrate, there will be a small amount of enzyme, but as the level of substrate increases, the level of enzyme increases. In this experiment, the speed at which enzymes and substrates react was observed.

Materials and Methods

For the first experiment, different enzyme concentrations were tested to see how fast they reacted. A Spec20 spectrophotometer was first turned on and set to a wavelength of 340 nm. A reaction mixture consisting of 0.850 ml distilled water, 1.000 ml buffer stock, 0.100 ml 15 mM NAD+, and 1.000 ml ethanol was then put into the test tube using a micropipettor. This mixture was agitated to mix it together, and the spectrophotometer was then calibrated using this test tube. Once calibrated, 0.050 ml ADH enzyme was added to the test tube mixture and agitated. The test tube was then quickly put into the spectrophotometer and its absorbance readings were recorded. The readings were recorded every 15 seconds for 3 minutes. This process was repeated for 2 more reaction mixtures with adjustments to the amount of distilled water and AHD enzyme included. The second reaction mixture had 0.100 ml ADH enzyme and 0.800 ml distilled water, and the third one had 0.025 ml ADH enzyme and 0.875 ml distilled water.

For the second experiment, different substrate concentrations were tested to see their reaction rates. Once again, a Spec20 spectrophotometer was turned on and set to a wavelength of 340 nm. A reaction mixture consisting of 1.85 ml distilled water, 1.0 ml buffer stock, 100 ul NAD+, and 0 ul ethanol stock was put into a test tube using a micropipettor. The mixture was agitated, then put into the spectrophotometer and the spectrophotometer was calibrated. Once 50 ul of ADH was added to the reaction mixture, the test tube was agitated and quickly put into the spectrophotometer. Absorbance readings were then recorded every 15 seconds for 2 minutes. Another reaction mixture with the same amounts of each substance was made, observed, and recorded. This process was repeated for five more reaction mixtures with differences in the amount of distilled water and ethanol stock used. The amounts of distilled water and ethanol stock for these reaction mixtures were 1.80 ml distilled water and 50 ul ethanol stock, 1.75 ml distilled water and 100 ul ethanol stock, 1.65 ml distilled water and 200 ul ethanol stock, 1.35 ml distilled water and 500 ul ethanol stock, and 0.85 ml distilled water and ethanol stock.

Results

Time (min:sec) Absorbance (50 ul enzyme) Absorbance (100 ul enzyme) Absorbance (25 ul enzyme)
0:00 n/a n/a n/a
0:15 n/a n/a n/a
0:30 n/a .15 .035
0:45 .14 .22 .07
1:00 .175 .28 .095
1:15 .21 .3325 .12
1:30 .24 .38 .145
1:45 .27 .42 .1675
2:00 .2975 .455 .19
2:15 .32 .485 .2125
2:30 .3425 .505 .235
2:45 .36 .525 .255
3:00 .3825 .55 .2725
3:15 .4 .57 .29
Amount of Enzyme (ul) [E] (ug/ml) Initial Slope (A/min) Velocity (umol/min)
50 0.00333 1.60 0.772
100 0.00667 2.29 1.10
25 0.00167 0.889 0.429

The concentration of enzyme had an effect of the reaction rate. As the concentration of the enzyme went up, the velocity went up (Table II, Figure 2). The initial slope also became steeper as the concentration of the enzyme increased (Table II, Figure 1).

The concentration of substrate affected the rate of reaction. As the concentration of substrate went up, the velocity increased steeply then evened out (Figure 4). From Figure 4 (v vs. [S]), the estimated Vmax is 0.0620 umol/min and the estimated Km is 0.0140 mM. From Figure 3 (1/v vs. 1/[S]), the estimated Vmax is 0.0568 umol/min and the estimated Km is 10 mM. The actual Vmax and Km recorded from SigmaPlot were 0.06566 umol/min and 0.01522 mM. The Vmax obtained from each graph are fairly close, but the Km are not very close at all. The Vmax and Km from the v vs. [S] graph are very similar to the actual Vmax and Km obtained from SigmaPlot.

Discussion

Shono and co-workers (1995) observed the rate of reaction versus the concentration of a substrate different from the one used in this experiment. Their graphs resulting from their experiment are very similar to the graph resulting from this experiment. The “velocity vs. substrate concentration” graphs follow the almost exact same curve, but the levels of concentration were higher in Shono’s experiment, resulting in higher rates of reaction. In the “1/v vs. 1/[S]” graphs, Shono’s line seems to have about the same slope, but it crosses the x-axis at a much lower value than it did in the graphs for this experiment.

This can be attributed to error in procedure of the experiment, which caused outlier values the line to be skewed. It was not known these outlier values should have been disregarded until computing calculations for Km and Vmax using the “1/v vs. 1/[S]” graph and comparing to other similar graphs. The Vmax recorded from Figure 3 was fairly close to the Vmax found from Figure 4. The Km from Figure 3, 10 mM, is not very close, however. That value is way larger than the Km found from Figure 4, 0.0140 mM, and the actual Km calculated by SigmaPlot, 0.01522 mM.

From the experiment, the results showed that with a higher concentration of enzyme, the higher rate of reaction. When catalyzing a reaction, the enzyme binds to the substrate (Bolsover et al., 1997). If there is a higher concentration of enzyme, this means that there will be more enzyme to bind to substrate at once, therefore making the turnover rate of substrate to product higher. This is why a higher concentration of enzyme produces a higher turnover rate. The turnover rate begins to slow down and stop as the amount of substrate runs out, and that is why the absorbance rates began to even out in Figure 1. There was a limited amount of substrate that could be converted into product.

