1. First, keep in mind the first two laws of thermodynamics. One, the total energy of a system and its surroundings is constant. Two, the total entropy of a system plus its surroundings is always increasing. The fundamental thermodynamic relation can be described with the equation , where G is Gibbs free energy, H is enthalpy, S is entropy, and T is temperature (in kelvins). A reaction will be spontaneous if ΔG is less than 0, non-spontaneous if ΔG is greater than 0, and at equilibrium if ΔG is = 0. A reaction is favorable if ΔH is negative and/or ΔS is positive. When ΔH is negative and ΔS is positive, the reaction will be spontaneous. When ΔH is positive and ΔS is negative, the reaction will not be spontaneous. In any other situation, the result will depend on the values of H and S.

Sign of Entropy +H -H

+S

Depends on H and S values

Spontaneous

-S

Not spontaneous

Depends on H and S values

2. Several steps in glycolysis have positive ΔG’s, but the pathway still occurs spontaneously in

cells. The first student hypothesizes that this is due to enzymes lowering the free energy of the equation. However, the ΔG listed for the steps of glycolysis take the enzymes into account, and so the enzymes themselves cannot account for the spontaneity of the overall pathway. Student B thinks that food energy is the solution, and that eating will give the energy needed to drive glycolysis. However, glycolysis is the process through which the body breaks down glucose, which is how the body gains energy from starchy food, so eating more wouldn’t drive the reaction forward. It would instead put more glucose into the system, requiring even more glycolysis action. Student C is correct, reaction coupling will drive the entire pathway forward. Some steps in glycolysis have positive ΔG’s, but other steps have very negative ΔG’s. Those steps require the products of previous steps. Because the reactions are linked by the product of one reaction providing the substrates of the following reactions, the very negative ΔG’s will pull the pathway forward, even though the positive ΔG’s are earlier in the pathway than the very negative ΔG’s.

3. ΔG°, or standard free energy, is the ΔG of a reaction at “standard” conditions, when each reactant is at a concentration of 1.0 M. In the body, this is almost never the case. Further, reactions in the body are typically at near-equilibrium, when ΔG is very close to 0. ΔG° of a single reaction in the body is typically not very useful, as reactions are often coupled or regulated in some way such that something may appear to be non-spontaneous, but in the body it occurs spontaneously.

4. A) True. Reaction X, ΔG = -30.78 kJ. Reaction Y, ΔG = 22.5. Reaction Y is closer to equilibrium.

The sign of ΔG tells us the direction in which the reaction needs to proceed to reach equilibrium and the magnitude tells us the relative distance from equilibrium. If the magnitude is zero, ΔG° = 0, then we know the reaction is a equilibrium. B) False. ΔG°’ = 0 does not represent equilibrium. At equilibrium, because and at equilibrium, ΔG = 0 C) True. ΔG = 0 at equilibrium D) False. ΔG°’ is independent of the concentrations of reactants and products. It is a set standardized value taken from a table for the given reaction under standard conditions, which include 1M concentrations of each species. In other words, the ΔG°’ value is SPECIFIC to 1M concentrations of products and reactants (and other standard states such as temperature and pressure). If a reaction with a certain ΔG°’ value is NOT at standard conditions (say the ratios of products and reactants are quite different from 1M each) then the reaction has some ΔG value (Note: NOT ΔG°’) associated with those particular non-standard conditions. However, the ΔG°’ for that reaction would still be whatever is listed in the table. Some texts refer to ΔG as ΔGactual to emphasize that it is Gibbs Free Energy for a reaction at any actual set of conditions other than standard conditions. E) True. ΔG is not associated with standard conditions. ΔG can be calculated for a reaction at any point in time for any concentration of products and/or reactants. F) False. We have noticed a considerable number of students who have this misconception and found it repeated on a professor’s website which shows up in a Google search. The value ΔG°’ is NOT the free energy change for the completion of a reaction to its endpoint. Recall that almost no reaction ever converts all reactants to products because equilibrium is reached beforehand. Rather, the value ΔG°’ represents the free energy associated with proceeding from a standard state concentration (1M of each reactant) to equilibrium. For example, a negative value for ΔG°’ indicates that the reactants have more free energy than the products. Thus, running the reaction toward the products results in a lower energy state and is favored. However, as the reaction approaches equilibrium this driving force gradually disappears. At equilibrium, there is no driving force to move in either direction, and thus ΔG = 0 (Note: NOT ΔG°’).

