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Biochemistry exam 3

by: Grace Gerhards

Biochemistry exam 3 BC 351

Grace Gerhards
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Biochemisty is not easy. This study guide will make your life easier! Great notes, easy grades.
Paul Laybourn PH.D.
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Biology, Math, Science
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This 26 page Study Guide was uploaded by Grace Gerhards on Thursday July 21, 2016. The Study Guide belongs to BC 351 at Colorado State University taught by Paul Laybourn PH.D. in Summer 2016. Since its upload, it has received 17 views. For similar materials see Biochemistry in Biochemistry & Molecular Biology at Colorado State University.

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Date Created: 07/21/16
Biochemistry Exam 3 study guide lectures 29-43 1. Lipid anchored proteins are covalently attached to membrane lipids or hydrophobic molecules (fatty acids or prenyl groups) stuck into the lipid bilayer, thus anchoring them to the membrane. Integral membrane proteins have hydrophobic regions/surfaces that interact well with the hydrophobic core of the lipid by layer alloying them to insert into or more often span the membrane. Peripheral membrane proteins associate with the membrane through noncovalent interactions with lipid anchored or integral membrane proteins or with membrane lipids, thus they can be removed with high salt, urea, etc. washed. 2. A hydrophobic α-helix is composed solely of amino acids whose R groups are hydrophobic. Thus, its entire external surface is hydrophobic. In contrast, an amphipathic α- helix contains a specific pattern of hydrophobic and polar amino acids so that one face of the α-helix is hydrophobic and the other is polar. Hydrophobic α-helices are found in transmembrane proteins (often single-pass), in which the entire α-helix is interacting with the membrane. This type of protein has extramembrane portions that stick out from the inner and outer sides. These portions are hydrophilic, at least on the surface. Therefore, these are amphipathic proteins with hydrophobic α-helices. Multipass transmembrane proteins forming channels have amphipathic α-helices, with the side facing the membrane lipids being hydrophobic and the side forming a portion of the inner wall of the channel being hydrophilic. 3. In an isolated membrane-spanning domain, it is essential for the entire surface of the α- helix to contain hydrophobic residues to optimize its interactions with the hydrophobic side chains of the lipid molecules. This is accomplished with a hydrophobic α-helix. 4. Multiple amphipathic α-helixes can orient to form an external surface that is entirely hydrophobic and an internal surface that is polar. The former can interact effectively with the hydrophobic side chains of the lipid molecules while the latter can create a polar pore or channel across the membrane. Amphipathic α-helix-based pores can form a greater range of pore sizes and have far greater flexibility for conformational changes, etc. than β- barrelbased pores. 5. Non-ionic detergents (mild) form micelles on their own in aqueous solutions due to amphipathic structures and single non polar tails. When added to membranes (lipid bilayers) from cells the can form end-capping micelles on the ends of lipid bilayer fragments and around the hydrophobic portions (for example helices) and shield them from interaction with water or other polar molecules. This dissipates the driving hydrophobic forces “dissolving” the membrane proteins out of the membrane and allows the integral membrane proteins to remain in their native structures outside of the lipid bilayer hydrophobic environment. The solubilized membrane proteins can then be separated from each other and the membrane fragments by various standard isolation methods. Please see tutorial 11C for diagrams depicting this process. 6. Passive transport proceeds down a concentration gradient with the release of free energy. In contrast, active transport proceeds against a concentration gradient and requires the input of energy. Primary active transport uses an energy source, like ATP or light, directly. In addition, primary active transport often functions to create an electrochemical gradient. Secondary active transport uses an electrochemical gradient created by a primary active transporter. 