As there was a higher concentration of substrate, the rate of reaction increased and then leveled off, as shown in Figure 4. The Michaelis-Menten graph assumes that the enzyme and substrate are in equilibrium (Zubay et al., 1995). Therefore, as there is a higher concentration of substrate, there will be an equally high concentration of enzyme to react with the substrate. As there is a higher concentration of each, the rate of reaction increases. The curve evens off, however. This is because each substrate has a maximum velocity at which it can convert from substrate to product. The enzyme can not catalyze the substrate to turnover faster than this.

The affinity for the enzyme and substrate in this experiment was fairly high. The Km, 0.01522 mM, is a low number. This means the enzyme and substrate reacted very quickly to produce product.

Literature Cited

Bolsover, S.R., J.S. Hyams, S. Jones, E.A. Shepard, and H.A. White. 1997. From Genes to Cells . (Wiley-Liss, NY). 424 p.

Shono, M., M. Wada, T. Fujii, 1995. Partial Purification of a Na+ -ATPase from the Plasma Membrane of the Marine Alga Heterosigma akashiwo. Plant Physiol 108: 1615-1615.

Zubay, Geoffrey, William M. Parson, and Dennis E. Vance. 1995. Principles of Biochemistry . (Wm. C. Brown, Dubuque, Iowa) 863 p.


Enzyme Activity Changes as the Enzyme Concentration Decreases

Enzymes are a very important class of biocatalysts. Due to the action of the enzyme, the chemical reaction in the living body can be carried out efficiently and specifically under extremely mild conditions. The life processes of every creature on the earth, from small to large, proliferating, living and dying, and metabolism, are all related to enzymes. Without enzyme catalysis, the most basic food digestion, oxygen breathing cannot be carried out, let alone other life activities. In fact, almost all kinds of reactions occurring in living organisms are carried out by enzyme catalysis.

Enzymes are very important to life. The reason for this is that the rate of enzyme catalysis is very fast, which is thousands of times faster than the rate of reaction catalyzed by a chemical catalyst. For example, glucose in food reacts with oxygen to become carbon dioxide and water, and the energy released is the energy that maintains the body's body temperature and all life activities. If there is no catalyst, the catalytic process takes several years or longer under normal temperature and pressure conditions. To speed up the reaction, it must be carried out above three hundred degrees Celsius, burning and oxidizing to release energy. In the living body, under the catalysis of a series of enzymes, it can be completed instantaneously at normal temperature and pressure at an unimaginable speed.

Enzymes assist in chemical reactions by allowing molecules to reduce the energy required to initiate the reaction. This energy, called activation energy, is provided by the environment. For example, ambient thermal energy associated with ambient temperature can be used as activation energy. The rate of chemical reactions in a biological environment is usually limited by a limited amount of environmental energy, but enzymes overcome this limitation because they enable a smaller amount of energy to activate more reactions.

In most cases, reducing the enzyme concentration has a direct effect on the enzyme activity. An enzyme only catalyzes a reaction of one substance, or the same reaction of a chemically similar substance, and will never catalyze other substances and reactions. Most enzymes catalyze only one reaction of a substance, even if the structure is very similar. This property is called the specificity of enzyme catalysis. The enzyme-bound molecule is called a substrate. Typically, an enzyme is combined with a substrate to reduce the activation energy of a chemical reaction. If all of the enzymes in the system bind to the substrate, the additional substrate molecules must wait for the enzyme to become available after the reaction is complete. This means that as the enzyme concentration decreases, the reaction rate will decrease.

In most biological environments, the concentration of the enzyme is lower than the concentration of the substrate. The relationship between enzyme concentration and enzyme activity is directly proportional. On a graph showing reaction rate versus enzyme concentration, this direct proportional relationship looks like a straight line with a slope of one. In other words, one additional enzyme increases the rate by one reaction per unit time, and one removed enzyme reduces the rate by one reaction per unit time. An exception to the direct proportional relationship is that if the substrate concentration is below the enzyme concentration, lowering the enzyme concentration does not result in a decrease in enzyme activity. In this case, the removed enzyme has no effect because the system still has enough enzyme to bind to all available substrates. Thus, as the enzyme concentration increases to a level similar to the substrate concentration, the map of enzyme activity versus enzyme concentration will eventually settle to a flat line.


One of the properties of enzymes that makes them so important as diagnostic and research tools is the specificity they exhibit relative to the reactions they catalyze. A few enzymes exhibit absolute specificity that is, they will catalyze only one particular reaction. Other enzymes will be specific for a particular type of chemical bond or functional group. In general, there are four distinct types of specificity:

  • Absolute specificity - the enzyme will catalyze only one reaction.
  • Group specificity - the enzyme will act only on molecules that have specific functional groups, such as amino, phosphate and methyl groups.
  • Linkage specificity - the enzyme will act on a particular type of chemical bond regardless of the rest of the molecular structure.
  • Stereochemical specificity - the enzyme will act on a particular steric or optical isomer.

Though enzymes exhibit great degrees of specificity, cofactors may serve many apoenzymes. For example, nicotinamide adenine dinucleotide (NAD) is a coenzyme for a great number of dehydrogenase reactions in which it acts as a hydrogen acceptor. Among them are the alcohol dehydrogenase, malate dehydrogenase and lactate dehydrogenase reactions.


Results

Coevolution Employing a Common Substrate.

The CL1 and DSL enzymes (Fig. 1) first were challenged to undergo continuous coevolution within a common reaction mixture, employing limited amounts of a single oligonucleotide substrate. This process quickly led to the extinction of one or the other enzyme, depending on which substrate was used. A substrate similar to that used in the evolutionary development of the CL1 enzyme was used more efficiently by that enzyme, and likewise for the DSL enzyme and its preferred substrate. This advantage could not be overcome by evolution before extinction of the disadvantaged enzyme occurred. By staggering the starting concentration of the 2 enzymes, it was possible to delay the onset of extinction. It became clear, however, that additional constraints would be required to prevent runaway growth of the more advantageous species.

