Effects of Enzymes on Chemical Reaction Rates
|✅ Paper Type: Free Essay||✅ Subject: Biology|
|✅ Wordcount: 1018 words||✅ Published: 24th May 2018|
Enzymes increases the rates of chemical reactions by providing a reaction pathway with a transition state free energy barrier lower than the uncatalyzed reaction (i.e. by lowering the free energy barrier that separates the reactants and products). The diagram below shows the reaction coordinate.
Enzymes catalyzed oxidation-reduction reactions only in relationship with small cofactors, which act like their chemical teeth. Cofactors may be ions like Cu2, Fe3+ or Zn2. The essential of cofactors determine why organisms only require trace amount of certain elements in their diets. Cofactors may also be organic molecules known as coenzymes. Cofactors are only temporary associates with the given enzyme molecules to function like cosubstrates. Nicotinamide adenine dinucleotide (NAD+) and nicotinamide adenine dinucleotide phosphate (NADP) are examples of cosubstrates. Enzymes have an optimum temperature in which they perform best. An enzyme assay measures the yield of products from substrates under specific conditions like; cofactors, pH and temperature of the enzyme at its optimum. In the initial reaction, high substrates concentrations were used, for the rate to be proportional to the enzyme concentration. By using spectrophotometer usually using coenzymes reduced nicotinamide adenine dinucleotide (NADH) and reduced nicotinamide adenine dinucleotide phosphate (NADHP), which absorb light at 340nm, are frequently used to monitor the progress of the reaction. Then either the rate of appearance of the product or the rates of the disappearance of the substrate are measured by monitoring the change in absorbance of the wavelength. In case of fluorescence substrates, change in the concentration can be measured by monitoring the change in fluorescence using a fluorimeter. As absorbance (or fluorescence) is proportional to concentration, the change in absorbance or fluorescence rate is proportional to the rate of enzymes activity in moles of substrates used or product formed per unit time. E.g. the activity of lactate dehydrogenase with lactate as substrate can be assayed by monitoring the increase in absorbance at 340 nm. In the following equation: CH3CH(OH)COO- + NAD+ ↔ CH3COCOO- + NADH + H+ In a coupling or linking reaction assay, where the second enzyme does not have any characteristic rate in absorbance change, the catalyze enzyme coupling or linking the reaction can be resolute. In this analyse, the catalyzed enzyme must be in excess for the progress of the assay to continue. This will ensure the rate of production is proportional to enzyme activity. Question (b):
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An enzyme activity is often articulated by the initial rate (V0) of the reaction been catalyzed. V0 are µmol min-1; and can also be represented by the enzyme unit (U) or Katal (Kat), where 1µmol min-1 equals 1unite (U) equals 16.67 nanokatal (Kat). An enzyme unit is the sum of the enzyme that catalyzed the transformation of 1µmole of substrate per minute under specified conditions of temperature, pH and substrate concentration. The Michaelis-Menten model of enzyme catalysis:
E + S â‡‹K1K2 ES fiK3 E + P
Where the rate constants K1; K2; and K3 describe association rates with each catalytic step processes. At low [S], V0 is directly proportional to [S], while at high [S] the velocity tends towards the maximum velocity (Vmax). The Michaelis-Menten equation Vo = (Vmax. [S])/ (Km + [S]), and this describe the hyperbolic curve on the graph shown in fig 2.
Km is defined by Michaelis and Menten as; a sum of stability of the ES complex equals the sum of the rates of the products [ES] over its rate of formation.
Km = (K2 +K3)/K1
For many enzymes, K2 is much greater than K3. Under these situations, Km is taking as a measure of affinity of enzyme’s substrate as its value is dependent of the relative values of K1 and K2 for ES formation and dissociation, correspondingly. Km is determined experimentally by substrate concentration as this equivalent to Km value at which the velocity is equal to half of Vmax. The Lineweaver-Burk plot:
As Vmax and Km is impossible to estimate them due to their values being achieved at infinite substrate concentration, they can be experimentally determined by measuring V0 at different substrate concentrations; (see fig 2). Although, the Linewearver-Burk plot is a derivation of Michaelis-Menten equation:
1/V0 = (1/Vmax + Km)/ (Vmax .1/[S])
Which gives a straight line, with the interception on the Y-axis equals 1/Vmax and that of the X-axis equals -1/Km, it is also useful in determining how an inhibitor binds to enzyme. The Km and Vmax can also be determined using Eadie-Hofstee plot of V0/[S] against V0, where intercept on X-axis equals Vmax and the slop of the line are equal to -1/Km. Question (C):
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Inhibitors lower the catalytic rate of an enzyme activity. There are two main types of inhibitors: Irreversible or reversible. Reversible inhibitors can be subdivided into competitive and non-competitive. An irreversible inhibitor covalently binds tightly to amino acid residues at the active site of the enzyme and permanently in-activate the enzyme active site. A competitive inhibitor caused a conformation change to enzyme by binding on to enzyme other than its active site and therefore, decreases enzymes catalytic rate. This can be determined using Lineweaver-Burk plot where non-competitive inhibitors were seen to decrease Vmax without any change to Km. Whereby competitive inhibitors decrease the enzyme catalytic rate by binding onto the enzyme active site and overcome substrates. At high substrate concentration competitive inhibitors can be overcome. The Lineweaver-Burk plot of competitive inhibitors shows an increase of Km without any change to Vmax. So by using Lineweaver-Burk plots, the effects of inhibitors of catalytic activity can be determined. For example:
Non-competitive inhibitor effects cannot be overcome by increasing substrates, as this does not bind to the active site.
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