Biochemistry Notes PDF

32 Dental Student Raghad Ibrahim Nafez Abutarboush Kinetic of enzymatic reactions K1 K2 K-1 • We know from the previous lectures that Velocity=constant*concentration • The maximum rate, Vmax, is equal to the product of k 2, also known as kcat, and the total concentration of enzyme, that’s beca...

32 Dental Student Raghad Ibrahim Nafez Abutarboush Kinetic of enzymatic reactions K1 K2 K-1 • We know from the previous lectures that Velocity=constant*concentration • The maximum rate, Vmax, is equal to the product of k 2, also known as kcat, and the total concentration of enzyme, that’s because the maximum velocity is reached when all the active sites of the enzyme are saturated, that makes the concentration of the ES the same for the enzyme ([E]Total). Vmax = k2 *[E]T • In other words, the maximum rate, Vmax, reveals the turnover number of an enzyme if the total concentration of active sites [E] T is known. • K2 is the same as Kcat and the same as the turnover number. Example: a 10^-6 M solution of carbonic anhydrase catalyzes the formation of 0.6 M H2CO3 per second when it is fully saturated with substrate Hence, Kcat is 6 X 10^5 s-1 (Kcat = Vmax / [E]T —> 0.6/ 10^-6) 3.6x 10^7/min • Each catalyzed reaction takes place in a time equal to 1/k2, which is 1.7 μs for carbonic anhydrase • The turnover numbers of most enzymes with their physiological substrates fall in the range from 1 to 104 per second Specificity and Efficiency By rearranging the equation V=(Kcat/Km)[E][S] ▪ Specificity constant (Kcat/Km) it tells how much the enzyme is specific to its constant, when you increase Kcat you increase how much product you make, when you increase Km the affinity decreases, which decreases the specificity Enzyme’s catalytic efficiency: the higher the ratio, the more efficient the enzyme. ▪ The importance of the specificity constant is that it allows for easy comparison between different enzymes, Kcat and Km ranges are very big which makes the comparison between different enzymes difficult but when it’s K cat / Km the range is only 4. ▪ Reaction rate: measures the concentration of substrate consumed (or product produced) per unit time (mol/ {L.s} or M/s) ▪ Enzyme activity: measures the number of moles of substrate consumed (or product produced) per unit time (mol/s). ➢ Enzyme activity = rate of reaction × reaction volume. ▪ Specific activity: measures moles of substrate converted per unit time per unit mass of enzyme (mol/ {s.g}), it’s used for determining the purity of the enzyme after purification (for example, u took a blood sample which have RBCs, plasma, buffy coat and u want an enzyme that is present in white blood cells so you’ll have to remove everything else in order to get it and once u do, you’ll need to check how pure that enzyme is so u use the specific activity) ➢ Specific activity = enzyme activity/ actual mass of enzyme ▪ Turnover number: measures moles of substrate converted per unit time per moles of enzyme (min-1 or s-1) ➢ Turnover number = specific activity × molecular weight of enzyme -Remember: moles= mass / molecular weight Example: A solution contains initially 25X10-4 mol L-1 of peptide substrate and 1.5 μg chymotrypsin in 2.5 ml volume. After 10 minutes, 18.6X10-4 mol L-1 of peptide substrate remain. Molar mass of chymotrypsin is 25,000 g mol-1. A. How much is the rate of the reaction? (conc./time) B. How much is the enzyme activity? (mol./time) C. How much is the specific activity? (enz. Act. / enz. Mass) D. How much is the turnover number? (sp. Act. X enz. Molar mass) You can solve it on your own, always pay attention to the units. Disadvantages of Michaelis-Menten equation & LineweaverBurk or double-reciprocal plot. • The problem with the Michaelis-Menten equation that it’s not linear, we can’t do a lot of experiments which makes it hard to draw the plot, also to know the vmax you have to keep slightly increasing the concentration of the substrate to very high amounts and do a lot of experiments, and that’s expensive. Lineweaver-Burk solved this problem by doing some math. Take this equation and divide over 1, it will become this new equation: We can see here that the new variables are 1/v and 1/[S], where Km/Vmax is the slope, 1/ Vmax is the y intercept and -1/Km is the x intercept, this is a linear equation which made the equation much easier to plot because we don’t need to do many experiments to plot it. Important example: A biochemist obtains the following set of data for an enzyme that is known to follow Michaelis-Menten kinetics. Approximately, Vmax of this enzyme is … & Km is …? A. 5000 & 699 B. 699 & 5000 C. 621 & 50 D. 94 & 1 E. 700 & 8 We can see how the velocity increases in smaller gaps when we have higher concentrations, the approximate Vmax is 699, half of it is almost 349 which makes Km 8, the answer is clearly E . But if we had the choice E as 690 & 8 and one answer is none of the above, what should you choose? 690 & 8 because we approximate the 699 to 690. ➢ When you read approximately you should understand that you don’t have to to calculations. Another example (not important but try to solve it): You are working on the enzyme “Medicine” which has a molecular weight of 50,000 g/mol. You have used 10 μg of the enzyme in an experiment and the results show that the enzyme converts 9.6 μmol per min at 25C̊. the turn-over number (kcat) for the enzyme is: A. 9.6 s-1 B. 48 s-1 C. 800 s-1 D. 960 s-1 E. 1920 s-1 Enzyme regulation Enzymes should be highly regulated because they’re highly processive (can catalyze millions of reactions per second) so the effects of any mistake in its work would be very damaging. Modes of regulations: ➢ All enzymes are affected by multiple modes, but no enzyme is affected by all modes of regulation. 1-Isozymes or isoenzymes • This mode changes the conformation of the enzyme (the shape) which influences its efficiency. • What are isozymes? Same substrate & product, different gene, different localization, different parameters (Km, Vmax, kcat). • For example, hexokinases are found in the RBCS and liver and pancreas, the one found in RBCS is hexokinase I, it’s Km is around 0.1 mM. The one found in liver and pancreas is hexokinase IV (AKA glucokinase) it’s Km is around 10 mM. It’s different because RBCS can synthesize energy only by glycolysis, but the liver can use amino acids, glucose and fats. The need for glucose in RBCS is higher than the liver so we need higher affinity of hexokinase I for glucose. • When blood glucose falls below its normal fasting level (≈ 5 mM), RBCs could still phosphorylate glucose at rates near Vmax. • High KM of hepatic glucokinase promotes storage of glucose, and the Pancreas works as a sensor. • Also, aldehyde dehydrogenase (ALDH) which oxidizes acetyl aldehyde to acetate, has four tetrameric isozymes (I-IV). • ALDH I has low Km and it’s mitochondrial, while ALDH II has high K m and it’s cytosolic. • Half of the Chinese and Japanese population are unable to produce ALDH I (not observed in Caucasian & Negroid populations), which causes the effects of Acetyl aldehyde toxicity which are tachycardia and flushing response. End of sheet 32