7. A symport system transports two solutes in the same direction while an antiport system transports two solutes in the opposite direction; i.e., it acts as an exchange system. 8. The Na+/K+ ATPase maintains a large electrochemical gradient of Na+ ions across the plasma membrane. Thus, a symport system that transports Na+ ions and a second solute (in the same direction) will derive energy from the movement of Na+ ions down an electrochemical gradient. This system can be used to accomplish the active accumulation of the second solute against a concentration gradient. The negative ΔG of the flow of Na+ ions down the gradient must be more negative than the positive ΔG of the pumping of the second solute, for example glucose, against the concentration gradient is positive. In other words, the net ΔG of the two coupled reactions must be negative. 9. The E1 form of the Na+ pump has the Na+ and K+ binding sites facing the cytoplasm and has a high affinity for Na+ and a low affinity for K+. When E1 is bound by Na+ in the cytoplasm, this triggers its self-phosphorylation using ATP. The phosphorylated E1 changes conformation (phosphorylation is an allosteric regulator) to become the E2 form. The E2 form has the binding sites facing the exterior of the cell and has a low affinity for Na+ so these ions are released outside the cell and has a high affinity for K+ so it binds K+ ions. K+ binding triggers dephosphorylation, which leads to an allosteric BC 351: Principles of Biochemistry Page 3 of 5 change back to the E1 form. The E1 form has the binding sites facing the cytoplasm and has a low affinity for K+ so these ions are released and a high affinity for Na+ starting the cycle over. t10. (B) The binding of Na+ ions allosterically converts the yransporter to a form with ahigh affinity for glucose so that it binds glucose even at the lower concentration found outside the cell. Glucose binding causes a second conformational change so that the binding sites are now facing the cytoplasm and the lower Na+ concentration, leading to its release and a second conformation change to a form with a very low affinity for glucose. This allows glucose release at the relatively high concentration in the cytoplasm. Thus, Na+ ions flowing down their concentration gradient can lead to and power conformational changes that result in pumping of glucose against a concentration gradient. (C) The overall process to go must and does have an over all negative free energy change. The negative free energy change of two Na+ ions flowing across the plasma membrane down the electrochemical gradient is more negative than the positive free energy change of transporting one glucose molecule into the cell against the concentration gradient. (D) Probably not, since the conformational changes that occur do not include changes in the affinity of the transporter for Na+. Otherwise, if they did, any enzyme should theoretically be able to catalyze the forward or reverse reaction and which reaction catalyzes is only dictated by the overall free energy change. 11. (A), (B), (C) and (D) 12. The Warburg Effect or aerobic glycolysis refers to cells exhibiting increased glucose uptake and lactate production in cancer cells. This has been determined to be due to a high rate of glycolysis, converting glucose to pyruvate. However, pyruvate conversion to acetyl CoA is inhibited and its conversion to lactate by lactate dehydrogenase is up regulated. Since Warburg’s group first described this phenomenon cancer cells have also been shown to be highly dependent on glutamine as a precursor for nitrogen containing compounds and for TCA cycle anaplerotic (filling up) functions. This is the basis for PET scans with radiolabeled glucose for detecting tumors in patients. It has only recently begun to be exploited as an “Achilles heel” for anticancer therapies. 13. The two examples of “normal” aerobic glycolysis are cell in early mammalian embryos up to the 16-cell stage and in brain cells. Early embryonic cells are dividing rapidly, suggesting that aerobic glycolysis may be an attribute of rapidly dividing cells and perhaps providing an important function useful to cancer cells. In the brain, astrocytes and neuron cells form a commensal metabolic relationship. Astrocytes carry out aerobic glycolysis producing lactate. Nerve cells take up the lactate, convert it back to pyruvate and oxidize it through the TCA cycle and produce ATP by oxidative phosphorylation. Astrocytes also function to take up glutamate and convert it back to glutamine, which nerve cells release as a neurotransmitter. Glutamate is toxic to nerve cells at high levels and glutamine is an important nutrient. 