Sequence and secondary structure of the evolved CL1 (A) and DSL (B) enzymes, shown with substrates S4 and S5, respectively. Open rectangles indicate primer binding sites at the 5′ end of the substrate and 3′ end of the enzyme. Curved arrow indicates the site of ligation. Filled circles highlight mutations present in typical clones isolated after 50 transfers of coevolution (with 5 substrates) relative to the starting enzymes.

A serial transfer experiment was initiated, during which a 14-mer antisense oligodeoxynucleotide was used to inhibit the growth of the dominant ligase. Antisense oligos were designed to hybridize to a critical region of each of the 2 RNA enzymes, thus impairing their catalytic activity. Beginning with a population of 1 pmol each of randomized variants of the CL1 and DSL enzymes, 40 serial transfers of continuous coevolution were carried out, monitoring the concentration of each ligase (based on cDNA) before and after each transfer [supporting information (SI) Fig. S1]. To maintain diversity within the evolving population, error-prone PCR was performed on both enzymes after transfers 10, 20, and 30. Extinction was narrowly avoided during this coevolutionary process by employing the antisense oligos, but it required knowledge by the experimenter of which sequence to target and manipulation of the reaction conditions based on which enzyme required antisense inhibition. Furthermore, the enzymes began to develop resistance to the antisense oligos by acquiring escape mutations that destabilized the relevant base-pairing interactions. Prolonged use of this approach would require that the oligos be redesigned to maintain the desired effect.

Coevolution Employing 5 Different Substrates.

A second continuous coevolution experiment was initiated, employing a mixture of 5 different substrates (Table S1). Each substrate (S1–S5) contained a single nucleotide change within the promoter sequence compared with the substrate that was used previously (S0). This change reduced the efficiency of the promoter to 9%–63% relative to that of the wild-type promoter and created a single-nucleotide mismatch for each combination of enzyme and substrate.

Continuous coevolution was initiated with 10 pmol of the CL1 and 1 pmol of the DSL enzymes, which were transcribed from templates that had been subject to mutagenic PCR. The concentration of each ligase was measured before and after each transfer (Fig. 2), and the time between transfers was adjusted periodically to allow the enzymes to exhaust the supply of their corresponding substrates (Fig. S2). The concentrations of the CL1 and DSL enzymes were again staggered after transfers 30, 32, 37, 42, and 46 to compensate for their differential growth rates. Diversity was maintained within the evolving population by performing mutagenic PCR after transfers 5, 7, 10, 15, 20, 25, 30, 32, 37, 42, and 46.

Time course of 50 transfers of continuous coevolution in the presence of 5 different substrates. The concentration of the CL1 (orange) and DSL (blue) enzymes was determined (based on their respective cDNAs) before and after each transfer. Paired bar graph at the Top indicates the substrate preference of each enzyme (CL1 at Left, DSL at Right) after various transfers when provided 1 μM S1 (red), S2 (green), S3 (purple), S4 (orange), or S5 (blue). Stepped graph at the Bottom indicates the concentration of S5 present during the coevolution experiment all other substrates were present at 1 μM concentration throughout.

The initial reaction mixture contained 1 μM each of the 5 substrates. Assays were conducted before the start of the experiment and periodically thereafter to measure the ability of each enzyme to amplify in the presence of 1 μM of each substrate, tested individually (Fig. 2). Both enzymes showed a strong tendency to adapt to utilization of S5, and accordingly, the concentration of S5 was reduced to 0.1 μM or eliminated completely for intervals during the course of the experiment to provide selective advantage favoring the use of one of the other substrates.

After 50 transfers of continuous coevolution, with an overall amplification of ≈10 100 -fold, individual CL1 and DSL molecules were cloned from the population and sequenced. A typical clone of each enzyme was chosen for more detailed analysis (Fig. 1). The evolved CL1 enzyme contained 5 mutations relative to the starting enzyme, whereas the evolved DSL enzyme contained 11 mutations. These mutations included compensatory changes within the portion of the enzyme that binds the substrate: A13→G for CL1 and U22→A for DSL, providing perfect complementarity with substrates S4 and S5, respectively. Amplification profiles were obtained for each of the evolved enzymes, demonstrating mutually exclusive use of S4 by CL1 and of S5 by DSL. The exponential growth rate of DSL appreciably exceeded that of CL1 when each enzyme was allowed to operate in the presence of its preferred substrate. After 30-min incubation, the DSL enzyme exhibited 150-fold amplification, whereas the CL1 enzyme exhibited only 13-fold amplification. In addition, DSL achieved a maximum extent of growth that was approximately 3-fold greater than that of CL1 (Fig. 3).

Amplification profiles of the CL1 (circles) and DSL (squares) enzymes operating in the presence of 1 μM of either S4 or S5. Filled symbols indicate behavior in the presence of the cognate substrate (CL1 with S4, DSL with S5) open symbols indicate behavior in the presence of the noncognate substrate. The concentration of RNA enzyme was determined at various times, and the data were fit to the logistic equation: [enzyme] = a/(1 + be −ct ), where a is the maximum extent of amplification and c is the exponential growth rate. Curvilinear regression coefficients were 0.995 for CL1 with S4 and 0.998 for DSL with S5. Inset shows behavior during the linear phase of growth.

