Enzymes: basic concepts & kinetics

1. Harini Chandra Affiliations Enzymes: basic concepts & kinetics Enzymes are highly specific, powerful catalysts of biological systems that help to accelerate reactions taking place in organisms by factors of millions or more.

2. Master Layout (Part 1) 1 This animation consists of 5 parts: Part 1 – Enzymes & their components Part 2 – Energetics of enzymatic reactions Part 3 – Models for enzyme-substrate binding Part 4 – Kinetics of enzymatic reactions Part 5 – Enzyme inhibition 2 Cofactor Apoenzyme Holoenzyme Enzyme classification 3 4 5

3. Definitions of the components:Part 1 – Enzymes & their components 1 1. Enzymes: Majority of enzymes are proteins (some are RNAs) that increase the rates of biochemical reactions in living organisms. They are highly specific molecules that catalyze only a particular group of reactions and like other catalysts, they are regenerated at the end of a reaction. The rates of enzyme catalyzed reactions are often more than million times faster than those of un-catalyzed ones. Molecular weights of enzymes range from 12,000 to more than 1 million daltons. Enzymes offer the advantages of high reaction rates, substrate specificity, capacity for regulation and mild reaction conditions when compared to any other chemical catalysts. 2. Apoenzyme: The protein part of an enzyme that requires other additional components in order to make it catalytically functional is known as the apoenzyme or apoprotein. 3. Cofactor: Many enzymes require the presence of an additional component known as a cofactor for their catalytic activity. These cofactors can either be metal ions such as Fe2+, Mg2+, Mn2+ etc. or organic molecules known as coenzymes. A coenzyme or metal ion that is very tightly or covalently bound to the apoenzyme is known as the prosthetic group. 4. Holoenzyme: The complete, catalytically functional enzyme containing both the protein part as well as its cofactor is known as the holoenzyme. 5. Enzyme classification: Enzymes are classified on the basis of the reactions that they catalyze. Most enzymes are named by adding the suffix ‘ase’ either to their substrate or the type of activity they carry out. Classification by international organizations has led to six enzyme classes with many subgroups within each class, depending upon the type of reaction being catalyzed. Every enzyme has a unique, four-part classification number known as the Enzyme Commission number (E.C. number). 2 3 4 5

4. Part 1, Step 1: 1 Coenzyme Example: Pyruvate dehydrogenase complex Thiamine pyrophosphate (TPP) 2 Apoenzyme Holoenzyme Example: Carbonic anhydrase Fe2+ 3 Metal cofactor Fe2+ Zinc ion Holoenzyme Apoenzyme 4 Action Description of the action Audio Narration Most enzymes are made up of a protein part known as the apoenzyme as well as a cofactor which can either be an organic molecule known as a coenzyme or a metal ion. These cofactors are essential for the enzyme to be catalytically functional and the complete functional enzyme is referred to as the holoenzyme. Pyruvate dehydrogenase is a complex enzyme which uses Thiamine pyrophosphate as its coenzyme while carbonic anhydrase uses zinc ion as its cofactor. (Please redraw figures.) First show the blue figure and the green triangle with their labels. The triangle must then enter its groove on the blue figure giving rise to the figure on right followed by the example figure in green. Similarly the orange figure and red circle must then be shown with their respective labels. The red circle must then fit into its groove and then the example figure in blue on the left must be shown. As shown in animation. 5 Source: http://www.biochemj.org/bj/361/0437/3610437.pdf; http://www.rcsb.org/pdb/explore/images.do?structureId=1NI4

5. Part 1, Step 2: 1 Enzyme classification 2 3 4 Action Description of the action Audio Narration Enzymes are classified on the basis of the reactions that they catalyze. Most enzymes are named by adding the suffix ‘ase’ either to their substrate or the type of activity they carry out. However as more enzymes came to be known, it became increasingly difficult to name them in this manner. Classification by international organizations has therefore led to six enzyme classes with many subgroups within each class, depending upon the type of reaction being catalyzed. Every enzyme has a unique, four-part classification number known as the Enzyme Commission number (E.C. number), in which subclass number gives finer details about that particular enzyme reaction. Display the table one row at a time. Each row of the table must be displayed one at a time. 5

