f) Adding an enzyme does not change the Gibbs free energy. It only increases the rate of the reaction by lowering the activation energy.

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1 Problem Set 2-Answer Key BILD1 SP16 1) How does an enzyme catalyze a chemical reaction? Define the terms and substrate and active site. An enzyme lowers the energy of activation so the reaction proceeds at a faster rate than if it were uncatalyzed. The enzyme accomplishes this by bringing reactants into closer contact within the active site and facilitating bond breakage or union. The substrate is the molecule bound by the active site of the enzyme and converted into product. The active site is that portion of the enzyme that binds with the substrate molecule to initiate the formation of product. 2) Label the terms A thru E on the energy diagram curve at the right. A = products; B = G; C = transition state C or unstable intermediate; D = energy of activation (EA); E = substrates or reactants a) Endergonic b) No, the reaction requires an infusion of E energy. c) G > 0 as G = GProducts GReactants. D A B d) Yes, this reaction is reversible. The reaction can go in the forward manner where the reactant goes to form the product and it can also proceed in the reversed manner where the product is used to form the reactant. The energy diagram suggests that the reverse reaction may be more favorable than the forward. e) Adding an enzyme will lower the activation energy, the amount of energy needed to overcome the energy barrier, and therefore decrease D. An enzyme, by definition, will speed up the rate of the reaction in both the forward and the reverse reactions, as a result does not change the Gibbs free energy of the reactants (E) and the products (A). f) Adding an enzyme does not change the Gibbs free energy. It only increases the rate of the reaction by lowering the activation energy. 1

2 3) At the right is a temperature sensitivity curve for a particular enzyme. a) The Gibbs Free Energy Equation states that G = H T S. The activity of the enzyme increases, and the reaction becomes more favorable, because the Gibbs Free Energy decreases at (T) increases. Enzyme Activity TEMPERATURE b) Beyond an optimal temperature, too much heat will cause noncovalent, ionic, and hydrogen bonds to break. The enzyme will unfold and become inactive, in a process called denaturation. c) A curve of enzyme activity over varying ph s will look similar to the curve for temperature. This is because ph affects the ionization of the side chains the amino acids, which in turn will affect the structure of the protein. 4) Is protein synthesis an endergonic or exergonic reaction? Is S positive or negative? Protein synthesis is an endergonic process (a more complex molecule is being made from smaller and less complex molecules and G is positive). S is negative because there is less disorder (more order) in a polypeptide than in its individual amino acids. 5) When NaOH is dissolved in water, the solution becomes hot. When CsCl is dissolved in water the solution becomes cold. Indicate for each reaction whether it is exergonic or endergonic and whether G, H, and S are positive or negative (i.e. + or -). For NaOH: G is negative exergonic; H is negative exothermic; and S is positive. That is why the solution becomes very hot. For CsCL: G is negative exergonic; H is positive endothermic; and S is positive. The dissociation of the CsCl salt into solvated ions in solution is an endothermic process ( H > 0). Therefore it takes up heat. However, it is strongly favored entropically because the ions are dispersed randomly in solution. In this case, T x S is greater than H so the reaction is spontaneous even though it is endothermic. Thus, the beaker gets cold. 2

3 6) What are the three principle ways RNA differs structurally from DNA? RNA is usually single stranded whereas DNA is double stranded. RNA contains the nucleotides ACGU whereas DNA contains ACGT. RNA contains a ribose sugar whereas DNA contains deoxyribose. 7) THINKING ABOUT THERMODYNAMICS (a) Draw a phospholipid bilayer. Using what you know about the two laws of thermodynamics, explain why the formation of this ordered structure occurs spontaneously? The second law of thermodynamics, states that every process increases entropy in the universe. Given ONE phospholipid, water is attracted to and hydrogen bonds with the hydrophilic head of a phospholipid and forms an ordered cage around the hydrophobic tails of the phospholipid. Therefore, in this situation water is ordered. When MANY phospholipids are present, a bilayer is formed with the hydrophobic tails all placed together and the hydrophilic heads exposed to the water. Although the phospholipids become ordered by the formation of a bilayer, the water becomes highly disordered. (b) ATP is hydrolyzed to form ADP and Pi. Draw the reaction schematically. What are the reactants? Speculate on why you think that ATP hydrolysis has a negative Gibbs Free Energy? ATP + H2O ADP + Pi + energy ΔG= -7.3 kcal reactants ATP hydrolysis has a negative Gibbs free energy because the phosphoester bond is broken, releasing energy ΔH<0. The reaction is also increasing entropy by creating two molecules from one ΔS>0. Also, the negative charges accompanying the phosphate groups repel each other and want to separate. 3

