Specific Co Factor Determining Reaction Rate Biology Essay

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Experiments were put forward to determine how enzyme kinetics functions under normal conditions and when the enzyme is subject to the effects of an inhibitor. Along with this procedure the enzyme is subject to a range of pH's to determine the optimum rate and also subject to a range of temperatures separately to deduce the optimum. A method used to determine the optimum rate is known as spectrophotometry and this is important. In basic terms it measures the substrates colour conversion and shows how much light will pass through, thus doing so may or may not obey beer lamberts law. Using a calibration graph is essential to convert absorbance to [V] rate of reaction per minute and is useful to help construct a michaelis-menten and lineweaver-burk graph to more accurately deduct the optimum. A similar set of graphs using a curve and linear approach can deduce how pH and temperature affect the enzyme's activity. Some key results related to the enzymes activity are Vmax (an estimated velocity rate maximum) and Km (concentration of substrate that gives half the Vmax).Without inhibitor the Vmax = 66.6 micromoles/min this is determined by the overall velocity and determination of the flattening of the curve at its peak. Km = 0.0947mM which is determined after the Vmax is estimated. With Vmax= 116.7 and Km = 0.833. The enzyme subject to pH has a value of 0.00204 micromoles/min at pH 5. Temperature shows that 50 degrees Celsius is an optimum revealing 0.004420 micromoles/min. This results can be interpreted to show that when an enzyme is the subject to change It generally follows a trend depicting the enzymes optimum where more energy = higher [V] per/min and higher amounts of substrate equate to a higher velocity (until the active sites are saturated).

Introduction:

General enzyme kinetics

"The use of enzymes in the diagnosis of disease is one of the important benefits derived from the intensive research in biochemistry since the 1940's."(Worthington Biochem corp, 2012, http://www.worthington-biochem.com/introbiochem/default.html, accessed 15/11/12) Enzymes have provided the basis for the field of clinical chemistry. Catalysis is an important factor in an enzymes function, it is an enzyme efficiently used to increase a specific reaction e.g. - Hydrolases are enzymes that will catalyse a hydrolytic cleavage. Enzyme concentration corresponds with the correlation of the substrate, which means more active sites are available to become saturated; this will increase the rate of the reaction. i.e. - E + S <-> ES -> EP -> E + P. Substrate concentration may be increased up until the enzyme's active sites are completely saturated so that adding anymore substrate to the enzyme substrate complex will not increase or alter the maximum rate of reaction (Vmax). A measurement known as (Km) which is half the (Vmax) can demonstrate how tightly the substrate is bound, meaning a low value of (Km) corresponding to a strong bond and vice versa. This also can demonstrate how fast the enzyme is catalysing the reaction; meaning the lower the (Km) the faster the enzyme reaches its (Vmax) also showing that this happens in a lower concentration of substrate. (B. Alberts et al., 2007, Chapter 3: Proteins)

Data from enzyme kinetics can be interpreted by a Michaelis-menten graph (shown Figure 1). This graph demonstrates a typical conclusive enzyme reaction; where a curve is shown representing reaction rate corresponding to substrate concentration. It outlines that a typical enzyme reaction will reach a peak and flatten out i.e.- Vmax. This example also demonstrates the Km and by viewing this point, can determine the speed of reaction rate. If the graph shows this progression then it is obeying Michaelis-menten kinetics.

Figure 1 (CHEMWIKI. A.Nath University of Washington 2011)http://chemwiki.ucdavis.edu/@api/deki/files/13014/=mm1.gif

Michaelis-menten can be determined by using similar estimated values (substrate [S]) and resubmitting them into the formulae V = Vmax x[S]/Km + [S], This will show a curve and dependant on the correlation compared to the initial results, it could indicate that the Vmax was over estimated; i.e. - the theoretical curve will initially increase at a faster rate but will peak at a higher Vmax.

Using the same data a graph known as a Lineweaver-burk plot can be used to more accurately determine a Vmax for the enzyme reaction rate. Since the Michaelis-menten graph does not represent the data accurately and the Vmax may have been over estimated. A plot of a Lineweaver-burk graph is taken by the reciprocal

1/V = 1/(Vmax x [S])/(Km + [S]) this data can be plotted to show the (1/Vmax) which can be used to determine the accurate Vmax rather than the estimated value. (-1/Km) shows a more accurate value for (Km) and so relating this to the enzyme reaction rate a higher (Km) means there is a lower affinity for the enzyme meaning as previously discussed, there would be a strong bond in this case.