14. The metabolic enzymes and their inhibitors tested for their potential as anti cancer drugs: 15. Targeting of metabolic changes in cancer cells with metabolic enzyme inhibitors, etc. like those described in the table above is promising for anticancer therapies. They may be particularly useful in combination with drugs exploiting other changes found in cancer cells. However, just like other changes in cancer cells, metabolic changes are often specific to cancer types (lung, skin, pancreatic, colon, etc.) and specific cancers among these types. Therefore, it will be important to develop means to determine the metabolic profiles typical of cancer types and for specific cancers to effectively target them through their metabolic changes. Enzyme Metabolic reaction catalyzed Inhibitor Hexokinase 2 (and Glyceraldehyde 3P Dehydrogenase) Steps 1 and 6 of glycolysis 3-bromopyruvate Phosphofructokinase-1 through Phosphofructokinase-2 production of F2,6 bisP levels Step 3 of glycolysis 3PO nPyruvate Kinase Step 10 of glycolysis Testing inhibitors Lactate Dehydrogenase NAD+ regeneration through conversion of pyruvate to lactate FX11 Pyruvate Dehydrogenase Kinase 1 Phosphorylates pyruvate dehydrogenase and inhibits it, blocking pyruvate conversion to Acetyl CoA for oxidation in the TCA cycle Dichloroacetate (DCA) Glutaminase Converts glutamine to glutamate, which is converted to the TCA cycle intermediate alpha ketoglutarate 968 and BPTES Isocitrate dehydrogenase Converts isocitrate to alpha ketoglutarate, step 3 of the TCA cycle Some cancers have mutations in this enzyme, which make them more sensitive to glutaminase inhibitors Glutamate Dehydrogenase and Glutamate Pyruvate Transaminase Convert glutamate to alpha ketoglutarate EGCG (in green tea) and AOA BC 351: Principles of Biochemistry Page 5 of 5 16. The three main types of metabolic pathways are catabolic (oxidation), anabolic (reduction) and amphibolic. Catabolic pathways couple the oxidation of nutrients to the production of ATP. Anabolic pathways couple to hydrolysis of ATP to the biosynthesis of macromolecules. Amphibolic pathways can do both. The catabolic pathways are connected with the anabolic pathways by way of ATP and NADH, which catabolic pathways produce and regenerate, for the anabolic pathways to use. 17. The intrinsic property of ATP hydrolysis is the biochemical standard free energy change that is quite negative primarily due to the neutral pH in cells. The system property is the [ATP]/[ADP] ratio in cells being at least 10/1, causing K = to be at lease 0.1. The two properties combine to give ATP hydrolysis a very negative change in free energy in cells. This negative free energy change can then be coupled to unfavorable reactions in the cell, like biosynthesis, to make the overall reaction still have a negative free energy change. The other two “energy currency” molecules in the cell are NADH and Acetyl Co Enzyme A. 18. The energy can be captured in the formation of multiple energy currency molecules rather than a large release of heat energy. 19. Entropy is a measure of disorder. Thus, if there is an increase in order there is a decrease in entropy. The greater the entropy of a system, the smaller its free energy. Thus, an increase in entropy during a reaction will result in a decrease in free energy. 20. ΔG'° = –RT ln Keq'. If Keq' is a large (positive) number, the term –RT ln Keq' (and therefore ΔG'°) has a relatively large, negative value. 21. ΔG'° is the difference in free energies of the products and reactants starting at standard conditions and going to equilibrium. I could be thought of as the difference between the intrinsic G of the reactants and the G of the products. Thus, it is a physical constant, characteristic of and intrinsic to each chemical reaction. ΔG is a variable that depends on ΔG'°, the temperature, and the concentrations of all reactants and products: ΔG = ΔG'° + RT ln [product] [reactant] Question 1 0.5 / 0.5 pts An integral membrane protein can be extracted with:    a buffer of alkaline or acid pH.      a chelating agent that removes divalent cations.   Correct!    a solution containing detergent.      a solution of high ionic strength.      hot water.     Question 2 0.5 / 0.5 pts Amphipathic α­helical structures    are composed of only hydrophobic amino acids.   Correct!    