Biochem Only

Southern blot

A Southern blot is exactly like a northern blot except it involves DNA instead of RNA. After using gel electrophoresis to separate pieces of DNA, the DNA is still double-stranded. We want to use a probe like the one used in Northern blot, but we'll first need to denature the double-stranded DNA into single-stranded DNA. That way, a probe with a complementary DNA sequence can actually bind. The rest of our steps are exactly the same!

A bond formed between which of the following amino acids is disrupted in reducing SDS-PAGE gel electrophoresis? A) Histidine B) Cysteine C) Leucine D) Methionine

B - Cysteine forms disulfide bonds, which are broken in reducing SDS-PAGE gel electrophoresis (choice B is correct; choices A, C, and D are incorrect).

Where does fatty acid synthesis take place in the cell? A) Mitochondria B) Ribosomes C) Cytoplasm D) Nucleus

C - Although certain preliminary steps to fatty acid synthesis take place in the mitochondria, such as acetyl-CoA generation, the actual biosynthesis of palmitic acid occurs in the cytoplasm (choice A is incorrect). Ribosomes are the site of protein synthesis (choice B is incorrect). The nucleus stores DNA and is the site of DNA replication and transcription (choice D is incorrect).

Assuming that the pea of the side chain of histidine is approx 6, which of the following is the closest to the isoelectric point of this amino acid? A) 4 B) 5.7 C) 7.5 D) 9

C - To calculate PI, you must consider the pI values of the amino acid in question. Histidine has a carboxylic acid group (pKa ~ 2) and an amino group (pKa ~ 9), and its side chain (pKa ~ 6), you only average the most most basic peas because histidine is a basic amino acid. (6 +9)/2

Allosteric Enzyme

The passage states that in wild-type Tcd proteins, IP6 binding induces a conformational change that moves the flap region away from the active site. This change is an example of allostery, a phenomenon in which the binding of an allosteric effector (eg, IP6) at an allosteric site (eg, the binding pocket) alters the activity at the catalytic site. Allosteric regulation occurs due to conformational changes induced in the protein upon binding of the effector and can include changes in the conformation of the active site itself, obstruction of the active site or, as in this case, exposure of the active site when the flap region moves. Unlike wild-type Tcd, autoprocessing occurs in H757A and H759A mutants even in the absence of IP6. This behavior supports the conclusion that H757 and H759 mediate an interaction necessary for the flap region to obstruct the active site in the absence of IP6. Conversion of histidine to alanine abolishes this interaction, so the flap region cannot fully block the active site in mutant proteins, resulting in a loss of allosteric control.

Sandwich ELISA

1. Antibody fixing: the antibody of interest is immobilized on a surface 2, Antigen binding: a liquid containing the antigen is washed over the immobilized antibody 3. Secondary antibody binding: a secondary antibody that is specific to the primary body is introduced 4. Reporting: the reporter enzyme conjugated to the secondary antibody creates a color change if binding occurs If we get a large signal or a strong color change (high color saturation), we have a high concentration of bound antibody-reporter, which also indicates a high concentration of antigen in our original sample. If we get a small signal or a weak color change, our original sample had a low concentration of antigen.

Step 6: G3P dehydrogenase

G3P dehydrogenase catalyzes the reversible conversion of glyceraldehyde 3-phosphate into 1,3-bisphosphoglycerate, which generates one molecule NADH. However, one molecule of glucose (a 6-carbon structure) generates 2 molecules of glyceraldehyde 3-phosphate—so this step yields two molecules of NADH per glucose molecule.

Direct ELISA

1. Antigen fixing: the antigen of interest is immobilized on a surface 2. Primary antibody binding: a liquid containing the primary antibody is washed over the immobilized antigen 3. Reporting: the reporter enzyme conjugated to the primary antibody creates a color change if binding occurs

Indirect ELISA

1. Antigen fixing: the antigen of interest is immobilized on a surface 2. Primary antibody binding: a liquid containing the primary antibody is washed over the immobilized antigen 3. Secondary antibody binding: a secondary antibody that is specific to the primary antibody is introduced 4. Reporting: the reporter enzyme conjugated to the secondary antibody creates a color change if binding occurs

The intermediates of which metabolic pathway were shown to increase in muscles due to exercise? A) The citric acid cycle B) Fermentation C) Glycolysis D) Glycogenolysis

A - A metabolic pathway consists of a specified set of molecules, called intermediates, and the enzymes that catalyze the conversion of one intermediate to the next in the pathway. The citric acid cycle is the set of reactions that converts citrate to oxaloacetate, which then reacts with acetyl-CoA to regenerate citrate. The final two steps in the cycle prior to formation of oxaloacetate are the conversion of succinate to fumarate, and conversion of fumarate to malate. Figure 1 shows that the levels of both fumarate and malate are significantly increased in muscles after exercise. Therefore, in muscle, exercise causes an increase in the levels of intermediates of the citric acid cycle. (Choice B) Fermentation is the reduction of pyruvate to generate NAD+ for continued glycolysis under anaerobic conditions. In mammals, fermentation is carried out by the conversion of pyruvate to lactate. Figure 1 shows that lactate levels increase in the hippocampus after exercise, but are depleted in muscles. (Choice C) Glycolysis is the conversion of glucose to pyruvate through several intermediates. Figure 1 shows that the glycolysis intermediate G6P is depleted in both muscles and the hippocampus after exercise. (Choice D) Glycogenolysis is the degradation of glycogen for use in other metabolic pathways. Figure 1 shows that glycogen is depleted in both muscles and the hippocampus after exercise.

Amino acid catabolism releases nitrogen in the form of ammonia. In the liver, the urea cycle prepares ammonia for excretion. Which amino acid could undergo deamidation to produce ammonia for the urea cycle? A) Glutamine B) Arginine C) Glycine D) Aspartate

A - Amino acids all have a common backbone structure that consists of an amino group bonded to a carbon atom (the α-carbon), which is bonded to a carboxylic acid. The α-carbon is also bonded to a side chain, called an R-group, which gives the amino acid its identity. Amino acid catabolism (degradation) involves the removal of nitrogen from backbones and certain R-groups in the form of ammonia, which is detoxified and prepared for excretion by the urea cycle: 1. The nitrogen of amino acid backbones is removed by transamination and deamination (removal of an amine to produce a ketone), producing α-keto acids in the process. 2. The side chains of asparagine and glutamine contain amides, which are catabolized by deamidation rather than deamination. Deamidation releases ammonia from an amide instead of an amine, producing a carboxylic acid instead of a ketone. Of the choices given, only glutamine can produce ammonia through deamidation. (Choices B, C, and D) Arginine, glycine, and aspartate each contain an amino group in their backbones, but they do not have any amido groups in their R-groups. Therefore, these amino acids cannot undergo deamidation.

Which type of inhibitor binds only to the enzyme-substrate complex, and how does this form of inhibition affect the kinetics of the enzyme in question? A) Uncompetitive; both Km and Vmax are reduced B) Competitive; Km is reduced, while Vmax is not changed C) Noncompetitive; Vmax is reduced, while Km is not changed D) Uncompetitive; Vmax is reduced, while Km is not changed

A - An uncompetitive inhibitor is an inhibitor of the enzyme-substrate complex. This reduces the ability for the enzyme to catalyze the reaction, lowering Vmax. Simultaneously, it seemingly increases the enzymes affinity for its substrate, as ES complexes are removed via inhibition. Thus, uncompetitive inhibition also reduces Km.

Researchers isolate a protein that runs as a single band on a nonreducing SDS-PAGE gel but as two distinct bands on a reducing gel. The protein is most likely a: A) Heteromultimer that formed in the oxidizing environment of the endoplasmic reticulum B) Heteromultimer that formed in the reducing environment of the cytosol C) Homomultimer that formed in the oxidizing environment of the endoplasmic reticulum D) Homomultimer that formed in the reducing environment of the cytosol

A - Multimers are typically bound together by noncovalent interactions but may also have covalent bonds between cysteine residues called disulfide bonds. Disulfide bonds form when two cysteine residues are oxidized, and therefore the presence of disulfide bonds requires an oxidizing environment. These bonds can be broken by reducing agents such as β-mercaptoethanol or dithiothreitol. The question states that the protein forms one band in a normal (nonreducing) SDS-PAGE gel but in a reducing gel, it forms two bands. The formation of two distinct bands when the multimer is broken suggests that the protein contained nonidentical subunits, making it a heteromultimer. The fact that a reducing gel is required to break the multimer indicates that disulfide bonds are present. They most likely formed in the oxidizing environment of the endoplasmic reticulum. (Choices B and D) In reducing environments such as the cytosol, proteins are unlikely to form disulfide bonds. Therefore, such proteins are unlikely to be affected by a reducing gel, as they are already reduced. (Choice C) A homomultimer consists of two or more identical subunits. The individual subunits will run faster on a gel than the intact multimer would, but they will not run differently from each other. Therefore, only one band would form on a reducing gel if the protein were a homomultimer.

Which of the following is not used in PCR? A) ddNTPs B) Heat-resistant DNA polymerase C) Double-stranded DNA D) Two primers

A - PCR is used to amplify a double-stranded piece of DNA (choice C is incorrect). ddNTPs, which have an H instead of an OH at the C'3 carbon, prevent DNA polymerase from adding additional nucleotides (choice A is correct). A heat-resistant DNA polymerase is needed to synthesize new DNA (choice B is incorrect). Two primers—one complementary to the 3' end of each DNA strand—are needed (choice D is incorrect).

What is the fate of the propionyl-CoA molecule generated by the oxidation of a saturated fatty acid with an odd number of carbon atoms? A) It undergoes a series of reactions and is ultimately converted to succinyl-CoA B) It undergoes a series of reactions and is ultimately converted to acetyl-CoA C) It is marked for destruction by ubiquitin and is degraded by a proteasome D) It is shuttled to the cytoplasm where it is degraded by a lysosome

A - Recall that in the oxidation of a saturated fatty acid with an even number of carbon atoms, nearly all of the carbon atoms are converted into acetyl-CoA. With an odd-numbered fatty acid, the bulk of the fatty acid is also converted into acetyl-CoA. However, the last 3 remaining carbon atoms on the chain cannot be easily converted into the 2-carbon acetyl-CoA molecule. Instead, the remaining 3-carbon group will undergo a series of reactions to become succinyl-CoA, which can be used in the citric acid cycle (choice A is correct).

Thermolysin is a protease that can cleave peptides at the N-terminal side of branched alkyl residues. Which of the following peptides has the most thermolysin cleavage sites? A) HQICSVGPKLNT B) DNKGSTVYQRWC C) GHPTLDQIWCER D) RTHCQGYNESDK

A - The amino acid residues in a peptide can be classified by their general properties. Proteases cleave peptide bonds at specific sites by recognizing residues with specific properties. They may cleave the bond on the N-terminal side of the residue (the side that is closest to the N-terminus of the peptide), or they may cleave the bond on the C-terminal side of the recognized residue (the side that is closest to the C-terminus). For example, trypsin cleaves bonds on the C-terminal side of positively charged residues (R, K), and chymotrypsin cleaves bonds on the C-terminal side of large aromatic residues (F, Y, W). The question states that thermolysin can cleave peptide bonds at the N-terminal side of branched alkyl residues. Leucine (L), isoleucine (I), and valine (V) are the only amino acids with branched alkyl side chains, and the peptide with the most L, I, and V residues will have the most cleavage sites. The peptide HQICSVGPKLNT has three thermolysin cleavage sites, one each of L, I, and V, and more than any of the other choices given.

If DoRA aggregation is caused by charge neutralization, then aggregates should form as pH: A) Increases from 7 to 9 B) Decreases from 7 to 5 C) Increases from 9 to 11 D) Decreases from 9 to 7

A - The isoelectric point (pI) of a protein is the pH at which the net charge of the protein is neutral. If the ambient pH is lower than a protein's pI, the protein gains protons and becomes positively charged. If the pH exceeds the pI, the protein loses protons and becomes negatively charged. The pI can be determined by isoelectric focusing, in which proteins migrate through a pH gradient until they become neutrally charged and are no longer influenced by an electric field. The pH at which a protein stops migrating is its pI. The scientists in the passage proposed that reflectins aggregate as their net charges are neutralized. If this is correct, then DoRA should have the greatest tendency to aggregate when the ambient pH is equal to its pI. Figure 2 shows that the pI of DoRA is approximately 9, so DoRA aggregates should form as the pH increases from 7 to 9. (Choice B) Decreasing the pH from 7 to 5 would move the ambient pH far below the pI of 9 and increase the positive charge on DoRA. As a result, the tendency of DoRA to aggregate would be greatly reduced. (Choices C and D) DoRA should have the greatest tendency to aggregate at a pH of 9, where it has a neutral charge. Moving the pH away from 9, either by increasing it or decreasing it, would reduce the tendency of DoRA to aggregate.

Titration Curves

Acidic and basic chemical groups, also known as ionizable groups, have intrinsic tendencies to exchange protons with water. This tendency is described by the pKa, or the pH at which half of the ionizable groups of interest are protonated. pKa values can be determined by titration, which is typically performed by gradually adding a base to an acidic solution of the molecule of interest. As the pH of the solution approaches the pKa of a chemical group, the system acts as a buffer and resists changes in pH by accepting protons from water or donating protons to it. On a titration curve, the region near a pKa (called the buffering region) is relatively flat because the addition of base changes the pH by only a small amount. All amino acids have ionizable carboxyl and amino groups with pKa values near 2 and 9.5, respectively. Therefore, all amino acid titration curves have two buffering regions, one near each pKa. Seven of the standard amino acids (R, K, Y, C, H, E, and D) also have ionizable side chains, each with a unique pKa. The titration curve for each of these amino acids has a third buffering region associated with the side chain pKa. Because the titration curve in the question shows only two buffering regions, the amino acid has only two pKa values. Therefore, its side chain is not ionizable and cannot exchange protons with water.

Affinity chromatography

Affinity chromatography separates proteins based on their specific noncovalent-binding interactions with antibodies or other ligands attached to a column. Negatively charged amino acids such as aspartate and glutamate interact with positively charged amino acids such as arginine, histidine, and lysine to form ionic bonds.

Protein Metabolism

Amino acids are classified as glucogenic or ketogenic depending on the metabolic intermediate to which they are converted. Glucogenic amino acids are converted to pyruvate or citric acid cycle intermediates, which can be converted to glucose. Ketogenic amino acids are converted directly to acetyl-CoA, which can enter the citric acid cycle or be used to form ketone bodies. The passage uses this classification to place amino acids into the three categories in Table 1. Amino acids in Category I are converted exclusively to pyruvate, and those in Category II become acetyl-CoA. Category III consists of amino acids used to make various citric acid cycle intermediates. Glycine and cysteine are Category I amino acids, which are converted directly into pyruvate. Patients with ME at rest have sufficient ATP to convert pyruvate into citric acid cycle intermediates (via oxaloacetate) during aerobic respiration. However, the shortage of oxygen caused by prolonged physical exertion inhibits aerobic respiration by preventing NAD+and FAD, which are used by the citric acid cycle, from being regenerated. To maintain ATP production, cells switch from aerobic to anaerobic respiration, which does not require oxygen and consists of glycolysis and fermentation. Glycolysis contributes two molecules of ATP by breaking down glucose into two molecules of pyruvate. During fermentation, the NAD+ required for glycolysis is generated by reducing pyruvate to lactate with the enzyme lactate dehydrogenase. Therefore, glycine and cysteine are ultimately converted to lactate during periods of physical exertion.

Antibodies

Antibodies are proteins that function in the immune system to recognize foreign particles known as antigens. Specifically, antibodies mark foreign antigens for destruction (eg, phagocytosis) by other immune cells. Each antibody contains a variable region that binds to a specific chemical structure within an antigen, known as an epitope. Typically, the epitope is a sequence of amino acids within a protein or peptide. Antibodies bind their epitopes through noncovalent interactions such as hydrogen bonding and electrostatic attractions. The passage states that cell lysates were incubated with α-flag, an antibody that binds the flag peptide. Therefore, flag was the epitope, and it was this portion of flag-IRS4 to which the antibody noncovalently bound. Because the antibody and flag-IRS4 are both proteins, the binding interaction can be classified as a noncovalent protein-protein interaction.

The metabolism of which biomolecule is least likely to directly increase aerobic respiration in the cell? A) Proline B) Cytosine C) A short chain fatty acid D) Glycogen

B - Aerobic respiration in the cell consists the CAC, which is fed by glycolysis and the ETC. The pyrimidines cytosine and uracil can be metabolized, but doing so they are converted into beta alanine, and later to malonyl-CoA, which is needed for fatty acid synthesis. A) If glycogen is exhausted, glucose can be made from some amino acids via gluconeogensis; proline is a gluconeogenic amino acid. C) Fatty acids such as 3-ketoacyl-CoA can be broken down via beta oxidation to produce two carbon units which are fed into the CAC via acetyl CoA D_ Glycogen can be broken down into glucose, which can directly enter glycolysis.

Which parameter would NOT be affected by the addition of phosphoglycerate mutase? A) The stability of the transition state between 3-phosphoglycerate and 2-phosphoglycerate B) The equilibrium concentrations of 2-phosphoglycerate and 3-phosphoglycerate C) The activation energy barrier between 3-phosphoglycerate and 2-phosphoglycerate D) The rate constant for conversion of 2-phosphglycerate to 3-phosphoglycerate

B - Catalysts increase reaction rates by facilitating formation of a more stable transition state between reactants and products. This enhanced stability decreases the activation energy Ea necessary for the molecules to react, allowing them to reach the energy threshold more often at any given temperature. As a result, the reaction rate increases in both the forward and reverse directions, resulting in an increased rate constant as calculated by the Arrhenius equation (Choices A, C, and D). Catalysts only stabilize the transition state of a chemical reaction. They do not stabilize the reactants or products themselves, and therefore the difference in energy between them (ΔG°) is not affected. The equilibrium constant of a reaction Keq is the ratio of products to reactants at equilibrium, and is related to the standard free energy change by the equation ΔG° = −RT ln(Keq). Because ΔG° does not change in the presence of a catalyst, the equilibrium concentrations of products and reactants also do not change. Therefore, the addition of phosphoglycerate mutase does not change the equilibrium concentrations of 2-PG or 3-PG.

SGLT proteins found in the plasma membrane of the same cells import glucose against its concentration from the tubular filtrate into the cell while exporting sodium down its concentration gradient. This is an example of which form of solute transport? A) Primary active transport B) Symport C) Secondary active transport D) Facilitated diffusion

C - Secondary active transport requires the input of energy to transport molecules across a membrane, but unlike in primary active transport, there is no direct coupling of ATP. The electrochemical potential difference created by pumping ions from the cell via primary active transport, thereby establishing a concentration gradient, allowing the molecule to flow down its concentration gradient.

An enzyme is subjected to an inhibitor with an unknown mechanism. Upon graphing the Lineweaver-Burk plots of the inhibited and uninhibited enzymes, researchers notice that the two plots share an x-intercept. The unknown inhibitor is most likely: A) Competitive B) Noncompetitive C) Uncompetitive D) Either noncompetitive or uncompetitive

B - On a Lineweaver Burk plot, the x-intercept represents -1/Km, if the inhibited and uninhibited plots have the same x-intercept, then they must have the same Km. Noncompetitive inhibitors do not alter Km, so the unknown inhibitor is most likely noncompetitive. Competitive inhibitors increase Km, while uncompetitive inhibitors decrease it.

Which of the following DNA sequences is most likely to be recognized by a restriction endonuclease? A) 5'-AACCAA-3' B) 5'-ACATGT-3' C) 5'-ATATCGC-3' D) 5'-GACGAC-3'

B - Restriction endonucleases recognize palindromic sequences of double-stranded DNA, generally between 4 and 8 bases long. A palindromic sequence of DNA reads the same way from 5' to 3' on both strands of DNA. You need to determine which one of these sequences has a complementary sequence like this. While A looks like a "palindrome," it is not a palindromic sequence of DNA. The complementary sequence read 5' to 3' is read TTGGTT. The complementary sequence of B is read TGTACA from 3' to 5', but it is read ACATGT from 5' to 3'.

Which of the following molecules could decrease the activity of one or more Krebs cycle enzymes? A) Glucagon B) NADH C) Acetyl-CoA D) ADP

B - The Krebs cycle, also known as the citric acid cycle, is tightly regulated to produce ATP at the optimal rate for the needs of the cell. The cycle includes three irreversible steps, catalyzed by the enzymes citrate synthase, isocitrate dehydrogenase, and α-ketoglutarate dehydrogenase. The activity of each of these enzymes can be increased or decreased when specific molecules bind to the enzymes and alter their shape (allostery). Each of these molecules serves as an indicator of the energy state of the cell. Molecules that are associated with depleted energy (eg, ADP) and muscle contractions (eg, calcium) indicate a need for additional ATP production. These molecules allosterically activate Krebs cycle enzymes to meet the demand for energy. Conversely, molecules that indicate that the Krebs cycle is producing sufficient energy allosterically inhibit the enzymes. These molecules include several products of the Krebs cycle and the electron transport chain such as ATP, citrate, succinyl-CoA, and NADH. NADH decreases the activity of all three of the regulated Krebs cycle enzymes. (Choice A) Glucagon is a hormone that upregulates glucose production in the liver through glycogen breakdown (glycogenolysis) and gluconeogenesis. The enzymes of the citric acid cycle are not under hormonal control. (Choice C) Acetyl-CoA reacts with oxaloacetate to form citrate at the beginning of the Krebs cycle. Its presence increases the overall rate of the cycle by increasing substrate concentration. (Choice D) ADP is produced when ATP is hydrolyzed, and it signals to the cell that energy has been depleted. It allosterically activates the Krebs cycle enzymes citrate synthase and isocitrate dehydrogenase, thereby increasing their activity.

What is the rate-limiting step of fatty acid synthesis? A) Polymerization via fatty acid synthase B) Acetyl-CoA shuttling via the citrate shuttle C) Reduction of NADP+ during the reaction catalyzed by malic enzyme D) Conversion of acetyl-CoA into malonyl-CoA via acetyl-CoA carboxylase

D - Acetyl-CoA carboxylase is an enzyme that catalyzes the rate-limiting step of fatty acid synthesis (choice D is correct). Fatty acid synthase and malic enzyme are both enzymes involved in fatty acid synthesis but are not rate-limiting (choices A and C are incorrect). Similarly, the shuttling of acetyl-CoA is a step in fatty acid synthesis but is also not rate-limiting (choice B is incorrect).

An infant who experiences seizures and has lactic acid buildup is shown to have deficient pyruvate dehydrogenase complex (PDHC) activity. Genetic analysis shows that both the maternal and paternal alleles for all PDHC subunits are normal. These results could indicate: A) Overexpression of one or more PDHC subunits B) Impaired ability to synthesize lipoic acid C) Upregulated acetyl-CoA entry into the Krebs cycle D) An unusually low rate of pyruvate fermentation

B - The pyruvate dehydrogenase complex (PDHC) is an enzyme composed of three subunits (E1, E2, and E3). It catalyzes the oxidative decarboxylation of pyruvate to form acetyl-CoA, along with the reduction of NAD+ to form NADH. During this process, electrons are passed from one subunit in the complex to the next until they can be transferred to NAD+. This transfer is facilitated in part by the cofactor lipoic acid. Cofactors are non-amino acid groups of proteins that are necessary for the protein to function. Because they are not amino acids, cofactors are not directly encoded by DNA. The question states that PDHC has decreased activity, yet the alleles (DNA) encoding it are normal. Therefore, some component of PDHC activity that is not directly encoded in the DNA must be altered. Because lipoic acid is a necessary cofactor of the PDHC, impaired lipoic acid production would result in decreased activity of the PDHC complex. (Choice A) The question states that alleles for all PDHC subunits are normal. Underexpression, not overexpression, could account for decreased activity of the normal proteins. (Choice C) The PDHC synthesizes acetyl-CoA, which enters the Krebs cycle (also known as the citric acid cycle). A deficient PDHC would decrease acetyl-CoA concentration and downregulate its entry into the Krebs cycle. (Choice D) Lactic acid is exclusively produced by pyruvate fermentation, so the fact that the infant experienced lactic acid buildup indicates that the rate of fermentation is high.