6. Master Layout (Part 2) 1 This animation consists of 5 parts: Part 1 – Enzymes & their components Part 2 – Energetics of enzymatic reactions Part 3 – Models for enzyme-substrate binding Part 4 – Kinetics of enzymatic reactions Part 5 – Enzyme inhibition Transition state [ES] ∆G‡ uncatalyzed reaction Free energy ∆G‡ catalyzed reaction Substrate [S] 2 Product [P] Reaction progress Transition state 3 Enzyme Substrate Product Enzyme 4 Spontaneous reaction – Negative ∆G [E] + [S] [ES] [E] + [P] 5

7. Definitions of the components:Part 2 – Energetics of enzymatic reactions 1 1. Enzyme: The biocatalyst responsible for bringing about an increase in the rate of reaction for conversion of substrate to product. 2. Substrate(s): The molecules present at the beginning of a reaction that are modified by means of the enzyme are known as substrates. An enzymatic reaction may have one or more substrates depending upon the reaction. 3. Product(s): The molecules produced as a result of an enzymatic reaction are known as the products. A reaction may yield one or more products with the enzyme being regenerated at the end of the reaction. 4. Transition state: Enzymatic reactions proceed through formation of a transition state i.e. an intermediate state between substrate and product having higher free energy than that of either the substrate or product. The transition state is the least stable species of the reaction pathway due to its high free energy and is therefore the most seldom occupied. 5. Free energy: The Gibbs free energy is a useful thermodynamic property that can be used to understand enzymatic reaction mechanisms. The free energy difference between the reactants and products as well as the energy required to activate conversion of reactants to products can provide information about the spontaneity and rate of a reaction. The free energy change (∆G) of a reaction is only dependent on the free energy of the reactants and products and not on the path of the reaction. A reaction can occur spontaneously only if the ∆G is negative. 2 3 4 5

8. Definitions of the components:Part 2 – Energetics of enzymatic reactions 1 6. Free energy of activation (∆G‡): The rate of any reaction is dependent on the Gibbs free energy of activation (∆G‡) i.e. the energy difference between the substrate and the transition state. Enzymes function to lower this energy gap by forming favourable interactions with the substrate, thereby facilitating formation of the transition state and allowing a larger number of molecules to overcome this energy barrier. Enzymes do not function by modifying the reaction equilibrium but only alter the reaction rate. 2 3 4 5

9. Part 2, Step 1: 1 Uncatalyzed reaction: Transition state [ES] ∆G‡ uncatalyzed reaction Free energy ∆G‡ catalyzed reaction Product [P] Substrate [S] Transition state 2 Substrate [S] Enzyme catalyzed reaction: Substrate [S] Product [P] 3 Reaction progress Transition state [ES] Enzyme [E] Product [P] Enzyme [E] 4 Action Description of the action Audio Narration Conversion of substrate to product proceeds through formation of a transition state. The free energy of activation of an uncatalyzed reaction is very high. Enzymes form favorable interactions with the substrate and facilitate formation of the transition state by lowering the free energy of activation. The transition state then dissociates to give the product and regenerates free enzyme. For a reaction to be spontaneous, the ∆G must be negative. It must be emphasized that enzymes do not alter the equilibrium of a reaction. First show the blue circle on top being converted into the red circle in the centre with simultaneous appearance of upward black curve in the graph. Next, show the red circle being converted into the violet circle with simultaneous appearance of the black downward curve of the graph. Then show the green shape and blue circle, which must bind in the groove of the green shape along with appearance of red upward curve of graph. Next show the blue circle changing color and then dissociating along with appearance of red downward curve of the graph. As shown in animation. 5

10. Master Layout (Part 3) 1 This animation consists of 5 parts: Part 1 – Enzymes & their components Part 2 – Energetics of enzymatic reactions Part 3 – Models for enzyme-substrate binding Part 4 – Kinetics of enzymatic reactions Part 5 – Enzyme inhibition 2 Fischer’s Lock-and-Key hypothesis Active site 3 Enzyme-substrate [ES] complex Enzyme Substrate Koshland’s induced fit hypothesis: Active site 4 Enzyme-substrate [ES] complex Substrate Enzyme 5 Source: Modified from Biochemistry by Lubert Stryer et al., 5th edition (ebook)