The compensatory mutations alone were installed in the CL1 and DSL enzymes. For CL1, the matched pairing between the enzyme and substrate resulted in growth characteristics similar to those of the fully evolved enzyme, suggesting that the other 4 mutations in the evolved enzyme are near-neutral with regard to fitness. For DSL, in contrast, the compensatory mutation alone did not provide the full growth rate of the evolved enzyme. DSL is a younger enzyme compared with CL1, having been subject to many fewer rounds of in vitro evolution, and thus is more likely to acquire mutations that enhance its fitness within the context of continuous in vitro evolution.

It is possible to balance the disparate growth rates of the evolved CL1 and DSL enzymes by adjusting the concentrations of their respective substrates (Fig. S3). The CL1 enzyme operating in the presence of 1 μM S4 has a nearly identical rate and maximum extent of growth compared with the DSL enzyme operating in the presence of 0.02 μM S5. Under these conditions, it is possible to carry out the sustained coevolution of CL1 and DSL within a common reaction mixture over the course of a serial transfer experiment (Fig. 4). In the absence of further evolutionary change, this behavior could be maintained indefinitely, demonstrating occupancy of distinct niches by the 2 different RNA enzymes.

Sustained coevolution of the CL1 (circles) and DSL (squares) enzymes. Five successive rounds of amplification and 100-fold dilution were performed over a period of 2.5 h. The concentration of each enzyme was determined based on incorporation of [α- 32 P]ATP into newly synthesized RNAs. The concentrations of S4 and S5 were 1 and 0.02 μM, respectively.

Kinetic and Transcriptional Analyses.

The catalytic activity of the evolved CL1 and DSL enzymes with their preferred substrates was examined under the conditions of continuous evolution (Fig. 5 A and B). CL1 demonstrated a kcat of 290 ± 30 min −1 and KM of 17 ± 4 μM. This catalytic rate is among the fastest ever measured for an RNA enzyme. In contrast, DSL exhibited a kcat of 0.17 ± 0.01 min −1 and KM of 0.24 ± 0.04 μM. Thus, the 2 enzymes exhibit markedly different kinetic parameters, with a ≈100-fold difference in observed rates in the presence of 1 μM of their respective substrates. Yet both enzymes are viable within the continuous evolution system.

Catalytic activity and transcription rate of the evolved CL1 (circles) and DSL (squares) enzymes. (A) Reaction of CL1 with S4. (B) Reaction of DSL with S5. Data were fit to the Michaelis–Menten equation curvilinear regression coefficients were 0.984 and 0.981 for CL1 and DSL, respectively. (C) Transcription rates were measured by using various concentrations of RNA/cDNA heteroduplex templates. Linear regression coefficients were 0.994 and 0.983 for CL1 and DSL, respectively.

Interestingly, the slower DSL enzyme amplified more efficiently in the continuous evolution mixture, warranting investigation into other factors that might affect amplification rate. The rates of reverse transcription of the 2 ligated enzymes were measured and found to be 0.2 and 0.3 min −1 for CL1 and DSL, respectively. However, the rates of forward transcription (starting from cDNA) were more disparate, with DSL exhibiting a 2.2-fold faster rate of transcription over a range of cDNA concentrations equivalent to those present during continuous evolution. Synthetic DNA templates were prepared to validate the observed difference in transcription rates. The rate of transcription was linear over a range of DNA template concentrations, with a rate constant of 0.40 and 1.0 min −1 for the CL1 and DSL enzymes, respectively (Fig. 5C). This 2.5-fold difference is consistent with the measurements obtained starting from cDNAs that had been generated in the continuous evolution mixture.


Relative concentration of enzyme vs reaction product - Biology

KINETICS OF MONOMERIC ENZYMES

Enzyme kinetics is a high-yield topic that can score us several points on Test Day. Just as the relief our student derives from squeezing a stress ball depends on a number of factors, such as size and shape of the ball and his or her baseline level of stress, enzyme kinetics are dependent on factors like environmental conditions and concentrations of substrate and enzyme.

The concentrations of the substrate, [S], and enzyme, [E], greatly affect how quickly a reaction will occur. Let's say that we have 100 stress balls (enzymes) and only 10 frustrated students (substrates) to derive stress relief from them (high enzyme concentration relative to substrate). Because there are many active sites available, we will quickly form products (students letting go and feeling relaxed) in a chemical sense, we would reach equilibrium quickly. As we slowly add more substrate (students), the rate of the reaction will increase that is, more people will relax in the same amount of time because we have plenty of available stress balls for them to squeeze. However, as we add more and more people (and start approaching 100 students), we begin to level off and reach a maximal rate of relaxation. There are fewer and fewer available stress balls until finally all active sites are occupied. Unlike before, inviting more students into the room will not change the rate of the reaction. It cannot go any faster once it has reached saturation. At this rate, the enzyme is working at maximum velocity, denoted by vmax. The only way to increase vmax is by increasing the enzyme concentration. In the cell, this can be accomplished by inducing the expression of the gene encoding the enzyme. These concepts are represented graphically in Figure 2.4.

Figure 2.4. Michaelis–Menten Plot of Enzyme Kinetics As the amount of substrate increases, the enzyme is able to increase its rate of reaction until it reaches a maximum enzymatic reaction rate (vmax) once vmax is reached, adding more substrate will not increase the rate of reaction.

For most enzymes, the Michaelis–Menten equation describes how the rate of the reaction, v, depends on the concentration of both the enzyme, [E], and the substrate, [S], which forms product, [P]. Enzyme–substrate complexes form at a rate k1. The ES complex can either dissociate at a ratek2 or turn into E + P at a rate k3:

Equation 2.1

Note that in either case, the enzyme is again available. On Test Day, the concentration of enzyme will be kept constant. Under these conditions, we can relate the velocity of the enzyme to substrate concentration using the Michaelis–Menten equation:

Equation 2.2

Some important and Test Day-relevant math can be derived from this equation. When the reaction rate is equal to half of vmax, Km = [S]:

Km can therefore be understood to be the substrate concentration at which half of the enzyme's active sites are full (half the stress balls are in use). Km is the Michaelis constant, and is often used to compare enzymes. Under certain conditions, Km is a measure of the affinity of the enzyme for its substrate. When comparing two enzymes, the one with the higher Km has the lower affinity for its substrate because it requires a higher substrate concentration to be half-saturated. The Km value is an intrinsic property of the enzyme–substrate system and cannot be altered by changing the concentration of substrate or enzyme.