can form a hydrophilic pore within a lipid bilayer.     are positioned in the membrane with the polar amino acid R­groups in contact with the fatty acid tails of the lipid bilayer.    are destabilized by the hydrophobic environment of the lipid bilayer.     Question 3 0.5 / 0.5 pts Which of these statements is generally true of integral membrane proteins? Correct!   The secondary structure in the transmembrane region consists solely of α­helices or β­sheets.   The domains that protrude on the cytoplasmic face of the plasma membrane nearly always have  covalently attached oligosaccharides.    They are unusually susceptible to degradation by trypsin.     They can be removed from the membrane with high salt or mild denaturing agents.   They undergo constant rotational motion that moves a given domain from the outer face of a  membrane to the inner face and then back to the outer.   Question 4 0.5 / 0.5 pts A hydropathy plot indicates    the amino acids forming an amphipathic β­barrel.      why membranes are self sealing.   Correct!   a stretch of amino acids forming a single­pass transmembrane domain.     regions of proteins consisting of multipass transmembrane α­helices forming a pore or channel.   Question 5 0 / 0.5 pts Sodium (Na+) transport across a membrane    has linear kinetics in response to increased Na+ concentration.   You Answered    often uses secondary active transport.   Correct Answer    uses ~30% of the ATP hydrolyzed in mammalian cells.      has no role in glucose transport in intestinal epithelial cells.     Question 6 0.5 / 0.5 pts Which of these statements about facilitated diffusion across a membrane is true?  Correct!   A specific membrane protein lowers the activation energy for movement of the solute through the  membrane.   It can increase the size of a transmembrane concentration gradient of the diffusing solute.   It is impeded by the solubility of the transported solute in the nonpolar interior of the lipid bilayer.   It is responsible for the transport of gases such as O2, N2, and CH4 across biological membranes.    The rate is not saturable by the transported substrate.     Question 7 0.5 / 0.5 pts Aerobic glycolysis or the Warburg Effect    is only seen in cancer cells.    produces acetyl CoA.   is characterized by higher than normal pyruvate dehydrogenase activity.    provides cancer cells the ability to live on glucose alone. Correct!    can be visualized through PET scans.   Question 8 0 / 0.5 pts Adenosine 5'­triphosphate (ATP) You Answered    is usually produced by anabolic pathways.     has a biochemical standard free energy (ΔG˚’) of hydrolysis of approximately +2.4 kcal/mol. Correct Answer   can be used as a cosubstrate by enzymes to drive unfavorable reactions.    is primarily used for long term energy storage in cells.     Question 9 0.5 / 0.5 pts Biological oxidation­reduction reactions always involve:    direct participation of oxygen.      formation of water.      mitochondria.   Correct!    transfer of electron(s).      transfer of hydrogens.     Question 10 0.5 / 0.5 pts The coenzyme NAD+    functions in reduction reactions.     donates electrons directly to molecular oxygen in the electron transport chain. Correct!    carries only one hydride anion (1 H+ and 2 e­s).      is worth approximately 5 ATP equivalents.   MODULE 10 1. A reaction for which ΔG'° is positive can proceed under conditions in which ΔG is negative. From the relationship ΔG = ΔG'° + RT ln [product] [reactant] it is clear that if the concentration of product is kept very low (by its subsequent metabolic removal, for instance), the logarithmic term becomes negative and ΔG can then have a negative value. 2. Reaction 3 is the sum of reaction 1 and the reversal of reaction 2. Because of the additivity of free energy changes, the overall ΔG'° for reaction 3 is the sum of the free energy changes for reaction 1 and the reversal of reaction 2: 1. ATP à ADP + Pi ΔG1'° = –30.5 kJ/mol 2. glucose + Pi à glucose 6-phosphate ΔG2'° = +13.8 kJ/mol 3. ATP + glucose à ADP + glucose 6-phosphate ΔG3'° = ΔG1'° + ΔG2'° = (–30.5 + 13.8) kJ/mol = –16.7 kJ/mol 3. Hydrolysis of the ATP would result in the loss of most of the free energy as heat. In a transfer reaction, the gamma (third) phosphate of ATP is transferred to the reaction substrate to produce a high-energy phosphorylated intermediate, which can then form the product in an exergonic reaction. 