11. Definitions of the components:Part 3 – Energetics of enzymatic reactions 1 1. Fischer’s Lock-and-Key hypothesis: Emil Fischer in 1894 postulated that enzymes and their substrates possess specific complementary geometric shapes that fit exactly into each other like a key fits in a lock. This model accounts for the specificity of enzymes but fails to account for stabilization of the transition state. 2. Koshland’s induced fit model: Daniel Koshland in 1958 suggested that the active site of an enzyme gets continually reformed and reshaped based on the interactions that it forms with the substrate molecule. Therefore, rather than assuming a rigid shape, the enzyme’s active site gets molded into the exact shape and position required to carry out catalysis of the substrate. This accounts for both the enzyme specificity and the stabilization of the transition state. 3. Active site: All enzymes possess an active site that is lined with around 4-5 suitable amino acid residues that are responsible for catalysis of the reaction. Cofactors that are also involved in the reaction are usually bound at or near the active site. 4. Enzyme-substrate [ES] complex: Enzymatic reactions proceed through formation of a transition state i.e. an intermediate state between substrate and product having higher free energy than that of either the substrate or product. This state results from binding of the substrate to the active site of the enzyme resulting in an [ES] complex. 2 3 4 5

12. Analogy / Scenario / Action 1 Fischer’s Lock-and-Key hypothesis 2 Different keys – analogous to different substrates available Lock – analogous to enzyme 3 4 Action Description of the action Audio Narration Fischer’s hypothesis is aptly defined as the ‘lock-and-key’ hypothesis. Any lock, which is analogous to an enzyme, can have only one suitable key of appropriate shape and size to open it. The various available keys, which are analogous to the thousands of substrates available, can attempt to open the lock but only one will be the perfect fit that is capable of opening the lock. Similarly only one particular substrate will fit into the active site of the enzyme and the enzymatic reaction can occur. (Please redraw figures.) First show the lock and the various keys. Each key must move to the lock one at a time attempting to open it. The first three keys must be wrong and therefore the cross sign must appear after each attempt. The final key must be the correct one, a tick mark should appear and the lock must open. As shown in animation. 5

13. Part 3, Step 1: 1 Fischer’s Lock-and-Key hypothesis Substrate Active site 2 Enzyme-substrate [ES] complex Enzyme Koshland’s induced fit hypothesis: 3 Substrate Substrate Active site Enzyme-substrate [ES] complex Active site modified upon substrate approach 4 Enzyme Action Description of the action Audio Narration (Please redraw figures.) First show the light green figure ‘enzyme’ followed by the darker green ‘substrate’. The substrate must be shown to fit into its complementary cavity on the enzyme as shown and the complex on the right must be formed. Next, the pale violet ‘enzyme’ on the bottom must be shown followed by the darker violet ‘substrate’. In this, when the substrate moves down to bind to the groove on the enzyme below, the groove must slowly be modified to take the complimentary shape as shown in the middle panel. And once this happens the complex on the right is formed. According to the Fischer’s hypothesis, enzymes and their substrates possess specific complementary geometric shapes that fit exactly into each other. This model accounts for the specificity of enzymes but fails to account for stabilization of the transition state. Koshland modified this hypothesis and suggested that the active site of an enzyme gets continually reformed based on the interactions that it establishes with the substrate molecule. This accounts for both the enzyme specificity and the stabilization of the transition state since the enzyme is not considered to be a rigid molecule. As shown in animation. 5

14. Master Layout (Part 4) 1 This animation consists of 5 parts: Part 1 – Enzymes & their components Part 2 – Energetics of enzymatic reactions Part 3 – Models for enzyme-substrate binding Part 4 – Kinetics of enzymatic reactions Part 5 – Enzyme inhibition Reaction velocity (Vo) 2 Vmax Michaelis-Menten enzyme kinetics Rate constants k2 k1 Vmax/2 [E] + [S] [ES] [E] + [P] k-1 k-2 Km 3 Substrate concentration [S] Substrate [S] Turnover number 4 Product [P] Enzyme activity Enzyme [E] Enzyme-substrate complex [ES] 5