5. At equilibrium, Q = Keq and ΔG = 0. Starting with ΔG = ΔG°’ + RTlnQ we have ΔG = ΔG°’ + RTlnKeq and then 0 = ΔG°’ + RTlnKeq. Subtracting RTlnKeq from both sides, we get ΔG°’ = -RTlnKeq

6. . ΔG ≠ ΔG°’, as explained in part B of Italicized Question #4. The one

exception is when Keq = exactly 1.0, as explained in part A of Italicized Question #7.

7. A) True. If Keq = 1 then ΔG° = 0, since ΔG° = -RTlnKeq, and the ln(1) = 0. However, do not confuse this with the fact that ΔG = 0 at equilibrium in every case. For all other cases, ΔG° does NOT equal zero at equilibrium (but ΔG does). ΔG° = 0 at equilibrium ONLY if Keq for that reaction = exactly 1.0. In this rare case, ΔG = ΔG°. This is proven by the equation: ΔG = ΔG°’ + RTlnKeq. If Keq = 1, then the second term is zero and ΔG = ΔG°. B) False. ΔG = ΔG° + RTln(Q), so ΔG = -RTln(Keq) + RTln(Q), and if Keq = 1 then ΔG° = 0 and ΔG = RTln(Q), which is not 0. C) True. If Keq = Q, the reaction is at equilibrium, and ΔG = 0.

D) False. The reaction is at equilibrium, but ΔG° is not 0 at equilibrium, as discussed in Italicized Question #4. E) False. ΔG = ΔG° + RTln(Q), and if Q is equal to 1 then ΔG = ΔG° because ln(1)=0 F) False. ΔG = ΔG° + RTln(Q), so ΔG = -RTln(Keq) + RTln(Q), and if Keq = 1 then ΔG° = 0 and ΔG = RTln(Q), which are not equal. G) True. ΔG = ΔG° + RTln(Q). Since Q = 1 and Ln(1) = 0, ΔG = ΔG°. H) True. ΔG° = -RTlnKeq

I) False. ΔG = ΔG° + RTln(Q), and various concentrations of reactants and products (Q) can affect ΔG, regardless of what the Keq and ΔG° values are.

8.

FIGURE 1

9. The phosphoanhydride bonds that connect the phosphate groups are what make ATP an

effective energy store for the cell. This type of bond is highly energetic. At physiological pH, the phosphate groups lose their protons and become negatively charged. The negative charges repel one another, and with three charged phosphate groups, there is a much greater repulsion occurring. Also, ADP and Pi are much more stable than ATP, due to less repulsion and resonance stabilization, so the ΔG of converting ATP to ADP and Pi is very negative.

10. In the reaction, GADP is the reduced form, NAD+ is the oxidized form 1,3 BPG is the oxidized form, and NADH is the reduced form. Thus NAD+ is being reduced (it is an oxidizing agent) and GAPD is being oxidized (it is a reducing agent). NAD+ + H+ + 2e- → NADH GADP + Pi → 1,3-BPG + 2e- + H+ Together, those equal NAD+ + Pi + GADP → NADH + 1,3-BPG

11. Respiration is a process in which an in organic compound serves as the ultimate electron acceptor in order to generate ATP. Aerobic respiration uses oxygen as the final electron acceptor, while anaerobic respiration uses a molecule other than oxygen. For question purposes, aerobic respiration involves all the reactions involved in the citric acid cycle and electron transport. Anaerobic respiration will typically refer to fermentation, using glycolysis in the absence of oxygen, or the lactic acid cycle in muscles. Humans use aerobic respiration to generate the vast majority of our ATP. However, we use anaerobic respiration in our muscles during exercise that results in a buildup of lactic acid. Many bacteria and yeast use anaerobic respiration, including during fermentation.

12. The term “obligate” implies that there is no other option, so obligate aerobes must use aerobic respiration and cannot survive without oxygen, while obligate anaerobes must use anaerobic respiration and cannot survive in the presence of oxygen. “Facultative” implies that the organism will use whichever respiration is available. So if oxygen is present, the organism will use aerobic respiration, and if oxygen is absent, the organism will use anaerobic respiration. Facultative anaerobes prefer anaerobic conditions while facultative aerobes prefer aerobic.