The researchers observed that the last three amino acids added to the Tat monomer (RKKRRQRRRPPQ) during solid-phase peptide synthesis required more time for linkage than the first several amino acids added. Which of the following best explains this observation? A) The amino groups of the last three amino acids are less able to react with the carboxyl group of the growing peptide B) The side chains of the last three amino acids experience stronger electrostatic repulsions with the growing peptide than the first several amino acids C) Proline residues add kinks to the peptide, resulting in steric repulsions as proline is added to the end of the chain D) Amino acids are generally less likely to collide with longer peptides than they are with shorter peptides

B - The side chains of the amino acids in a peptide interact with each other through electrostatic interactions, including ionic attractions and repulsions, hydrogen bonding, London dispersion forces, and steric interference. These interactions can also influence the ability of amino acids and peptides to react with each other. Amino acids with similar charges repel each other, reducing the probability that they will collide in a way that results in a reaction (ie, a productive collision). The passage states that solid-phase peptide synthesis occurs from C-terminus to N-terminus. However, by convention the sequence of a peptide or protein is always written from N-terminus to C-terminus. Therefore, the last three amino acids added to the Tat monomer (ie, those at the N-terminus) were a lysine (K), another lysine, and an arginine (R) residue. These residues are all positively charged and will repel each other, resulting in fewer productive collisions. In contrast, the first several amino acids added were glutamine (Q), proline (P), and another proline, which are neutral and will not experience the same electrostatic repulsion. These amino acids are more likely to collide productively and react more quickly.

What is the purpose of the urea cycle? A) To catabolize proteins for energy use during extreme conditions B) To process ammonium to be eliminated from the body C) To regulate blood pH D) To generate NADPH for fatty acid synthesis

B - The urea cycle processes the toxic substance ammonium into a safer form, urea, in order to eliminate it from the body (choice B is correct). Protein catabolism, the breakdown of proteins, is used to provide energy during extreme starvation (choice A is incorrect). The bicarbonate buffer system is responsible for regulating blood pH (choice C is incorrect). Malic enzyme catalyzes the reaction that generates NADPH for fatty acid synthesis (choice D is incorrect).

The pKa pf the side chain of histidine is approximately 6. In a solution with a pH of 8, the side chain of histidine is most likely to be: A) Protonated and charged B) Deprotonated and uncharged C) Deprotonated and charged D) Protonated and uncharged

B - When pH > pKa, the group in question will be deprotonated. Since the side chain of histidine contains and NH ,it will be positively charged when it is protonated and uncharged when it is deprotonated.

Activation

Before their breakdown, fatty acids undergo key preliminary processes. First, they are activated in the cytoplasm by being attached to CoA, creating a compound referred to as acyl-CoA. For short-chain and medium-chain fatty acids, this attachment is sufficient to allow the fatty acid to enter the mitochondria and undergo oxidation. Long-chain fatty acids (those containing 14 - 20 carbons) must be transported into the mitochondria using a carnitine shuttle. The fatty acid is modified by carnitine acyltransferase I, the rate-limiting step of β-Oxidation that adds carnitine to the fatty acid. Upon entry into the mitochondria, the carnitine is removed from the fatty acid.

Beta oxidation

Beta oxidation (β-oxidation) refers to the breakdown of fatty acids. This process is called β-oxidation because the breakdown occurs at the β-carbon of the fatty acid chain. 1. Lipid absorption 2. Activation 3. Oxidation

Secondary Protein Structure

Beta-sheets can be oriented in either a parallel or an antiparallel manner. The passage states that the sheets in ataxin-3 are in the parallel orientation. Parallel strands run in the same direction, like lanes on a track, so the N-terminal portion of one strand aligns with the N-terminal portions of the others (Number II). In antiparallel sheets, on the other hand, the individual strands run in opposite directions from each other, like switchbacks on a hiking trail, so that the N-terminal region of one strand lines up with the C-terminal regions of neighboring strands. Secondary structure always includes hydrogen bonds between carbonyl and amino groups of the polypeptide backbone (Number III). Parallel and antiparallel sheets differ in their hydrogen bond geometries, with bond pairs directly aligned in antiparallel sheets and slightly offset in parallel sheets. Nevertheless, both types of beta-sheets have hydrogen bonds between adjacent backbones. (Number I) Because antiparallel strands run in opposite directions, they may be linked by a short sequence of amino acids called a beta-turn that induces a 180° bend in the polypeptide chain. Parallel beta-strands do not reverse directionality, so neighboring strands must instead be linked by longer loops that make 360° turns to align the N-terminal regions of neighboring strands. Parallel strands such as those in ataxin-3 can never be linked by beta-turns.

Glucose Production

Blood metabolite and pH levels are normally tightly controlled. As the passage states, however, glucose 6-phosphatase (G6Pase) inactivation leads to decreased blood sugar and increased blood acidity. Group B mice were given tetracycline, which inactivated G6Pase 10 days into the experiment. Therefore, group B mice should maintain nearly constant glucose and pH levels prior to day 10, and should demonstrate decreased blood glucose and increased acidity after day 10. Because increased acidity means a decrease in pH, the correct graph should show a decrease in both glucose and pH levels, starting at day 10.

Biotin is a key coenzyme for which of the following enzymes? I. Propionyl-CoA carboxylase II. Phosphoenolpyruvate carboxykinase III. Pyruvate carboxylase A) I only B) I and II C) I and III D) I, II, and III

C - Carboxylase enzymes in humans are biotin-dependent (choices I and III are correct). Phosphoenolpyruvate carboxykinase is an enzyme used during gluconeogenesis to convert oxaloacetate into phosphoenolpyruvate. It is not biotin-dependent (choice II is incorrect).

Scientists could confirm that an allosteric effector increases the catalytic efficiency of an enzyme if it has what effect on kcat and Km? A) kcat decreases, Km increases B) kcat decreases, Km remains unchanged C) kcat increases, Km decreases D) kcat remains unchanged, Km increases

C - Catalytic efficiency is a measure of how well an enzyme facilitates reactions at low substrate concentrations. When enzyme concentration is held constant, catalytic efficiency is proportional to the initial slope of a Michaelis-Menten curve (when [S] << Km). This slope is governed by both the catalytic turnover kcat and the Michaelis constant Km. Mathematically, catalytic efficiency is calculated as the ratio of kcat to Km. An increase in kcat increases the catalytic efficiency, and a decrease in Km also increases the efficiency. Therefore, an allosteric activator that both increases kcatand decreases Km will increase the catalytic efficiency of an enzyme. Conceptually, a large (steep) slope indicates an efficient enzyme because a small amount of substrate yields a high reaction rate. Catalytic turnover (kcat) measures the number of substrate molecules converted to product per enzyme per second when the enzyme is saturated; therefore, a high kcat corresponds to a high maximum rate (Vmax) and, by extension, a larger slope at lower substrate concentrations. The Michaelis constant (Km) is an indicator of binding affinity. A small Km indicates high affinity, allowing the reaction to approach Vmax more rapidly (higher slope) with smaller amounts of substrate. Therefore, a small Km contributes to high catalytic efficiency. (Choice A) A decrease in kcat and an increase in Km would decrease catalytic efficiency. (Choice B) If Km is unchanged, a decrease in kcat will cause a decrease in catalytic efficiency. A decrease in kcat must be overcome by a proportionally greater decrease in Km for catalytic efficiency to increase. (Choice D) If kcat is unchanged, an increase in Km will decrease catalytic efficiency. An increase in Km must be overcome by a proportionally greater increase in kcat for catalytic efficiency to increase.

Acetylation of lysine residues in histones increases gene expression because: A) DNA is tightly bound to negatively charged amino acids on histones B) The carboxyl oxygen atoms in acetyl groups form hydrogen bonds with nitrogenous bases. C) The salt bridges between charged amino acids and phosphate groups are disrupted D) Lysine residues in histones associate with positively charged phosphate groups in DNA

C - DNA winds tightly around proteins referred to as histones to form structural units known as nucleosomes. Gene expression depends partially on the association of histones with DNA. Genes that are actively transcribed are found on unwound stretches of chromatin called euchromatin, which transcription machinery can easily access. Inactive genes are usually in highly condensed DNA known as heterochromatin, which is much less accessible. Covalent post-translational modifications such as methylation, acetylation, and phosphorylation alter the association of histones with DNA. DNA is predominantly negatively charged due to the phosphate groups on the backbone. Histones associate with DNA by forming salt bridges between positively charged amino acid residues and negatively charged phosphate groups. These ionic interactions allow histones to bind tightly to DNA and prevent genes from being transcribed. Acetylation of histones involves the transfer of acetyl groups from acetyl coenzyme A to positively charged amino groups on lysine or arginine residues. This modification disrupts salt bridges by reducing the positive charge on histones, which allows DNA to unwind and become more accessible to transcription machinery. As a result, the acetylation of histones causes nucleosomes to relax and increases gene expression.

A researcher designing an upcoming immumosorbent assay presents a list of molecules to use as a label for their experiments. Which molecule was most likely rejected for use as a label? a) Fluorescein isothyocyanate B) 131 I C) Albumin D) Horseradish peroxidase

C - Detection of the antigen of interest in immunoassay may require labelling or conjugation of the antibodies with another molecule, these are commonly flurophores (a), enzymes (d), radioisotopes (b), or biotin. These labeling agents can be conjugated to either a primary antibody (direct assay), or to a secondary antibody (secondary assay). While small polypeptides can be used, a large protein like albumin is not likely to be chosen.

In the DNA polymerase active site, which nucleotide base would provide the most Watson-Crick donors for interactions with amino acid residues? A) Adenine B) Cytosine C) Guanine D) Thymine

C - Hydrogen bonds form between donor hydrogen atoms and acceptor oxygen and nitrogen atoms. Donor hydrogen atoms must be covalently bonded to electronegative atoms such as nitrogen or oxygen to confer a partial positive charge, and acceptor atoms must have at least one lone pair of electrons. In Watson-Crick base pairing, hydrogen bonds form between particular acceptor atoms of one base and donor atoms of the complementary base, and vice versa. Adenine (A) and thymine (T), which each have 1 donor and 1 acceptor atom, form two hydrogen bonds between them (note that thymine has two potential acceptors in the form of carbonyls, but only one participates in Watson-Crick pairing). Guanine (G) and cytosine (C) pairs form three hydrogen bonds: Guanine contributes 1 acceptor and 2 donorswhereas cytosine contributes 2 acceptors and 1 donor. The passage states that amino acids in the active site of DNA polymerase can interact with nucleotides by forming hydrogen bonds with Watson-Crick donor and acceptor atoms. Guanine has two Watson-Crick donors; therefore, it contributes more donors for bonding to active site amino acids than any other nucleotide base.

The addition of AMPK phosphatases to skeletal muscles exposed to leptin would result in which of the following? A) Increased shuttling of fatty acids from the cytosol into mitochondria B) Increased funneling of electrons to carriers NAD+ and FAD C) Upregulated activity of the acyl carrier protein in fatty acid synthase D) Upregulated phosphorylation of acetyl-coenzyme A carboxylase

C - Phosphatases remove phosphate groups from enzymes/substrates (ie, dephosphorylation), whereas kinases add phosphate groups (ie, phosphorylation). In this scenario, AMPK phosphatases in leptin-exposed muscles would dephosphorylate AMPK and thereby inactivate it. Inactive AMPK would be unable to inhibit (phosphorylate) ACC, which carries out the first and rate-limiting step of fatty acid synthesis by carboxylating acetyl-CoA to malonyl-CoA. As a result, unphosphorylated ACC would promote fatty acid synthesis, preventing β-oxidation (Choice D). Fatty acid synthase carries out the main reactions of fatty acid synthesis and contains an acyl carrier protein (ACP)responsible for transferring acyl intermediates. Acetyl-CoA and malonyl-CoA are activated to acetyl-ACP and malonyl-ACP, which are condensed into acetoacetyl-ACP. Subsequently, acetoacetyl-ACP is reduced, dehydrated, and further reduced to generate butyryl-ACP. The coenzyme NADPH is used for both reductive steps. The reactions are repeated until a 16-carbon fatty acid (palmitate) is formed via addition of two carbon atoms per cycle. (Choice A) In β-oxidation, a carnitine transporter moves 14-16 carbon fatty acids from the cytosol into the mitochondria. However, β-oxidation is inhibited in this setting. Instead, fatty acid synthesis requires shuttling of citrate from the mitochondria into the cytosol to yield the acetyl-CoA precursor. (Choice B) NAD+ and FAD are reduced to NADH and FADH2 (electrons gained) in β-oxidation. However, fatty acid synthesis requires NADPH (as an electron donor) to reduce the acyl groups added to the fatty acid.

Atpenins are succinate-ubiquinone reductase inhibitors with antifungal properties. Atpenins illustrate the fact that: A) Succinate oxidation at Complex II decreases the pH of the intermembrane space B) Protons from FADH2 flow back into the matrix against their concentration gradient C) Electron transfer from Complex II to III is facilitated by ubiquinone in the inner membrane D) Ubiquinone has a higher redox potential than cytochrome c reductase in the cell membrane

C - Succinate-ubiquinone reductase, also known as Complex II or succinate dehydrogenase, is an oxidoreductase found in the inner membrane of the mitochondria in eukaryotes. In the electron transport chain, Complex II facilitates the transport of electrons from succinate in the citric acid cycle to Complex III. Electrons move through Complex II by three consecutive reactions. First, electrons from succinate are transferred to flavin adenine dinucleotide (FAD) to produce FADH2 as succinate is oxidized to fumarate. Then, FADH2 is reoxidized to FAD when electrons are shuttled to the iron-sulfur centers in Complex II. Finally, a small mobile electron carrier called ubiquinone, or coenzyme Q, accepts electrons from the iron-sulfur centers and transports them to Complex III. Electrons transferred by Complex II contribute to the overall production of ATP, which provides cells with the energy to grow. The question indicates that atpenins are effective antifungal agents that inhibit succinate-ubiquinone reductase (ie, Complex II). Consequently, the correct answer choice should illustrate that atpenins disrupt at least one of the three reactions of Complex II. The transfer of electrons from Complex II to III through ubiquinone is the only reaction accurately described. (Choices A and B) Complexes I, III, and IV pump protons into the intermembrane space against their concentration gradient, but Complex II does not pump protons against their gradient, and therefore does not contribute to the decrease in pH in the intermembrane space. (Choice D) In the electron transport chain, electrons are transferred to carriers with higher redox potentials at each step. Electrons are transferred from ubiquinone to cytochrome c reductase (Complex III) because ubiquinone has a lower, not a higher, redox potential.

Which of the following enzymes is involved with the digestion of proteins? A) Lipase B) Amylase C) Trypsin D) Aromatase

C - Trypsin is a key enzyme released by the pancreas to digest proteins. Lipase is involved with the digestion of fats (choice A is incorrect). Salivary and pancreatic amylases work to digest carbohydrates (choice B is incorrect). Aromatase is essential for estrogen synthesis and is not involved with digestion (choice D is incorrect).

An allosteric enzyme changes its conformation when it binds an effector molecule at a site distinct from its active site. These enzymes are most likely to be prevalent in which pathways? A) Digestive pathways B) Metabolic regulatory pathways C) Feedback loops D) Cell growth and division pathways

C - While allosteric enzymes can play roles in each of the mentioned pathways, they are particularly important in feedback loops. Downstream products often inhibit an enzyme in an earlier step to prevent overproduction of the final product

Protein absorption

Carbohydrates, lipids, and proteins each have unique pathways to digestion. In the mouth, salivary amylases initiate the digestion of carbohydrates. Lipids, as we discussed earlier, are primarily digested in the small intestine. Proteins start to undergo proteolysis in the stomach via pepsin. Additional enzymes, such as trypsin, chymotrypsin, and pancreatic juices further break down proteins in the lower digestive tract. In the small intestine, brush-border enzymes finish digestion. These enzymes all digest enormous proteins into much smaller amino acids, dipeptides, and tripeptides.

Kcat and Catalytic efficiency

Catalytic turnover refers to the rate at which an enzyme-bound substrate is converted to product. It is given by the rate constant kcat and is proportional to the maximum reaction rate Vmax. On a Michaelis-Menten plot, catalytic turnover is reflected in the height of the plateau that occurs at high substrate concentrations. A high kcat (or a high plateau) corresponds to high catalytic turnover. Catalytic efficiency is defined by the ratio kcatKmkcatKm and is a measure of enzyme specificity. It indicates the rate of catalysis at low substrate concentrations and is proportional to the initial slope of the Michaelis-Menten plot. An enzyme that quickly creates product at low substrate concentrations (high initial slope) is said to be catalytically efficient because only a small amount of substrate is needed to generate substantial activity. A high kcat, a low Km, or both may contribute to catalytic efficiency. An enzyme-substrate system may have any combination of high and low efficiency and turnover. In this case, CSL-173 has the highest kcat (turnover) and CSL-176 has the highest kcatKmkcatKm ratio (efficiency).

Electron Transport Chain

Complex III facilitates consumption of UQH2 and cyt-Cox, producing UQ and cyt-Cred. If this complex is inhibited, the concentration of UQH2 will build as complexes I and II continue to produce it but Complex III fails to consume it. Because UQ is no longer being regenerated, any additional NADH and FADH2 will have nothing to pass their electrons to, so their concentrations will build up as well. Finally, because cytochrome C is no longer being reduced, the reduced form will be depleted over time as complex IV continues to oxidize it. Reduced electron carriers upstream of the inhibited enzyme build up, and reduced electron carriers downstream are depleted. (Choice A) Inhibition of complex I would result in the buildup of NADH and the depletion of the downstream electron carriers. Although complex II is not downstream of complex I, FADH2 is still depleted because NADH buildup shuts down the citric acid cycle, and therefore prevents FADH2 regeneration. (Choice B) Inhibition of complex II would result in the buildup of FADH2 and the depletion of the downstream electron carriers. Although complex I is not downstream of complex II, NADH is still depleted because FADH2 buildup shuts down the citric acid cycle, and therefore prevents NADH regeneration. (Choice D) If complex IV were inhibited, all of the reduced electron carriers would eventually build up because each would ultimately lose the acceptor that takes its electrons.

During the Gabriel synthesis of an amino acid, which of the steps listed below occurs? A) A Masonic ester undergoes an Sn1 reaction B) An unprotected amine attacks the carbonyl carbon of an ester C) The oxygen atom of an aldehyde is protonated D) Heat is used to remove an extra COOH group

D - A final step in the Gabriel synthesis of an amino acid is the administration of heat to decarboxylate the molecule, removing an extra COOH. The Gabriel synthesis involves an Sn2 mechanism, and the amine in a Gabriel synthesis is protected, and choice C is something that happens during the Strecker synthesis.

Which step is NOT required to generate the western blot shown in Figure 3? A) Gel electrophoresis of BAZ2B proteins recovered from affinity columns B) Transfer of proteins to a protein-binding membrane C) Blocking of nonspecific antibody binding D) Incubation with antibodies against histone subunit H3

D - A western blot is a technique used to detect the presence of a specific protein and to compare its relative abundance in one set of conditions to that obtained under other conditions. It is performed in the following steps: Samples containing the protein of interest are loaded onto a gel and subjected to electrophoresis, causing the protein to migrate through the gel. In this case, the protein of interest is BAZ2B, which must be present in the gel for detection to occur (Choice A). Proteins in the gel are transferred to a protein-binding membrane such as nitrocellulose, where they become immobilized (Choice B). Transfer is usually carried out by electrophoresis. The portions of the membrane to which protein was not transferred are blocked by protein-rich mixtures such as bovine serum albumin (BSA) or low-fat milk. This prevents antibodies, which are themselves proteins, from nonspecifically binding to the nitrocellulose in the next step (Choice C). The membrane is incubated with antibodies that specifically bind the protein of interest. These antibodies may be labeled with a marker and detected directly, but more commonly they are subjected to a labeled secondary antibody that specifically binds the first (primary) antibody. Use of secondary antibodies helps improve detection. Antibodies are detected by techniques such as fluorescence or chemiluminescence.

The oxidation state of critical cofactors play an important role in regulating cell metabolism. Which of the following correctly describes the relationship between oxidation state and cellular process? A) An increased NAD/NADH ratio will drive fatty-acid synthesis B) An increased FADH2/FAD ratio will drive ATP hydrolysis C) A decreased FAD2/FAD ratio will increase the activity of ATP synthase D) A decreased NADH/NAD ratio will drive fatty-acid oxidation

D - Catabolism of molecules like fatty acids is used to produce energy. Anabolism describes construction of molecules, which requires energy. A lower NADH/NAD means energy is needed, and fatty acids will be oxidized to eventually provide ATP.

Which of the following techniques would you most likely use to isolate a protein with a net positive charge of +35? A) Size exclusion chromatography B) Gel filtration chromatography C) Anion-exchange chromatography D) Cation-exchange chromatography

D - Cation-exchange chromatography is used to isolate positively charged substances from the mobile phase (choice D is correct; choices A, B, and C are incorrect).

In order to transport long-chain fatty acids from the cytosol to the mitochondria, three of the following molecules are required. Which one is NOT necessary? A) Coenzyme A B) ATP C) Carnitine D) Oxaloacetate

D - Entry into the mitochondrial matrix is tightly regulated by the inner mitochondrial membrane. Crossing the membrane generally requires the aid of transport proteins, which only recognize specific molecules. Some short-chain fatty acids can cross the inner mitochondrial membrane directly, but long-chain fatty acids can only cross the outer mitochondrial membrane. Additionally, long-chain fatty acids are not recognized by any transport proteins; they must be modified or "activated" to enter the matrix. Fatty acid activation and transport includes the following steps: In the cytosol, the enzyme acyl-CoA synthetase catalyzes the reaction of fatty acids with coenzyme A to form acyl-CoA molecules. This reaction is thermodynamically unfavorable and requires ATP hydrolysis to proceed (Choices A and B). Acyl-CoA molecules then migrate to the intermembrane space, where they can react with carnitine to form acylcarnitine. The transport protein acylcarnitine translocase, located on the inner mitochondrial membrane, recognizes acylcarnitine and carries it into the mitochondrial matrix (Choice C). Acylcarnitine is then converted back to acyl-CoA and carnitine. Acyl-CoA is subsequently digested by β-oxidation, and carnitine is transported out of the matrix where it can pick up and carry a new fatty acid into the mitochondrial matrix. Although oxaloacetate is an intermediate in the citric acid cycle, gluconeogenesis, and transport of acetyl-CoAfrom the mitochondria into the cytosol for fatty acid synthesis, it is not involved in transport of fatty acids from the cytosol intothe mitochondria.