Inhibition

Figure 2 (Berg et al., 2011)

Inhibition is a process in which an ion or molecule will affect the activity of an enzyme. Inhibitors play a vital role because they are used in the body for regulation. E.g.-"The regulation of allosteric enzymes." (Berg et al., 2011) Figure 2 shows a diagram of a substrate binding to an enzyme active site; this is expected of the enzyme under normal conditions. A competitive inhibitor will do as it suggests, it will attempt to get to the active site and bind before the substrate preventing catalysis. A competitive inhibitor can be reversed by increasing the ratio of substrate:inhibitor i.e.- higher concentration of substrate overrules the inhibitor to bind to the enzyme creating an enzyme - inhibitor complex. An uncompetitive inhibitor will only bind to the enzyme when it is in an enzyme-substrate complex this is because the binding site is only available to the inhibitor in an ES complex - it is created when the substrate binds. A non-competitive inhibitor is where the inhibitor will bind independently to the enzyme whether it is an ES complex or not. It will decrease the overall function of the enzyme rather than prevent the binding of the substrate, this is because of a conformational change to the active site preventing it from binding, and this is why increasing the concentration of the substrate will have no effect.

A Michaelis-menten plot can also be used to show that there is still an initial increase in reaction rate because not all the enzyme is inhibited; gradually the enzyme becomes an EI or ES complex. On its own the graph plot of an inhibitor cannot show essentially how much it has affected the reaction rate. To show this it is plotted along the plot of what would be expected if the substrate can effectively bind.

Figure 3 is an example of what is expected when an inhibitory experiment on an enzyme is compared alongside the same experiment with the enzyme under normal conditions. The competitive inhibitor will reach the same Vmax because the substrate can still bind with the active site, while competing. The non-competitive inhibitor will have a lower Vmax due to it binding outside the active site; as shown on the graph it is obvious how a non-competitive inhibitor can affect the Vmax and the Km. http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/M/Michaelis_Menten2.gif

Figure 3: Enzyme kinetics, http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/E/EnzymeKinetics.html, 2011, accessed 12/11/12.

As explained before a michaelis-menten plot does not determine accurately the Vmax and it could most likely have been overestimated for the inhibitory curve. A lineweaver-burk plot is used to show how affective the inhibitor has been upon the Vmax in comparison to normal conditions.

Figure 4 shows the lineweaver-burk plot of an inhibitor/s compared to the enzyme reaction under normal conditions (shows as the red line). As shown in figure 3 the non-competitive inhibitor has a lower Vmax and this graph also shows more effectively that the active site isn't affected by the inhibitor. This is shown by the (Y) intercept. http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/L/Lineweaver_Burk2.gif

Figure 4: Enzyme Kinetics, http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/E/EnzymeKinetics.html, 2011, accessed 12/11/12.

Temperature

Temperature is an important co-factor that can alter the reactive state of an enzyme. It can efficiently speed up the reaction rate up until the denaturation of the enzyme. Denaturation occurs when the temperature exceeds the enzymes optimum and therefore the hydrogen, inter and intra-molecular bonds are no longer holding the enzymes conformation together (Average kinetic energy has disrupted the molecules in the conformation). This indefinitely means that a substrate cannot bind to the active site because it has lost its form due to the denaturation. (RSC, http://www.rsc.org/Education/Teachers/Resources/cfb/enzymes.htm, n/a, accessed 13/11/12)

Typically as the temperature raises the reaction rate increases up until a peak known as the optimum temperature at which the enzyme will function. This can be shown by a graph.

Figure 5: This curve represents the effectiveness of temperature on an enzymes activity. Essentially the temperature increase will give the enzyme more kinetic energy up until that energy is strong enough to denature the conformation. Notice that the line does not intercept at "0" this is because the reaction is taking place very slowly beforehand at lower temperatures, meaning less kinetic energy for the substrate and enzyme. Graph of enzyme activity verses temperature

Figure 5: Enzymes, http://www.rsc.org/Education/Teachers/Resources/cfb/enzymes.htm, RSC, 2011, accessed 13/11/12

A much easier way to view the optimal temperature for an enzyme (i.e. - when it denatures) is to plot Log10 of the rate alongside 1/T (which is temperature in kelvin). Arrhenius equation is used to plot this: K = A -Ea/RT.