15. Definitions of the components:Part 4 – Kinetics of enzymatic reactions 1 1. Michaelis-Menten enzyme kinetics: A simple model to explain the kinetic characteristics of enzymatic reactions was proposed by Leonor Michaelis and Maud Menten in 1913. According to this model, formation of the [ES] complex intermediate is essential for the enzymatic catalysis reaction to take place. A group of enzymes that does not obey the Michaelis-Menten enzyme kinetics is comprised of the allosteric enzymes. These enzymes have multiple active sites and binding of substrate to one site affects, positively or negatively, the binding of substrate to the remaining sites. 2. Rate constants: An enzyme [E] combines with its substrate [S] to form the enzyme-substrate complex [ES] with a rate constant of k1. This complex can either form the product [P] with rate constant k2 or can dissociate to undergo a reverse reaction with rate constant k-1 which results in release of the substrate. In the Michaelis-Menten model for enzyme kinetics, it has been assumed that dissociation of the [ES] complex to give back the substrate is negligible. This condition applies for the initial stages of a reaction when the equilibrium favours product formation. 3. Reaction velocity (Vo): The rate of increase in product concentration [P] with time when the concentration is low is known as the reaction velocity (Vo). It is also often defined as the number of moles of product formed per second. Vo is linearly proportional to substrate concentration [S] when [S] is low but becomes independent of [S] when [S] is high. 4. Vmax: The maximal rate that can be achieved by an enzyme when all its catalytic sites are occupied by the substrate is referred to as Vmax or maximum velocity. 2 3 4 5

16. Definitions of the components:Part 4 – Kinetics of enzymatic reactions 1 5. Km: The substrate concentration at which the reaction rate is half its maximal value is known as Km or the Michaelis constant. Km value of an enzyme is an indicator of the affinity that the enzyme has for its substrate or in other words, it is a measure of the strength of the [ES] complex. A high Km indicates weak affinity between enzyme and substrate while a low Km is indicative of strong affinity. 6. Enzyme activity: The activity of an enzyme can be defined either in International Units (U) or Katal. 1 U of enzyme activity is defined as the amount of enzyme that catalyzes the conversion of 1 mmole of substrate into product at 25oC under the specified assay conditions. 1 Katal is the amount of enzyme that catalyzes the conversion of 1 mole of substrate into product per second. (1 U = 16.67 nanokatal) 7. Specific activity: This can be defined as the enzyme activity per milligram of protein. It is a measure of enzyme purity and increases continuously during a purification process until it achieves a maximum, constant value indicating the pure enzyme state. 8. Turnover number (kcat): The turnover number of an enzyme can be defined as the maximum number of substrate molecules that can be converted into product per active site of the enzyme per unit time. Unit of kcat is s-1 or min-1 and can obtained from parameters of the Michaelis-Menten equation (Vmax/[E]). 2 3 4 5

17. Part 4, Step 1: 1 k2 k1 [E] + [S] [ES] [E] + [P] k-1 k-2 Substrate [S] 2 Product [P] Reaction velocity (Vo) – number of moles of product formed per second during the initial stages of the reaction. 3 Enzyme [E] 4 Action Description of the action Audio Narration (Please redraw figures.) First show the brown ‘enzyme’ shape and the green ‘substrate’ circles along with the digital stopwatch set to 1 minute and the reaction on top. The stopwatch must then be shown to count down from 1 minute and the green circles must enter the groove on the brown molecule, change color and then be released as depicted in the animation. This must occur for all the green circles within 1 minute at the end of which the red start must appear with the text. Enzymes catalyze the formation of product from its substrate via an enzyme-substrate intermediate complex. During the initial stages of the reaction, the equilibrium favors product formation rather than dissociation of the [ES] complex to give back the substrate. The number of moles of product formed per second during these stages determines the reaction velocity for that particular enzyme. Vo has an almost linear relation with substrate concentration when the substrate concentration is low but becomes independent at higher concentrations. As shown in animation. 5