A single-stranded DNA oligonucleotide composed of which of the following would move most slowly down an alkaline agarose gel during electrophoresis? A) Deoxyadenosine monophosphate B) Deoxythymidine monophosphate C) Deoxycytidine monophosphate D) Deoxyguanosine monophosphate

D - Gel electrophoresis separates molecules by molecular weight, with larger molecules migrating more slowly than smaller molecules. DNA is composed of the four deoxyribonucleotides (dNMPs), each of which has a different molecular weight. The purine deoxyguanosine monophosphate (dGMP) is the largest dNMP, followed by deoxyadenosine monophosphate (dAMP), deoxythymidine monophosphate (dTMP), and deoxycytidine monophosphate (dCMP). An oligonucleotide is a short strand of DNA, and its molecular weight is determined by its composition of dNMPs. An oligonucleotide composed entirely of dGMP will be larger than any other oligonucleotide of the same length and therefore would move the most slowly through an agarose gel. (Choices A, B, and C) Deoxyguanosine has the largest molecular weight of any nucleotide, so all other nucleotides will move faster through an agarose gel.

Which of the following regulatory mechanisms helps increase net glucose catabolism in the liver after a meal? A) Inhibition of hexokinase by glucose-6-phosphate B) Allosteric suppression of phosphofructokinase-1 by ATP binding C) Hormonal suppression of fructose-2,6-bisphosphate synthesis D) Allosteric inhibition of fructose-1,6-bisphosphatase catalysis

D - Glucose uptake into cells and its subsequent metabolism are controlled primarily by the hormones insulin and glucagon. The binding of insulin and glucagon at the cell membrane stimulates or suppresses the biochemical pathways involved in the uptake, breakdown, and storage of glucose. Insulin stimulates glycolysis, which consists of a series of reactions that break down glucose into pyruvate to produce ATP. Glucagon acts in reverse by activating gluconeogenesis in the liver, which results in the synthesis of glucose from pyruvate and other precursor molecules. The well-fed state is characterized by the release of insulin in response to high levels of glucose in the blood. Insulin stimulates glucose catabolism (glycolysis) by activating synthesis of fructose-2,6-bisphosphate (F2,6BP). F2,6BP allosterically activates phosphofructokinase-1 (PFK-1), which catalyzes the rate-limiting step of glycolysis. In the liver, F2,6BP also inhibits fructose-1,6-bisphosphatase, which catalyzes the dephosphorylation of F1,6BP in gluconeogenesis. Inhibiting gluconeogenesis in the liver leads to increased net glucose catabolism. (Choice A) Hexokinase converts glucose into glucose-6-phosphate in the first step of glycolysis. Inhibition by glucose-6-phosphate would reduce glucose catabolism. (Choice B) Phosphofructokinase-1 catalyzes the conversion of fructose 6-phosphate to fructose-1,6-bisphosphate in glycolysis. Inhibiting this process with ATP would slow down glucose catabolism. (Choice C) Activation, not suppression, of F2,6BP synthesis increases glucose catabolism.

Medium-chain acyl-CoA dehydrogenase (MCAD) is an enzyme involved in β-oxidation, and is encoded by the autosomal ACADM gene. MCAD deficiency is caused by a mutation in ACADM, and results in impaired ability to digest medium-chain fatty acids for energy, and in their subsequent accumulation. Which of the following interventions would best relieve the negative effects of MCAD deficiency? A) Mitochondrial gene therapy B) Supplementation with long-chain fatty acids C) Prolonged periods of fasting D) A diet rich in glucose

D - Inborn errors of metabolism are genetic disorders caused by deleterious mutations in genes that code for metabolic enzymes. These disorders result in the accumulation of metabolites upstream of the affected enzyme and decreased levels of the downstream products in the pathway. The resulting changes in metabolite levels frequently have negative clinical effects such as lethargy, impaired development, and toxicity. According to the question, medium-chain acyl-CoA dehydrogenase (MCAD) is an enzyme involved in β-oxidation, a metabolic process that hydrolyzes fatty acids into acetyl-CoA and generates reduced electron carriers for ATP production. β-oxidation is an important source of ATP during periods of fasting, after glycogen stores are exhausted. Because mutations within the gene that encodes MCAD result in impaired β-oxidation, therapies for MCAD deficiency should be aimed at restoring energy production either by supplementing the deficient enzyme with a functional version or providing an alternate fuel source that follows a different metabolic pathway. A diet rich in glucose is an effective therapy because, like β-oxidation, digestion of glucose leads to the production of acetyl-CoA, reduced electron carriers, and ultimately ATP. (Choice A) Although β-oxidation occurs in the mitochondria, the gene that encodes MCAD (ACADM) is autosomal. Autosomal genes are found in the nucleus, so mitochondrial gene therapy would neither affect functional MCAD levels nor relieve the negative effects of MCAD deficiency. (Choice B) Long-chain fatty acids are digested into medium-chain fatty acids, which must be processed by MCAD. Supplementation with long-chain fatty acids in an MCAD-deficient patient would lead to an increased buildup of medium-chain fatty acids and yield little ATP. (Choice C) β-oxidation is an important energy source during periods of fasting. MCAD deficiency inhibits β-oxidation, so fasting is harmful to MCAD-deficient individuals.

Which of the following correctly describes the stereospecific formation of a citric acid cycle intermediate? A) Citrate synthase consumes ATP to generate citrate from oxaloacetate and acetyl-CoA B) Fumarase converts fumarate into both L- and D-malate through a condensation reaction C) Succinyl-CoA synthetase consume GTP to transform succinyl-CoA to succinate D) Aconitase converts citrate into isocitrate through an intermediate

D - The first reaction in the cycle is the conversion of citrate to isocitrate by the aconitase enzyme through an intermediate called cis-aconitate. Aconitase catalyzes consecutive dehydration and hydration reactions that exchange the hydrogen atom and hydroxyl group of the second and third carbon. This isomerase activity results in the transformation of citrate, which is achiral (no chiral centers), into isocitrate, which contains two new chiral centers. Similarly, fumarate is converted to malate by fumarase in a stereospecific reaction. The enzyme referred to as fumarase splits a molecule of water and adds the hydroxyl group and hydrogen atom across the C=C double bond in fumarate. This reaction forms a new chiral at the second carbon, and generates L-malate only (Choice B). Furthermore, this is a hydration reaction, not a condensation reaction. (Choice A) Synthases such as citrate synthase do not require energy to break or form new bonds, and do not consume ATP. (Choice C) The energy from cleavage of the thioester bond in succinyl-CoA is used by succinyl-CoA synthetase to produce GTP.

The absence of which of the following lipids is most likely to affect the assembly of viral envelopes? A) Palmitic acid B) Triacylglycerides C) Sphingomyelin D) Phosphatidylserine

D - The plasma membrane is composed mainly of proteins, phospholipids, and cholesterol arranged in a bilayer with hydrophilic surfaces enclosing a hydrophobic interior. Phospholipids have a wide range of structures but generally consist of one or more fatty acid tails linked to a polar head group that contains a phosphate. The charges of phospholipids depend on the head groups attached to the phosphate. Head groups can be either neutral or negatively charged. Influenza A obtains its viral envelope from the plasma membrane of its host cell. However, the passage states that the composition of the envelope differs significantly from the plasma membrane. Unlike human membranes, the viral envelope contains mostly phosphatidylethanolamine and negatively charged phospholipids. Therefore, phospholipids with ethanolamine and negatively charged head groups such as serine, inositol, and diphosphatidylglycerol are needed for envelope formation. Of the choices presented, phosphatidylserine is the only negatively charged phospholipid. Therefore, its absence is most likely to affect the assembly of the viral envelope. (Choice A) Palmitic acid is the only free fatty acid synthesized by cells. Although it can be incorporated into phospholipids, the passage does not indicate that it is required. (Choice B) Triglycerides are storage lipids found in adipocytes (fat cells) and consist of three fatty acid tails bound to a glycerol backbone by ester linkages. They are not found in cell membranes or viral envelopes. (Choice C) Sphingomyelin is a phospholipid with a sphingosine backbone. The phospholipids required for the viral envelope (phosphatidylethanolamine and negatively charged glycerophospholipids) do not contain sphingosine, but instead contain a glycerol backbone. Sphingomyelins are not part of the viral envelope.

DNA repair

DNA polymerases are capable of detecting and removing incorrect nucleotides at the ends of a growing DNA strand. This function, referred to as proofreading, prevents mutations or errors in DNA replication from accumulating. Normally, DNA polymerases are equipped with both 5′-3′ and 3′-5′ exonuclease activity that allows them to remove and replace incorrect nucleotides at either end of a DNA strand. However, the passage states that the Klenow fragment (KF) enzyme described in the experiment does not have 5′-3′ exonuclease activity. KF can only proofread DNA in the 3′-5′ direction on the template strand, so only errors at the 3′ end of the growing strand can be repaired.

Protein Denaturation

Electrostatic interactions such as ionic bonds are noncovalent forces that help maintain the tertiary and quaternary structure of a protein. They occur mainly between the acidic and basic side chains of amino acids on the polypeptide chain, and are commonly referred to as salt bridges. Neutral salts such as NaCl, which dissociate in solution, can disrupt salt bridges by forming ionic bonds with amino acids. Positively charged sodium ions (Na+) associate with the negatively charged side chains of acidic amino acids. Chloride ions (Cl−), which are negatively charged, associate with the positively charged side chains of basic amino acids. High concentrations of salt can cause a protein to unfold through a process known as denaturation. The accompanying loss of three-dimensional structure also results in a loss or reduction of function. Under controlled physiological conditions, ALDH2 produces NADH during oxidation-reduction reactions. However, ALDH2 proteins exposed to increasing concentrations of salt in the eluate could become denatured, resulting in a decrease in the rate of NADH production. (Choice A) NaCl can interfere with hydrogen bonding in the secondary structure of a protein but cannot enhance it. As with ionic bonds, interrupting hydrogen bonds within ALDH2 could also lead to denaturation and a reduction in NADH production. (Choice B) NaCl does not alter the hydrophobicity of a protein. Hydrophobicity is determined by the number and location of hydrophobic amino acids in the protein's primary structure. (Choice C) NaCl would not increase the basicity because it is a neutral salt that does not contribute to a net change in hydrogen or hydroxide ion concentration.

Enzyme catalysts

Enzymes are biological catalysts that lower the activation energy of a reaction, allowing the reaction to proceed at a faster rate. Substrate molecules bind to the enzyme's active site, where catalytic activity occurs. The enzyme and substrate undergo conformational changes that make the transition state more stable and favorable, resulting in lower activation energy. Enzyme function is sensitive to changes in pH and temperature as well as deleterious mutations, particularly those involving the active site. A missense mutation at the histidine in position 57 of the NSP4 sequence (ie, His57) would result in a different amino acid at this site and alter the interaction between the viral peptide (substrate) and the NSP4 active site. This mutation would be expected to decrease the binding affinity of the enzyme for its substrate, resulting in an increased dissociation constant Kd for the enzyme-substrate complex (Number III). As a result, the mutation would interfere with normal enzymatic function and lead to a decreased reaction rate (ie, increasing the time the reaction takes to reach equilibrium) (Number I). (Number II) Enzymes do not alter the free energy change ΔG or the equilibrium constant Keq of a reaction. The overall ΔG depends only on the energy difference between products and reactants (ΔG = Gproducts − Greactants), and can be visualized in a free energy diagram. The equilibrium constant Keq of a reaction is the ratio of products to reactants for a given reaction at equilibrium at a specific temperature and is related to ΔG (ie, ΔG = −RT ln Keq). Although a functional enzyme would decrease the time taken for the reaction to reach equilibrium (ie, ΔG = 0 for the reaction) by increasing the reaction rate, it would not alter Keq.

ATP Hydrolysis

Fatty acid synthesis produces long hydrocarbon chains that are linked to a carboxylic acid. This process is facilitated by the enzyme fatty acid synthase, which successively links 2-carbon units (acetyl-CoA) to each other. The first acetyl-CoA molecule attaches directly to fatty acid synthase. Subsequent acetyl-CoA molecules are activated by combining with CO2 to form malonyl-CoA in a reaction that requires the hydrolysis of 1 ATP molecule to form 1 ADP molecule. Malonyl-CoA is then linked to the growing fatty acid chain on fatty acid synthase, releasing CO2 and CoA in the process. A carbonyl carbon in the chain is then reduced to the alkyl form in a process that requires oxidation of 2 NADPH molecules to form 2 NADP+ molecules. This process is repeated until a carbon chain of the desired length is formed. The question asks how many ADP and NADP+ molecules will form during synthesis of a 16-carbon fatty acid chain. The reaction scheme shows that for n + 1 acetyl-CoA molecules used, n ADP and 2n NADP+ molecules form. The scheme also shows that the resulting fatty acid contains 2n + 2 carbons. If the total number of carbons in the fatty acid chain (2n + 2) is 16, then 2n = 14 and n = 7. Therefore, 7 ADP molecules and 14 NADP+ molecules form in this process.

Fatty acid synthesis

Fatty acids are long hydrocarbon chains that serve as great sources of energy for the body. The only fatty acid that the human body can synthesize by itself is palmitic acid, a 16-carbon fatty acid. Each of the stages of this synthesis: 1. The citrate shuttle 2. The oxaloacetate shuttle 3. Palmitic acid synthesis

Gel electrophoresis

Gel electrophoresis is an experiment used to separate different components of a mixture based on their size and charge. Normally, these components are strands of DNA, RNA, or different proteins. Let's say you have a test tube containing 5 different proteins, but you only want one of them. If the 5 proteins each have a different size (or different amino acids), you can separate them with gel electrophoresis. During an experiment, you begin by placing your mixture on the gel, which you can think of as a large sheet of Jell-O. You then apply an electric field across the gel using a negatively charged side (cathode) and a positively charged side (anode). The molecules will travel through the Jell-O, and you can think of the system as a molecule swim meet through a Jell-O pool. (Note: many gel electrophoresis experiments are referred to with the acronym PAGE, or polyacrylamide gel electrophoresis. Polyacrylamide just refers to the type of gel that is used!) All gel electrophoresis experiments work by taking advantage of three properties: size, charge, and shape.

Glycogen metabolism pt II

Glycogen is primarily composed of several glucose units linked to each other through α-1,4-glycosidic bonds. Degradation of glycogen begins with a process called phosphorolysis, which removes individual glucose subunits. The enzyme glycogen phosphorylase, which catalyzes this reaction, breaks the α-1,4 bond by adding inorganic phosphate at carbon 1 of the terminal glucose unit. As a result, glucose in the form of glucose-1-phosphate (G1P) is released from glycogen stores. Figure 2A shows that after exercise to exhaustion, glycogen stores in the hippocampus are partially depleted. When DAB is injected, glycogen is depleted to a lesser extent. This suggests that DAB prevents glycogen degradation. Of the choices presented, inhibiting phosphorolysis is the only option that will lead to decreased glycogen degradation. (Choice A) Phosphorylated glucose in the form of glucose-6-phosphate (G6P) is generated during glycogen degradation by phosphorolysis and isomerization, or during glycolysis by phosphorylation. Inhibiting phosphorylation of glucose (catalyzed by hexokinase) would deplete the amount of G6P available. As a consequence, glycogen degradation would most likely increase to replenish the G6P needed for glycolysis and fermentation. Figure 2A shows that DAB causes glycogen degradation to decrease instead. (Choice C) Inhibition of lactate export would cause an increase in lactate concentration. Figure 2B shows that lactate is present at decreased levels in rats treated with DAB. This is most likely because less glycogen was degraded for use in lactate synthesis, as indicated by Figure 2A. (Choice D) Figure 2 indicates that DAB inhibits glycogenolysis. However, the first step in glycogenolysis breaks a bond in glycogen by adding inorganic phosphate across it (phosphorolysis) rather than by adding water across it (hydrolysis).

Glycogen Metabolism

Glycogen serves as a form of energy storage in several tissues, including muscle, liver, and in glial cells such as astrocytes. The passage presents the hypothesis that during exercise, astrocytes convert glycogen to lactate, which is then sent to neurons for use as a fuel. This conversion requires three metabolic pathways: glycogenolysis, glycolysis, and fermentation. Glycogenolysis is the degradation of glycogen into individual glucose-6-phosphate units. The first step in glycogenolysis produces glucose-1-phosphate, which is then converted to glucose-6-phosphate units through the enzyme phosphoglucomutase (Number I). Glucose-6-phosphate then enters glycolysis, which involves the conversion of 3-phosphoglycerate to 2-phosphoglycerate by the enzyme phosphoglycerate mutase (Number III). The final step of glycolysis produces pyruvate, which is then fermented to lactate by lactate dehydrogenase (Number V). (Number II) In glycogenolysis, glycogen is converted to glucose-1-phosphate and then to glucose-6-phosphate (G6P). G6P can then enter glycolysis directly, and free glucose is not converted to G6P in this process. (Number IV) Pyruvate is converted into acetyl-CoA (a 2-carbon molecule) prior to its entry into the citric acid cycle. This step includes the irreversible loss of a carbon atom as CO2. The fuel that astrocytes produce for neurons is lactate (a 3-carbon molecule), but once acetyl-CoA is formed, it cannot be converted to lactate.

Aliphatic Amino Acids

Hydroxylation reactions are involved in the oxygen-induced degradation of organic compounds and entail the additionof a hydroxyl group (-OH), usually in place of a hydrogen (H) atom. In the passage, HIF-PHD hydroxylates HIF at aliphatic residues. Aliphatic amino acids are nonpolar and hydrophobic ("water-fearing") because their R-groups are composed mostly of carbon atoms that can have single, branched, or cyclic (nonaromatic) carbon backbones. Hydrophobicity increases as the number of carbon atoms in the side chain increases. For this reason, glycine, which is nonpolar but has an H atom as its R-group, is often excluded from the list of aliphatic amino acids. However, methionine is often included due to the inert nature of its sulfur atom. Of the choices listed, proline (Choice A) is the only amino acid with an aliphatic side chain; therefore, proline is most susceptible to hydroxylation by HIF-PHDs.

Polar Amino Acids

In each strand of a beta-sheet, the amino acid side chains project out from and perpendicular to the backbone in alternating opposite directions. As beta-sheets form, the R groups become aligned. Glutamine is a polar uncharged amino acid with an amide side chain. Amides contain both a carbonyl oxygen and an amide nitrogen group, which can act as a hydrogen bond acceptor and donor, respectively. Because the side chains are aligned in beta-sheets and because glutamine can simultaneously act as an acceptor and a donor, polyglutamine beta-sheets can form networks of hydrogen bonds. These networks add stability to the beta-sheet conformation, so increased polyglutamine length increases hydrogen bond networks and stabilizes the beta-conformation. (Choice A) Stacking interactions can indeed stabilize secondary structures. However, they only occur between aromatic side chains such as phenylalanine, tyrosine, and tryptophan. (Choice B) It is glycine, not glutamine, whose flexibility is favorable for beta-turns. (Choice C) The rigidity of proline, along with its tendency to adopt the cis-conformation, is conducive to tight turns. However, glutamine is not particularly rigid.

Protein Metabolism

In eukaryotes, different metabolic processes are carried out in different compartments of the cell. For example, glycolysis and the pentose phosphate pathway occur in the cytosol, whereas the citric acid cycle and fatty acid oxidation occur in the mitochondria. The enzymes required to facilitate each process must be localized to the appropriate compartments. Transportation of proteins synthesized in the cytosol is prompted by a short amino acid sequence at the N-terminus, which serves as a signal for cellular machinery to move the protein to its correct destination. The pyruvate dehydrogenase complex (PDHC) catalyzes the decarboxylation of pyruvate to form acetyl-CoA, which then enters the citric acid cycle. This process occurs in the mitochondrial matrix, so the PDHC is found in this organelle. According to the passage, SIRT4 colocalizes with the PDHC and must also be found in the mitochondrial matrix. Therefore, SIRT4 most likely has a mitochondrial targeting signal to induce its transport to the mitochondria. (Choice A) Nuclear localization sequences signal for the transport of proteins such as transcription factors to the nucleus. According to the passage, SIRT4 colocalizes with the PDHC, which is found in the mitochondria. (Choice B) Polyubiquitin tags target defective or unnecessary proteins for destruction by the proteasome. SIRT4 is used routinely in the cell, so it would have a polyubiquitin tag only if only it were defective. (Choice C) N-linked carbohydrate chains are added to certain asparagine residues of proteins in the endoplasmic reticulum. Mitochondrial proteins such as SIRT4 are translated completely in the cytosol and are not modified in the endoplasmic reticulum.

Nucleophilicity

In general, amino acids containing thiol (-SH) or hydroxyl (-OH) groups can act as nucleophiles in biological reactions. Nucleophiles are attracted to the positive charge of atomic nuclei, and the greater the negative charge possessed by a molecule, the greater its nucleophilicity. Deprotonation of the cysteine thiol or the hydroxyl group of serine results in negatively charged sulfur or oxygen atoms and enhances their otherwise weak nucleophilicity. The question states that the mechanism begins with the modification of a thiol group. Cysteine is the only amino acid with a thiol (-SH) in its R group. The question further states that the thiol is modified by an amino acid (histidine, in this case) acting as a general base. Bases are proton acceptors; therefore, histidine must be deprotonating, not protonating, cysteine. The deprotonation of cysteine would likely enhance the rate of hydrolysis by converting a cysteine residue in the active site into an anionic—and more nucleophilic—sulfide residue. This sulfide can donate electrons to the carbonyl carbon of the peptide in a process called nucleophilic attack, ultimately leading to hydrolysis. This class of enzyme, known as a cysteine protease, is commonly employed in protein degradation.

RT-qPCR

In quantitative PCR, you want a more accurate quantification of how much DNA you are producing. This can be determined using a fluorescent probe that binds to double-stranded DNA. You can then measure the amount of fluorescence, which is directly related to the amount of double-stranded DNA. To complete RT-qPCR, you follow several steps: 1. Convert RNA to cDNA using reverse transcriptase 2. Magnify the amount of cDNA by completing rounds of PCR 3. Determine how much cDNA you have produced using a probe

Lipid absorption

In the small intestine, lipids undergo chemical digestion by bile and various enzymes, such as lipases. Fatty acids with short chains end up being absorbed into the blood after they cross the intestine. Long-chain fatty acids, however, will form micelles for their absorption and then be assembled into chylomicrons. Chylomicrons are a specific type of lipoproteins that transport dietary lipids, such as triacylglycerols, to tissues. They enter the lymphatic system via lacteals, lymphatic vessels in the small intestine, and re-enter the blood through the thoracic duct.

Agonist and Antagonist

In the study described in the passage, the control (baseline) was established as normal AMPK activity (at 100%). The activity of AMPK was assessed when PH was present or absent during leptin exposure (i.v. and i.h.p. at various time points). Muscles exposed to leptin only (no PH) showed a substantial increase in AMPK activity at i.v. 0.5 hr, i.v. 6 hr, and i.h.p. 1 hr. However, AMPK activity decreased in muscles exposed to both PH and leptin at i.v. 6 hr and i.h.p. 1 hr (due to faster drug delivery). This result signified that PH was blocking the leptin receptor and preventing AMPK activity from being increased by leptin, causing AMPK activity to return to a value close to baseline. PH can therefore be an antagonist, a compound that alters receptor-ligand binding affinity or binds a receptor directly to decrease the cell's response to the ligand. In addition, the increased amount of time taken for PH to lower AMPK activity (ie, inactive at 0.5 hr but active at i.v. 6 hr and 1 hr i.h.p) suggests that PH acts in a time-dependent manner. (Choices A and D) An agonist enhances or duplicates the effect of a ligand binding to a receptor. For PH to act as an agonist, it would have to bind to leptin-specific receptors and induce a biochemical response similar to that of leptin (ie, increase AMPK activity). (Choice B) AMPK activity was still increasing at i.v. 0.5 hr due to leptin exposure in the presence of PH. This increase implies that PH does not act immediately, which is supported by the fact that AMPK activity returned to a value close to baseline only with leptin (plus PH) at i.v. 6 hr.