Where K = Rate constant, A = Pre-exponential factor; representing number of collisions Log10 C (where C = proportionality constant) Ea = Activation energy of rate of reaction, R = Gas Constant (8.31JK-1mol-1) T = Temperature (kelvin) (G.Sibert, http://www.files.chem.vt.edu/RVGS/ACT/notes/temp_effects.html, accessed 13/11/12)

Figure 6 shows a linear fashion using the Arrhenius plot and can determine the denaturation point more easily and accurately. The denaturation at optimum temperature is indicated by the anomaly or stray point. This graph is definitely clearer to read and determine. Cold temperature will slow down the enzyme activity by decreasing molecular motion and kinetic energy. (G E.Kaiser, 2001, http://student.ccbcmd.edu/~gkaiser/biotutorials/proteins/enzyme.html, accessed 13/11/12)

Effect of pH

The pH is a measure of H ions, there for a high pH is high concentration of H+ and low pH is low concentration. An enzyme can be very delicate when subject to pH change. If the pH extends beyond the optimum the enzyme will denature. This can affect the ionisation of amino acids which affects the conformation, leading to a change in the active site and function of the enzyme. These changes may also have a great effect on the weaker charges that hold the enzyme structure together; it may also alter the substrate complex so that it is unable to bind to the active site. All enzymes are unique to a function and some can function in pH levels which others cannot e.g. - in highly acidic conditions such as the stomach, pepsin functions in the stomach and its optimum pH is very low because of these conditions. An enzyme can also function in the complete opposite pH such as trypsin, which operates at a higher pH (alkaline). Below is some examples, assuming the environment pH is 7.

Figure 7a is an example of a substrate binding to the active site of an enzyme. Normal conditions mean the Carboxyl (COO-) group can bind to the amino (NH3+) group creating a substrate enzyme complex.

(a)(b)(c)http://www.chemguide.co.uk/organicprops/aminoacids/asiteph1.gif

Figure 7a, J.Clark(a), http://www.chemguide.co.uk/organicprops/aminoacids/enzymes2.html(b), accessed 13/11/12 (c)

Figure 7b, is an example of what happens when the pH in the environment lowers, because this is an increase in H+ ions, the Carboxyl group will gain an H+. This means that ionic bonds cannot form and the substrate will not bind to the active site. Also new bonds are formed in other positions changing the conformation of the enzyme and its active site. Overall charge = Negative

(a)(b)(c)

http://www.chemguide.co.uk/organicprops/aminoacids/asiteph2.gif

Figure 7b, (a), (b), (c)

Figure 7c, This is an example of when the pH increases so that the environment becomes alkaline, because this is a low H+ concentration it doesn't affect the carboxyl group, instead the amino group will lose a H+ ion effectively disrupting the active site bonding. Overall charge = Positive (a) (b) (c)

Figure 7c, (a), (b), (c)http://www.chemguide.co.uk/organicprops/aminoacids/asiteph3.gif

Results:

PNP Calibration Graph

Table 1

Tube

Amount of PNP (µ/mol)

Absorbance (400nm)

1

0.00

Table 1 represents known concentrations of P-nitrophenol (PNP), which is the product formed after the enzyme is introduced. 0

2

0.15

0.380

3

0.20

0.530

4

0.25

0.646

5

0.30

0.791

6

0.35

0.929

7

0.40

1.045

8

0.45

1.204

9

0.50

1.320

Graph 1is plotted in accordance to the results of table 1, so the graph represents how the concentration can affect the absorbance more clearly than reading it alone. Simply the absorbance represents how much light travels through the solution, in this the case the solution has a dedicated amount of PNP, and as represented by the graph the higher the concentration of PNP the higher the absorbance, i.e. - because less light can travel through the PNP molecules.

Tube Number

Substrate(PNPP) Concentration (mM)[S]

Absorbance of Sample @400nm

0

0

1

0.03

0.114

3

0.05

0.144

5

0.08

0.173

7

0.16

0.290

9

0.33

0.369

11

0.66

0.423

13

1.66

0.481

15

3.33

0.580

Absorbance

[V] per 15 min

[V] per min (μmoles/min)

0

0

0

0.114/2.6387 =

0.0432

0.00288

0.144

0.0545

0.00363

0.173

0.0655

0.00437

0.290

0.1099

0.00733

0.369

0.1398

0.00932

0.423

0.1603

0.01069

0.481

0.1822

0.01215

0.580

0.2198

0.01465

n/2.6387= [V]p/15min

[V]p/15min/15 = [V]p/min

Experiment A Effect of a substrate (Without inhibitor)

Table 2

This table represents the data collected from experiment A where absorbance is relevant to determine the rate of reaction. The first part of table 2 indicates how the increase of substrate concentration correlates to absorbance i.e. - Higher concentration is equal to an increase in absorbance. The absorbance collected from this experiment is used and converted into rate per minute or velocity of the reaction. This is worked out by dividing the absorbance by the Y intercept (2.6387) from the calibration graph and then dividing by 15 (since the experiment was timed over a period of 15 minutes).