4 8) Match the terms on the left with the properties on the right. (there may be more than one answer) A. Competitive inhibitor 1 C Modifies an amino acid at the active site B. Noncompetitive inhibitor 2_B, D, C_ Prevents conversion of an enzyme to its active form. C. Irreversible Inhibitor 3 A, C Resembles the substrate in shape. D. Allosteric Inhibitor 4_B,C,D, E_ Functions at a location on the enzyme distant from the active site. E. Allosteric Activator 5 E Increases activity of the enzyme 9) Enzymes are remarkably specific. Below are two very related hormones. What mechanisms could be used by an enzyme to allow it to specifically recognize Cortisol and its complex ring structure, but preclude binding to Testosterone An enzyme is specific for one substrate because of the side chains on the amino acids that make up the active site. Both Cortisol and Testosterone could participate in Van der Waals attractions with the active site through interactions with the hydrophobic ring structure at #4. Also, potentially, They could both hydrogen bond with the active site amino acid side chains at #5. For an enzyme to specifically recognize Cortisol and not Testosterone, there would need to be polar amino acids present in the active site that would need to hydrogen bond at #1, #2, and #3. Testosterone is unable to hydrogen bond at these sites and would therefore be rejected by the enzyme. 4

5 10) A metabolic pathway converts A into B, then B into C, then C into D, and finally D into E. E is a molecule that is needed by the cell, and used for many purposes. (a) Draw this simple metabolic pathway. A B C D E (b) How many distinct enzymes would you expect to participate in this metabolic pathway? Four different enzymes would participate in this pathway, one for each step. (c) EBC (the enzyme that catalyzes B into C) is regulated in an allosteric fashion. Both an allosteric activator and an allosteric inhibitor of EBC exist among the pathway molecules. What could you guess about the possible identities of the positive and negative regulators, if you assume that they "make sense" in terms of controlling the pathway to provide more product when it is needed, and less product when it is not needed. Molecule A would be an allosteric activator and molecule E would be an allosteric inhibitor. When more product is needed molecule A is in excess and will speed up EBC by activating it to produce more of molecule E. When there is too much product, molecule E will act as an inhibitor to slow down EBC and less product will be made. (d) Plot a curve showing the affects the allosteric activator and the allosteric inhibitor on the reaction rate of enzyme, EBC. activator reaction rate normal inhibitor [substrate] 11) An enzyme, Cutase, cuts the dipeptide GLY-GLY, separating it into two amino acids. a) Cutase uses a hydrolysis reaction. Water is added to split the peptide bond between Gly and Gly. 5

6 b) The 2 nd Law of Thermodynamics states that all processes contribute to the increasing entropy of the universe. Cutase cuts the dipeptide Gly Gly into two amino acids and thereby increasing entropy. An increase in entropy done by Cutase will increase T S, which will make G < 0. c) A competitive inhibitor will have a similar shape as the substrate and will compete with the substrate for access to the binding sites. Molecule I is Alanine (Ala), which has a similar shape to amino acid, Glycine (Gly). Molecule I is most likely the competitive inhibitor. CH 3 NH 2 - C-COOH H Molecule I CH 3 -CH 3 -CH 2 -CH-CH 3 CH 3 Molecule II 12) Briefly described how a competitive inhibitor decreases activity. How can you distinguish it from a non-competitive inhibitor? Plot the rate vs. substrate curves for both in your answer. The Km (amount of substrate to reach half of Vmax(max rate)) rate will increase because adding a competitive inhibitor will compete with the substrate for binding to active sites. At higher substrate concentrations, reaction curve approaches the same end point as a reaction curve without a competitive inhibitor. See curve below on the left, line a). Adding large amounts of the competitive inhibitor will reduce the reaction curve even further. See curve below on the left, line b). Competitive Non-competitive 6