Electrophoresis

In their native (ie, properly folded) form, many proteins consist of multiple polypeptide chains known as subunits. These proteins, called multimers, may consist of two subunits (dimers), three subunits (trimers), four subunits (tetramers), or more. When a multimer runs through a reducing SDS gel, the sodium dodecyl sulfate (SDS) causes proteins to unfold, and the reducing agent breaks any disulfide bonds that may link subunits together. Each subunit migrates through the gel by itself, and subunits of different sizes are separated. For multimers in which all subunits are identical (homomultimers), all subunits migrate the same distance and appear as a single band. If the subunits are not identical (heteromultimers), they migrate different distances according to molecular weight and appear as separate bands. According to the question, each protein in the gel is a dimer. Lanes A and C each show single bands, indicating that these proteins are homodimers. In contrast, Proteins B and D separate into two distinct bands, indicating that they are heterodimers. In their native (biologically active) form, the dimers have a molecular weight equal to the sum of the weights of their subunits. For heterodimers, the weights of each band (indicated by the marker lane) can be added. For homodimers, the weight of the single band can be doubled to account for both subunits. From largest to smallest, the order of the proteins is Protein C (80 kDa) → Protein D (65 kDa) → Protein A (60 kDa) → Protein B (55 kDa).

Protein Binding

In vitro assays can be used to measure the parameters of purified biomolecules such as proteins. They are tightly controlled experiments performed outside of living organisms that allow variables to be isolated because only a few types of molecules are present. However, in vitro assays are limited in their usefulness because they may not accurately reflect the behavior of a complex biological system presented by in vivo assays, which are conducted in whole living organisms with thousands of different molecules present. Therefore, results from in vitro assays must be confirmed by in vivoassays. The passage illustrates a case in which in vitro and in vivo results do not fully agree. In the ITC experiment (an in vitrotechnique that uses purified proteins), the dissociation constant Kd is used to determine the binding affinity of proteins. The low Kd of the 10M+ BAZ2B interaction with H3 shows that it has a higher binding affinity with H3 than with H3K14ac. In the in vivo experiment, in which proteins were extracted directly from cells, the results are the opposite. 10M+ BAZ2B has a higher affinity for H3K14ac than it has for unmodified H3, as shown by the thicker band in the H3K14ac lane of the western blot, which indicates increased binding.

Protein folding

Molecular chaperones are defined as proteins that facilitate the proper folding of other proteins. They usually accomplish this by binding hydrophobic regions of nascent, misfolded, or aggregated proteins, preventing exposure of those regions to aqueous solvent. This precludes incorrect interactions between hydrophobic residues and can prevent or reverse aggregation by blocking interactions between hydrophobic regions of separate polypeptides.

ATP Phosphate Transfer

Kinases are enzymes that catalyze the transfer of phosphate groups from nucleotide triphosphates (ie, ATP, GTP) to other molecules or from phosphorylated molecules to nucleotide diphosphates (ie, ADP, GDP). According to the passage, εPKC is a kinase that phosphorylates proteins such as ALDH2 at certain serine and threonine residues. Attaching phosphate to a molecule is usually a thermodynamically unfavorable process and must be coupled to a thermodynamically favorable process to provide the necessary energy. Hydrolysis of ATP or GTP is thermodynamically favorable. For this reason, kinases use ATP or GTP as a source of phosphate in the reactions they catalyze. Therefore, in addition to εPKC, ATP must also be present for phosphorylation of ALDH2 to occur.

Structural Lipids

Lipids are broadly classified as either hydrolyzable or nonhydrolyzable. They can also be grouped by function as structural lipids, energy-storing lipids, and signaling lipids. Structural lipids are found in the membranes surrounding cells and organelles and include hydrolyzable glycerophospholipids and sphingolipids as well as nonhydrolyzable cholesterol. The properties of individual lipids such as the number of fatty acid tails, double bonds, and head group structures influence both fluidity and curvature of biological membranes. The sphingolipids in Table 3 contain a sphingosine backbone attached to a single fatty acyl chain and a polar head group. In red blood cells and other cell types these sphingolipids are essential structural components of the plasma membrane (Number I). (Number II) The lipids that are primarily responsible for energy storage are triacylglycerides, which are metabolized through hydrolysis and beta-oxidation. The fatty acyl groups of sphingolipids may be hydrolyzed and used for energy production, but this is not their primary function. (Number III) Glycerophospholipids contain two hydrolyzable fatty acyl chains, but sphingolipids contain only one. The other long chain hydrocarbon in a sphingolipid is part of the sphingosine backbone and cannot be hydrolyzed. Therefore, sphingolipids produce one fatty acid upon hydrolysis, not two.

Glucose Production

Maintaining blood glucose levels is critical for proper body function, so several biological processes exist to maintain glucose levels between meals. They take place predominantly in the liver, and all produce glucose 6-phosphate (G6P). G6P must be dephosphorylated by G6Pase to be released as glucose into the bloodstream, so all of these processes will be directly impaired by the inactivation of G6Pase. They include the following: Glycogenolysis is the breakdown of glycogen into glucose. It is the first source of glucose during fasting and requires G6Pase in its final step (Number I). Gluconeogenesis is the synthesis of glucose from pyruvate and other precursors. It is essentially glycolysis in reverse with a few differences, including the conversion of pyruvate to oxaloacetate and the use of phosphatases such as G6Pase instead of kinases to remove phosphate groups (Number II). Gluconeogenesis and glycolysis are connected by the Cori cycle in which lactate, produced in the muscle during anaerobic glycolysis, is sent to the liver. There it is converted to pyruvate and undergoes gluconeogenesis. The newly synthesized glucose is then returned to the muscles through the bloodstream. (Number III) The pentose phosphate pathway is required for nucleotide and NADPH synthesis. Although the pathway includes G6P as a substrate, it does not require G6Pase activity, so it will not be directly impaired by von Gierke disease.

Oxidation of saturated fatty acids

Our body can most easily begin to break down saturated fatty acids. Recall that saturated fatty acids possess single bonds between each carbon of the alkyl chain and do not contain any double or triple bonds. The breakdown of saturated fatty acids, β-oxidation, can be broken down into four basic steps: 1. Oxidation: A dehydrogenase forms a double bond between the alpha and beta carbons at the end of the fatty acid tail, resulting in the formation of an FADH2 molecule. 2. Hydration: A hydroxyl group is added to the beta carbon, breaking the double bond with the alpha carbon. 3. Oxidation: The hydroxyl group on the beta carbon is oxidized, resulting in the formation of a carbonyl group and the production of 1 NADH molecule. 4. Thiolysis: Cleavage with CoA-SH results in the formation of acetyl-CoA and the shortened acyl-CoA. This process is repeated until we are left with one acetyl-CoA, a two-carbon molecule.

O2 Consumption in ETC

Oxygen consumption takes place in complex IV together with consumption of reduced cytochrome C. ETC reactions are connected such that inhibition of one complex affects the other complexes. Figure 1 shows that reduced cytochrome C is depleted in the presence of drug X. Because complex IV can reduce oxygen only when sufficient cyt-Cred is available to donate electrons, adding more reduced cytochrome C is the only way to allow complex IV to temporarily resume oxygen consumption. (Choice B) Oxidized cytochrome C is a product of oxygen consumption. It cannot be used to reduce oxygen to water. (Choice C) Ubiquinol provides the electrons necessary for complex III to produce cyt-Cred, which in turn can reduce O2. However, in the presence of drug X, complex III will not be able to use electrons from ubiquinol to produce cyt-Cred, no matter how much ubiquinol is added. (Choice D) Ubiquinone can be reduced to ubiquinol by complexes I and II. The generated ubiquinol would not be able to reduce cytochrome C because complex III is inactive in the presence of drug X.

Peptide Bond Hydrolysis

Peptides are linear sequences of amino acids linked by amide bonds (-CO-NH-) that join the α-carboxyl group of one amino acid to the α-amino group of another amino acid. Proteins that have completed their biological purpose must be degraded by proteases so that their constituent amino acids can be reused to generate new peptides. To cleave an amide bond during protein degradation, a water molecule (H2O) must be added using acids (H+) or bases (OH−) in a reaction known as amide hydrolysis. Figure 1 shows the acylation process (R-C=O addition) of amide hydrolysis by NSP4, which cleaves the amide bond of the viral peptide by adding a hydrogen atom to the nitrogen. The hydrogen atom displaces the C-N single bond when electrons from the oxygen on the carbonyl carbon reform the double bond. The acyl group (R1-C=O) attached to NSP4 is subsequently converted into carboxylic acid when electrons from an OH− group attack the electrophilic (electron-attracting) carbonyl carbon. (Choices A, C and D) Hydrolysis reactions only involve consumption of water to break bonds. They do not involve water production, which occurs in condensation reactions such as peptide bond formation.

Protein catabolism

Protein catabolism is the breakdown of proteins and is used only in extreme conditions as a supplement to gluconeogenesis when the body lacks sufficient energy supply. This process generally occurs in the liver and muscle. To begin the process, amino acids first undergo transamination or deamination reactions, which result in the loss of amino groups—either by transferring amino acids to another group or losing it completely. After the amino acid has lost its amino group, its carbon skeleton can now be converted into glucose or acetyl-CoA. Glucogenic amino acids are converted into pyruvate and will feed into gluconeogenesis to produce glucose. Ketogenic amino acids will be converted into acetyl-CoA, the precursor to ketone bodies.

SDS-PAGE

Since the charge density throughout a protein can vary, separating proteins is a little more complex than separating DNA or RNA. While smaller DNA and RNA strands will almost always travel faster than larger strands, proteins may break this general rule of thumb if they have different charge densities. For example, you might have a large protein with a lot of negative side chains and a slightly smaller protein with fewer negative side chains. Will the larger protein travel faster because it is more negatively charged? Or will the smaller protein travel faster because it is smaller? Even for similarly sized and charged proteins, the 3D structure of the protein may vary a lot, meaning aerodynamics are another factor we might have to deal with. In order to eliminate the effects of the differences in charge distribution and 3D shape for proteins that we mentioned above, researchers use SDS-PAGE. In other words, if you want to separate proteins just by their size (number of amino acids), use SDS-PAGE! In SDS-PAGE, researchers add sodium dodecyl sulfate (SDS) to their proteins before running them on the gel. SDS denatures the protein and adds a number of negative charges that are proportional to the size of the protein, thereby creating an equal charge distribution (just like we see in DNA and RNA).

Cofactors and Coenzymes

Some proteins, including some enzymes, can only function with the help of non-amino acid groups called cofactors. These groups may be organic (often called coenzymes) or inorganic, and they may be tightly or loosely bound to the proteins they help. Tightly bound cofactors are called prosthetic groups. The passage states that a magnesium ion is required on each subunit for rubisco activity. Therefore, Mg2+ is a cofactor of rubisco (Number II). The passage also states that Mg2+ is a prosthetic group, meaning that it is bound tightly to rubisco. The dissociation constant Kd measures the tendency of a protein-ligand complex (ie, rubisco and magnesium) to separate into its components. A small Kd indicates a low tendency to dissociate and therefore indicates the tight binding expected of a prosthetic group (Number I). (Number III) Allosteric effectors bind proteins at positions other than the active site and either increase (activators) or decrease (inhibitors) their activity. The passage says that magnesium ions bind rubisco at the active site; so although magnesium activates rubisco, it does not do so allosterically.

Native-PAGE

Sometimes you want to analyze a protein in its natural state, but once a protein is denatured, it is often impossible to return it back to its normal or native shape (imagine untying a massive knot and then being told to re-tie that exact knot!). If you want to isolate a protein and conduct a study that would require the protein to be in its native shape (e.g., studying your protein's activity after adding an inhibitor), you need to use native-PAGE. In native-PAGE, you do not add SDS or reducing agents, and the gel is non-denaturing, so the protein can remain in its native shape and maintain its secondary, tertiary, and quaternary structure.

Stereochemistry

Stereocenters are designated as having R or S configurations based on the arrangement of their substituents and the priority ranking of each. Carbohydrates with multiple stereocenters are given an L or D designation based on the configuration of the highest-numbered stereocenter. Almost all carbohydrates found in nature are in the D configuration.

Oxidative phosphorylation

The ETC produces a pH difference across the inner mitochondrial membrane as protons are pumped from the matrix to the intermembrane space. Transporting protons from high- to low-concentration areas is energetically favorable but prohibited by the membrane. Normally, the only path by which protons can cross the membrane is through ATP synthase, which can couple the released energy to ATP synthesis from ADP and inorganic phosphate (Pi). The addition of FCCP provides an alternate pathway to carry protons across the membrane from high concentration to low concentration. However, protons that enter the mitochondrial matrix by this means will not interact with ATP synthase, and therefore will not drive ATP synthesis. This phenomenon is called "decoupling." Because fewer protons are being used to drive ATP synthase, the rate of ATP synthesis will decrease. (Choice B) Carrying protons from the matrix to the intermembrane space would increase the rate of ATP synthesis, as more protons would be available to flow through ATP synthase. FCCP causes a decrease in ATP synthesis. (Choice C) FCCP will bring the proton gradient closer to equilibrium because it only carries protons down the concentration gradient. Protons can only move against the gradient (further from equilibrium) with the aid of energy input, as in the ETC. (Choice D) The passage states that the citric acid cycle is fully active in the presence of FCCP, and that reduced electron carriers are neither depleted nor accumulated. Therefore, the ETC must be able to consume the reduced electron carriers that the citric acid cycle produces, and must be fully active.

DNA sequencing via Sanger method

The Sanger method can be used to determine the sequence of a strand of DNA. What do we need to carry out Sanger sequencing? The method requires 5 basic ingredients: single-stranded DNA (template strand), a primer, DNA polymerase, deoxynucleotide triphosphates (dNTPs), and labeled dideoxynucleotide triphosphates (ddNTPs). You'll also need four reaction tubes. The single-stranded DNA is the DNA you are trying to sequence, and you can follow this description by looking at Figure 6. First, you must denature double-stranded DNA by heating in order to isolate the single-stranded DNA. The primer is a small piece of DNA that is complementary to (and binds to) the 5' end of this single-stranded DNA. DNA polymerase then binds to the double-stranded region formed by the interaction between the single-stranded DNA and the primer, and it begins extending the primer by adding dNTPs complementary to the nucleotides on the template strand.

CAC

The citric acid cycle is an oxidative process in which a six-carbon molecule (citrate) is converted into a four-carbon molecule (oxaloacetate). Reducing the number of carbons is achieved by two successive oxidative decarboxylation events in which carbon dioxide (CO2) is released. The second event, referenced in the passage as the oxidation of α-ketoglutarate, requires α-ketoglutarate dehydrogenase (αKDH). This enzyme removes a CO2 from α-ketoglutarate (five carbons) and forms a bond between a carbon atom and coenzyme A to produce succinyl-CoA, a four-carbon molecule. Figure 1 shows that α-ketoglutarate is oxidized at a significantly lower rate in KO mice than in WT mice. Because all livers were treated with the same amount of α-ketoglutarate, this indicates that αKDH works more slowly in KO mice than in WT mice under identical experimental conditions. Therefore, the product of α-ketoglutarate oxidation, succinyl-CoA, will most likely be depleted in KO livers. (Choice B) Figure 2 shows that oxygen consumption is not affected by SIRT4, indicating that the electron transport chain and oxidative phosphorylation are fully functional. Because this is the primary means of ATP generation under aerobic conditions, it is unlikely that ATP levels are significantly reduced in KO mice. (Choice C) Isocitrate is a precursor to α-ketoglutarate in the citric acid cycle. Because α-ketoglutarate is consumed more slowly in KO livers, isocitrate is also likely to be consumed more slowly, and will accumulate. (Choice D) Figure 2 shows that KO mouse livers consume oxygen at the same rate as WT livers, indicating that the Cyt-C necessary to reduce oxygen is present at the same level in both genotypes.

CAC

The electron transport chain (ETC) depends on the production of NADH and FADH2 by the citric acid cycle and other metabolic pathways. Complex II is a flavoprotein (an oxidoreductase that contains an FAD or FMN prosthetic group) that is also known as succinate dehydrogenase. It is part of both the ETC and the citric acid cycle. In Complex II, electrons are transferred from succinate to an FAD molecule that is covalently attached to the protein complex, producing fumarate and FADH2. The FADH2 then transfers its electrons to ubiquinone, forming ubiquinol and regenerating FAD. Oxygen is the final electron acceptor in the ETC and is consumed only if sufficient NADH and FADH2 are present. Increasing the rate of FADH2 synthesis increases the rate of electron transfer and therefore increases the rate of oxygen consumption. In Figure 2, researchers measured oxygen consumption to assess ETC output under various conditions. They most likely added succinate to stimulate FADH2 synthesis (and subsequent electron transfer) by Complex II. (Choice A) The addition of succinate can indirectly stimulate NADH synthesis at the downstream step catalyzed by malate dehydrogenase. However, Complex I only consumes NADH and does not synthesize it. (Choice B) Succinyl-CoA synthetase produces GTP only when it consumes succinyl-CoA to produce succinate. Adding succinate would shift the equilibrium to favor the reverse reaction and would not stimulate GTP synthesis. (Choice D) Ubiquinol is synthesized in Complex I and Complex II of the ETC. It is not synthesized by fumarase.

Membrane Fluidity

The fluidity and rigidity of plasma membranes is partially dependent on the structure of fatty acyl tails in the lipid bilayers. Increasing hydrophobic interactions between fatty acyl tails leads to decreased motion and more rigidity. The length and number of double bonds in a fatty acyl tail influence the number of hydrophobic interactions possible. Short tails participate in fewer hydrophobic interactions than long tails and therefore result in increased fluidity. Carbon-carbon double bonds in fatty acyl tails are typically in the cis-conformation and introduce a bend in the tail, preventing adjacent tails from interacting efficiently with each other. A higher number of double bonds leads to decreased interaction and greater fluidity. The lipid in Table 3 that has the shortest fatty acyl tail length and the most double bonds will make the greatest contribution to membrane fluidity. GL6 and GL3 have the shortest fatty acyl tails (20 carbons), but GL6 has more double bonds (one instead of zero). Therefore, GL6 makes the greatest contribution to membrane fluidity.

can

The passage indicates that lactate is transported into neurons through MCT2 channels to serve as a fuel source. Inside the cell, it is then used to generate pyruvate and enters the mitochondria for further processing. In the mitochondria, pyruvate is converted to acetyl-CoA by decarboxylation (release of a carbon atom as CO2). Acetyl-CoA then reacts with oxaloacetate to enter the citric acid cycle. Within this cycle, another carbon atom is released as CO2 when isocitrate is converted to alpha-ketoglutarate, and a third is released when alpha-ketoglutarate becomes succinyl-CoA. Therefore, the three carbon atoms in lactate are ultimately removed by three decarboxylation events, and their metabolic fate is to be released as CO2. (Choice A) Triacylglylcerides are composed of three fatty acid chains connected to a glycerol backbone. The fatty acids are synthesized in the cytosol of cells, not in the mitochondria. (Choice B) Conversion of lactate to glucose requires gluconeogenesis, which primarily occurs in the cytosol rather than the mitochondria. Gluconeogenesis requires energy input, so if lactate enters gluconeogenesis, it would not be acting as a fuel. (Choice C) Glycogen is produced by the formation of bonds between glucose units. Glycogen production occurs in the cytosol rather than the mitochondria. In addition, lactate in neurons cannot be converted to the G6P needed for glycogen synthesis because neurons lack the enzymes necessary for gluconeogenesis.

Gluconeogenesis

The physiological effects of von Gierke disease include low blood glucose and high lactate levels as G6Pase inactivation shuts down the pathways that control them. One of these pathways, gluconeogenesis, consumes lactate to produce glucose, and blocking any enzyme of gluconeogenesis would therefore likely lead to effects similar to those of von Gierke disease. Phosphoenolpyruvate carboxykinase (PEPCK) catalyzes the second step in gluconeogenesis, the conversion of oxaloacetate to phosphoenolpyruvate, and its inactivation can indeed lead to both lactate buildup and glucose depletion in the blood. (Choice A) Glucose 6-phosphate dehydrogenase (G6PDH) is an enzyme of the pentose phosphate pathway. Although G6PDH uses glucose 6-phosphate as a substrate, it is not involved in gluconeogenesis. (Choice B) Pyruvate decarboxylase converts pyruvate to acetaldehyde as the first step in the production of ethanol by yeast. Humans do not have this enzyme. (Choice D) Acyl-CoA dehydrogenase catalyzes the first step in fatty acid catabolism and does not affect blood glucose or lactate levels.

Western Blot

The protein concentrations detected are represented by bands, which may vary in width/intensity (corresponding to greater protein abundance). Proteins that are ubiquitously expressed (ie, they have consistent concentration levels across all cell/tissue types) are used as loading controls to show that all lanes in the gel had the same amount of sample loaded into them. Properly loaded gels have the same amount of loading control in every lane. The most common loading controls are proteins necessary for baseline cellular function that are transcribed from housekeeping genes (eg, α- and β-tubulin proteins are structural/mobile cytoskeleton components). Based on this fact, α-tubulin was used as a loading control as it can be detected at consistent levels in all cell types and has a molecular weight different from HIF-1α. The consistent expression of α-tubulin in the gel under all conditions confirms that differences in HIF-1α expression were not a result of experimental errors in protein loading or detection.

Glucose

This is an aldose sugar, and in its cyclic form, it is classified as a pyranose, meaning it is a six-membered ring.

Fatty Acid Metabolism Regulation

The reactions involved in lipid synthesis are sensitive to hormones, such as insulin and glucagon, that control blood sugar (glucose) levels. Insulin reduces blood sugar levels by increasing glucose uptake and storage, whereas glucagon increases the production of glucose and its release into the bloodstream. Ongoing digestion after a meal leads to high blood sugar levels that induce the release of insulin and the suppression of glucagon production. Liver cells, in particular, respond to insulin by increasing their rates of lipid synthesis and mobilization. In the mitochondria, insulin activates pyruvate dehydrogenase (PDH), which converts pyruvate into acetyl-CoA. The accumulated acetyl-CoA is transported to the cytoplasm by combining with oxaloacetate to form citrate, which can exit the mitochondria. In the cytoplasm, citrate is converted back to acetyl-CoA and oxaloacetate. Insulin activates both acetyl-CoA carboxylase (ACC) and fatty acid synthase (FAS), which convert acetyl-CoA to malonyl-CoA and catalyze the synthesis of 16-carbon fatty acid chains from malonyl-CoA, respectively.

Isoelectric focusing

There is a stationary pH gradient (ranging from about 0 to 14) inside the gel in addition to the electric field. The lowest pH is found near the positive side of the gel (anode), and the highest pH is found near the negative side of the gel (cathode). If you add components to this gel, they will migrate until reaching a region where the pH is equal to their isoelectric point, which is known as the pI. The isoelectric point is the pH at which the protein has a completely neutral charge. How do we determine the isoelectric point of a protein? Let's go back to those acidic and basic side chains. Acidic side chains will be deprotonated and negatively charged at a high pH. As you decrease in pH, however, more and more of these side chains will be protonated. In fact, most acidic side chains will be protonated by the time you get to a pH of about 2. Basic side chains, on the other hand, will be protonated and positively charged at low pH. As you increase pH, more and more of these side chains will become deprotonated. Most basic side chains will be deprotonated by a pH of about 12. So, as you increase the pH, negative charges are gained whereas decreasing pH results in gaining positive charges. The isoelectric point is the pH at which the number of negative charges is equal to the number of positive charges and your protein has a net neutral charge. Once your protein is neutrally charged, it will stop moving, and it won't be attracted to the positive or negative poles of the gel!