KEY CONCEPT

We can assess an enzyme's affinity for a substrate by noting the Km. A low Km reflects a high affinity for the substrate (low [S] required for 50% enzyme saturation). Conversely, a high Km reflects a low affinity of the enzyme for the substrate.

For a given concentration of enzyme, the Michaelis–Menten relationship generally graphs as a hyperbola, as seen in the Michaelis–Menten plot in Figure 2.4. When substrate concentration is less than Km, changes in substrate concentration will greatly affect the reaction rate. At high substrate concentrations exceeding Km, the reaction rate increases much more slowly as it approaches vmax, where it becomes independent of substrate concentration.

The Lineweaver–Burk plot is a double reciprocal graph of the Michaelis–Menten equation. The same data graphed in this way yield a straight line as shown in Figure 2.5. The actual data are represented by the portion of the graph to the right of the y-axis, but the line is extrapolated into the upper left quadrant to determine its intercept with the x-axis. The intercept of the line with the x-axis gives the value of . The intercept of the line with the y-axis gives the value of . The Lineweaver–Burk plot is especially useful when determining the type of inhibition that an enzyme is experiencing because vmax and Km can be compared without estimation.

Figure 2.5. Experimentally Determined Lineweaver–Burk (Double Reciprocal) Plot Used to Calculate the Values of Km and vmax

Certain enzymes do not show the normal hyperbola when graphed on a Michaelis–Menten plot (v vs. [S]), but rather show sigmoidal (S-shaped) kinetics owing to cooperativity among substrate binding sites, as shown in Figure 2.6. Cooperative enzymes have multiple subunits and multiple active sites. Subunits and enzymes may exist in one of two states: a low-affinity tense state (T) or a high-affinity relaxed state (R). Binding of the substrate encourages the transition of other subunits from the T state to the R state, which increases the likelihood of substrate binding by these other subunits. Conversely, loss of substrate can encourage the transition from the R state to the T state, and promote dissociation of substrate from the remaining subunits. Think of cooperative enzyme kinetics like a party. As more people start arriving, the atmosphere becomes more relaxed and the party seems more appealing, but as people start going home the party dies down and more people are encouraged to leave so the tense hosts can clean up. Enzymes showing cooperative kinetics are often regulatory enzymes in pathways, like phosphofructokinase-1 in glycolysis. Cooperative enzymes are also subject to activation and inhibition, both competitively and through allosteric sites.

Figure 2.6. Cooperative Enzyme Kinetics

MCAT EXPERTISE

This cooperative binding of hemoglobin, which acts as a transport protein rather than an enzyme, results in a characteristic sigmoidal binding curve that is an MCAT favorite.

MCAT Concept Check 2.3:

Before you move on, assess your understanding of the material with these questions.

1. What are the effects of increasing [S] on enzyme kinetics? What about increasing [E]?

2. How are the Michaelis–Menten and Lineweaver–Burk plots similar? How are they different?

3. What does Km represent? What would an increase in Km signify?

4. What do the x- and y-intercepts in a Lineweaver–Burk plot represent?


The Effect of Concentration, Ph and Temperature on Enzyme Activity Essay Example

Many important processes in the body involve the work of enzymes, including the digestion of nutrients such as carbohydrates, proteins and fats (Raven 45). Enzymes are also organic catalysts. A catalyst is a chemical that controls the rate of a reaction, but is itself not used up in the process. Reactions that are accelerated due to the presence of enzymes are known as enzyme-catalyzed reactions (Raven 112). The substrate is the reactant within the reaction that fits with the enzyme. Once the substrate enters the enzyme’s active site, the enzyme’s shape changes to form an enzyme-substrate complex.

The substrate is then metabolized or broken down, resulting in a product, which can be utilized to energize cells. Once the product is released from the active site, the enzyme returns to its original shape (Raven 117). The three factors that can affect the activity of an enzyme include temperature, pH, and concentration. The temperature effects enzyme activity. As the temperature increases, enzyme stability decreases. Therefore there are more collisions of the substrate with the active site and the formation of activated complex’s and product (Raven 52). The rate of reaction is increasing.

The optimal temperature is the highest rate of reaction. Greater temperature raises the kinetic energy of the enzyme atoms, resulting in the breakdown of bonds, and an alteration of shape of the active site. Most enzymes present in living tissue have their secondary protein structure denatured, at the temperature 40°C or above (Raven 53). Effects of enzyme concentration on enzyme activity were evident with the level of substrate concentration increased, so did enzyme activity. Conversely, this activity had a stop point where activity ceased suddenly (Raven 114).

Without enzymes, reactions may take place very slowly, but with enzymes the rate was directly proportional to the amount of enzyme that was present. Cells control the production of these enzymes, and if the breakdown of these enzymes exceeds synthesis, the rate slows down. But, if the synthesis exceeds the breakdown, the reaction rate increases (Silverthorn 91). Low substrate concentration is the least productive (Silverthorn 98). In the effects of pH on enzyme activity, the way a protein folds can be changed in the presence of various.