13.

Figure 2

FIGURE 2

14. All the radiolabeled molecules will be dihydroxyacetone phosphate. This is because when fructose 1,6-bisphosphate is cleaved, it is cleaved the same way every time. Carbons 1, 2, and 3 become dihydroxyacetone phosphate, and carbons 4, 5, and 6 become glyceraldehyde 3-phosphate. If carbon 2 is labeled, the labeled molecule will always be DHAP. If the isomerase is active, some of the DHAP will be converted to GAP, so the label will be distributed between the two. But without the isomerase, all labeled molecules will be DHAP.

15. Mature erythrocytes lack cellular organelles, they are basically sacks of hemoglobin. This means they also lack mitochondria, the normal location of the citric acid cycle, ETC, and oxidative phosphorylation. Red blood cells use fermentation—even in the presence of oxygen—because they lack the cellular machinery found in other cells and therefore fermentation is their only route to produce energy.

16. During glycolysis, the transfer of phosphate groups is done through substrate level phosphorylation. Substrate level phosphorylation is the transfer of a phosphate group from a phosphorylated substrate (such as phosphenolpyruvate) to ADP to make ATP. Oxidative phosphorylation occurs in the electron transport chain. ATP synthase, or complex 5, catalyzes the conversion of ADP and Pi into ATP. This process uses electron transfer to build up a proton gradient to drive the function of ATP synthase. A basic gluconeogenesis chart showing the level of detail required for MCAT-2015 is shown on the following page. The enzymes and cofactors in red are different in gluconeogenesis than in glycolysis. Those in black are common between the two.

FIGURE 3

17. The four enzymes specific to gluconeogenesis replace three glycolytic enzymes which all

catalyze irreversible reactions. Those three steps that are replaced are all phosphorylation reactions.

18. For each glucose 6-phosphate, two NADPH molecules are generated. Both come from the first reaction, glucose 6-phosphate converting to ribulose 5-phosphate. The pentose phosphate pathway does not directly generate glutathione molecules. However, the NADPH generated in the PPP can be used to reduce the oxidized form of glutathione to protect the cell from reactive oxygen species. To regenerate glutathione, the reaction is 1 glutathione disulfide + NADPH → 2 glutathione + NADP+. This means that for each glucose 6-phosphate, four glutathiones are generated, because two NADPHs are generated.

19. The Citric Acid Cycle is shown below. Acetyl-CoA is the end-product of glycolysis after passing through the PDH complex.

FIGURE 4

20. Substrate level phosphorylation is the process by which a phosphate group is transferred to ADP or GDP from a phosphorylated intermediate. The Citric Acid Cycle uses substrate level phosphorylation. Oxidative phosphorylation occurs in ATP synthase in the electron transport chain. This process combines ADP and inorganic phosphate to generate ATP through the generation of a proton gradient by transporting electrons.

21. The Electron Transport Chain is shown.

Figure 5

FIGURE 5

22. Complex I pumps 4 protons. Complex II pumps 0 protons (because it does not traverse the membrane, it can’t pump protons). Complex III pumps 4 protons. Complex IV pumps 2 protons. Utilizing electrons from an NADH molecule will result in 10 protons pumped because the electrons go through Complex I, III, and IV (So 4 + 4 + 2). Utilizing electrons from FADH2 will result in 6 protons pumped because the electrons go through Complex II, III, and IV (so 0 + 4 + 2). To generate one ATP molecule, 3 protons are needed. This means that every NADH generated will result in 3 ATPs, and every FADH2 will generate 2 ATPs.

23. When NADH is oxidized, the electrons flow through Complexes I, III, and IV. When FADH2 is oxidized, the electrons flow through Complexes II, III, and IV. Because Complex I pumps 4 electrons and Complex II pumps 0 electrons, oxidation of FADH2 results in fewer protons being pumped. ATP synthase requires the proton gradient to function, and so fewer protons pumped will result in fewer ATPs being generated.