Reducing SDS-PAGE

There's one small thing that we haven't told you about SDS-PAGE yet: adding SDS completely denatures (or straightens) the protein except at places where there are disulfide bonds. Disulfide bonds are formed by the connection between two different cysteine side chains of a protein, and you can think of them as taping two points on a string together. This might cause the protein to travel more slowly than if these two points weren't attached. In order to break these disulfide bonds, you have to add a reducing agent that reduces the single disulfide S-S bond to two S-H bonds. This is known as reducing SDS-PAGE. By using reducing SDS-PAGE, you ensure that all of the higher structure of a protein has been eliminated, including any disulfide bonds.

Metabolism Mechanisms

To determine whether a metabolic pathway proceeds by a proposed mechanism, a critical component of the proposed mechanism should be blocked and the effects observed. The passage presents the hypothesis that during exercise, neurons receive fuel from astrocytes in the form of lactate, which was generated by degrading glycogen. If this prediction is correct, then neurons should use lactate from astrocytes to produce ATP. The passage also states that lactate enters neurons through MCT2 channels. Inhibiting these channels will prevent lactate generated by astrocytes from entering neurons. If neurons rely on lactate from astrocytes as a fuel source, but lactate is blocked from entering neurons, the level of ATP production would be expected to decrease. Therefore, measuring ATP production in the hippocampus in both the presence and absence of MCT inhibitors would help determine whether lactate from astrocytes can serve as a fuel source in neurons.

Hydrophillic and Hydrophobic interactions

Typically, hydrophilic amino acids (polar and charged) are found on the outside of proteins, where they interact favorably with water molecules. Hydrophobic amino acids are usually shielded from the solvation layer in the protein interior or found at protein-protein interfaces. This arrangement is entropically favorable, as water molecules cannot form hydrogen bonds with hydrophobic residues, and therefore must organize around these residues. Leucine is hydrophobic, and leucine zipper layers act as a watertight seal around the zero ionic layer, in which a positively charged, basic arginine (R) residue interacts with polar glutamine (Q) residues at the interface of the complexed proteins. NSF destabilizes the cis-complex by breaking the leucine zipper seal (unzipping). Without the watertight seal provided by the leucine zipper, the charged and polar residues of the zero ionic layer are exposed to water. Because water is polar, exposure of the charged and polar residues results in preferential binding to water and resultant destabilization of the cis-complex.

Triacylglyceride Metabolism

When lipids must be degraded (catabolized) for fuel, triglycerides stored in adipocytes are first hydrolyzed by hormonally controlled lipases, yielding three fatty acids and one glycerol molecule. The fatty acids are released into the bloodstream and delivered to fuel-dependent tissues by albumin. Subsequently, the fatty acids are activated and enter the mitochondria for β-oxidation, which is their major catabolic pathway. Glycerol cannot be metabolized further by adipocytes; it is secreted into circulation and transported to the liver, where it is phosphorylated to glycerol-3-phosphate by glycerol kinase. Glycerol-3-phosphate is then converted to dihydroxyacetone phosphate (DHAP) by glycerol phosphate dehydrogenase. DHAP is isomerized to glycerladehyde-3-phosphate, an intermediate compound of glycolysis. (Choice B) Although glycerol (part A) can be incorporated into gluconeogenesis, it is an anabolic, not a catabolic pathway. Furthermore, free fatty acids (part B) are the final products of fatty acid synthesis, not the initial reagents. (Choice C) The reagents for fatty acid synthesis are acetyl-CoA and malonyl-CoA. Fatty acid chains with 16 carbon atoms (part B) do not enter fatty acid synthesis. Glycerol (part A) is not involved in fatty acid synthesis but is converted to DHAP for glycolysis. (Choice D) Glycogenolysis is a separate catabolic pathway that breaks down glycogen stored in the liver into glucose to increase ATP production. Free fatty acids (part B) are not glycogenolysis reagents, but glycerol (part A) can enter glycolysis.

Western blot

When you're looking for a specific protein, you use a Western blot. The only difference between this technique and the technique used for Northern and Southern blotting is the type of probe used to search for a specific protein. Rather than using complementary strands of nucleic acids as we did in the other blotting techniques (there's no such thing as a complementary strand to a protein), you need to use antibodies that recognize single proteins with high specificity. In Western blotting, you generally use two antibodies. Your first antibody specifically binds to your target protein and is known as the primary antibody. Rather than labeling this antibody with a reporter, you use a labeled secondary antibody that recognizes your primary antibody. The secondary antibody is often labeled with a light-producing substance or radioactive substance as is the case for Northern and Southern blots. We conduct Western blotting as follows: gel electrophoresis -> immobilization of proteins -> primary antibody wash -> rinse away unbound primary antibodies -> secondary antibody wash -> rinse away unbound secondary antibodies -> look for signal.

Northern blot

ou can use a Northern blot to identify the presence of a specific RNA strand in your sample. After using gel electrophoresis to separate RNA strands by size, you must transfer the RNA strands to another surface (nitrocellulose) and immobilize them to this surface using UV light. After the RNA is immobilized to the surface with UV light, you can introduce a labeling strand or probe to the already separated RNA strands, and the probe will specifically bind to the single RNA strand you are searching for. Note: the probe is often just an RNA strand that has a complementary sequence to the RNA strand you are looking for and a label so that you can visualize it! In order to visualize the area where the probe has bound, the probe is generally labeled with a reporter. This reporter, which can take many forms, must be able to emit some sort of signal that can indicate where the probe is bound. In many cases, this reporter gives off colored light or a radioactive signal.

Fatty Acid Oxidation

β-oxidation is the breakdown of fatty acid chains into acetyl-CoA through a series of oxidation reactions in the mitochondrial matrix that produce ATP. Short- and medium-chain fatty acids can diffuse through the inner and outer membranes of the mitochondria, but long-chain fatty acids obtained from lipid stores must undergo a set of reactions to enter the mitochondrial matrix. First, fatty acids in the cytoplasm are activated when the enzyme acyl synthetase attaches them to the carrier molecule CoA using ATP. Subsequently, in the rate-limiting step of the reaction, carnitine palmitoyltransferase I (CPTI) converts fatty acyl-CoA molecules into fatty acylcarnitine, which enters the intermembrane space. Fatty acylcarnitine is then moved by acylcarnitine translocase across the inner membrane and into the matrix. Lastly, to begin the oxidation reactions, carnitine palmitoyltransferase II (CPTII) on the inner membrane reconverts fatty acylcarnitine into fatty acyl-CoA, which can be broken down into acetyl-CoA. Therefore, researchers interested in confirming that fatty acid oxidation is not impaired would need to ensure that translocation of acylcarnitine into the mitochondrial matrix is functional. (Choice A) During fatty acid synthesis, acetyl-CoA in the mitochondria is converted into citrate and then transported into the cytoplasm, where it is reconverted into acetyl-CoA and oxaloacetate. This does not occur during β-oxidation. (Choice B) CO2 is released from malonyl-CoA during fatty acid synthesis but not during fatty acid oxidation. (Choice D) β-oxidation reactions transfer electrons from fatty acids to the electron carriers NAD and FAD to produce their reduced forms. Therefore, reduced electron carriers are created, rather than depleted, by β-oxidation.

Isoelectric Focusing

A peptide's isoelectric point (pI) is the pH at which it has no net charge and therefore does not migrate in an electric field. This point is determined by the ionizable amino acid side chains present in the peptide. Seven side chains (R, K, H, Y, C, E, and D) are "ionizable," meaning their charges can be altered as protons are added or removed with varying pH: Ionizable side chains that contain oxygen or sulfur (Y, C, E, and D) are negatively charged when deprotonated and tend to decrease pI. For side chains that can be negatively charged, a lower pKa correlates with a stronger effect on pI. Therefore, aspartate and glutamate cause the greatest decreases in pI. Ionizable side chains that contain nitrogen (R, K, and H) are positively charged when protonated and tend to increase pI. Side chains that can be positively charged have a greater effect if the pKa is high, and therefore arginine and lysine cause the greatest increases in pI. All other side chains are nonionizable and do not affect pI.

How much ATP would be produced by 10 NADH and 2 FADH2 molecules? A) 11 B) 7 C) 32 D) 28

D - For every molecule of NADH, 2.5 ATP is eventually produced. For every molecule of FADH2, 1.5 ATP will be produced. 10 NADH molecules would result in 25 ATP, and 2 FADH2 would result in 3 ATP. This gives us a total of 28 ATP (choices A, B, and C are incorrect).

Step 3: Phosphofructokinase 1 (PFK-1)

Phosphofructokinase 1, also known as PFK-1, catalyzes the rate-limiting step of glycolysis. It uses ATP to catalyze the irreversible conversion of fructose 6-phosphate into fructose 1,6-bisphosphate. This step is highly regulated. Citrate (a metabolic product of aerobic respiration) and ATP have a negative feedback effect on PFK-1. Why would this be? The presence of citrate and/or ATP indicates that the cell's energy needs are being met, and thus signals that the glycolysis pathway is not immediately needed. Since this step is an irreversible conversion—and thus requires energy to be performed—shutting down PFK-1 when it is not needed allows the cell to conserve valuable energy. On the other hand, the presence of AMP (adenosine monophosphate) indicates low energy in the cell and activates PFK-1.

Protein Binding

The affinity of a protein for its ligand is measured by its dissociation constantKd, which quantifies the tendency of the protein-ligand complex to degrade into protein and ligand. Conceptually, Kd represents the concentration of ligand at which half of the proteins in the solution are bound to ligands and half are not. A smallKd indicates a low tendency for the complex to break apart and corresponds to a high affinity because only a small amount of ligand is required to achieve a high amount of binding.

ETC

The electron transport chain consists of the reactions catalyzed by four membrane-bound complexes located in the mitochondrion's inner membrane. Several of these complexes belong to a class of enzymes called flavoproteins, which use FAD embedded within their structure to accept and transfer electrons. We've listed these first four complexes below. While you won't need to memorize these complexes, it may be useful to be familiar with their names. Note that each complex pumps protons from the mitochondrial matrix into the intermembrane space—with the exception of complex II. Complex I: NADH-CoQ oxidoreductase Complex II: Succinate-CoQ oxidoreductase Complex III: CoQH2-cytochrome C oxidoreductase Complex III is also the site of the Q cycle. Complex IV: Cytochrome C oxidase Here, cytochrome C is oxidized and an oxygen molecule is reduced, resulting in the formation of two water molecules.

Step 10: Pyruvate kinase

The final enzyme of glycolysis, pyruvate kinase, catalyzes the irreversible conversion of phosphoenolpyruvate into pyruvate, or the removal of a phosphate group from phosphoenolpyruvate. This generates one ATP per molecule of phosphoenolpyruvate (or 2 ATP per glucose molecule).

Pentose Phosphate Pathway

The pentose phosphate pathway (PPP) ultimately produces ribose-5-phosphate: a molecule that is critical for nucleotide synthesis. This pathway takes place in the cytoplasm of the cell. While the MCAT will not test you on the specific details of nucleotide synthesis, it's important to know the byproducts of the PPP and why they are important. The PPP begins with the first step of glycolysis: the phosphorylation of glucose into glucose-6-phosphate. The first phase of the pathway, known as the oxidative phase, results in the production of two NADPH molecules and a molecule known as ribulose-5-phosphate. The second phase of the pathway, known as the non-oxidative phase, can result in multiple possible outcomes depending on the energy needs of the cell. The pathway can produce glycolytic intermediates to feed into glycolysis or ribose-5-phosphate to feed into nucleotide synthesis. It's important to note that the PPP does not produce any NADH—the type of electron carriers required for oxidative phosphorylation and ATP synthase. Rather, it produces NADPH, an important electron carrier in fatty acid synthesis.

Fructose

This monosaccharide is a ketose sugar, and in its cyclic form, it is classified as a furanose, meaning it is a five-membered ring. (You may also see certain sugars referred to as hexoses and pentoses, meaning they have six and five carbon atoms, respectively. Note that it's possible for a sugar to be both a hexose and a furanose—the sugar would simply have five atoms within the ring and at least one carbon atom outside of the ring.)

Phospholipid Bilayer

To speed up the process of switching phospholipids between layers (or "leaflets"), special enzymes can be used. Flippases move a phospholipid from an outer leaflet to an inner leaflet. Floppases move a phospholipid from an inner leaflet to an outer leaflet. Since these are energetically demanding processes, ATP is required for both of these enzymes to function.

Lactic acid fermentation

Under anaerobic conditions, or when there is a lack of oxygen, the pyruvate molecules generated by glycolysis will undergo fermentation. During this process, lactate dehydrogenase catalyzes the conversion of pyruvate into lactate (another 3-carbon molecule) and generates NAD+ as a byproduct. Since it is the only enzyme in the process, it is the rate-determining step. Why would our cells perform lactic acid fermentation, if it does not yield any ATP? The primary purpose of lactic acid fermentation is to replenish the NAD+ that was converted into NADH during glycolysis by glyceraldehyde 3-phosphate dehydrogenase. This makes additional NAD+ available to glycolytic enzymes, so our cells can continue producing 2 ATP at a time through glycolysis. Lactic acid fermentation is part of a larger pathway known as the lactic acid cycle, or Cori cycle. Lactate generated by the muscles is sent to the liver through the bloodstream. The liver has specialized enzymes that can convert lactate into glucose, which is then sent back to the muscles.

Which type of bond does debranching enzyme hydrolyze? A) α-1,4 glycosidic linkage B) α-1,6 glycosidic linkage C) β-1,4 glycosidic linkage D) Both A and B

D

Step 1: Hexokinase/Glucokinase

Glucokinase is found in hepatocytes (liver cells) and pancreatic β-islet cells. It is activated by insulin. Hexokinase, on the other hand, is a bit more universal and found in most tissues. Both enzymes serve the same function: to use ATP to catalyze the irreversible phosphorylation of glucose. The product of this reaction, glucose 6-phosphate, is now unable to spontaneously diffuse out of the cell. Glucose 6-phosphate also has an inhibitory effect on the hexokinase enzyme.

Insulin

Is released by β-cells when blood glucose levels are high. It promotes catabolic pathways, such as glycolysis, to derive energy from glucose. It also promotes anabolic pathways, such as glycogenesis, fatty acid synthesis, and peptide synthesis.

Somatostatin

Is released by δ-cells in response to high blood glucose and amino acid levels. It inhibits the secretion of insulin and glucagon.

The affinity of a protein for its ligand was measured by isothermal titration calorimetry. The experiment was repeated multiple times to mitigate error by: A) Assessing the precision of the results B) Verifying the calibration of the instruments C) Assessing the validity of the experiment D) Verifying the accuracy of the results

A - A measurement is said to be precise (reliable) if similar results are obtained on repeated trials. It is accurate if it correctly represents the real-world value as assessed by comparison with a "gold standard," the best available benchmark test that is considered accurate. One or more gold standards are used to calibrate the instrument being used for measurement, allowing researchers to be confident that future measurements are accurate. If experimental results are both precise and accurate, they are said to be valid, as they reliably represent the correct value. In this scenario, an experiment was repeated multiple times. This allowed researchers to determine whether repeated trials yield similar results. Therefore, the purpose of the repetition was to assess the precision of the results. (Choices B and D) To assess accuracy, the instrument used for measurements must be calibrated against a gold standard. Because no gold standard calibration is mentioned, accuracy of the results cannot be determined. (Choice C) Valid experiments must have results that are both precise and accurate because they must correctly measure the real-world value of the phenomenon of interest. Because the results were not compared to a gold standard, accuracy cannot be determined.

Replacing native gel electrophoresis with nonreducing sodium dodecyl sulfate (SDS) polyacrylamide gel electrophoresis (PAGE) during the purification step results in eluates with no enzymatic activity. The best explanation of this observation would be: A) ALDH2 must retain its quaternary structure to catalyze reactions B) The SDS-PAGE breaks disulfide bonds between ALDH2 subunits C) ALDH2 has the mobility of a 55-kD protein in native PAGE D) The primary structure of the ALDH2 active site has been modified

A - ALDH2 proteins were initially isolated by native polyacrylamide gel electrophoresis (PAGE), which is used to separate proteins by charge and molecular weight without affecting their three-dimensinal structure. In this procedure, an electric field is applied to a gel and proteins migrate toward one of the electrodes based on their mass and charge. Under these conditions, proteins maintain their biological function because they retain their tertiary and quaternary structure during native PAGE purification. In contrast, SDS-PAGE separates proteins only on the basis of molecular weight. To accomplish this, proteins are coated with sodium dodecyl sulfate (SDS), a detergent that confers a negative charge. Under these conditions, each protein in the gel has a similar charge and migrates toward the anode. Although this technique is superior for estimating the molecular weight of a protein, detergents such as SDS disrupt noncovalent interactions and cause proteins to unfold, or become denatured. As a result, proteins exist only in their primary structure (linear sequence of amino acids) and cannot perform their biological function. Replacing native PAGE with SDS-PAGE results in the denaturation of active ALDH2 tetramers. The lack of enzymatic activity in eluates obtained from SDS-PAGE purification indicates that ALDH2 proteins lost their quaternary structure, preventing them from functioning. (Choice B) Reducing agents can separate multisubunit proteins into individual subunits by cleaving disulfide bonds, but nonreducing SDS-PAGE does not contain reducing agents. (Choice C) ALDH2 would migrate as a 220-kD tetramer (four subunits) in native PAGE. (Choice D) SDS does not alter the primary structure (amino acid sequence) of proteins.

A protein consisting of domains A, B, and C binds a ligand. Which experimental result would provide the best evidence that domain A is involved in binding and domains B and C are not? A) A mutant containing only domain A binds the ligand with the same Kd as the wild-type protein B) A mutant containing only domains B and C binds the ligand with a lower Kd than the wild-type protein C) A mutant containing only domain A binds the ligand with a higher Kd than the wild-type protein D) A mutant containing only domains B and C cannot bind the ligand

A - In this question, the protein consists of domains A, B, and C. If domain A is the only domain involved in ligand binding, then binding affinity will be unaffected in the absence of domains B and C. Therefore, if a mutant that contains only domain A can bind the ligand with the same Kd as the wild-type protein, this would suggest that only domain A is involved in ligand binding, and that the other domains are not involved. In contrast, if domains B or C are involved in binding, removing them would affect Kd.

Why might ataxin-3 aggregates become insoluble? A) The change in conformation from alpha-helix to beta-sheet exposes hydrophobic residues B) The conformational change from alpha-helix to beta-sheet disrupts side-chain ionic interactions C) Interactions between glutamine side chains exclude them from interacting with water D) Beta-sheets expose more of the hydrophobic protein backbone than do alpha-helices

A - Disruption of a protein's native conformation may force exposure of hydrophobic residues to the aqueous environment. Hydrophobic residues are unable to form hydrogen bonds with water, so when they are exposed to an aqueous environment, water molecules are forced to form a rigid network to satisfy hydrogen bond requirements. These interactions are thermodynamically unfavorable. To avoid these unfavorable interactions, hydrophobic molecules aggregate to decrease the total surface area exposed to water. This phenomenon, known as the hydrophobic effect, is a driving force in protein folding. Hydrophobic residues are normally buried in the protein core to minimize exposure to the surrounding aqueous environment. As the alpha-helices of ataxin-3 are converted to beta-sheets, the protein must undergo an overall conformational change. This change may prohibit hydrophobic residues from being buried, and may expose them to water. This in turn can cause the hydrophobic residues of multiple proteins to aggregate, to the exclusion of water. By definition, the inability to interact with a solvent (water, in this case) is insolubility. (Choice B) Glutamine has an uncharged polar side chain. It has no ionic interactions to disrupt. (Choice C) The interaction of glutamine side chains with each other does not prevent their interaction with water. Hydrogen bonds between side chains may transiently break to bond with water, and vice versa. (Choice D) Peptide backbones are not very hydrophobic because they have both a carbonyl and an amino group that can form hydrogen bonds.

Which cofactor is regenerated by lactate synthesis? A) NAD+ B) NADH C) NADP+ D) NADPH

A - During glycolysis, NAD+ is reduced to NADH as glyceraldehyde 3-phosphate is oxidized to 1,3-bisphosphoglycerate; therefore, NAD+ is required for glycolysis to continue. Under aerobic conditions, NAD+ can be regenerated by oxidation of NADH in the electron transport chain (ETC). Under anaerobic conditions, the ETC cannot regenerate enough NAD+, and another source is required. Under these conditions, lactate synthesis takes place and oxidizes NADH to NAD+ as pyruvate (the final product of glycolysis) is reduced to lactate. (Choice B) Lactate synthesis consumes NADH to regenerate NAD+. (Choices C and D) NADP+ and NADPH are used in the pentose phosphate pathway and anabolic pathways such as fatty acid synthesis, but not in glycolysis. They therefore do not need to be regenerated or consumed by lactate synthesis.

An increase in which of the following would lead to decreased ethanol production in yeast? A) Partial pressure of oxygen B) Pyruvate kinase activity C) Concentration of glucose D) Krebs cycle inhibitors

A - Glycolysis (the breakdown of glucose) consumes NAD+ to produce NADH, and for glycolysis to continue, NAD+ must be regenerated. Under aerobic (high-oxygen) conditions, the NAD+ used in glycolysis is regenerated by the electron transport chain (ETC) in a process known as respiration. Because oxygen is required as the final electron acceptor in the ETC, insufficient oxygen (in anaerobic conditions) will prevent the ETC from functioning and inhibit the regeneration of NAD+. Accordingly, in anaerobic conditions, an alternate process called fermentation is necessary to regenerate NAD+. In higher eukaryotes, fermentation occurs by reduction of pyruvate to lactate. However, in bacteria and lower eukaryotes such as yeast, fermentation occurs through decarboxylation of pyruvate to form acetaldehyde, followed by reduction of acetaldehyde to form ethanol. If the partial pressure of oxygen in the environment were increased, yeast cells would exhibit increased function of the ETC and the need for fermentation to regenerate NAD+ would decrease. Because fermentation in yeast generatesethanol, increased partial pressure of oxygen (and decreased fermentation) would most likely result in reduced ethanol production. (Choice B) Pyruvate kinase catalyzes the transfer of a phosphate group from phosphoenolpyruvate to ADP, forming pyruvate and ATP. If pyruvate kinase activity were increased, more pyruvate would be produced. At a given O2 partial pressure, fermentation of this pyruvate would likely increase (not decrease) ethanol production. (Choice C) The carbon atoms in ethanol originate in glucose. Therefore, increased glucose concentration would make more carbon available, likely increasing ethanol production. (Choice D) The Krebs cycle (citric acid cycle) uses acetyl-CoA, synthesized from pyruvate, to produce NADH and FADH2 for the ETC. Pyruvate that is converted to acetyl-CoA cannot undergo fermentation. Krebs cycle inhibition would increase the amount of pyruvate available for fermentation and would most likely increase ethanol production.

Where in the cell does glycolysis occur? A) The cytoplasm B) The inner membrane of the mitochondria C) The mitochondrial matrix D) The intermembrane space

A - Glycolysis occurs in the cytoplasm (choice A is correct). The electron transport chain occurs along the inner membrane of the mitochondria (choice B is incorrect). Pyruvate oxidation and the citric acid cycle occur in the mitochondrial matrix (choice C is incorrect). The intermembrane space is where hydrogen ions are pumped during the electron transport chain, resulting in its acidification (choice D is incorrect).