The pH also affects the rate of reaction of an enzyme catalyzed reaction at about or below the optimal pH the rate decreases (Raven 116). The change in rate is because bonds are made and broken, which changes the shape of the active site and therefore decreases the rate of reaction. As the substrate concentration is increased, the rate of reaction increases. Further increases in substrate also increase the rate, but proportionately the rate is constant (Raven 117). Enzyme activity can be changed by enzyme concentration, pH, and temperature changes. Materials and Methods There are three parts to test concentration, pH, and temperature.

To begin, concentration is tested by collecting five test tubes and labeling each one 1-5 and placing them in a test tube rack. Next, a beaker is filled half way full of water and placed on a hotplate until it begins to boil. Afterward, different solutions are filled in each test tube: Test Tube 1: 2 ml of 1% sucrose and then add 2ml of pH buffer 4. 4. Test Tube 2: 2 ml of 1% sucrose and then add 2 ml of pH buffer 4. 4. Test Tube 3: 2 ml of 1% sucrose and then add 2 ml of pH buffer 4. 4. Test Tube 4: 2 ml of 1% sucrose and then add 2 ml of pH buffer 4. 4. Test Tube 5: 2 ml of 1% sucrose and then add 2 ml of pH buffer 4.

Consequently, predetermined amounts of enzyme and distilled water are added to each tube at room temperature: Test Tube 1: 0 ml sucrase and then add 3ml of water. Test Tube 2: 0. 5 ml sucrase and then add 2. 5 ml of water. Test Tube 3: 1 ml of sucrase and then add 2 ml of water. Test Tube 4: 1. 5 ml of 1% sucrase and then add 1. 5ml of water. Test Tube 5: 3 ml of sucrase and 0 ml of water. Each test tube is allowed to sit for five minutes. This stage is imperative to allow the sucrose to act upon the sucrose. Afterwards, 3 mL of Benedict’s reagent is added to each tube.

Each tube is placed in the boiling water for three minutes, and each tube is observed for color changes which are compared to the provided Benedict’s test for simple sugars, with results recorded. The second part of the experiment was targeted to determine the effects of temperature on the rate of enzyme activity. Four test tubes are labeled 1A-4A along with four additional tubes labeled 1B-4B. A beaker is filled ? full with water and brought to a boil. Test tubes 1A-4A is filled with 2 mL of both sucrose and a pH buffer. These four tubes are placed in the boiling water for fifteen minutes. Test Tube 1A: 0 C

Test Tube 2A: 30 C Test Tube 3A: 30 C Test Tube 4A: 100 C Measure and fill test tubes 1B-4B with 3 mL of enzyme. Test tube 1B-4B placed to water for 15 minutes. Test Tube 1B: 0 C Test Tube 2B: 30 C Test Tube 3B: 30 C Test Tube 4B: 100 C Put enzyme from B tubes into A tubes after fifteen minutes and place back into water for 5 minutes. Remove each tube from the water and add 3 mL of Benedict’s solution to each tube. Return each tube back into water and boil for 3 minutes. Results calculated and recorded. Lastly, part three of this experiment strives to understand the effects of pH on the rate of enzyme activity.


Question: Describe The Definitions Of Substrate, Enzyme Active Site And Its General Characteristics, And Apoand Holo-enzymes. · Describe The Fact That Enzyme Catalysis Is Specific In Terms Of The Type Of Reaction And The Exact Substrate Structure. · Explain The Difference In The “lock-and-key’ Model And The Induced Fit Model Of Enzyme Substrate Interaction. .

· describe that reaction rate constant thus reaction rate is dependent on the activation energy.

· describe the relationship between the energy of the transition state and the activation energy.

· describe the relative binding affinity of a substrate and its transition state to the enzyme.

· Explain how enzymes speed up reaction rate, i.e., how do enzymes lower the activation

energy. Note: Factors other than preferential stabilization of transition state also contribute to increased

reaction rate. They include 1) enzymes placing one substrate (reactant) next to the other substrate (reactant)

so that the reaction is no longer dependent on the collision rate of the two substrates 2) enzymes sequester the

substrate(s), minimizing solvent competition thus speeding up the reaction 3) in enzyme complexes with each

enzyme catalyzing one step in a multi-step reaction sequence, the product of one reaction is fed to the active site

of another enzyme for the next step without the product diffusing away from the enzyme complex thereby

· describe the Michaelis-Menten equation relating the initial velocity to total substrate

· describe the assumptions that went into the derivation of the Michaelis-Menten equation. ([E]

is constant, steady state assumption, etc)

· describe the relationship between Vmax and kcat and the fact that kcat is a constant for a given

enzyme and substrate at a given temperature and solution condition whereas Vmax can differ

for a given enzyme and substrate at a given temperature and solution condition, depending

on the enzyme concentration.

· describe the definition of Kd (equilibrium dissociation constant for ES into E and S).

· describe the relationship between Kd and the affinity of substrate for the enzyme, i.e., the

higher the Kd, the weaker the binding between E and S.

· describe the relationship between Kd and Km, i.e., in general, the higher the Kd, the higher theKm.

· Use the Michaelis-Menten equation and the relationship between Vmax and kcat to do

· describe the other names for the catalytic rate constant, kcat (i.e., the turn over number also

called krelease) and why it is so called (kcat reports on per unit time per enzyme molecule the

number of product molecules formed/released, thus the name turnover number or krelease).

· Determine Vmax and Km using both the Michaelis-Menten plot (initial velocity versus [S]) and

the Lineweaver-Burk Plot (the double reciprocal plot), given the data of initial velocity and

· Draw the Michaelis-Menten plot AND the double reciprocal plot (Lineweaver Burk), given

· describe the fact that the term kcat/Km is the best parameter for measuring the catalytic

capability (efficiency) of an enzyme.