24. The overall reaction for glycolysis is:

glucose + 2 NAD+ + 2Pi + 2ADP 2 pyruvate + 2ATP + 2NADH + 2H+ The overall reaction for the citric acid cycle (including pyruvate dehydrogenase) is:

pyruvate + 4 NAD+ + FAD + GDP + Pi + 2 H20 3 CO2 + 4NADH + 4H+ + GTP + FADH2 Keep in mind that you’ll need to double the citric acid cycle equation because glycolysis results in two pyruvates. This means that from complete oxidation of one glucose, we get:

NADH FADH2 ATP

Glycolysis 2 (4 ATPs) 0 (0 ATPs) 2

Citric Acid Cycle 8 (24 ATPs) 2 (4 ATPs) 2 (GTP)

Total 10 (28 ATPs) 2 (4 ATPs) 4

NADH = 3 ATP FADH2 = 2 ATP NADH from glycolysis = 2 ATP (because it costs 1 ATP to transport it in)

Total, we have (2 x 2) + (8 x 3) + (2 x 2) + 4 = 36 ATPs

NOTE: Some Biochemistry Textbooks state that NADH yields 2.5 ATP and FADH2 , 1.5 ATP. In this case, a total of 30 ATPs are produced. Students should know that either answer may be correct.

25. A decrease in the free energy released by a reaction will, in general, decrease the rate of the reaction. In this case, decreasing the free energy released as a result of translocating a proton through ATP synthase will A) decrease ATP production, because the equilibrium will be shifted such that the phosphorylation of ADP to ATP will happen at a slower rate. B) Electron transport chain function will decrease as well because there won’t be as many protons to pump out by Complexes I, III, and IV if they aren’t being pumped in at the same rate by ATP synthase. ETC function will not decrease as much or as quickly as ATP production because the enzymes involved are not directly affected by the free energy decrease. C) The electrochemical gradient will increase slightly because protons will be pumped out by Complexes I, III, and IV, but ATP synthase won’t be pumping them back in as quickly. D) The citric acid cycle will still function normally because it does not depend on ATP synthase.

26. A) In glycolysis, three reactions are regulated. Keep in mind that glycolysis is needed when energy in the cell is low, so regulation will be such that enzymes are activated by a lack of energy and inhibited when energy is abundant. Phosphofructokinase is inhibited by high levels of ATP and is the main source of regulation for glycolysis. AMP reverses the inhibition of ATP, and so the ratio of ATP to AMP is crucial to determining the activity of glycolysis. Glycolysis is stimulated when energy available to the cell falls, and it is reduced when energy increases. Hexokinase is another way to regulate glycolysis. It is inhibited by glucose 6-phosphate, its product. Because glucose and glycogen are both converted to glucose 6-phosphate, when this molecule is at high concentration, hexokinase is inhibited because the cell has plenty of access to energy. Glucose will stay at higher concentrations in the blood or be converted to glycogen for storage. The third regulated enzyme is pyruvate kinase. ATP inhibits pyruvate kinase to slow glycolysis. Alanine also inhibits pyruvate kinase because pyruvate is used as a building block for amino acids, and a high concentration of alanine signals that building blocks aren’t needed.

B) Gluconeogenesis is needed when energy levels are high and glucose levels are low, the opposite of when glycolysis is needed. As such, the enzymes are regulated oppositely to those in glycolysis such that when one pathway is activated, the other is being actively inhibited, and vice versa. Fructose 1,6-bisphosphatase is inhibited by AMP and stimulated by ATP, exactly opposite of its glycolysis counterpart, phosphofructokinase. Both pyruvate carboxylase and phosphoenolpyruvate are inhibited by ADP. These enzymes are the counterpart to pyruvate kinase in glycolysis. C) Much as glycolysis and gluconeogenesis are reciprocally regulated, glycogenolysis and glycogen synthesis are regulated in opposite ways. Glycogenolysis, the breakdown of glycogen into glucose 6-phosphate via glucose 1-phosphate, is stimulated by the presence of glucagon and epinephrine in the bloodstream. In both cases, the hormones stimulate a cAMP cascade, which ultimately activates protein kinase A. This phosphorylates phosphorylase kinase, which activates glycogen phosphorylase. Glycogenolysis is shut down when the stimulating hormones are gone from the blood stream. Without the stimulating hormones, the cAMP cascade is withdrawn, and protein phosphatase I dephosphorylates glycogen phosphorylase, rendering it inactive.