The first step of the autoprocessing reaction mechanism described in the passage involves modification of a thiol in the active site by an amino acid functioning as a general base. This action is likely to enhance protease activity by: A) The deprotonation of cysteine, which results in an anionic sulfur with increased nucleophilicity B) The protonation of serine, which results in a structure with improved capacity to act as a leaving group C) The deprotonation of serine, which results in an anionic oxygen with increased nucleophilicity D) The protonation of cysteine, which results in a structure with improved capacity to act as a leaving group.(13%)

A - In general, amino acids containing thiol (-SH) or hydroxyl (-OH) groups can act as nucleophiles in biological reactions. Nucleophiles are attracted to the positive charge of atomic nuclei, and the greater the negative charge possessed by a molecule, the greater its nucleophilicity. Deprotonation of the cysteine thiol or the hydroxyl group of serine results in negatively charged sulfur or oxygen atoms and enhances their otherwise weak nucleophilicity. The question states that the mechanism begins with the modification of a thiol group. Cysteine is the only amino acid with a thiol (-SH) in its R group. The question further states that the thiol is modified by an amino acid (histidine, in this case) acting as a general base. Bases are proton acceptors; therefore, histidine must be deprotonating, not protonating, cysteine. The deprotonation of cysteine would likely enhance the rate of hydrolysis by converting a cysteine residue in the active site into an anionic—and more nucleophilic—sulfide residue. This sulfide can donate electrons to the carbonyl carbon of the peptide in a process called nucleophilic attack, ultimately leading to hydrolysis. This class of enzyme, known as a cysteine protease, is commonly employed in protein degradation. (Choices B and D) A general base will deprotonate, rather than protonate, a reaction partner. Although protonation would produce a good leaving group, it would not aid in attacking the carbonyl of a peptide bond. (Choice C) Deprotonation of serine would produce a good nucleophile, and serine proteases are common in nature. However, serine does not have a thiol group.

Is an arginine or a valine more likely to be a catalytically important amino acid based on what you know about its structure and chemical properties? A) Arginine B) Valine

A - In general, charged, polar, or nitrogen-containing amino acids are more likely to be involved in the catalytic steps of an enzyme. Why? These amino acids have interesting properties that allow them to carry out chemistry with the substrate. For example, a lone pair on a terminal nitrogen in lysine can serve as a nucleophile.

The fact that glycogenolysis in muscle cells is activated by signal molecules binding to G protein-coupled receptors suggests that glycogenolysis can be induced by which of the following? A) Activation of adenylate cyclase B) Inhibition of glycogen phosphorylase C) Increased blood glucose concentration D) Depletion of cytosolic cAMP

A - Intracellular glycogen levels are controlled by two opposing pathways, glycogenesis (synthesis) and glycogenolysis(degradation). In muscle cells, glycogenesis occurs when sufficient blood glucose induces insulin synthesis, and glycogenolysis is induced when hormones such as epinephrine bind G protein-coupled receptors and activate an intracellular signal cascade. Upon epinephrine binding, G protein-coupled receptors activate the enzyme adenylate cyclase, which converts ATP into cyclic AMP (cAMP) in the cytosol. Newly synthesized cAMP then allosterically activates protein kinase A, initiating a phosphorylation cascade that ultimately activates glycogen phosphorylase, the first enzyme in glycogenolysis. (Choice B) Because glycogen phosphorylase is the first enzyme in the glycogenolysis pathway, inhibiting it would lead to decreased glycogenolysis. (Choice C) Increased blood glucose levels lead to increased insulin expression. Insulin induces glycogenesis (synthesis of glycogen), the opposite of glycogenolysis. (Choice D) Cytosolic cAMP is required for activation of protein kinase A and, ultimately, glycogen phosphorylase. Depletion of cAMP would lead to decreased glycogenolysis.

Which of the following processes would be directly impaired in von Gierke disease? I. Glycogenolysis II. Gluconeogenesis III. Pentose phosphate pathway A) I and II only B) I and III only C) II and III only D) I, II, and III

A - Maintaining blood glucose levels is critical for proper body function, so several biological processes exist to maintain glucose levels between meals. They take place predominantly in the liver, and all produce glucose 6-phosphate (G6P). G6P must be dephosphorylated by G6Pase to be released as glucose into the bloodstream, so all of these processes will be directly impaired by the inactivation of G6Pase. They include the following: Glycogenolysis is the breakdown of glycogen into glucose. It is the first source of glucose during fasting and requires G6Pase in its final step (Number I). Gluconeogenesis is the synthesis of glucose from pyruvate and other precursors. It is essentially glycolysis in reverse with a few differences, including the conversion of pyruvate to oxaloacetate and the use of phosphatases such as G6Pase instead of kinases to remove phosphate groups (Number II). Gluconeogenesis and glycolysis are connected by the Cori cycle in which lactate, produced in the muscle during anaerobic glycolysis, is sent to the liver. There it is converted to pyruvate and undergoes gluconeogenesis. The newly synthesized glucose is then returned to the muscles through the bloodstream. (Number III) The pentose phosphate pathway is required for nucleotide and NADPH synthesis. Although the pathway includes G6P as a substrate, it does not require G6Pase activity, so it will not be directly impaired by von Gierke disease.

Which of the following most accurately describes the ribose component of a nucleotide triphosphate? A) It is a pentofuranose B) It is a pentopyranose C) It is a hexofuranose D) It is a hexopyranose

A - Nucleotide triphosphates (NTPs) consist of a nitrogenous base (adenine, guanine, cytosine, or uracil), a sugar (ribose), and a triphosphate. Ribose is a pentose (five-carbon sugar) that links to the nitrogenous base through a glycosidic bond. A phosphoester bond links the 5′ carbon of ribose to the triphosphate group. Carbohydrates can participate in glycosidic linkages only when in cyclic form (furanose or pyranose), so ribose must be cyclic in NTPs. In the furanose form, the 5′ hydroxyl group can readily participate in phosphoester bonds. Because the ribose in an NTP is a pentose in the furanose form, it is classified as a pentofuranose.

All of the following could confirm apoptosis due to increased oxidative stress in cells with an L156P mutation in Complex V EXCEPT: A) Inhibition of caspase activity B) Increased cytosolic cytochrome C C) Activation of proteolysis D) Increased mitochondrial superoxide concentration

A - Oxidative stress occurs when there is an overabundance of reactive oxygen species (ROS) in cells, which can cause damage when they react with cell membranes and other biological molecules. The ETC in the mitochondria produces ROS such as superoxide, hydrogen peroxide, and hydroxyl radicals when it fails to fully reduce oxygen to water. Normal levels of ROS (about 4% of oxygen consumed) are removed by enzymes such as peroxidase and superoxide dismutase. However, defects in Complex V and other ETC complexes can cause oxidative stress by increasing the mitochondrial concentrations of superoxide and other ROS beyond the ability of the cell to control (Choice D). Apoptosis is the controlled, programmed death of cells. It occurs in response to a variety of circumstances, including oxidative stress as well as DNA damage and certain developmental events. Apoptosis is initiated when the mitochondrial membrane is permeabilized in response to these environmental cues. This allows the release of cytochrome C from the mitochondria, resulting in increased cytosolic cytochrome C (Choice B). Cytochrome C induces proteolysis (Choice C) and other degradative pathways by activating caspase proteases (inactive caspase is found in healthy cells).

Which glycolysis intermediate could serve as the sole starting material for the synthesis of R5P in muscle tissue? A) Fructose 6-phosphate B) Fructose 1,6-bisphosphate C) Glyceraldehyde 3-phosphate D) Dihydroxyacetone phosphate

A - Ribose 5-phosphate (R5P) is a phosphorylated five-carbon sugar (pentose phosphate) that is produced in the pentose phosphate pathway (PPP). The PPP consists of two major parts: an irreversible oxidative phase and a reversible nonoxidative phase. Fructose 6-phosphate (F6P) can enter either phase by different means. F6P is readily converted to G6P by the enzyme phosphoglucose isomerase. G6P can then enter the oxidative phase of the PPP, where it is decarboxylated (loss of CO2) and converted to R5P. The nonoxidative phase recycles R5P by rearranging its carbon atoms to produce F6P and glyceraldehyde 3-phosphate (GAP). Because this phase is reversible, F6P and GAP can also combine and rearrange to become R5P. F6P can be converted to GAP, so F6P alone is sufficient as a starting material to enter this pathway. (Choice B) Fructose 1,6-bisphosphate (F1,6BP) must be converted to F6P before it can enter the PPP. This can be accomplished in liver cells by the gluconeogenic enzyme fructose 1,6-bisphosphatase, but this enzyme is not present in muscle cells. (Choices C and D) GAP can only enter the PPP by combining with F6P. GAP can be converted to dihydroxyacetone phosphate (DHAP), and GAP and DHAP can combine to form F1,6BP. However, muscle cells lack the enzyme necessary to convert F1,6BP to F6P. These molecules are not sufficient for R5P synthesis in muscle.

During hypoxia, skeletal muscles have been shown to accumulate succinate and ADP molecules. Which of the following explains this result? A) Anaerobic conditions prevent mitochondrial regeneration of NAD+ and FAD B) Succinate dehydrogenase kcat is decreased after prolonged hypoxia C) Excess fumarate is converted back into succinate in ATP-rich conditions D) Overabundance of NADH stimulates succinyl-CoA production

A - The accumulation of succinate and ADP during hypoxia would occur because anaerobic conditions prevent the normal regeneration of the NAD+ and FAD needed for progression of the citric acid cycle. This cycle involves reactions that occur within the mitochondrial matrix and perform aerobic processing (oxidation) of pyruvate to generate CO2 and ATP, the cell's energy currency. The dehydrogenase enzymes in this cycle transfer hydride ions to an electron acceptor (eg, NAD+ or FAD), creating the high-energy electron carriers NADH and FADH2. NADH and FADH2 are then shuttled to the electron transport chain (ETC) in the inner mitochondrial membrane where proteins transfer the released electrons to oxygen, the final electron acceptor. NAD+ and FAD are also regenerated, and the electron movement between ETC proteins generates a proton motive force that drives ATP synthase to phosphorylate ADP to ATP. During hypoxia, less oxygen is available to accept electrons from the citric acid cycle; therefore, NAD+ and FAD are not regenerated in the mitochondria. Consequently, ADP concentrations increase (due to the absent proton motive force) as ATP production is significantly reduced. In addition, there is an accumulation of succinate and other citric acid cycle intermediates that require NAD+ or FAD for their oxidation reactions. (Choice B) Succinate dehydrogenase remains fully functional during hypoxia, so its kcat is unchanged. However, because FAD is a substrate of succinate dehydrogenase, hypoxia-induced depletion of FAD causes a decrease in the reaction rate. (Choice C) Fumarate is not converted back to succinate in ATP-rich conditions. Instead, excess citric acid cycle intermediates such as fumarate are used as amino acid precursor molecules. (Choice D) High concentrations of NADH inhibit the α-ketoglutarate dehydrogenase that generates succinyl-CoA from α-ketoglutarate. As a result, succinyl-CoA production is reduced, not stimulated.

If the pathological effects observed in patients with the L156P mutation are caused by decreased Complex V activity, the effects could be best counteracted by a drug that increases which of the following? A) Malate-aspartate shuttle activity to transfer cytosolic NADH to the mitochondria B) Potassium concentration in the mitochondrial matrix to decrease the charge gradient C) The pH of the intermembrane space to equalize it with the pH of the matrix D) Electron transfer from NADH to FAD via reduction of glycerol 3-phosphate to DHAP

A - The activity of Complex V depends on the number of protons available to flow through it. A decrease in activity could be counteracted by an increase in proton concentration in the intermembrane space, which is dependent upon the number of NADH and FADH2 molecules that enter the ETC. NADH pumps more protons into the intermembrane space than FADH2 does, so an increase in the number of NADH molecules entering the ETC would yield the greatest increase in Complex V activity. Glycolysis produces two cytosolic NADH molecules per glucose consumed, but these are unable to enter the mitochondrial matrix directly. Instead, the cytosolic NADH must transfer its electrons to the matrix through an intermediate. This can be achieved most efficiently by reducing oxaloacetate to malate, which can enter the mitochondrial matrix (malate-aspartate shuttle). In the matrix, malate can be oxidized back to oxaloacetate as part of the citric acid cycle, generating mitochondrial NADH in the process. This process effectively transfers cytosolic NADH to the mitochondria, increasing the total proton output by the ETC. Therefore, increased malate-aspartate shuttle activity would stimulate Complex V. (Choice B) The energy released when protons flow through Complex V is partially due to the difference in charge across the inner membrane. Decreasing the charge difference by adding potassium to the matrix would cause a reduction in ATP synthesis. (Choice C) ATP synthase is driven by protons flowing from a high concentration in the intermembrane space toward a low concentration in the matrix. Equalizing the pH of the intermembrane space and the mitochondria by raising the pH of the intermembrane space would destroy the proton gradient and stop ATP synthesis. (Choice D) The glycerol 3-phosphate shuttle effectively converts cytosolic NADH to mitochondrial FADH2 by oxidizingglycerol 3-phosphate to DHAP, not by reducing it. FADH2 pumps fewer protons than NADH does, so this system is less efficient than malate translocation at stimulating Complex V activity.

The amino acid methionine can be converted to succinyl-CoA to enter the citric acid cycle. How many NADH molecules can be produced by each methionine molecule that enters the citric acid cycle by this pathway? A) 1 B) 2 C) 3 D) 4

A - The citric acid cycle is a metabolic process that oxidizes a series of molecules to produce the reduced electron carriers NADH and FADH2 as well as the high-energy molecule GTP. The reduced electron carriers then enter the electron transport chain and contribute the electrons necessary for ATP production. Various molecules can enter the citric acid cycle at different points, and each produces a distinct number of reduced electron carriers, depending on the number of oxidative steps that remain in the cycle at the point of entry. According to the question, methionine enters the citric acid cycle by conversion to succinyl-CoA. At this point in the cycle, two oxidative steps remain. The first step is the conversion of succinate to fumarate, which produces one FADH2molecule. The second step is conversion of malate to oxaloacetate, which produces one NADH molecule. All other oxidative steps in the cycle require precursors to succinyl-CoA, which do not form in the metabolism of methionine. Based on this, entry of methionine into the citric acid cycle will yield one NADH molecule.

Based on information given in the passage, stabilizing interactions are most likely to occur between phosphate groups on IP6 and which of the following amino acids lining the Tcd binding pocket? A) Lys704 via electrostatic interactions B) Gly705 via hydrophobic interactions C) Cys707 via covalent bonding D) Asp709 via ionic interactions

A - The passage states that IP6 is stabilized in the binding pocket through salt bridging. In protein chemistry, salt bridging describes an electrostatic interaction that occurs between charged residues and ligands, stabilizing the protein conformation or strengthening protein-ligand interactions. Electrostatic interactions combine elements of both hydrogen bonding and ionic bonding. In proteins, it most commonly occurs between the anionic carboxylate of aspartic or glutamic acids and the cationic ε-amino group of lysine or the guanidinium group of arginine, but can potentially involve any pair of oppositely charged chemical groups. IP6 contains six phosphate groups, each bearing a −2 charge at physiological pH, so IP6 can only provide the anionic component of salt bridges; therefore, it must form salt bridges with cationic amino acids. Of the amino acids given, IP6 could potentially form a salt bridge with a cationic lysine residue. (Choice B) Glycine is neutral and nonpolar, and therefore cannot participate in salt bridge formation. Furthermore, salt bridging is not a hydrophobic interaction. (Choice C) Although cysteine residues can form covalent disulfide bonds with other cysteine residues, they do not participate in salt bridge formation. Salt bridging is a noncovalent interaction. (Choice D) Aspartate residues can participate in ionic interactions such as salt bridges, but are negatively charged, and IP6 can form a salt bridge only with a positively charged residue.

Scientists have been studying the actin-myosin crossbridge cycle. In one experiment, it is noted that myosin successfully binds ATP and releases ADP, but there is no apparent movement of actin microfilaments. Which of the following is a plausible explanation for this phenomenon? A) A mutation in the myosin causes it to attach to actin microfilaments directly instead of swinging the myosin head backward B) Myosin is unable to hydrolyze ATP C) Multiple myosin heads are moving in opposite directions, causing no net movement. D) Not enough information presented to answer the question

A - The question stem states that myosin is able to go through the entire cycle but unable to produce movement. If myosin is not able to swing its head backward, it will not be able to correctly perform its power stroke and move the actin microfilaments despite the presence of ATP hydrolysis (choice A is correct). Without the hydrolysis of ATP, myosin would not be able to release from actin and complete the crossbridge cycle (choice B is incorrect). Due to the polarity of actin, myosin heads can only move in one direction along a microfilament (choice C is incorrect).

One of the challenges of antimicrobial peptides is their short physiological half-life. To solve this problem, some researchers have begun investigating synthetic antimicrobial peptides composed of D-amino acids. Why might these peptides have an increased half-life in the body relative to natural antimicrobial peptides? A) Proteases cannot act on peptides made of D-amino acids B) D-amino acids are more soluble at physiological pH C) Peptide bonds between D-amino acids require more energy to break D) Hydrolysis of D-peptides is catalyzed by acids instead of bases

A - With the exception of glycine, which has no chiral center, amino acids are chiral molecules that can exist in two enantiomeric conformations, designated L and D forms. In nature, amino acids are almost always found in the L-conformation, so almost all biomolecules are designed to work with L-amino acids. Proteases typically recognize specific amino acid side chains and cleave the peptide bond nearest that side chain. When D-amino acids are used to make peptides, the side chain protrudes from the peptide at a different angle than it does in the L-conformation and is unrecognized by proteases. D-peptides therefore remain uncleaved and have a longer physiological half-life than their L counterparts. (Choice B) D- and L-amino acids have essentially identical solubility. (Choice C) A peptide bond is an amide bond between two amino acids. Amide bonds require approximately the same amount of energy to break, regardless of the chemical groups on either side of the bond. (Choice D) Hydrolysis of both L- and D-peptides can be catalyzed by strong acids or strong bases. Because physiological pH is nearly neutral, these modes of catalysis are essentially absent in the body.

Researchers hope to identify the antigen-binding region of the nanobody. Based on the information provided in Figure 2, (shows binding around ph pf 7) which of the following amino acids is most likely in this region? A) His B) Arg C) Lys D) Asp

A- According to Figure 2, nanobody binding peaks at pH=7-8. Its affinity gradually decreases from pH 8-12 and decreases very quickly from pH 5-6. This rapid decrease in affinity from pH 5-6 is likely due to protonation of some residue. Recall that the pKa of histidine is approximately 6, resulting in protonation at this pH level (choice A is correct). Both arginine and lysine are basic residues and have higher pKa values (choices B and C are incorrect). Aspartate, an acidic residue, has pKa~3 and so would likely be deprotonated throughout this entire range of values (choice D is incorrect).

Tertiary Protein Structure

According to the passage, the globular domains of CI-MPR are on the luminal side of the Golgi apparatus and vesicles. To optimize interactions between these domains and the aqueous environment of the lumen, polar and charged residuestend to reside on the surface, where they can interact with water. Nonpolar residues tend to be found buried within the protein or in the hydrophobic environment of a membrane, where they can avoid water. Therefore, nonpolar aliphatic residues are the least likely to be found on the surface of a CI-MPR globular domain.

The citrate shuttle

After the consumption of excess carbohydrates, acetyl-CoA begins to accumulate in the mitochondrial matrix. Recall that glycolysis produces pyruvate, which is converted into acetyl-CoA via the pyruvate dehydrogenase complex. Citrate synthase then catalyzes the formation of citrate from acetyl-CoA and oxaloacetate. Typically, this is how acetyl-CoA enters the tricarboxylic acid cycle (TCA). However, since an excess of carbohydrates has been consumed, regulatory measures are taken to slow the TCA cycle, and citrate begins to accumulate. Recall that the TCA cycle's rate-limiting step is isocitrate dehydrogenase, which acts downstream of citrate synthase—hence causing a build-up of citrate. To remedy this, citrate is shuttled to the cytoplasm via a citrate shuttle. An enzyme in the cytoplasm catalyzes the reverse reaction of citrate synthase, by splitting citrate into acetyl-CoA and oxaloacetate.

Aspirin inhibits the enzyme cyclooxygenase. Researchers discovered a new drug that inhibits cyclooxygenase more effectively. If they report that the difference between aspirin and the new drug is statistically significant, it can be assumed that they: A) Rejected a true null hypothesis as there was no real difference between the groups B) Rejected a false null hypothesis as there was a real difference between the groups C) Failed to reject a false null hypothesis as there was no real difference between the groups D) Failed to reject a true null hypothesis as there was a real difference between the groups

B - A hypothesis test is performed to assess the validity of a claim made about 2 groups. The null hypothesis (H0) theorizes that there is no difference between these 2 groups. In contrast, the alternative hypothesis (HA) rivals the null by claiming that a difference does exist. p-values evaluate how well the data support H0, assuming H0 is true. The statistical significance of the results is determined by whether the p-value is equal to or less than α, the predetermined level of significance. In general, α is set to equal 0.05, which means that p-values ≤0.05 are considered statistically significant. The researchers found that the new drug they discovered inhibits cyclooxygenase more effectively than aspirin does. To arrive at this result, they first established the following hypotheses: H0: The inhibition of cyclooxygenase by the new drug will not differ from that of aspirin. HA: The inhibition of cyclooxygenase by the new drug will differ from that of aspirin. The researchers' claim of statistical significance means that the resultant p-value they obtained was ≤α. Therefore, H0 is false. They rejected a false H0 as there was a true difference between the groups (HA). (Choice A) Rejecting a true null hypothesis (type I error) leads to the incorrect conclusion that there is a difference between groups. In reality, there is no difference between groups. (Choice C) Failing to reject a false null hypothesis (type II error) leads to the erroneous conclusion that there is no difference between groups. In reality, there is a difference between groups. (Choice D) Failing to reject a true null hypothesis leads to the correct conclusion that there is no difference between groups. However, the null hypothesis is false in this case and was rejected as a result.

Which of the following is a covalent bond that could stabilize the tertiary structure of CI-MPR domains? A) Peptide B) Disulfide C) Electrostatic D) Thioester

B - A protein's tertiary structure is stabilized by interactions between amino acid side chains. Most of these interactions are noncovalent, including hydrogen bonding and electrostatic attractions. In oxidizing environments such as the lumen of vesicles and the extracellular space, covalent bonds can form between cysteine residues. Because these bonds form between two sulfur atoms, they are known as disulfide bonds. The passage states that each domain of CI-MPR has between 6 and 12 cysteine residues. Because the globular domains are found in the oxidizing environment of the Golgi apparatus and its vesicles, their tertiary structure is likely stabilized by disulfide bonds between these cysteine residues. (Choice A) Peptide bonds are covalent bonds that form between amino acids to make proteins and peptides. They contribute to a protein's primary structure but not its tertiary structure. (Choice C) Electrostatic interactions are frequently involved in tertiary protein structure, but they do not involve shared electrons and therefore are not covalent bonds. (Choice D) Thioester bonds form between a sulfur atom and a carbonyl carbon. They are important for coenzymes and metabolites such as acetyl-CoA and succinyl-CoA, but they do not participate in tertiary protein structure.