7 Adding a non-competitive inhibitor will bind reversibly to second site and the enzymes will change its 3-D conformation blocking its ability to bind the substrate. Km will BE UNCHANGED and Vmax will decrease. Note: although reversible both Vmax and Km are affected why? It is a non-competitive inhibitor and that also reflects the inability of the substrate to compete for the inhibitors binding site. Can a Non-competitive inhibitor bind irreversibly? 13) Why is hemoglobin (binds O 2 ) regulated by cooperative regulation? Draw the rate vs. substrate curve. Hemoglobin uses Cooperative regulation - a type of allosteric regulation where the binding of a substrate (in this case O2) to a subunit of the enzyme will result in a conformational change in the other subunits. The reaction curve for a cooperative enzyme would look sigmoidal. 14) For the reaction shown below, which form is oxidized and which form is reduced? Does this reaction occur in animals or in ferns? Explain. NADP+ + 2H + NADPH + H NADP + is oxidized, whereas NADPH gains electrons and is therefore reduced. NADPH is a coenzyme electron carrier in plants, so this reaction occurs in ferns, NOT animals 15) If the inner mitochondrial membrane was permeable to H +, how would it affect the production of ATP? A gradient of H+ could not be produced and therefore the efficiency of the ATP synthase to produce ATP would be severely diminished. Note the directionality of the H+ gradient is important that s why in the normal case H+ conc. increase in the inner mitochondrial space so that they can flow 7

8 down their conc. Gradient through the ATP synthase into the Matrix providing the energy to make the ATP key to Chemiosmosis! 16) ATP can be synthesized 2 ways in the cell. The first way is by substrate level phosphorylation. The second is by oxidative phosphorylation. (a) Describe the key differences and similarities between these two methods of ATP synthesis. Substrate level phosphorylation is when the enzyme kinase transfers a Pi from an high energy intermediate to ADP, creating ATP. Substrate level phosphorylation occurs in glycolysis and the Krebs cycle of respiration. Oxidative phosphorylation also creates ATP from ADP and Pi, however, in this case the energy comes originally from NADH, a product of glycolysis and the Krebs cycle. NADH is oxidized and the H+ and e- removed. The e- are delivered to the electron transport chain, which couples movement of the e- to pumping of the H+ across the inner mitochondrial membrane. This buildup of charged H+ in the intermembrane space provides the proton motive force for ATP synthase. ATP synthase couples movement of the H+ from high concentration in the intermembrane space to low concentration in the matrix. This movement releases energy, which is harvested by ATP synthase to join ADP and Pi to create ATP. (b) Define Chemiosmosis. What is a proton motive force and what does it have to do with chemiosmosis? Chemiosmosis is a process that couples H + transport to ATP synthesis. As mentioned above, the intermembrane space of mitochondria has a high concentration of H + that want to cross the membrane due to both high concentration and charges repelling each other, but they cannot because the membrane is IMPERMEABLE. ATP synthase uses the energy released by the H+ s movement across the membrane to create ATP. (c) Why do we need NADH? During respiration NADH functions as an electron carrier and the oxidation of NADH by oxygen releases 53 kcal of energy. NADH is produced during the glycolysis and Krebs cycle of respiration. NADH is oxidized to NAD+ and the freed electrons (e-) and protons (H+) are used as described above to make ATP. 8

9 17) There are two types of fermentation, one that occurs in muscle and other tissues and another that is seen commonly in yeast and bacteria. (a) How do these two processes differ? Fermentation follows glycolysis under anaerobic conditions. The process creates NAD + and allows glycolysis to continue making ATP in the absence of oxygen. Fermentation in muscles and other tissues produces lactic acid, while fermentation in yeast and bacteria produces ethanol. (b) What would be the consequence if organisms were unable to undergo fermentation and could only undergo respiration? An organism unable to undergo fermentation would die as there would be no more NAD+ to continue Glycolysis in the absence of ) In photosynthesis there are stages termed light reactions and dark reactions. (a) What are the products of the light reactions? The light reactions produce oxygen, NADPH, and ATP. (b) What are the products of the dark reactions? The dark reactions produce NADP, ADP, and sugar. (c) Why can t plants produce glucose in the dark? Plants cannot produce glucose in the dark because the dark reactions require the products of the light reactions (ATP and NADPH). (d) Which reactions involve chemiosmosis, light or dark? List two MAJOR differences between this chemiosmosis process when compared to that seen in respiration. The light reactions involve chemiosmosis. The electron carriers in respiration are NAD + and FAD, but in photosynthesis, NADP + carries the electrons for chemiosmosis. In photosynthesis, there are two different systems of electron flow, noncyclic and cyclic, but in respiration, there is only one. Chemiosmosis occurs across the inner membrane of the mitochondria in respiration and in photosynthesis, it takes place across the thylakoid membrane of the chloroplast. 9