· describe the existence of irreversible enzyme inhibition and the general mechanisms of

· describe the different types of reversible inhibitions, namely the competitive, uncompetitive,

and mixed inhibitions (noncompetitive inhibition is a special case in which the inhibitor

binds E and ES with equal affinity therefore α is equal to α’ – Km is not affected).

· Describe each type of inhibition in terms of inhibitor binding site (active site or other than

the active site), how the inhibitor affects vmax and Km and the quantitative relationship of

each parameter with and without the inhibitor present and their relationship with [I] and KI

· Show the characteristics of each type of inhibition on a Michaelis-Menten plot (initial

velocity versus total substrate concentration), i.e. draw the curves with and without the

· Show the characteristics of each type of inhibition on a double reciprocal plot (LineweaverBurk

Plot) by drawing the lines with and without the competitive inhibitor.

· Given the Michaelis-Menten or double reciprocal plot with and without the inhibitor,

determine the type of inhibitor.

· describe the fact that transition state analogs are potent inhibitors.

· Explain why transition state analogs are potent inhibitors.

· describe that not all enzymes obey Michaelis-Menten kinetics.

· describe situations when an enzyme will not obey Michaelis-Menten kinetics (when the enzyme

catalyzes reactions with multiple substrates or when the enzyme is allosteric, i.e. binding of

the substrate at one active site affects the other active site(s) for the same substrate).

· Name the four types of catalytic mechanisms. (electrostatic catalysis, general/specific acid

base catalysis, metal ion catalysis, covalent catalysis)

· Contrast acid/base catalysis to general acid/base catalysis, i.e., comment on their similarity

· Name the eight common general acid/base pairs in proteins and write the formula of the

acid and base form for each pair (hint: consider the eight amino acids that have ionizable

· describe the meaning of covalent catalysis.

· describe the reaction catalyzed by chymotrypsin and the three catalytic residues in

· Draw the interactions among the three catalytic residues (catalytic triad).

· State the catalytic role of each of the catalytic residues and how they accomplish that.

· describe the steps in the formation of the acyl-enzyme intermediate, and the substrate and

products in this first part of the chymotrypsin mechanism.

· Draw the transition state structure for the formation of the acyl-enzyme intermediate and

indicate by drawing how it is preferentially stabilized.

· describe the steps in the regeneration of chymotrypsin (deacylation), the substrates and

product of this second part of the chymotrypsin mechanism.

· Draw the transition state for deacylation and explain how it is preferentially bound to or

· Explain what general acid and base catalysis refers to and what groups act as the general acid

and base in the chymotrypsin mechanism.

· Explain what covalent catalysis refers to in the chymotrypsin mechanism.

· Explain the basis for the substrate specificity of chymotrypsin (consider the composition of

the active site – chymotrypsin specifically cleaves peptide bonds adjacent to aromatic

· Explain the term “serine protease”. Answer: these are proteases (enzymes that

hydrolyze peptide bonds) that have serine as a catalytic residue (that acts as a

nucleophile, which can attack the carbonyl carbon of the peptide bond).

· Key features of Hexokinase, enolase and lysozyme mechanisms that were discussed in class

(specific methods of catalysis utilized by these enzymes general description of overall

· describe the four regulatory strategies of enzyme catalysis

· describe the definition of feedback inhibition (mechanism of cellular regulation that regards an

enzyme that catalyzes the production of a given substance that then inhibits the cell when

concentration of that substance has reached an appropriate level – this balances/regulates

amount of product provided with amount needed)

· describe the most common type of covalent modification and the types of enzymes

responsible for the addition and removal of the chemical group in this covalent

· Explain the general mechanism of phosphorylation in regulating enzyme activity.

· describe the general naming scheme for active and inactive proteases.

· describe that protease inhibitors exist to inhibit protease.

· describe that enzymes are regulated through covalent and non-covalent processes.

o Describe noncovalent modification in general terms. Elaborate on the relation

among enzyme activity, allostery, effector molecules (small molecules) and control

o Concerning allosteric regulation, some enzymes have multiple active sites for the

same substrate and the binding of the substrate increases the activity of the enzyme

by stabilizing the R (more active) state over the T (relatively inactive) state. (Lecture

13: “Noncovalent Modification: Allosteric Regulators”)

ß Draw a curve for the initial reaction velocity versus substrate concentration.

ß describe and explain the specific shape of the v0 vs. [S] plot.

ß Draw on the same graph above a v0 vs. [S] curve for the enzyme in the

absence of allostery. and how th i.e., in the absence of conformational change

upon substrate binding. Comment on the shape of this curve.

ß Comment on the benefit of allosteric activation.

ß When an inhibitor is present for such an allosteric enzyme, the v0 versus [S]

plot will shift. Draw a v0 vs. [S] curve with and without the inhibitor.

ß When an activator is present for such an allosteric enzyme, the v0 versus [S]

plot will shift. Draw a v0 vs. [S] curve with and without the inhibitor.

ß When an activator is present for such an allosteric enzyme, the v0 versus.

ß describe the relationship between feedback inhibition and allosteric regulation

and the reason behind the term “feedback inhibition”.

o describe the most common covalent modification, the enzymes that catalyze the

modification and the removal of that modification, respectively and the amino acid

residues that tend to be the site of this modification.

o Write the structure of the added group bonded to those residues and the reaction for

this covalent modification.

o Rationalize the effectiveness of phosphorylation on altering the activity of an

o describe the mechanism of chymotrypsinogen activation to chymotrypsin (cleavage of


Relative concentration of enzyme vs reaction product - Biology

The Discussion should be written after the Results section so that you have a good idea of what the experiment has demonstrated. The discussion section should definitely have a statement of your expected findings. (Pechenik, 86) This should include your hypothesis and a brief statement about why these types of results are expected. There should also be a comparison of how your actual results related to your expected findings. (Pechenik, 86) Here, you should state whether or not your results supported or didn't support your hypothesis. In addition, the degree to which the evidence supported your hypothesis should be stated. For example, were the results completely supportive, or were there variances?