When pyruvate enters the mitochondria, it is converted to acetyl-CoA and further oxidized in a series of reactions that produces four NADH molecules and one FADH2 molecule. If all of the NADH and FADH2 produced by 2 moles of pyruvate digestion in the mitochondria enter the ETC, how many moles of ATP can wild-type Complex V produce (Every NADH molecule that enters the electron transport chain (ETC) pumps 10 protons into the intermembrane space for use by Complex V whereas only 6 protons are pumped per FADH2 molecule)? A) 18 B) 23 C) 28 D) 32

B - According to the passage, 10 protons are pumped into the intermembrane space per NADH molecule that enters the ETC, and 6 protons are pumped per FADH2 molecule. Therefore, each pyruvate can cause the pumping of 46 protons(40 from the four NADH molecules and six from FADH2). The passage also states that Complex V produces three ATP for every twelve protons (or one ATP per four protons) that flow through it; therefore, it produces 46/4 = 11.5 ATP molecules for every pyruvate molecule, or 11.5 moles of ATP for every mole of pyruvate. Because the question asks for the ATP yield of 2 moles of pyruvate, this result must be doubled, giving 23 moles of ATP for 2 moles of mitochondrial pyruvate.

Researchers quantified the amount of protein produced by cultured cells during a one-hour period by detecting the radioisotope 35S. Prior to this period, the cells were most likely provided with radiolabeled: A) Asparagine B) Methionine C) Threonine D) Serine

B - Cells continuously produce proteins by combining free amino acids into polypeptide chains. Once they are produced, proteins typically remain intact for hours to days before being degraded. Researchers interested in comparing the amounts of protein produced by cells in a given period must disregard any proteins that were already present before the assay began. One way to quantify the amount of protein produced during a specified period is to provide the cells with amino acidsthat contain a radioactive atom for the duration of the assay. Proteins that are synthesized during the specified time frame will incorporate the radiolabeled amino acids into their primary structure. These proteins can then be detected by measuring radioactive emissions, whereas proteins that were synthesized prior to the period of interest will contain no radioisotopes and will not be detected. Of the 20 standard amino acids, only cysteine and methionine contain sulfur atoms. Either of these amino acids could be labeled with 35S to facilitate detection. However, methionine is the only amino acid among the given choices that contains sulfur, so the cells were most likely provided with radiolabeled methionine.

Escherichia coli bacteria containing only 15N-labeled DNA were grown in media containing only 14N nucleotides. What percentage of double helices would be composed of one 15N strand and one 14N strand after three generations? A) 12.5% B) 25% C) 50% D) 100%

B - DNA replication is a semiconservative process in which DNA polymerase synthesizes each daughter strand using one of the parental strands as a template. Each new double helix contains one parental strand and one daughter strand. In the experiment described, Escherichia coli bacteria initially have the heavy isotope 15N in their DNA. Once the bacteria are transferred to 14N media and allowed to replicate, newly synthesized strands will contain 14N. In Generation 1, all new strands will be paired with parental strands that contain 15N. Therefore, 100% of double helices produced in Generation 1 will have one 15N strand and one 14N strand (Choice D). In Generation 2, both the 15N and 14N strands from Generation 1 are used as templates for the synthesis of new 14N strands. In the DNA that used a 14N strand as a template, both parental and daughter strands will contain 14N nucleotides. In DNA that used a 15N strand as a template, the parental strand will contain 15N and the daughter strand will contain 14N. Because half of the parental strands are 15N and half are 14N, 50% of the Generation 2 double helices will contain one 15N strand and one 14N strand (Choice C). The other 50% will contain two 14N strands. In each new generation, the percentage of double helices containing one 15N strand and one 14N strand will be half the percentage of the previous generation. Therefore, in Generation 3, 25% of double helices will contain a 15N strand and a 14N strand.

The method used to phosphorylate ALDH2 in Experiment 2 must include εPKC and which other component? A) Inorganic phosphate B) ATP C) Phosphoprotein D) Glucose-6-phosphate

B - Kinases are enzymes that catalyze the transfer of phosphate groups from nucleotide triphosphates (ie, ATP, GTP) to other molecules or from phosphorylated molecules to nucleotide diphosphates (ie, ADP, GDP). According to the passage, εPKC is a kinase that phosphorylates proteins such as ALDH2 at certain serine and threonine residues. Attaching phosphate to a molecule is usually a thermodynamically unfavorable process and must be coupled to a thermodynamically favorable process to provide the necessary energy. Hydrolysis of ATP or GTP is thermodynamically favorable. For this reason, kinases use ATP or GTP as a source of phosphate in the reactions they catalyze. Therefore, in addition to εPKC, ATP must also be present for phosphorylation of ALDH2 to occur. (Choice A) Kinases do not transfer inorganic phosphate to molecules. The phosphate must come from a high-energy nucleotide triphosphate such as ATP or GTP. (Choice C) ALDH2 becomes a phosphoprotein once phosphorylation occurs. Prior to phosphorylation, no phosphoproteins are present in this scenario. (Choice D) Glucose-6-phosphate is produced by hexokinase during glycolysis. It is not a requirement for εPKC activity.

Group A mice (who were not given tetracycline) , but not group B mice, could use which of the following dietary supplements to help maintain blood glucose levels after day 10? A) Cholesterol B) Alanine C) Acetyl-CoA D) Palmitic acid

B - Mice in group A were not given tetracycline at any point and so have functional G6Pase. They therefore will be able to use any metabolites that can enter gluconeogenesis to help sustain their blood glucose levels. Amino acids (except for leucine and lysine) can be converted into glucose through gluconeogenesis. For example, alanine can be converted to pyruvate by deamination. Mice in group A will be able to use alanine to make glucose in the liver, thereby maintaining blood glucose levels. Mice in group B were given tetracycline at day 10 and will be unable to carry out gluconeogenesis after that point. Therefore, glucogenic substrates such as alanine will not help maintain their blood glucose levels. (Choice A) Cholesterol is a lipid used as a structural molecule in cell membranes and in the synthesis of steroid hormones. It is not used in glucose synthesis. (Choice C) Although acetyl-CoA can be converted to the glucogenic molecule oxaloacetate through the citric acid cycle, it must first combine with oxaloacetate to do so. The two carbons gained by adding acetyl-CoA to oxaloacetate are effectively lost as CO2 during the cycle, so acetyl-CoA cannot provide additional carbon for glucose synthesis. (Choice D) Most fatty acids, including palmitic acid, are broken down to acetyl-CoA, which cannot be converted to glucose.

An increase in the activity of pancreatic β-cells levels would likely lead to: A) Direct activation of phosphofructokinase-1 B) Direct activation of phosphofructokinase-2 C) Direct activation of pyruvate kinase D) Direct activation of debranching enzyme

B - Pancreatic β-cells produce insulin. Insulin directly activates phosphofructokinase-2 (choice B is correct). This indirectly leads to an activation of phosphofructokinase-1 via the products of the reaction catalyzed by phosphofructokinase-2 (choice A is incorrect). Insulin does not activate pyruvate kinase (choice C is incorrect). Insulin promotes glycogen synthesis, and so would neither directly nor indirectly activate the debranching enzyme (choice D is incorrect).

Considering that NSPs are proteases, which of the following statements best describes the reaction catalyzed by NSPs? A) Water is produced by the reaction B) Water is consumed by the reaction C) Water is produced and then consumed by the reaction D) Water is consumed and then produced by the reaction

B - Peptides are linear sequences of amino acids linked by amide bonds (-CO-NH-) that join the α-carboxyl group of one amino acid to the α-amino group of another amino acid. Proteins that have completed their biological purpose must be degraded by proteases so that their constituent amino acids can be reused to generate new peptides. To cleave an amide bond during protein degradation, a water molecule (H2O) must be added using acids (H+) or bases (OH−) in a reaction known as amide hydrolysis. Figure 1 shows the acylation process (R-C=O addition) of amide hydrolysis by NSP4, which cleaves the amide bond of the viral peptide by adding a hydrogen atom to the nitrogen. The hydrogen atom displaces the C-N single bond when electrons from the oxygen on the carbonyl carbon reform the double bond. The acyl group (R1-C=O) attached to NSP4 is subsequently converted into carboxylic acid when electrons from an OH− group attack the electrophilic (electron-attracting) carbonyl carbon. (Choices A, C and D) Hydrolysis reactions only involve consumption of water to break bonds. They do not involve water production, which occurs in condensation reactions such as peptide bond formation.

Which of the following statements correctly characterizes the bond that is broken during the reaction catalyzed by the IP6-dependent Tcd protease? A) The bond rapidly degrades in the absence of enzyme B) The bond restricts molecular rotation C) The bond is part of a tetrahedral structure D) The nitrogen involved in the bond is negatively charged

B - Tcd exhibits protease activity. Proteases are a class of enzymes that catalyze the hydrolysis of peptide bonds in polypeptide chains. Peptide bonds are synthesized from a condensation reaction between the α-carboxyl group of one amino acid and the α-amino group of another. The amide nitrogen has a lone electron pair that it can donate to the peptide bond, allowing peptide bonds to exist in two resonance structures and giving them significant double-bond character. Just as a wheel cannot rotate about two axles, atoms in a molecule cannot rotate about two bonds. Therefore, peptide bonds restrict rotation and limit the conformations that a protein can adopt. (Choice A) Although peptide bond hydrolysis is thermodynamically favorable, it is kinetically slow in the absence of a protease, so peptide bonds do not degrade rapidly in the absence of an enzyme. (Choice C) The double-bond character of peptide bonds restricts each peptide unit to a single plane. Therefore, peptide units are planar rather than tetrahedral. (Choice D) The nitrogen atom in a peptide bond carries a partial positive charge due to resonance.

During prolonged fasting, which of the following liver enzymes has upregulated activity? A) Glycogen synthase B) Pyruvate carboxylase C) Glucokinase D) Phosphofructokinase

B - The liver is responsible for maintaining blood glucose levels during periods of fasting. During fasting, glucose consumption and storage are downregulated in the liver and glucose synthesis is upregulated. Newly formed glucose is exported into the blood to keep levels constant. In the early stages of fasting, the liver synthesizes glucose by degrading glycogen (glycogenolysis). However, if fasting is prolonged, glycogen stores are depleted, and the liver must upregulate the synthesis of glucose from smaller precursor molecules (gluconeogenesis). Many of the reactions in gluconeogenesis are glycolysis reactions in reverse, with both processes using the sameenzymes. However, a few glycolysis reactions are irreversible: phosphorylation of glucose by glucokinase, phosphorylation of fructose 6-phosphate by phosphofructokinase, and phosphate transfer from phosphoenolpyruvate by pyruvate kinase. The enzymes that catalyze these reactions are tightly regulated, and gluconeogenesis uses different enzymes to bypass irreversible steps. During prolonged fasting, the bypassing enzymes of gluconeogenesis are upregulated, whereas key glycolysis enzymes are downregulated. The first step in gluconeogenesis is carboxylation of pyruvate to form oxaloacetate. This step bypasses the irreversible pyruvate kinase reaction and is catalyzed by the enzyme pyruvate carboxylase. In the well-fed state, pyruvate carboxylase is downregulated so that pyruvate can be used in other pathways, but during prolonged fasting, pyruvate carboxylase activity is upregulated in the liver to increase the rate of gluconeogenesis. (Choice A) Glycogen synthase consumes glucose to produce glycogen. During fasting, the liver must synthesize glucose rather than consume it, so glycogen synthase activity is downregulated. (Choices C and D) Glucokinase and phosphofructokinase are tightly regulated enzymes of glycolysis that phosphorylate glucose and fructose 6-phosphate, respectively. During prolonged fasting, the activities of these enzymes are downregulated in the liver, whereas the gluconeogenesis enzymes—glucose 6-phosphatase and fructose 1,6-bisphosphatase—are upregulated to bypass these irreversible steps and increase gluconeogenesis.

Scientists predicted that gene copy number is proportional to protein expression for pyruvate kinase. If this hypothesis is correct, which kinetic parameter would be expected to double when the gene copy number doubles? A) Catalytic efficiency B) Maximum velocity C) Catalytic turnover D) Equilibrium constant

B - The rate of an enzyme-catalyzed reaction plateaus at high substrate concentrations ([S] >> Km) because the active sites of enzymes are completely saturated with substrate; adding more substrate fails to increase the rate of catalysis. Adding more enzyme, on the other hand, provides additional active sites, allowing the reaction to proceed more quickly as more substrate can be bound. The maximum velocity Vmax of the reaction is the product of the number of active sites (total enzyme concentration [Etot]) and the rate at which each site converts substrate to product (turnover number kcat). Mathematically, this is described as Vmax = kcat × [Etot]. Therefore, an increase in the [Etot] will increase Vmax. The question states that scientists predicted that the gene copy number is proportional to protein expression. If they are correct, doubling the gene copy number should result in double the total enzyme purified from a given number of cells. As a result, doubling pyruvate kinase's gene copy number should also double the Vmax of the pyruvate kinase reaction. (Choice A) Catalytic efficiency is the ratio of kcat to the Michaelis constant Km. Both are intrinsic properties of the enzyme and will not be altered by a change in enzyme concentration. (Choice C) Catalytic turnover kcat is an intrinsic property of the enzyme and will not be altered by a change in enzyme concentration. (Choice D) [Etot] (and therefore gene copy number) has no effect on the equilibrium constantKeq of a reaction. Enzymes increase the rate at which equilibrium is achieved but not the equilibrium itself.

Gluconeogenesis

Between energy stores available in glycogen and dietary intake, the glucose content in the body is usually sufficient to meet energy needs. However, these sources of energy can easily run out during exercise or periods of fasting, for example. How does energy continue to be supplied? Gluconeogenesis is a metabolic pathway that uses precursors from other sources, for instance, lipids or amino acids, to create glucose. The process can be thought of as the "reverse" of glycolysis: after converting these precursor molecules into pyruvate, several of the same enzymes used in glycolysis will run the reverse reaction to create glucose. Any reactions that are rate-limiting steps in glycolysis require an additional set of enzymes to catalyze the reverse reaction during gluconeogenesis. Each of these steps requires an additional ATP molecule to proceed spontaneously.

Each of the following molecules can be converted to oxaloacetate by a single enzymatic reaction EXCEPT: A) Pyruvate B) Malate C) Alanine D) Aspartate

C - A metabolic pathway is a series of enzymatic steps that interconvert metabolic intermediates to produce energy (catabolism) or structural molecules (anabolism). Many metabolic pathways intersect and have an intermediate in common. This intermediate may be synthesized from multiple precursors, depending on the pathway used in their synthesis. Oxaloacetate is an intermediate in several metabolic pathways, including gluconeogenesis, the citric acid cycle, and amino acid degradation. Each uses a different precursor and a different enzyme to catalyze the synthesis of oxaloacetate. In the first step of gluconeogenesis, the enzyme pyruvate carboxylase catalyzes the carboxylation of pyruvate to form oxaloacetate (Choice A). In this reaction, CO2 is added to pyruvate, and ATP is hydrolyzed to provide the necessary energy. In the final step of the citric acid cycle, oxaloacetate is synthesized by the oxidation of malate, a reaction catalyzed by malate dehydrogenase (Choice B). This reaction requires the simultaneous reduction of NAD+ to NADH. NADH can then enter the electron transport chain and provide electrons necessary for the production of ATP. In amino acid degradation, amino acids are converted to α-keto acids by transamination, in which an amino group is transferred from an amino acid to α-ketoglutarate to synthesize glutamate (Choice D). The α-keto acid derivative of aspartate is oxaloacetate. Alanine can be deaminated to yield pyruvate, which can then be converted to oxaloacetate by pyruvate carboxylase. This process requires two enzymatic steps instead of one.

Which of the following accurately describes the action of complex I of the electron transport chain? A) NAD+ is reduced to NADH and 4 protons are pumped into the intermembrane space B) NAD+ is reduced to NADH and no proton pumping occurs C) NADH is oxidized into NAD+ and 4 protons are pumped into the intermembrane space D) NADH is oxidized into NAD+ and no proton pumping occurs

C - At complex I, NADH is oxidized into NAD+ (choices A and B are incorrect). This results in four hydrogen ions being pumped into the intermembrane space (choice D is incorrect). While it is not necessary to memorize the action and effects of each complex of the electron transport chain, it is critical to know that the electron transport chain results in the oxidation of electron carriers and the pumping of hydrogen ions into the intermembrane space of the mitochondria.

Which statements regarding ataxin-3 parallel beta-sheets are true? I. Short sequences called beta-turns link one strand to the next. II. The N-termini of each strand are aligned. III. The adjacent backbones hydrogen bond with each other. A) I only B) I and II only C) II and III only D) I and III only

C - Beta-sheets can be oriented in either a parallel or an antiparallel manner. The passage states that the sheets in ataxin-3 are in the parallel orientation. Parallel strands run in the same direction, like lanes on a track, so the N-terminal portion of one strand aligns with the N-terminal portions of the others (Number II). In antiparallel sheets, on the other hand, the individual strands run in opposite directions from each other, like switchbacks on a hiking trail, so that the N-terminal region of one strand lines up with the C-terminal regions of neighboring strands. Secondary structure always includes hydrogen bonds between carbonyl and amino groups of the polypeptide backbone (Number III). Parallel and antiparallel sheets differ in their hydrogen bond geometries, with bond pairs directly aligned in antiparallel sheets and slightly offset in parallel sheets. Nevertheless, both types of beta-sheets have hydrogen bonds between adjacent backbones. (Number I) Because antiparallel strands run in opposite directions, they may be linked by a short sequence of amino acids called a beta-turn that induces a 180° bend in the polypeptide chain. Parallel beta-strands do not reverse directionality, so neighboring strands must instead be linked by longer loops that make 360° turns to align the N-terminal regions of neighboring strands. Parallel strands such as those in ataxin-3 can never be linked by beta-turns.

The physical properties of two isoforms of a protein, both of which bind the same ligand, are shown in the table: Isoform 1: 68.3kDa 4.7 pI Kd 3.1 X 10^-6 Isoform 2: 67.7kDa 5.9 pI Kd 2.9 X 10^-6 Which protocol would yield the best separation of the two isoforms? A) Run the mixture through a cation-exchange column starting at pH 10, then gradually increase the pH of the buffer B) Pass the mixture through a size-exclusion column with a pore size cutoff of 40 kDa C) Load the mixture onto an anion-exchange column at physiological pH, then gradually increase the NaCl concentration of the buffer D) Load the mixture onto an affinity column with the ligand covalently bound to the beads, followed by cleaving the covalent bond to elute

C - Both isoforms shown in the table have pI values below physiological pH (7.4), so at this pH, both are negatively chargedand will bind to an anion-exchange column. When NaCl is added, the chloride anions compete with the negatively charged proteins for binding. Because Isoform 1 has a lower pI than Isoform 2, it is more negatively charged and binds the column more tightly. Therefore, Isoform 2 will elute at a lower salt concentration than Isoform 1. Gradually increasing the salt concentration in the buffer allows each isoform to be collected separately. (Choice A) Cation-exchange columns have negatively charged beads that bind positively charged molecules and repel negatively charged ones. At pH 10, both isoforms are negatively charged and would be repelled by the beads, exiting the column together. (Choice B) Both isoforms are approximately the same size (~68 kDa), so they will elute from the size-exclusion column together. (Choice D) Both isoforms bind the same ligand with nearly identical affinity (~3 × 10−6 M), so both would bind to the affinity column. This technique would separate the two isoforms from other proteins but not from each other.

Glucosidases break down complex carbohydrates, such as glycogen. Which of the following enzymes is a glucosidase? A) Glycogen synthase B) Phosphofructokinase I C) Debranching enzyme D) Branching enzyme

C - Debranching enzyme is essential in the breakdown of glycogen (choice C is correct). Glycogen synthase catalyzes the polymerization of glucose monomers into linear chains of glycogen (choice A is incorrect). Phosphofructokinase I catalyzes the addition of a phosphate to fructose 6-phosphate during glycolysis (choice B is incorrect). Branching enzyme is a key enzyme in glycogenesis that creates the branches found in glycogen (choice D is incorrect).

Based on information in the passage, pyruvate in red blood (red blood cells rely almost exclusively on glycolysis to produce sufficient ATP for cellular functions.) cells can be further metabolized only by: A) Phosphorylation of ADP to form ATP B) Decarboxylation to form acetyl-CoA C) Reduction to form lactate and NAD+ D) Carboxylation to form oxaloacetate

C - Glycolysis breaks down glucose into pyruvate in a series of reactions that generate ATP and reduce NAD+ to NADH. NAD+ must be regenerated for glycolysis to continue, and this can occur either in the mitochondria (electron transport chain) or in the cytosol (fermentation). According to the passage, red blood cells lack mitochondria, and therefore they can regenerate NAD+ only by fermentation. Fermentation occurs when the enzyme lactate dehydrogenase catalyzes the reduction of pyruvate to lactate. The required electrons are provided by oxidation of NADH to form NAD+. (Choice A) Phosphorylation of ADP to form ATP occurs when pyruvate is synthesized by pyruvate kinase, not when pyruvate is subsequently metabolized. (Choice B) Pyruvate is decarboxylated in mitochondria to form acetyl-CoA. Because red blood cells do not have mitochondria, they cannot process pyruvate by this pathway. (Choice D) Carboxylation of pyruvate to form oxaloacetate is the first step in gluconeogenesis. Gluconeogenesis occurs in the liver, not in red blood cells.

Which of the following enzymes catalyzes an irreversible step in glycolysis? I. Phosphofructokinase-1 II. Glyceraldehyde 3-Phosphate Dehydrogenase III. Pyruvate Kinase A) I only B) II and III C) I and III D) I, II, and III

C - Hexokinase, phosphofructokinase-1, and pyruvate kinase catalyze the three irreversible steps for glycolysis (choices I and III are correct). Glyceraldehyde 3-phosphate dehydrogenase catalyzes a reversible step of glycolysis (choice II is incorrect).

POMT can use activated D-mannose, but not D-glucose, as a substrate. D-glucose and D-mannose are: A) Constitutional isomers B) Enantiomers C) Epimers D) Anomers

C - Molecules with the same molecular formula but different arrangements of the atoms are called isomers. Stereoisomersare molecules with the same kinds of bonds at any given position but with the bonds oriented differently in space. Stereoisomers have one or more stereocenters, also called chiral centers, which are atoms bonded to four different chemical groups. Molecules that have multiple stereocenters but differ at only one of them are called epimers. D-mannose and D-glucose each have several stereocenters but differ from each other only at the carbon 2 center; therefore, they are called C-2 epimers. Because POMT can use activated D-mannose but not D-glucose as a substrate, it must be able to distinguish between these epimers. (Choice A) Unlike stereoisomers, constitutional isomers have the same molecular formula but different bond types in at least one position (ie, different connectivity between atoms). For example, D-glucose and D-fructose are constitutional isomers, having a carbon with two bonds to oxygen at carbon 1 and carbon 2, respectively. (Choice B) Enantiomers are stereoisomers that differ at every stereocenter, not just one. The enantiomer of D-mannose is L-mannose. (Choice D) Anomers are a special kind of epimer that differ at the anomeric carbon and are designated as the α- and β-forms of the same sugar. This difference can only occur in the cyclical form of the carbohydrate because the anomeric carbon is not a stereocenter in the linear form.

Which of the following cellular metabolic processes would result in the highest net energy yield? A) The glycolysis of two molecules of glucose B) β-oxidation of an unsaturated 20 carbon fatty acid C) β-oxidation of a saturated 20 carbon fatty acid D) Both B and C

C - Saturated fats lack double bonds. Because they are more reduced, the oxidation of a saturated fatty acid will yield more energy than an unsaturated fatty acid of the same length (choice B is correct). Glycolysis only yields a small amount of energy (choice A is incorrect).