Michaelis-menten Graph (graph 2) Effect of a substrate (without inhibitor) Values from table 2.

Km

Maximum Rate (Vmax) = 0.015 µmoles/min Km (mM) = ½Vmax = 0.17mM

The Vmax shows the point at which the active site initially becomes saturated; this is an estimated value based on the plot of results therefore it is possible to overestimate it. Km shows the affinity for the enzyme; low Km = high affinity and high = low affinity.

Theoretical Values to determine curve reliability

[S]

Rate (V)

0

0.0000

0.04

Table 30.0029

0.07

0.0044

0.11

0.0059

0.18

0.0077

0.38

0.0103

0.52

0.0113

0.81

0.0123

Since the Vmax could have been overestimated some theoretical values can substituted into the Michaelis-menten equation: Rate (V) = Vmax x [S]/Km + [S] i.e. - Rate (V) = 0.015 x [S]/0.17 + [S]. Applying this to the values in table 3 will show a curve, this curve is plotted red on graph 2; the line can determine whether the Vmax was overestimated or underestimated.

Lineweaver-burk - Effect of a substrate (without inhibitor)

1/[V]

1/[S]

347.2222

(Table 4) This table represents the reciprocal of the values in table 3. These values are used to plot a lineweaver graph which shows the rate of the reaction in a linear fashion, making it easier to determine maximum rate. This is plotted on graph 3.33.33333333

275.2294

20

229.0076

12.5

136.4877

6.25

107.2961

3.03030303

93.57455

1.515151515

82.32711

0.602409639

68.24386

0.3003003

Graph 3 is a lineweaver-burk plot and it represents how clearly the linear fashion can show the progression of the rate of reaction. The smaller the number the bigger the reciprocal is therefore the higher values are smaller when plotted in a Michaelis-menten format. This graph can also be used to determine a more accurate Vmax and Km using the reciprocal of the Michaelis-menten formula;

i.e. - 1/V = 1/ (Vmax x[S]/ (Km + [S]) hence is in the format Y = mx + c.

The Km can be more accurately determined by rearranging the formula so that y is equal to 0, therefore:

0 = 8.636x + 83.728, thus 0 - 83.728/8.636 then 1/9.69 = 0.103mM = Km the reciprocal of the y intercept (83.728) is equal to 0.0119µmoles = Vmax

i.e. - means the initial Vmax was overestimated. Essentially since both graphs are plotted with the same data, it is obvious that the Vmax initially wasn't correct.

Effect of a substrate (With inhibitor)

Tube No

Substrate PNPP (mM)

(Table 5) This shows similar data to (table 2) but the rate of reaction is noticeable slower. This graph shows the absorbance of light for the substrate, determining rate. Absorbance(400nm)

0

0

1

0.03

0.017

3

0.16

0.081

5

0.33

0.133

7

0.66

0.222

9

1.66

0.284

11

3.33

0.334

Absorbance

[V] per 15min

(Graph 4) shows clearly the difference between how an enzyme is affected by an inhibitor. A- Representing the just substrate and enzyme, B representing the inhibited enzyme. Values determined by Michaelis-menten plot:

(With inhibitor) Vmax = 0.0086 μmoles/min Km = 0.47mM (low affinity)

(Without inhibitor) Vmax = 0.015 μmoles/min Km = 0.17mM (high affinity)

[V] per min

0

0

0

0.017

0.0064

0.00042

0.081

0.0306

0.00204

0.133

0.0504

0.00336

0.222

0.0841

0.00560

0.284

0.1076

0.00717

0.334

0.1265

0.00843

Lineweaver-burk (With Inhibitor)

1/[V]

(Table 6) This represents the reciprocal of the rate of reaction and amount of substrate relevant to the inhibitor. (Graph 4) This shows a more clear view of how an inhibitor affects enzyme reaction rate via a linear fashion.