18. Part 4, Step 2: 1 k2 k1 [E] + [S] Reaction velocity (Vo) [ES] [E] + [P] k-1 k-2 Vmax Substrate [S] 2 Vmax/2 Km Product [P] 3 Substrate concentration [S] Michaelis-Menten equation Vo = Vmax [S] Km + [S] Enzyme [E] 4 Action Description of the action Audio Narration The Michaelis-Menten model for enzyme kinetics assumes that the breakdown of [ES] complex to give back free substrate is negligible and also assumes steady-state conditions whereby the rates of formation and breakdown of the [ES] complex are equal. The reaction velocity increases linearly with substrate concentration when [S] is low but becomes independent at higher concentrations. The maximal velocity that can be achieved by an enzyme refers to the state in which all its catalytic sites are occupied. The substrate concentration at which the reaction velocity is equal to half its maximal value is known as the Michaelis constant, Km. As shown in animation. First show the brown ‘enzyme’ shape and the green ‘substrate’ circles along with the reaction on top and the axes of the graph. Next, show the green circles moving into the groove on the enzyme, changing colour and then getting released. This must continue for all the circles with the pace of movement slowing down after the first 3 circles. After the first 3, the upward slope of the graph must appear followed by the next 3 slower circles moving into the groove and then the second half of the enzyme slope as depicted in the animation. 5

19. Part 4, Step 3: 1 Lineweaver-Burk equation (Double reciprocal plot) – a useful tool for analyzing experimental data Michaelis-Menten equation Slope = Km / Vmax Vo = Vmax [S] 2 1 / Vo Km + [S] Taking reciprocal of both sides of the above equation and rearranging the terms, we get the Lineweaver-Burk equation: 3 1 / Vmax 1 = Km + 1 1 / [S] - 1 / Km Vo Vmax[S] Vmax 4 Action Description of the action Audio Narration The Lineweaver-Burk equation or the double reciprocal plot is a useful tool that can be plotted using simple experimental data from kinetics experiments. This equation is derived from the Michaelis-Menten equation by taking reciprocals on both sides and then plotting a graph of 1/Vo Vs 1/[S]. The Y-intercept on this graph can be used to deduce the value of Vmax while the X-intercept gives the value of Km. As shown in animation. First show the equation on top followed by the text in between and then the equation at the bottom in the green box. Finally show the graph on the right. 5

20. Master Layout (Part 5) 1 This animation consists of 5 parts: Part 1 – Enzymes & their components Part 2 – Energetics of enzymatic reactions Part 3 – Models for enzyme-substrate binding Part 4 – Kinetics of enzymatic reactions Part 5 – Enzyme inhibition 2 Enzyme inhibition 3 Reversible inhibition Irreversible inhibition 4 Competitive inhibition Uncompetitive inhibition Mixed inhibition 5

21. Definitions of the components:Part 5 – Enzyme inhibition 1 1. Enzyme inhibition: Several molecules are capable of binding at or near the active site of enzymes thereby decreasing or inhibiting their activity. It provides an important control mechanism in biological systems. Enzyme inhibition is also an important mechanism that is exploited during the manufacture of various drug molecules. 2. Reversible inhibition: Inhibition of an enzyme can be reversed if there is rapid dissociation of the enzyme-inhibitor complex. The inhibitor is associated to the enzyme molecule by relatively weaker interactions. 3. Irreversible inhibition: In case of irreversible inhibition, the inhibitor covalently modifies the enzyme and dissociates very slowly from the target enzyme because it is tightly bound. The action of penicillin, an important antibiotic, is through the irreversible inhibition of the enzyme transpeptidase that is essential for bacterial cell wall synthesis. Aspirin, the commonly used analgesic and anti-pyretic also functions by means of irreversible inhibition of the enzyme cyclooxygenase. 4. Competitive inhibition: This is a type of reversible inhibition wherein the inhibitor binds to the active site of the enzyme thereby preventing the substrate from binding to it. The inhibitor and substrate, in this case, are structural analogues and the reaction rate is decreased due to fewer enzyme molecules being bound to the substrate. This type of inhibition can be overcome by increasing the substrate concentration. 2 3 4 5