4. Oxidation is the loss of electrons; reduction is the gain of electrons. Free electrons are unstable (do not occur), so whenever an electron is released by oxidation of some species, an electron must be accepted by reduction of another species. 5. Negative. ΔG'° = –n ℑ ΔE'°. 6. ΔE'° = E'° (electron acceptor) + E'° (electron donor) = – 0.32 + 0.29 = - 0.03 V ΔG'° = –n ℑΔE'° = (–2)(96.48 kJ/V·mol)(–0.03 V) ΔG'° = +5.8 kJ/mol 7. (a) Co+; (b) glucose; (c) Fe2+; (d) acetate; (e) ethanol; (f) acetaldehyde. 8. The glycolytic pathway is so central to all of cellular metabolism that mutations in glycolytic enzymes are lethal; embryos with such mutations would not survive. Question 1 0 / 0.5 pts Metabolism involves only the oxidation of nutrients. consists of pathways that are either catabolic or anabolic, but not both. You Answered is made up of pathways of reactions coupled with 100% efficiency to production of energy carriers . generally is short pathways of reactions, some of which are catalyzed by enzymes. Correct Answer consists of metabolic pathways that are linear, cyclic and spiral. Question 2 0 / 0.5 pts Anabolic and catabolic pathways are related by You Answered the flow of energy between catabolic and anabolic pathways being reversible. the energy derived from catabolic pathways being used to drive the breakdown of organic molecules in anabolic pathways. the degradation of organic molecules by anabolic pathways providing the energy drive catabolic pathways. Correct Answer anabolic pathways synthesizing more complex organic molecules using the energy derived from catabolic pathways. catabolic pathways producing usable cellular energy by synthesizing more complex organic molecules. Question 3 0.5 / 0.5 pts Energy requiring metabolic pathways that yield complex molecules from simpler precursors are: amphibolic. Correct! anabolic. autotrophic. catabolic. heterotrophic. Question 4 0.5 / 0.5 pts Life is thermodynamically possible because living cells reproduce themselves. increase the degree of order in the universe. carry out energetically favorable reactions alone. Correct! release heat to the environment. Question 5 0.5 / 0.5 pts The standard free-energy changes for the reactions below are given. Phosphocreatine --> creatine + Pi ∆G'° = –43.0 kJ/mol ATP −−> ADP + Pi ∆G'° = –30.5 kJ/mol What is the overall ∆G'° for the following reaction? creatine + ATP −−> Phosphocreatine + ADP –73.5 kJ/mol –12.5 kJ/mol Correct! +12.5 kJ/mol +73.5 kJ/mol ∆G'° cannot be calculated without Keq' Question 6 0.5 / 0.5 pts Adenosine 5'-triphosphate (ATP) is usually produced in an anabolic pathway. has a biochemical standard free energy of hydrolysis (\DeltaG˚’) of approximately +2.4 kcal/mol. Correct! can be used as a cosubstrate to drive unfavorable reactions. is primarily used for long term energy storage in cells. Question 7 0.5 / 0.5 pts Regarding ATP, the hydrolysis of ATP supplies the energy needed for catabolic pathways. in cells, almost all of the free energy released by the hydrolysis of ATP is released as heat. Correct! the cycling between ATP and ADP + Pi provides an energy coupling between catabolic and anabolic pathways. None of the choices is correct. the energy released by hydrolysis of ATP is the result of breaking a high-energy bond. Question 8 0.5 / 0.5 pts The standard reduction potentials (E'°) for the following half reactions are given. Pyruvate- + 2H+ + 2e–→ lactate- E'° = -0.185 V NAD+ + H+ + 2e–→ NADH E'° = –0.320 V If you mixed pyruvate, lactate, NAD+, and NADH together, all at l M concentrations and in the presence of lactate dehydrogenase, which of the following would happen initially? pyruvate and lactate would become oxidized; NAD+ and NADH would become reduced. Correct! pyruvate would become reduced, NADH would become oxidized. No reaction would occur because all reactants and products are already at their standard concentrations. Lactate would become oxidized, NAD+ would become reduced. Lactate would become oxidized, NADH would be unchanged because it is a cofactor. Question 9 0.5 / 0.5 pts Which of the following is not true for the nicotinamide cofactors? The oxidized form is positively charged. The reduced form has a large extinction coefficient at 340 nm. Correct! The oxidized form provides reducing equivalents to other molecules. Oxidation-reduction reactions with nicotinamides usually involve hydride transfer. Enzymes transfer hydrides stereospecifically to one or the other side of the nicotinamide ring. Question 10 0.5 / 0.5 pts NADH reduces O2 in the first stage of oxidative phosphorylation to produce NAD+ and H2O. From table 13-7 (p. 515) and following the method of the worked example 13-3 (p 516) calculate the change in standard reduction potential (E) and change in free energy for the reduction of 1/2 mol of O2 by one mole of NADH. If synthesis of one mol of ATP from ADP and Pi has a free energy change of around +12 kcal per mol or +50.2 KJ per mol, what is the theoretical maximum number of ATP that could be synthesized by coupling the reaction to the reduction of 1/2 mol of O2 by one mol of NADH? 1 2 Correct! 