D) Glycogenesis, the synthesis of glygogen, is regulated oppositely to glycogenolysis. When glucagon and epinephrine are present in the bloodstream, the cAMP cascade is stimulated, as discussed in part C. Protein kinase A phosphorylates glycogen synthase, but this phosphorylation inhibits the enzyme (instead of activating, as it does with glycogen phosphorylase). When the cAMP cascade is withdrawn, protein phosphorylase I will dephosphorylate glycogen synthase, stimulating glycogen synthesis. This way, the same stimulus will simultaneously shut down one pathway and turn on the other.

E) The citric acid cycle is tightly regulated, because without regulation, large amounts of NADH and ATP could be wasted. The pyruvate dehydrogenase complex, the entry point to the CAC, is inhibited by its products, acetyl CoA and NADH. Pyruvate dehydrogenase is also inhibited by phosphorylation stimulated by high levels of NADH and ATP. When NADH and ATP are low, pyruvate dehydrogenase is dephosphorylated and activated. The citric acid cycle itself is regulated at many points. Isocitrate dehydrogenase is stimulated by ADP and inhibited by ATP and NADH. α-ketoglutarate dehydrogenase is inhibited by its products, succinyl CoA and NADH, and is also inhibited by ATP. F) The electron transport chain is stimulated by high levels of ADP and inhibited by high levels of ATP. Additionally, there are many poisons that inhibit ETC. Rotenone, an insecticide, blocks electron transfer in NADH-Q oxidoreductase. Cyanide, azide, and carbon monoxide block electron flow in cytochrome c oxidase. Oligomycin, an antifungal agent, prevents proton flow through ATP synthase by binding to the oligomycin sensitive subunit. G) The pentose phosphate pathway is primarily controlled by the levels of NADP+. The first enzyme, glucose 6-phosphate dehydrogenase, is inhibited by NADP+ because NADP+ is needed as the electron acceptor for the reaction.

27.

FIGURE 6

28.

FIGURE 7

29. To determine how many rounds of β-oxidation is required to oxidize an even numbered fatty

acid, simply divide the number of carbons by 2 and subtract 1. So for a 14-carbon fatty acid, 6 rounds of β-oxidation are needed. This is because every round cleaves 2 carbons off. At the end, a final round cleaves the 4-carbon fatty acid into two 2-carbon fatty acids, finishing the oxidation.

For an odd numbered fatty acid, subtract 1 to get to an even number. Then divide by 2, subtract 1. So for a 17-carbon fatty acid, 7 rounds of β-oxidation are needed (17-1=16. 16/2=8. 8-1=7). This is because every round cleaves off two carbons, but at the end of an odd numbered fatty acid, the final round cleaves the 5-carbon fatty acid into one 2-carbon fatty acid and one 3-carbon fatty acid.

If you’re unsure how many rounds are needed, simply draw it out quickly, marking through the fatty acid for every cleavage. In this diagram, every red line is a round of β-oxidation.

30. A ketogenic amino acid is degraded into acetyl CoA or acetoacetyl CoA (ketone bodies) through ketogenesis. The carbons of ketogenic amino acids are ultimately converted to CO2 in the citric acid cycle (because acetyl CoA carbons are converted to CO2). Leucine and lysine are both ketogenic. Glucogenic amino acids can be converted to glucose through gluconeogenesis. They are converted first to alpha keto acids and then to glucose in the liver. Some amino acids are both ketogenic and glucogenic: isoleucine, phenylalanine, tryptophan, tyrosine, and threonine are both. If you can remember those 5 and the 2 that are ketogenic, all the rest are glucogenic.

31. Ketogenesis is the process by which ketone bodies are produced through the breakdown of

fatty acids. This occurs during periods of starvation when blood glucose levels drop and no further source of carbohydrate fuel is available. Ketogenesis is able to provide energy by generating acetyl CoA to be fed into the citric acid cycle. This occurs in the liver. Ketolysis is the utilization of ketone bodies by converting them to acetyl CoA for energy. This occurs in organs other than the liver (mainly the heart and brain). The liver is lacking an essential enzyme for the utilization of ketone bodies for energy. When blood glucose is low, β-oxidation and ketogenesis occur in the liver. Ketone bodies are transported out of the liver to key tissues, where they can be used for energy through ketolysis. The brain (and CNS) in particular relies on ketone bodies when glucose is not abundant. Most other tissues and organs can use fatty acids for energy when glucose is low, but the CNS relies on glucose primarily, and ketone bodies during periods of starvation, for fuel.