22. Definitions of the components:Part 5 – Enzyme inhibition 1 5. Uncompetitive inhibition: In this case, the substrate and the inhibitor have distinct binding sites on the enzyme and the inhibitor binds only to the ES complex and not to the enzyme alone. Both Km and Vmax are found to decrease in this type of inhibition. 6. Mixed inhibition: A mixed inhibitor also binds to the enzyme at a site distinct from the substrate binding site with the difference being that it can bind either to E or ES. Binding of either one brings about conformational changes in the enzyme structure thereby affecting binding of the other. This inhibition can be reduced but not overcome by increasing substrate concentration. Non-competitive inhibition is a special case of mixed inhibition where inhibitor binding reduces catalytic activity but does not affect substrate binding. 2 3 4 5

23. Part 5, Step 1: 1 Enzyme inhibition 2 Reversible inhibition Irreversible inhibition 3 Competitive inhibition Uncompetitive inhibition Mixed inhibition 4 Action Description of the action Audio Narration Enzyme inhibition can either be reversible, where the inhibitor can dissociate quickly from the enzyme, or irreversible, where the inhibitor dissociates very slowly from the enzyme and can covalently modify the enzyme thereby rendering it unsuitable for further catalysis reactions. Reversible inhibition can be further classified as competitive, uncompetitive and mixed inhibition. Show each of the headings appearing sequentially followed by the three colored boxes at the bottom getting highlighted. As shown in animation. 5

24. Part 5, Step 2: 1 Competitive inhibition Product [P] Increasing [I] Substrate [S] ES complex 2 Enzyme [E] 1 / Vo Inhibitor [I] No reaction No inhibitor 3 EI complex 1 / Vmax Lineweaver Burk equation Slope = aKm / Vmax 1 = aKm + 1 1 / [S] Vo Vmax[S] Vmax 4 Action Description of the action Audio Narration In competitive inhibition, the inhibitor molecule is structurally similar to the substrate and therefore binds to the enzyme at the active site. Binding of inhibitor prevents substrate from binding, thereby decreasing the reaction rate. The Vmax in this type of inhibition remains the same and only the Km is altered. Competitive inhibition can be overcome by suitably increasing the substrate concentration, which allows the substrate to out-compete the inhibitor for the enzyme’s active site. First the blue triangle must bind to the red ‘enzyme’ followed by the green triangle. First show the blue triangle binding to the red ‘enzyme’ and then triangle changing color to purple followed by its dissociation. Next show the green triangle binding to another red ‘’enzyme’ but no reaction taking place as depicted in animation. This is followed by appearance of the graph on the right. 5

25. Part 5, Step 3: 1 Lineweaver Burk equation Uncompetitive inhibition 1 = Km + a’ P Vo Vmax[S] Vmax Product [P] Increasing [I] Substrate [S] 2 Enzyme-substrate [ES] Enzyme [E] Inhibitor [I] 1 / Vo No inhibitor 3 ESI complex 1 / [S] - 1 / Km No reaction 4 Action Description of the action Audio Narration In the case of uncompetitive inhibition, the substrate and inhibitor both have different binding sites on the enzyme . However, the inhibitor binds only to the enzyme-substrate complex and not to the enzyme alone. Binding of inhibitor to the ES complex prevents any further reaction and no product formation is observed. Both the Km and Vmax are found to decrease with this type of inhibition. (Please redraw all figures.) First show the blue ‘enzyme’ followed by the binding of hexagonal ‘substrate’ to one of its grooves. This is followed by two reaction arrows, one showing appearance of ‘product’ and the other showing binding of pentagonal ‘inhibitor’ to the second groove on the ‘enzyme’. There must be an arrow from this complex indicating no reaction. Finally graph shown on the right must appear. As shown in animation. 5