Without inhibitor: Vmax = 0.0119μmoles/min Km = 0.103mM

With inhibitor: Vmax = 0.0117μmoles/min Km = 0.833mM

1/[S]

0

0

2380.95

33.33

490.19

6.25

297.61

3.03

178.57

1.51

139.47

0.60

118.62

0.30

Temperature

(Table 7)

Tube

Temp (degrees C)

Absorbance (400nm)

Subtracted values

Control Tubes

(Abs [400nm])

2

0

4

0

6

0.002

8

0.004

10

0.038

12

0.221

(Abs [400nm])

1

5

0.011

0.011

0

3

20

0.051

0.051

0

5

25

0.056

0.054

0.002

7

37

0.107

0.103

0.004

9

50

0.213

0.175

0.038

11

75

0.235

0.0014

0.221

Subtracted values (400nm)

[V]/15min

[V]/min

0.011

0.0041

0.000273

0.051

0.0193

0.001287

0.054

0.0204

0.001360

0.103

0.039

0.002600

0.175

0.0663

0.004420

0.0014

0.0005

0.000033

(Table 7) Represents the experimental tubes used and the effects of temperature on absorbance by a curve. Graph 5 shows how the reaction rate increases up until where the enzyme denatures.

log(V)

1/t

1/T is 1/ temperature and the temperature is measured in kelvin thus 1/k are the values in column (1/T) i.e. - 1/278.2 = 0.00359K

-3.56

0.00359

278.2

-2.89

0.00341

293.2

-2.86

0.00335

298.2

-2.58

0.00322

310.2

-2.35

0.00309

323.2

-4.48

0.00287

348.2

Liner plot of temperature

(Table 8)

Graph 6

Table 8 shows the Log 10 values of the rate of reaction [V]/min and the 1/t or temperature in kelvin. Graph 6 shows a clearer view of the effect of temperature on the enzyme using a linear fashion; the anonymous result is denaturation.

Effects of pH on an enzyme

Table 9

Tube

pH

Abs (400nm)

[V]/15min

[V]/min

1

3

0.01

0.0037

0.0002467

3

4

0.019

0.0072

0.00048

5

4.5

0.028

0.0106

0.0007067

7

5

0.081

0.0306

0.00204

9

5.5

0.053

0.02

0.0013333

11

6

0.047

0.0178

0.0011867

13

7

0.007

0.0026

0.0001733

Table 9 shows the experimental values of the pH experiment, representing absorbance corresponding to pH thus incorporated into reaction rate. The experiment spanned over a period of 15minutes therefore each sample is [V]/15 = [V] min. This data is represented on graph 7 where a curve is shown to reach a peak; where at the optimum pH is for the enzyme activity rate. Estimated optimum of pH = 5.1

Question1:

This is because p-nitrophenylphosphate (PNPP) is chromogenic; in alkaline solutions it is colourless and changes pigment in acidic solutions therefore placing this initially into test tubes with in ranges of pH's of acidic to alkaline would change the pigment, making the experiment inaccurate because the measurements of absorbance are taken from product formed after the enzyme acid phosphatase is introduced. So the enzyme is left to incubate in the range of pH's because it isn't chromogenic and acid phosphatase, as it indicates functions well in acidic environments. The pH may start to alter the enzymes conformation as the solution reaches alkaline pH's, due to the fact it starts to neutralise the acid phosphatase.

Question2: The effects of pH on an enzyme are due to the increase in H+ when the pH is acidic or a decrease in H+ which is alkaline. Acidic amino acids have a carboxyl group in the side chain where alkaline amino acids have an amino group in their side chain. This means an increase or decrease in H+ ions can change the state of the bonds, meaning that the structure of the enzyme is altered. Also an increase or decrease in protons can alter the charge state of the substrate, preventing it from binding to the active site.

