Ali Yaghi. Gumana Ghashan. Mamoun Ahram

Transcription

1 21 Ali Yaghi Gumana Ghashan Mamoun Ahram

2 Kinetics The study of Kinetics deals with the rates of chemical reactions. Chemical kinetics is the study of the rate of chemical reactions. For the reaction (A P), The velocity (v) or rate of the reaction can be calculated by either: The appearance of products per unit of time. (Rate of formation) The disappearance of reactants per unit of time. (Rate of consumption) Or Rate law A mathematical equation describes the relationship between reaction rate and concentration of reactants. the mathematical equation describes how concentrations of reactants affect the rate of the reaction during a certain period of time. The higher the concentration of a substrate, the higher the rate of the reaction. Explanation: the enzyme s chance to bind to the substrate increases. At high level of a substrate, the rate of the reaction increases. Explanation: because the enzyme will collide with the substance and bind to it more frequently. When the substrate concentration decreases, the rate also decreases. K is a constant used to measure the velocity of a reaction, its unit is (time)-1 usually (sec)-1.

3 Hyperbolic plot, describes the rate of enzyme catalyzed reactions according to substrate concentration. Diagram Notes: 1- molecules are blind, they collide to each other randomly. When there are more substrates, the appropriate enzyme- substrate collision opportunity is greater, which means higher rate of reaction, and vice versa 2- At fixed concentrations of enzyme, V0 is almost linear proportional to [S] when [S] is small. 3- The rate of catalysis rises linearly as substrate concentration increases (conc. Is small) and then begins to level off and approach a maximum rate at higher substrate concentrations. 4- At low concentration: any small increase in the substrates concentration leads to great increase in the rate of the reaction. 5- At high concentration: increasing the substrates concentration (even if it is huge) leads to small increase in the rate of the reaction. 6- At low concentration, the limiting factor is the substrate, because the enzyme roughly "finds" the substrate. At this point, the active sites aren't saturated. 7- At high concentration, the limiting factor is the enzyme's active sites are saturated. 8- The maximal rate, Vmax, is achieved when the catalytic sites on the enzyme are saturated with the substrate.

4 Vmax: The number of substrate molecules converted into product by an enzyme molecule in a unit time when the enzyme is fully saturated with substrate. The Michaelis-Menten equation Km: it is the substrate concentration when V0 (velocity of the reaction) is half of Vmax. At very low substrate concentration, when [S] is much less than KM, V0= (KM)[S]; that is, the rate is directly proportional to the substrate concentration. At high substrate concentration, when [S] is much greater than KM, V0= Vmax; that is, the rate is maximal, independent of substrate concentration. One enzyme, two substrates a reaction is catalysed by an enzyme, it has high affinity for substrate S, but low affinity for substrate S. Vmax is the same for both substrates, but Km is higher for substrate S, the low affinity substrate Hexokinase is an enzyme that phosphorylates substances, so it must bind to ATP and to substrate (glucose or fructose).

5 Note: Hexokinase binds to both at different affinities, So Km describes the affinity. 1- Km is a property for the substrate not the enzyme. 2- Binding affinity for ATP and glucose is different, but Vmax of the enzyme won't change (Vmax is a property for the enzyme). Why the affinity of binding is different? It depends on shape, number of non-covalent interactions, type of noncovalent interactions, angle of binding, type of amino acids in the active site and many other factors. Example: find Vmax and Km. Vmax = 700 micro M/min Km= 8 micro M Explanation1: we reached a point that any increase in the concentration no matter how big it is, the speed won't increase any more. Explanation2: Km is the concentration when the velocity is half of maximum velocity.

6 The Michaelis constant (Km) It is related to the dissociation of the substrates from the enzyme divided by the association. Note: when the Km is lower, the affinity is higher. The reason is that when you need less concentration to reach the V1/2, the affinity is higher. remember P1/2 for hemoglobin and myoglobin. Myoglobin has P1/2 lower than hemoglobin, which means it has higher affinity. Notes: 1- Km describes the affinity of the enzyme for a substrate, but it is not an accurate measure of affinity (because the products are involved in the equation). 2- Dissociation Constant [KD] is the actual measure of the affinity. KD = (K-1/K1) 3- we can increase Vmax for the reaction by Increasing the enzyme concentration, because when we add more enzyme, the probability of collision between substrates and enzyme is higher. in this case, Km doesn't change. 4- If we increase the substrates concentration, we will reach Vmax, but we can't increase Vmax in this way.

7 *this figure shows that when you increase the enzyme conc. increases, but Km stays constant. Vmax If we have two enzymes, one has a Vmax of 10 micro moles/ minute, and the other one has a Vmax of 20 micro moles/minute, we can't determine which one of them is more efficient, because we don't have the concentrations of the enzymes. Measurement of enzyme efficiency(kcat)- (K2): K2= Kcat [E]t: total concentration of enzyme

8 What does it mean if K2 for an enzyme is 40 million? It means that this enzyme can make 40 million reactions per second. Or it can mean that each reaction needs 1/40 million second to be done. Note: Kcat unit is 1/s or s^ (-1). Example1 You are working on the enzyme X which has a molecular weight of 50,000 g/mol. You have used 10 μg of the enzyme in 1 liter in an experiment and the results show that the enzyme at best converts 9.6 μmol of the substrate per min at 25 C.What is the turnover number (kcat) for the enzyme? Answer: MW = 50,000 g/mol Weight = 10 μg Vmax = 9.6 μmol of the substrate per min Kcat = (9.6/60)/(10 μg /50,000) = 800 s-1 Remember to convert μmol/min to μmol/s.

9 Enzyme activity In order to measure enzyme activity, we measure the number of moles of substrate disappearing (or products appearing) per unit time (mol/sec) In other words: enzyme activity = rate of reaction reaction volume Specific activity (will be used later) Specific activity is a measure of enzyme purity and quality. It is calculated as moles of substrate converted per unit time per unit mass of enzyme (mol / (sec g)). In other words, Specific activity =enzyme activity / mass of enzyme (grams) This is useful in determining enzyme purity after purification. Turnover number Turnover number (kcat) is related to the specific activity of the enzyme Turnover number = specific activity molecular weight of enzyme It is expressed as moles of substrate converted per unit time (usually per second)/moles of enzyme Remember: kcat= Vmax/ [E]T

10 Disadvantage of Michaelis-Menten equation Determination of Km from hyperbolic plots is not accurate since a large amount of substrate is required in order to reach Vmax. This prevents the calculation of both Vmax and KM

11 The Lineweaver-Burk (double-reciprocal plot) A plot of 1/ V0 versus 1/[S], called a Lineweaver-Burk. It yields a straight line. Straight line has: 1- intercept of 1/Vmax 2- slope of KM/Vmax. 3- The intercept on the x-axis is -1/ KM Notes: 1- The equation is theoretical, because it is impossible to have a concentration of enzyme in negative. However, it is really useful. 2- you have to memorize the equations. 3- If x = 0, then y = b (x-axis is 0, then y-intercept = 1/Vmax) 4- If y = 0, then mx = -b (y-axis is 0, then x-intercept = -1/Km)