26. Part 5, Step 4: 1 Lineweaver Burk equation Mixed inhibition 1 = aKm + a’ P Vo Vmax[S] Vmax Product [P] Increasing [I] Substrate [S] 2 Enzyme-substrate [ES] Enzyme [E] Inhibitor [I] 1 / Vo No inhibitor 3 ESI complex Enzyme-inhibitor [EI] 1 / [S] - 1 / Km No reaction 4 Action Description of the action Audio Narration (Please redraw all figures.) First show the blue ‘enzyme’ followed by binding of hexagonal ‘substrate’ to the groove on it or binding of pentagonal ‘inhibitor’ next to the groove. Next the pentagonal ‘inhibitor’ must bind to the ‘ES’ complex and ‘substrate’ must bind to the ‘EI’ complex as depicted in animation. Finally arrows showing no reaction from ‘ESI’ complex and product formation from ‘ES’ complex must be shown followed by the graph on right. A mixed inhibitor also binds to the enzyme at a site distinct from the substrate binding site with the difference being that it can bind either to enzyme or ES complex. Binding of either one brings about conformational changes in the enzyme structure thereby affecting binding of the other. This inhibition can be reduced but not overcome by increasing substrate concentration. Both Km and Vmax are altered in this type of inhibition. As shown in animation. 5

27. Interactivity option 1:Step No:1 1 Methanol, which is commonly used as a solvent and is also a component of illicit liquor, is extremely toxic to the body. It commonly causes blindness and can even result in death. This is due to the conversion of methanol into toxic formaldehyde by the liver enzyme alcohol dehydrogenase. In order to effectively treat methanol poisoning, patients are often given slow intravenous infusions of ethanol. This ethanol is converted into harmless acetaldehyde by the same enzyme, alcohol dehydrogenase, while the methanol gets excreted by the kidneys. What principle does this therapy make use of? 2 3 A) Mixed inhibition B) Competitive inhibition C) Uncompetitive inhibition D) Non-competitive inhibition 4 Results Interacativity Type Options Boundary/limits (B) Is the correct answer. If user chooses this, it must light up in green with the comment ‘Correct answer’ displayed. If any of the other options are selected, it must turn red with the comment ‘Incorrect answer’. User must be allowed to choose any one of the four options. Choose the correct answer. 5

28. Interactivity option 2:Step No:1 1 Shown below are two enzymes and their substrates. Select the enzyme-substrate pair that exhibits Koshland’s induced fit hypothesis by dragging the substrate into the enzyme’s active site. Substrate Substrate 2 Koshland’s induced fit model 3 4 Enzyme Enzyme Results Interacativity Type Options Boundary/limits User must be able to drag the blue & green figures on top into the brown & violet figures below such that the projections fit into the grooves. When the user drags the green figure on the right top into the violet shape below, the grooves must be modified such that it resembles the brown shape on the left bottom and fits exactly into the grooves. Once the user has done this, the text ‘Koshland’s induced fit model’ must appear. Drag and drop. 5

29. Questionnaire 1 1. What is the cofactor for the enzyme carbonic anhydrase? Answers: a) Fe2+ b) Mg2+ c) TPP d)‏ Zn2+ 2. The enzyme fumarase falls under which of the following classes? Answers:a) Lyase b) Ligase c) Oxidoreductase d)‏ Hydrolase 3. In which of the following types of inhibition does the Vmax remain the same? Answers:a) Non-competitiveb)Competitivec) Mixed d) Uncompetitive‏ 4. Enzymes catalyze reactions by Answers: a) Shifting the reaction equilibrium b) Increasing the enthalpy c) Decreasing the free energy of activation d)‏ Making the net free energy change positive. 5. The protein part of an enzyme is known as? Answers: a) Coenzyme b) Apoenzyme c) Prosthetic group d)‏ Holoenzyme 6. In the Michaelis-Menten equation, what happens to the value of Vo when Km<< [S] and therefore can be ignored? Answers:a) Vo = [S]b)Vo = Vmaxc) Vo >> Vmax d) Vo << Vmax 2 3 4 5

30. Links for further reading Books: Biochemistry by Stryer et al., 5th & 6th edition Biochemistry by A.L.Lehninger et al., 4th edition Biochemistry by Voet & Voet, 3rd edition Fundamentals of Enzymology by Price & Stevens, 3rd edition