Discussion

Firstly an experiment took place in which the enzyme undergoes its catalytic process in normal conditions to consider and compare the results to further investigation and so it can be referred back to. Graph 2 is a Michealis-menten plot of the initial data and it shows that the effect of substrate concentration on the enzymes activity is determined by how efficient the active site is working. Therefore eventually the active site will become saturated and it will reach a peak. Graph 2 represents how the initial activity of the enzymes production accelerates at a sharp rate, and gradually peaks of to a flat gradient parallel to the x axis. This shows that the enzymes active site is gradually becoming saturated, until there is a point at where the substrate concentration won't affect the enzymes activity. The Vmax (0.015μmoles/min) from this graph demonstrates the point at which the enzyme is saturated; it may not be accurate due to estimation so a formula (Rate = Vmax x [S]/Km + [S]) can be used to plot some theoretical values to show whether the Vmax was over estimated. This can mean that increasing the substrate concentration further in the initial experiment could show a slow increase to a higher Vmax. It also shows that the line follows the predictions as so in the introduction. The Km (0.17mM) on graph 2 is interpreted by half of the Vmax read of at the x axis, thus graph 2 shows a low Km, so the enzyme substrate complex is strongly bonded together and there is rapid catalytic acceleration. Graph 3 can be used to show how the enzymes active site becomes saturated in a linear fashion. In this case small values become large and vice versa, so the values for the experiment: Vmax = 0.0119µmoles/min and Km = 0.103mM, thus the Vmax is how fast the substrate collides with the enzyme and binds to its active as demonstrated by a diagram (figure 2) in the introduction, using a typical lock and key method.

Secondly the same experiment was reproduced but with an inhibitor present, similar data was recorded as experiment one, but the values have a noticeable change when the tables are compared. When the data is plotted on a lineweaver-burk (graph 4) alongside the effects of the substrate alone, graph 4 shows the linear relationship between the two gradients and shows that the inhibitor indeed does follow prediction as stated in the introduction. Along with this graph 4 is showing that the inhibitor mostly affects the active site because it is a competitive inhibitor, which explains why the Vmax (0.0117µmoles/min) is closely related to when an inhibitor is not present, i.e. - the initial rate of reaction is predominately the same when compared but with an inhibitor present the gradient of the slope begins to flatten out earlier. The substrate can still bind to the active site even with the inhibitor present. The Km (0.833mM) when compared with out an inhibitor (0.103mM) present shows a very low affinity so this means that the there is a weak bonding appearance in the substrate-enzyme complex, which is underlined by the fact that inhibitors prevent the correct binding initial state of reaction.

The basics of a correlation correspond between [S] and [V] or the rate and this shows how the enzymes activity can correspond with a linear fashion.

Another experiment expands the evidence of enzyme activity with a substrate, graph 5 looks at how temperature and how an increase will affect the conformation of the active site and its binding with a substrate. Graph 5 shows that with general increase in temperature the enzymes-substrate complex binding becomes more efficient up to a peak; essentially the increase in temperature also increases the amount of collisions between substrate and enzyme molecules. At the peak on graph 5 which is at 49 degrees Celsius shows where the enzyme has reached its optimum efficiency, this means the maximum amount of substrate is binding with the active sites and potentially causing saturation. Naturally after the optimum the enzyme starts to denature because the high temperature breaks weak non-covalent forces such as hydrogen bonds that keep the conformation together for function, therefore average kinetic energy has disrupted the molecules in the conformation. This is shown by the steep slope after the peak on graph 5, whereas a temperature increase corresponds to a decrease in the rate of production. The data for this experiment can be interpreted a lot better by plotting it in a linear fashion which is represented by graph 6 using log10 of rate of reaction plotted against 1/T (1/K). The reason for this plot is it shows where the denaturing of the enzyme takes place according to temperature by the anomalous plot on the graph.

Temperature isn't the only factor, pH can change the environment and alter the conformation of the enzyme so it denatures. The measure of pH is H+ ions and these can alternate charges between the substrate and enzyme-complex, referring to the introduction figure 7b and 7c represent how the amino and carboxyl groups are affected by the change in positive charge. This can show how altering the substrate or/and enzyme active site can have an effect on the rate of binding. Since the enzyme is based to work in the kidney's and liver (acid phosphatase) it generally functions better at mid-low pH (5) (higher charge of H+ ions) and graph 7 shows the correlation of how the rate of product produced corresponds with the increase in pH. After the optimum point graph 7 shows that the enzyme has started to be altered by the change in charges due to increase in pH (decrease in H+ ions) this can be explained by the example shown in figure 7b where the causality is potential loss of hydrogen ions from the carboxyl group thus preventing the substrate from binding efficiently or at all.

Enzymes and substrates dependant on their environment have the potential to alter their activity between one another therefore, it concludes that enzyme-substrate complex is an important process and these set of experiment demonstrate how minute changes can influence a co-factor of the binding activity. When compared with background data the experiment has produced similar results concluding that the results are accurate for representation of a co-factor's effect on substrate-enzyme complexes. Carrying out the experiment more than once with different and more refined precise measurements i.e. - using a smaller range of pH or temperature with decimal values to determine a more accurate result; this will conclude the overall experiment for the enzyme to be more reliable.

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