32. a) cytosol, b) mitochondrial matrix, c) cytosol, d) mitochondrial matrix, e) cytosol, f) cytosol—primarily of liver cells, and in kidney cortex cells to a lesser degree g) mitochondrial matrix of liver cells/some kidney cells (NOTE: First, pyruvate carboxylase converts oxaloacetate to pyruvate in the matrix, gluconeogenesiss then continues in the cytosol), h) cytosol—primarily of liver cells, i) mitochondrial matrix, j) inner mitochondrial membrane (part of the ETC), k) inner mitochondrial membrane, l) inner mitochondrial membrane, m) mitochondrial matrix (fatty acids are activated in the cytosol, and very long chain fatty acids are oxidized first in peroxisomes, so you might find traces of radioactivity there, but the question asks for where it will be in abundance, which would be in the matrix), n) mitochondrial matrix, o) mitochondrial matrix of cells throughout the body during fasting or starvation, but NEVER In the liver—it lacks the necessary enzymes (ketogenesis occurs only in the matrix of liver cells, because the enzymes needed are only found in the matrix), p) cytosol—primarily of liver cells and to some extent in the kidneys (NOTE: This is ONLY true because we specified argininosuccinate, which is the product of a cytosolic enzyme that participates in the urea cycle. Of the five enzymes in the urea cycle, two are mitochondrial and three are cytosolic), q) mitochondrial matrix (from carbamoyl phosphate synthetase 1, one of the two mitochondrial urea cycle enzymes), r) mitochondrial matrix (product of the second mitochondrial urea cycle enzyme) AND the cytosol—citrulline is transported from the matrix to the cytosol as part of the urea cycle, s) BOTH the mitochondrial matrix and the cytosol (ornithine is transported from the cytosol to the mitochondrial matrix as part of the urea cycle), t) BOTH the cytosol and the matrix (malate is part of the TCA cycle in the matrix and participates in the Citrate shuttle in the cytosol).

33. In general, most metabolic substrates are able to move freely or be transported around the cell. This makes sense if you think about how many pathways are linked, which wouldn’t be possible if substrates were primarily held in one organelle or another. Enzymes, however, are often

“trapped” in one location. Consider the enzymes of the electron transport chain. All of them are membrane proteins, which means that they are always embedded in the inner mitochondrial membrane. Also trapped within the mitochondrial membrane are the cofactors of the ETC, such as ubiquinone and cytochrome c. Glycolysis occurs in the cytoplasm, and as such, the enzymes for glycolysis are located in the cytoplasm. The citric acid cycle occurs in the matrix of the mitochondria, where the enzymes are “trapped.” The products of glycolysis and the citric acid cycle can be transported across the membrane and are not trapped.

34. a) 9 kcal/g, b) 4 kcal/g, c) 4 kcal/g.

35. See the diagram of the Citrate shuttle below:

FIGURE 8

FIGURE CREDITS

1. Source: http://en.wikipedia.org/wiki/Adenosine_triphosphate#mediaviewer/File:Adenosintriphosphat_protoniert.svg http://en.wikipedia.org/wiki/Adenosine_diphosphate#mediaviewer/File:Adenosindiphosphat_protoniert.svg http://en.wikipedia.org/wiki/Adenosine_monophosphate#mediaviewer/File:Adenosinmonophosphat_protoniert.svg http://en.wikipedia.org/wiki/Cyclic_adenosine_monophosphate#mediaviewer/File:Cyclic-adenosine-monophosphate-2D-skeletal.png http://en.wikipedia.org/wiki/Phosphate#mediaviewer/File:Phosphat-Ion.svg

2. Source: Altius image 3. Source: Altius image 4. Source: Altius image 5. Source:

http://en.wikipedia.org/wiki/Oxidative_phosphorylation#mediaviewer/File:Mitochondrial_electron_transport_chain%E2%80%94Etc4.svg

6. Source : Altius image 7. Source: Altius image 8. Source: Altius image