4 8 12 Question 11 0.5 / 0.5 pts The conversion of 1 mol of fructose 1,6-bisphosphate to 2 mol of pyruvate by the glycolytic pathway results in a net formation of 1 mol of NAD+ and 2 mol of ATP. 1 mol of NADH and 1 mol of ATP. 2 mol of NAD+ and 4 mol of ATP. 2 mol of NADH and 2 mol of ATP. Correct! 2 mol of NADH and 4 mol of ATP. Question 12 0.5 / 0.5 pts When a mixture of glucose 6-phosphate and fructose 6-phosphate is incubated with the enzyme phosphohexose isomerase, the final mixture contains twice as much glucose 6-phosphate as fructose 6-phosphate. Which one of the following statements is most nearly correct, when applied to the reaction below (R = 8.315 J/mol·K and T = 298 K)? Glucose 6-phosphate ↔ fructose 6-phosphate Correct! ΔG'° is +1.7 kJ/mol. ΔG'° is –1.7 kJ/mol. ΔG'° is incalculably large and negative. ΔG'° is incalculably large and positive. ΔG'° is zero. 1. 2. There are at least three reasons (see p 547 or 531 for details). (a) So they will not cross the plasma membrane and be lost. (b) The resulting high-energy phosphoryl groups can be transferred directly to ADP to make ATP. (c) The high binding energy of the ionic interactions between the (-) charged phosphoryl groups and (+) charged enzyme side chains helps to stabilize the enzyme-substrate interaction. 3. 1; 4; 2; 6; 5; 7; 3; a. phosphofructokinase-1; b. pyruvate kinase; c. glyceraldehyde 3-phosphate dehydrogenase. 4. Cells often drive a thermodynamically unfavorable reaction in the forward direction by coupling it to a highly exergonic reaction through a common intermediate. In this example, to make glucose 6-phosphate formation thermodynamically favorable, cells transfer phosphoryl groups from ATP to glucose. ATP “hydrolysis” is highly exergonic, making the overall reaction exergonic. (Numerical solution below not required.) Glucose + Pi à glucose 6-phosphate + H2OΔG'° = +13.8 kJ/mol ATP + H2O à ADP + Pi ΔG'° = –30.5 kJ/mol –––––––––––––––––––––––––––––––––––––––––––––––––––––––––– Sum: ATP + glucose à ADP + glucose 6-phosphate ΔG'° = –16.7 kJ/mol See p. 498; see also Chapter 14. 5. Inorganic phosphate (Pi) is an essential substrate in the reaction catalyzed by glyceraldehyde 3-phosphate dehydrogenase. Pi OH C Pi H H C H C OH O ADP ATP C Pi H H C H C OH O 1,3-BisP-Glycerate 3-P-Glycerate Step 7 ΔG˚’ = +7.3 ΔG˚’ = -11.8 Step 7 ΔG˚’ = -4.5 Overall ΔG˚’ = -3.0 See also Fig. 2-76 in the PowerPoint slides. H Glyceraldehyde-3-P GA-3-P DHase NAD+ NADH + H+ Pi S Enz C Pi H H C H C OH O C Pi H H C H C OH O Step 6 ΔG˚’ = +1.5 PGlycerate Kinase BC 351: Principles of Biochemistry Page 2 of 4 6. Under aerobic conditions, pyruvate is oxidized to acetyl-CoA and passes through the citric acid cycle. Under anaerobic conditions, pyruvate is reduced to lactate to recycle NADH to NAD+, allowing the continuation of glycolysis. 7. Step 6 of glycolysis requires NAD+. Under aerobic conditions NAD+ is regenerated by donation of electrons to the ETC. Under anaerobic conditions it must be regenerated through fermentation. In muscle cells NADH is converted to NAD+ by lactate dehydrogenase reduction of pyruvate to lactate. Lactate produced in the muscle cells is carried through the blood to liver cells where it is converted to glucose by gluconeogenesis. The glucose is then transported by the blood back to the muscles for use in glycolysis. 8. Anaerobic glycolysis yields only 2 ATPs and 2 lactates per molecule of glucose. In contrast, aerobic catabolism of glucose completely oxidizes it to CO2 and H2O and yields 38 ATPs. Thus, the yield of "biologically useful energy" is 19 times greater. 