There should be an explanation of unexpected results. (Pechenik, 86) When looking for possible explanations, consider the following:

a. Was the equipment used adequate for the task? b. Was the experimental design valid? c. Were the working assumptions made correct?

A common mistake that many writers make is to blame themselves for the unexpected results. Unless you actually made a mistake following the methods of the experiment, and could not go back and correct it, do not make up such errors to explain the variances you observe. Think about and analyze the methods and equipment you used. Could something different have been done to obtain better results? Another possibility to consider is if the experiment was conducted under factors that were considerably different from those described in the manual. Be sure to include ideas on how to test these explanations. (Pechenik, p.86) Briefly explain a way to test these possible reasons for unexpected results. For example, if there is a problem with the methods, maybe the experiment should be reproduced with an added step. Also mention what kinds of experiments still need to be conducted in order to obtain more information.

Sample 1 : The results of the first experiment supported the hypothesis that the rate of conversion of the substrate would increase with increased amounts of enzyme. We observed that Tube 2, which had the highest concentration of enzyme, catecholase, also had the highest absorbance level. Since absorbance is used as a measure of reaction, the greatest rate of conversion of catechol and oxygen to benzoquinone was seen in Tube 2. The high ratio of enzyme to substrate caused the absorbance to grow rapidly and then level off (see Figure 1). The tubes with lower concentrations of enzyme had lower rates of conversion, as expected. However, there were some unexpected results in Tube 2. Between the time of around 6 minutes to 8 minutes there was decrease in the absorbance. One explanation of this observation is that the settling of the substrate to the bottom of the test tube caused the enzyme to become less efficient since it could not attack the substrate as well. The settling reduced the surface area of the substrate that could be attacked by the enzyme. The tube was inverted and the substrate was stirred up, which caused a rise in the absorbance. Further experiments, involving the constant stirring of the solution, could be performed to test this possibility.

The folding and combination of polypeptide chains forms the specific, three dimensional shape of an enzyme. This shape is extremely important to the enzyme's catalyzing efficiency and many environmental conditions can affect the shape of enzymes and thus their efficiency. A range of pH values exists for all enzymes, between which they reach their maximum catalyzing action. This range is usually between a pH of 6-8. pH levels outside this range can denature the enzyme, thereby decreasing its catalyzing ability. The results we obtained supported this assumption for the catecholase enzyme. The catecholase samples in tubes 3 and 4 had similar absorbance rates and, therefore, similar enzyme activities. However, the pH of 4 in tube 2 corresponded to low absorbance and low activity of the enzyme in that tube. This is due to the fact that the acidic environment is harmful to the enzyme, and denatures it. Catecholase, an enzyme found in fruits in nature, is well adapted for efficiency in nature. Its range of optimal pH levels, 6-8, allows it to function in the varying pH levels of soil and those caused by acid rain.

Sample 2: Enzymes catalyze reactions by lowering the activation energy of the reaction. Catecholase, an enzyme found in potatoes, converts catechol to benzoquinone in the presence of oxygen. It would be expected that more benzoquinone would be formed in the presence of a greater amount of catecholase. This hypothesis was supported by the results obtained. The most enzyme was placed in tube 2. The absorbance was also highest for this tube. This means that the most product was formed in this test tube. In accordance with this, tube four, which had the least amount enzyme, also had the least amount of absorption. There were some unexpected results, but this is most probably due to human error the absorbance levels were probably read wrong.

Enzymes are affected by the environment. The pH level of the environment is one factor that can alter enzymes. The rate at which the enzyme form product is slowed or sped up depends on how close to the norm the environment is. In the second experiment, the pH of the medium was different in each of the test tubes. The general trend seen in these reactions was that the more acid added to the test tubes, the less product formed. The more acidic solution caused the enzyme to work less efficiently.

Results : This author does a good job of answering the questions that should be addressed in a discussion. For example, in the very first sentence he stated what he expected to find and also whether or not the results he obtained supported or failed to support his hypothesis. This is a good, strong way to start a discussion section. It starts off with the facts of the experiment and then later on, the author can move on to his opinions. (return to Sample 1)

Absorbance : A good discussion includes good ideas and also exact and detailed support of these ideas. In addition to starting off well, the author also goes on to explain the specific results of the experiment that support his hypothesis. This is what defines the strength of his discussion section. (return to Sample 1)

Explanation : After his explanation he presents the unexpected results and discusses possible reasons for this data. The author's explanation of possible reasons for unexpected results is good because it shows that he thought about the problems. He does not blame himself for the unexpected. Instead, he considers the methods used, presents a possible explanation, and then justifies his ideas. (return to Sample 1)

Catalyze : This author does a good job outlining his discussion, however he is lacking the specifics to make a good discussion. The first two sentences are better placed in the introduction. However, he does state his expectations and whether or not his results supported these expectation. He could have made this part better by stating this more authoritatively, for example: "It was expected," and not, "It would be expected that." (return to Sample 2)

Unexpected results : The biggest problem this author had was explaining the unexpected results. He blamed himself, saying he read the equipment wrong and passed off the unexpected results as human error. (return to Sample 2)

Enzymes : This author does not develop his argument enough. One example of this, is the affects harsh environmental factors have on enzymes. He could have stated how the acidity caused the enzymes to denature, thus creating less efficiency. (return to Sample 2)

All citations from Pechenik, Jan A. A short guide to writing about Biology. pp. 54-102, Tufts University: Harper Collins College Publishers . 1993.


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