9. The dark reactions of photosynthesis require 12 NADPHs and 18 ATPs. Twelve NADPHs are equivalent to 36 ATPs (assuming ~3 ATP/NADPH). Thirty-six plus 18 ATPs used directly totals 54 ATP equivalents to produce one glucose molecule in the light independent reactions. In contrast, the complete (aerobic) oxidation of glucose generates 38 ATP equivalents. Thus, greater "biologically useful energy" is consumed in the biosynthesis of glucose than is generated during its catabolism. The difference between yield and input reflects the need to put additional energy into the anabolic pathways in order to make the all of the individual reactions and the overall process thermodynamically favorable. In addition, a level of inefficiency is required to allow the generation of heat, which is required for cells to obey the second law of thermodynamics. 10. The Warburg Effect is seen in cancer cells and is defined as the high level of anaerobic-like glycolysis (glucose to lactate) in the presence of sufficient oxygen. 11. Gluconeogenesis is the biosynthesis of glucose from simpler, noncarbohydrate precursors such as oxaloacetate or pyruvate. During periods of fasting, when carbohydrate reserves have been exhausted, gluconeogenesis provides glucose for metabolism in tissues (brain, erythrocytes) that derive their energy primarily from glucose metabolism. 12. Pyruvate is converted into phosphoenolpyruvate in two steps: (1) Pyruvate + HCO3 − + ATP à oxaloacetate + ADP + Pi (2) Oxaloacetate + GTP à CO2 + GDP + phosphoenolpyruvate BC 351: Principles of Biochemistry Page 3 of 4 The first reaction is catalyzed by pyruvate carboxylase, which requires biotin as a cofactor; the second, by phosphoenolpyruvate carboxykinase. [See Fig. 14-17, p. 554]. 13. Multimeric homopolymers of an enzyme can respond cooperatively to binding by a ligand through and all-or-none shift in conformation of the subunits. The ligand will have a higher affinity for the “R” state, stabilizing that conformation. Once one subunit reverts to the “R” state its neighbors are influenced to change to the “R” state, as well. This results in cooperative binding. Hexokinase is also stabilized in the “R” state by ATP or glucose binding, increasing the affinity of the enzyme for the second remaining substrate (glucose or ATP, respectively) and resulting in cooperativity. Generally, as seen when comparing myoglobin a monomeric protein with hemoglobin a four-subunit protein, the allostery or shape change resulting in cooperative binding requires multisubunit enzyme complexes. Myoglobin does go through a change in shape with oxygen binding, but it cannot result in cooperative binding of subsequent oxygens as it has only one subunit. 14. 15. Homoallostery is a change in shape leading in response to substrate binding (binding in the active site) leading to cooperative binding up subsequent substrate molecules and sigmoidal kinetics. Heteroallostery is a change in shape in response to non-substrate ligand binding. ATCase is an example of heteroallostery because the two regulator ligands, ATP and CTP bind elsewhere from the active site (on the regulatory subunits) and stabilize the “R” and “T” states, respectively. BC 351: Principles of Biochemistry Page 4 of 4 16. Heteroallosteric ligands can have positive or negative linkage with the substrate and the ligand can act as an activator or an inhibitor, respectively. This means that the ligands stabilize the same “R” state as the substrate (positive linkage) or they stabilize the “T” state (negative linkage), which the substrate does not bind and stabilize. A second type or meaning for linkage is the connection or influence one subunit of a multisubunit enzyme has on the shape (conformation) of its neighboring subunits. 17. Enzyme regulatory mechanisms other than ligand (small molecule binding) include posttranslational (covalent) modification. This is predominantly phosphorylation, but includes acetylation, methylation, ubiquitinylation, etc. Enzymes are also regulated through proteolytic cleavage (cutting off part of them), binding by other proteins (larger ligands), and GTP/GDP binding. Finally, they can be regulated by localization into specific sub- compartments in the cell. It is important to note that many enzymes are regulated by combinations of ligands, protein binding, and post-translational modifications. A great example is cyclin-dependent kinases, which you will learn about in cell biology courses. 18. The enzyme would show kinetics that do not fit the Michaelis-Menten equation; the plot of V vs. [S] would be sigmoidal, not hyperbolic. The enzyme kinetics would be affected by molecules other than the substrate(s).


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Refund Policy


All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.