Which of the following steps in the citric acid cycle does NOT produce reduced electron carriers? A) Conversion of isocitrate to α-ketoglutarate B) Conversion of α-ketoglutarate to succinyl-CoA C) Conversion of succinyl-CoA to succinate D) Conversion of succinate to fumarate

C - The citric acid cycle is an oxidative process. The oxidation state of carbon can be estimated by the number of oxygen or sulfur bonds and the number of C=C double bonds. Increasing the number of these bonds increases the overall oxidation state. When one molecule is oxidized, another molecule must be reduced, and vice versa. Therefore, each of the four oxidative steps in the cycle yield reduced electron carriers, either in the form of NADH or FADH2. The oxidative steps are: 1. Conversion of isocitrate to α-ketoglutarate (Choice A). In this reaction, a single bond to oxygen (hydroxyl group) is oxidized to a ketone with two bonds to oxygen, and a carboxyl group leaves as CO2. One NADH molecule is produced in this step. 2. Conversion of α-ketoglutarate to succinyl-CoA (Choice B). In this step, the ketone that was formed in the first step is further oxidized by formation of a bond to sulfur, and another carboxyl group leaves as CO2. Another NADH molecule is formed during this step. 3. Conversion of succinate to fumarate (Choice D). In this step, a carbon is oxidized by forming a double bond with another carbon. Oxidation steps of this nature typically use FAD as a cofactor, and this step produces one FADH2molecule as a reduced electron carrier. 4. Conversion of malate to oxaloacetate. In this step, a hydroxyl group is oxidized to a ketone, and a molecule of NADH is formed in the process. Conversion of succinyl-CoA to succinate does not involve oxidation because the C-S bond that breaks and the C-O bond that forms both have the same oxidation state. With no oxidation reaction, no reduction of electron carriers can occur. However, this conversion is still a high-energy step as it produces guanosine triphosphate (GTP), which is equivalent in energy to ATP.

Based on the molecular weight of a rubisco monomer (51kDa), approximately how many amino acids are present in a rubisco dimer? A) 460 B) 690 C) 920 D) 5100

C - The native form of a protein is the form in which the protein is biologically active, and it may consist of one or more polypeptide chains. The exact molecular weight of each chain is determined by the primary structure, or the sequence of amino acids that makes up the chain, with each amino acid contributing its own molecular weight to the total. The 20 standard amino acids have an average molecular weight of 110 Da, so the approximate number of amino acids in a protein can be estimated by dividing its molecular weight in daltons by 110. The passage states that the functional (native) form of rubisco is a dimer in which each subunit contributes 51 kDa (51,000 Da) of mass. This mass can be converted to the approximate number of amino acids (AA) by 51,000 Da110 Da/AA=463 AA≈460 AA51,000 However, this is the number of amino acids in a rubisco monomer (Choice A). Because the native form of rubisco is a homodimer (two identical subunits), the number of amino acids in the monomer must be doubled. Native, dimeric rubisco contains ~920 amino acids.

Inactivation of which enzyme is most likely to result in physiological effects similar to those of von Gierke disease ( low blood glucose and high lactate levels)? A) Glucose 6-phosphate dehydrogenase B) Pyruvate decarboxylase C) Phosphoenolpyruvate carboxykinase D) Acyl-CoA dehydrogenase

C - The physiological effects of von Gierke disease include low blood glucose and high lactate levels as G6Pase inactivation shuts down the pathways that control them. One of these pathways, gluconeogenesis, consumes lactate to produce glucose, and blocking any enzyme of gluconeogenesis would therefore likely lead to effects similar to those of von Gierke disease. Phosphoenolpyruvate carboxykinase (PEPCK) catalyzes the second step in gluconeogenesis, the conversion of oxaloacetate to phosphoenolpyruvate, and its inactivation can indeed lead to both lactate buildup and glucose depletion in the blood. (Choice A) Glucose 6-phosphate dehydrogenase (G6PDH) is an enzyme of the pentose phosphate pathway. Although G6PDH uses glucose 6-phosphate as a substrate, it is not involved in gluconeogenesis. (Choice B) Pyruvate decarboxylase converts pyruvate to acetaldehyde as the first step in the production of ethanol by yeast. Humans do not have this enzyme. (Choice D) Acyl-CoA dehydrogenase catalyzes the first step in fatty acid catabolism and does not affect blood glucose or lactate levels.

As NSP4 folds, the water molecules surrounding it become: A) More ordered; ΔS is negative B) More ordered; ΔS is positive C) More disordered; ΔS is positive D) More disordered; ΔS is negative

C - The polarity and charge of amino acid residues on the surface of a protein affect the order of the surrounding water molecules, as measured by entropy, ΔS. Increased order (decreased entropy) is expressed as a negative ΔS value whereas decreased order (increased entropy) corresponds to a positive ΔS value. Each water molecule must participate in hydrogen bonds, but the side chains of hydrophobic amino acids cannot hydrogen bond with water. Therefore, water molecules in the vicinity of hydrophobic residues must hydrogen bond with each other instead, and can only do so by forming a rigid, highly ordered network called a solvation layer around the hydrophobic molecule. Hydrophilic amino acids, on the other hand, readily hydrogen bond with water, so formation of a solvation layer around these residues is unnecessary. The water molecules surrounding hydrophilic residues are more disordered than those surrounding hydrophobic residues. Unfolded proteins expose their hydrophobic residues to the aqueous environment, causing the surrounding water molecules to form solvation layers. By contrast, folded proteins tend to hide hydrophobic residues in their interior, away from the surrounding water. Hydrophilic residues remain on the surface of the protein, where they can interact with water without a solvation layer. Therefore, as a protein folds, the water surrounding it becomes more disordered. This is reflected by a positive ΔS value. (Choices A and B) The protein becomes more ordered upon folding, but the question asks for the order of the water molecules surrounding the protein. The water molecules become more disordered when the protein folds. (Choice D) The water molecules become more disordered upon protein folding, but disorder is reflected as a positive ΔSrather than a negative ΔS.

Based on information in the passage and the data in Table 1, the 15 domains of CI-MPR are most likely similar to each other in which of the following ways? A) Post-translational modifications B) Affinity for their ligands C) Amino acid sequence D) Primary ligand

C - The three-dimensional folded form of a protein, including secondary and tertiary structure, is entirely determined by primary structure (amino acid sequence). Therefore, proteins with similar amino acid sequences often fold similarly, and those with similar folds most likely have similar amino acid sequences. Proteins with different sequences are likely to adopt distinct folds. Domains within a protein behave as individual units and typically fold independently from each other. The passage states that the 15 globular domains of CI-MPR fold similarly. Therefore, they most likely all have similar amino acid sequences. (Choice A) Post-translational modifications are non-amino acid additions to proteins, such as phosphorylation or glycosylation. Table 1 shows that some CI-MPR domains have no glycosylation sites, some have one, some two, and some three. Therefore, the domains are not similar in their post-translational modifications. (Choice B) The affinity of a protein for its ligand is described by the dissociation constant Kd. Table 1 shows that the different domains have Kd values that differ by several orders of magnitude. Therefore, each domain has a different affinity for its ligand. (Choice D) Table 1 shows that different domains bind different ligands, and at least one (domain 13) does not bind ligands on its own.

When the cell is in need of glucose, glycogenolysis is upregulated, beginning with the activation of glycogen phosphorylase. Glycogen phosphorylase catalyzes the production of which of the following molecules? A) UDP-glucose B) UDP-galactose C) Glucose 1-phosphate D) Glucose 6-phosphate

C- Glycogen phosphorylase breaks the bond between two adjacent glucose molecules in a glycogen chain by phosphorolysis, or the addition of a phosphate molecule across a bond. The bond broken by this reaction is an α(1→4) linkage, so the phosphate must be added to either carbon 1 or carbon 4 of a glucose unit. The glucose unit that is released was attached by carbon 1, and it is to this carbon atom that phosphate is added, producing glucose 1-phosphate (G1P). (Choice A) UDP-glucose is produced by the combination of G1P with UTP along with the release of pyrophosphate. This reaction, catalyzed by UDP-glucose pyrophosphorylase, is involved in glycogenesis but not glycogenolysis. (Choice B) UDP-galactose is an intermediate of galactose metabolism. It is converted to UDP glucose and later G1P, but is not itself a product of glycogenolysis. (Choice D) Glucose 6-phosphate (G6P) is the final product of glycogenolysis, produced when phosphoglucomutase moves the phosphate group from carbon 1 to carbon 6. It is not produced by glycogen phosphorylase.

O-linked D-mannose can form a glycosidic bond with any sugar that has a free anomeric carbon. Therefore it could theoretically form a glycosidic bond with all of the following sugars EXCEPT: A) Galactose B) Ribose C) Lactose D) Sucrose

D - A glycosidic bond is a bond between the anomeric carbon of a carbohydrate and any other biological molecule. The formation of glycosidic bonds requires at least one sugar with a free anomeric carbon, also known as a reducing end. A carbohydrate with a free reducing end is called a reducing sugar. Because O-linked D-mannose is connected to Dg through its anomeric carbon, it does not have a free reducing end and will only be able to form a glycosidic bond with a sugar that does have an available reducing end. Sucrose is a disaccharide in which the reducing end of D-glucose is linked to the reducing end of D-fructose. Because both reducing ends are already involved in a glycosidic bond, sucrose is not a reducing sugar. Therefore, O-linked mannose will not be able to form a glycosidic bond with sucrose. (Choices A and B) Galactose and ribose are monosaccharides (single carbohydrate units). All free monosaccharides are reducing sugars and can form glycosidic bonds. (Choice C) Lactose is a disaccharide formed by a β-1,4 linkage between galactose and glucose. The glucose unit has a free anomeric carbon, and therefore is a reducing sugar.

Under low ATP conditions, which of the following enzymes would be LEAST beneficial to the cell? A) Aldolase B) Enolase C) Glyceraldehyde 3-phosphate dehydrogenase D) 6-Phosphogluconate dehydrogenase

D - As stated in the passage, under low ATP conditions the G6P in a cell will enter glycolysis, which will produce ATP. The final ATP molecules of this pathway are produced by the final enzyme in the pathway, pyruvate kinase. Therefore, all glycolysis enzymes upstream of pyruvate kinase are required for maximum ATP synthesis, as their absence would prohibit pyruvate kinase function. These enzymes include the following: Aldolase catalyzes the cleavage of fructose 1,6-bisphosphate into glyceraldehyde 3-phosphate (GAP) and dihydroxyacetone phosphate (Choice A). Enolase catalyzes the conversion of 2-phosphoglycerate to phosphoenolpyruvate and is the second to last step in the pathway (Choice B). Glyceraldehyde 3-phosphate dehydrogenase, also known as GAPDH, catalyzes the conversion of GAP to 1,3-bisphosphoglycerate in a step that immediately precedes the first ATP-synthesis step of glycolysis (Choice C). 6-Phosphogluconate dehydrogenase (6PGD) is the enzyme in the pentose phosphate pathway that catalyzes the conversion of 6-phosphogluconate to R5P. It is in this step that a carbon atom is converted to CO2, decreasing the overall potential ATP output. Therefore, 6PGD would be less beneficial to a cell deficient in ATP than would any glycolytic enzyme.

Which of the following biomolecules can be found in the cell membrane? I. Carbohydrates II. Lipids III. Proteins IV. Nucleic Acids A) II only B) II and IV C) I, II, IV D) I, II, III

D - Carbohydrates can be found on the surface of the cell membrane and are involved in immune recognition (I). Lipids, such as cholesterol and phospholipids, are the most plentiful component of the membrane (II). Proteins, such as transmembrane proteins, are also very prevalent in the membrane (III). Nucleic acids, on the other hand, are virtually absent (IV).

Which of the following statements is not true? A) Cholesterol is a precursor for steroids B) Cholesterol is a precursor for vitamin D C) Cholesterol maintains the fluidity and stability of the cell membrane D) Cholesterol aids in immune and intracellular recognition

D - Carbohydrates, not cholesterol, can be found attached to proteins on the extracellular surface of a cell to aid in recognition (choice D is correct). The other statements regarding cholesterol are true (choices A, B, and C are incorrect).

The addition of which reagent would result in the greatest amount of dimerization of the Tat variants (Variants 1 and 2 contained a C-terminal and an N-terminal cysteine residue, respectively) ? A) A strong acid such as HCl B) A strong base such as NaOH C) A reducing agent such as dithiothreitol D) An oxidizing agent such as oxygen gas

D - Cysteine (C) residues in proteins and peptides can react with each other to form disulfide bonds. These bonds are commonly (although not always) present in dimers and multimers and help hold protein subunits together. The formation of a disulfide bond occurs through an oxidation-reduction reaction, in which the sulfur atoms of cysteine residues are initially in reduced form and become oxidized. For this reaction to occur, an oxidizing agent must be present. The passage indicates that the Tat variants dimerize by disulfide bond formation. Therefore, the addition of an oxidizing agent such as oxygen gas would help increase the amount of disulfide bond formation and, as a result, the amount of dimerization that occurs.

In humans, glycogen is stored in the form of: A) Triglycerides B) Chylomicrons C) Starch D) Granules

D - The glycogen chains produced during glycogenesis are stored as granules, which are spherical and contain a protein core (choice D is correct). Triglycerides are energy-storing lipids (choice A is incorrect). Chylomicrons are lipoproteins that transport lipids (choice B is incorrect). Starch is a polysaccharide made up of glucose monomers and is the primary form of energy storage used by plants (choice C is incorrect).

During anaerobic exercise, the Cori cycle connects which two metabolic pathways? A) Glycogenolysis and the citric acid cycle B) The urea cycle and gluconeogenesis C) The pentose phosphate pathway and glycogenesis D) Gluconeogenesis and glycolysis

D - During glycolysis, NAD+ is converted to NADH by the enzyme glyceraldehyde-3-phosphate (GAP) dehydrogenase. For glycolysis to continue, NAD+ must be regenerated. Under aerobic conditions, electrons from NADH can be transferred to the electron transport chain (ETC) and ultimately to oxygen. However, in anaerobic conditions, NAD+ cannot be regenerated by the ETC because there is insufficient oxygen to accept electrons. Consequently, NADH donates electrons to pyruvate, which is reduced to lactate in the process. Lactate that builds up from this mechanism must be removed from the system because it can lead to muscle pain and nausea. The lactate in muscles enters the bloodstream, which carries it to the liver. In the liver, lactate is converted to glucose during gluconeogenesis and is carried back to muscles by the blood. The process of carrying lactate from the muscle to the liver and moving regenerated glucose from the liver back to muscles is called the Cori cycle, which connects gluconeogenesis and glycolysis. (Choice A) Glycogenolysis is the consumption of glycogen to form glucose-6-phosphate, and the citric acid cycle is the consumption of acetyl-CoA (formed from pyruvate) to generate CO2, GTP, NADH, and FADH2. Neither pathway is part of the Cori cycle. (Choice B) The urea cycle processes nitrogen from amino acids for excretion, and gluconeogenesis generates glucose from pyruvate and other precursors. Both pathways occur primarily in the liver, and they are not connected by the Cori cycle. (Choice C) The pentose phosphate pathway produces ribose 5-phosphate and NADPH using glucose as a starting material. Glycogenesis is the production of glycogen from glucose. The two pathways diverge from each other and are not connected by the Cori cycle.

The Km of an enzyme is always decreased by which class of inhibitor? A) Mixed B) Noncompetitive C) Competitive D) Uncompetitive

D - Enzyme inhibitors are classified according to the mechanism by which they interfere with enzyme activity. Competitive inhibitors bind the free enzyme E exclusively and prevent substrate from binding. Uncompetitive inhibitors bind the enzyme-substrate complex ES exclusively, preventing the enzyme from converting substrate to product. Mixed inhibitors have characteristics of both competitive and uncompetitive inhibitors. All mixed inhibitors bind both E and ES but may favor one over the other. When a mixed inhibitor binds E and ES with equal affinity, it is a noncompetitive inhibitor. Different inhibitory mechanisms have different effects on an enzyme's properties. Specifically, inhibitor binding to E prevents substrate binding, so a higher substrate concentration is required to reach ½Vmax. This substrate concentration is the Km of the enzyme, so competitive inhibitors always cause an increased Km (Choice C). Conversely, inhibitor binding to ES effectively decreases the amount of ES in solution and shifts the equilibrium toward more ES formation. This causes a decrease in the amount of substrate needed to achieve ½Vmax (although Vmax itself is also decreased) and therefore decreases Km. Because uncompetitive inhibitors bind ES exclusively, they always cause a decrease in Km. (Choices A and B) Mixed inhibitors may cause a decrease, increase, or no change in Km, depending on whether they favor binding to E (increased Km) or ES (decreased Km). If the mixed inhibitor has equal affinity for E and ES, it is noncompetitive and does not alter Km.

If site-directed mutagenesis was used to create a missense mutation at His57, this would likely result in which of the following changes regarding the enzyme-catalyzed degradation of the viral peptide depicted in the active site? I. The time required for the reaction to achieve equilibrium would increase. II. The equilibrium constant Keq of the reaction would decrease. III. The enzyme-substrate dissociation constant Kd would increase. A I only B) III only C) I and II only D) I and III only

D - Enzymes are biological catalysts that lower the activation energy of a reaction, allowing the reaction to proceed at a faster rate. Substrate molecules bind to the enzyme's active site, where catalytic activity occurs. The enzyme and substrate undergo conformational changes that make the transition state more stable and favorable, resulting in lower activation energy. Enzyme function is sensitive to changes in pH and temperature as well as deleterious mutations, particularly those involving the active site. A missense mutation at the histidine in position 57 of the NSP4 sequence (ie, His57) would result in a different amino acid at this site and alter the interaction between the viral peptide (substrate) and the NSP4 active site. This mutation would be expected to decrease the binding affinity of the enzyme for its substrate, resulting in an increased dissociation constant Kd for the enzyme-substrate complex (Number III). As a result, the mutation would interfere with normal enzymatic function and lead to a decreased reaction rate (ie, increasing the time the reaction takes to reach equilibrium) (Number I). (Number II) Enzymes do not alter the free energy change ΔG or the equilibrium constant Keq of a reaction. The overall ΔG depends only on the energy difference between products and reactants (ΔG = Gproducts − Greactants), and can be visualized in a free energy diagram. The equilibrium constant Keq of a reaction is the ratio of products to reactants for a given reaction at equilibrium at a specific temperature and is related to ΔG (ie, ΔG = −RT ln Keq). Although a functional enzyme would decrease the time taken for the reaction to reach equilibrium (ie, ΔG = 0 for the reaction) by increasing the reaction rate, it would not alter Keq.

While maintaining the Lactococcus lactis cell cultures, Scientist A accidentally adds an unlabeled tube to Culture 6. Subsequent gel electrophoresis analysis of the culture's DNA shows its molecular weight to be less than half of that of the original sample. Which enzymes were most likely present in the tube? A) DNA polymerase and DNA ligase B) DNA helicase and DNA ligase C) DNA polymerase and DNA nuclease D) DNA helicase and DNA nuclease

D - Given the lower molecular weight, the enzymes in the unlabeled tube likely cleaved or degraded the DNA present in the cell culture. DNA polymerase synthesizes new strands of DNA, and DNA ligase joins fragments of DNA together (choices A, B, and C are incorrect). DNA helicase unwinds DNA, which would then allow DNA nuclease to degrade single-stranded DNA (choice D is correct).

The glycogen branching enzyme forms a glycosidic bond between carbon 1 on the incoming glucose monomer and which of the following? A) Carbon 3 on a glucose unit of the growing chain B) Carbon 4 on a glucose unit of the growing chain C) Carbon 5 on a glucose unit of the growing chain D) Carbon 6 on a glucose unit of the growing chain

D - Glycogen is composed of several linked glucose molecules. The majority of these units are connected to each other in an α(1→4) linkage, meaning carbon 1 of each incoming glucose molecule becomes linked in the α conformation to carbon 4 of the final carbon atom in the growing chain. This reaction is catalyzed by glycogen synthase. However, glycogen is a branched molecule rather than linear. The glycogen branching enzyme creates these branches by catalyzing the formation of α(1→6) linkages, acting on approximately every tenth glucose unit in the chain. In this case, carbon 1 of the incoming glucose molecule in the α conformation is linked to carbon 6 of a glucose molecule in the growing glycogen chain. This branch then begins a new series of α(1→4) linkages. (Choices A and C) Carbons 3 and 5 are not involved in glycosidic bonds in glycogen. (Choice B) Carbon 4 is involved in glycosidic bonds in glycogen, but glycogen synthase catalyzes the formation of these bonds, not the branching enzyme.

Which statement correctly describes the cleavage of protein disulfide bonds by the enzyme thioredoxin? A) A protease breaks the bond by catalyzing the addition of water across it B) An isomerase catalyzes a rearrangement of functional groups to break the bond C) A phosphorylase breaks the bond by catalyzing the addition of a phosphate group D) An oxidoreductase catalyzes the transfer of electrons to break the bond

D - In proteins, disulfide bonds form when the thiol groups of cysteine residues are oxidized, linking them together to form cystine. To break the disulfide bond in cystine, a reduction step that transfers electrons back to the sulfur atoms is required. Thioredoxin catalyzes the transfer of electrons from a donor molecule to cystine, which reduces the sulfur atoms and oxidizes the donor molecule. Consequently, thioredoxin can be classified as an oxidoreductase because it facilitates the cleavage (reduction) of disulfide bonds. (Choice A) Proteases are a type of hydrolase that break peptide bonds by hydrolysis (addition of water). They do not break disulfide bonds. (Choice B) Isomerases rearrange functional groups within a molecule but do not catalyze redox reactions. Therefore, they cannot catalyze cleavage of disulfide bonds. (Choice C) Phosphorylases are a type of transferase that break bonds by adding phosphate, as in glycogen degradation. They do not catalyze redox reactions and cannot break disulfide bonds.

In skeletal muscle cells, leptin resistance would lead to decreased concentrations of: A) Glycerol B) Fatty acids C) Malonyl-CoA D) Acetyl-CoA

D - Leptin resistance or leptin insensitivity refers to an inability to increase fatty acid oxidation in a cell despite adequate leptin production and normal binding of leptin to cell surface receptors. The passage states that leptin generally activates AMPK, which then phosphorylates/inhibits ACC. If leptin is administered to skeletal muscles with leptin resistance, the exogenous leptin would still fail to activate the AMPK signaling pathway. AMPK would therefore remain inactive and unable to block ACC activity. As a result, ACC would remain uninhibited and continue to carboxylate free acetyl-CoA molecules to produce malonyl-CoA, which would increase in concentration (Choice C). The levels of acetyl-CoA would be reduced due to its continued conversion to malonyl-CoA, resulting in increased fatty acid synthesis and decreased β-oxidation. (Choices A and B) Leptin resistance leads to accumulation of fatty acid precursors (eg, malonyl-CoA), resulting in an increased concentration of triglycerides (composed of glycerol and fatty acids) and reduced oxidation of fatty acids.

Which of the following statements is true and will enhance the function of the genetic circuit? A) If the promoter's cytosine and adenine groups are demethylated, the addition of DNA methylase will methylate the promoter region and enhance levels of transcription B) If the promoter's thymine and guanine groups are methylated, the addition of DNA demethylase will demethylate the promoter region and enhance levels of transcription C) If the promoter's thymine and guanine groups are demethylated, the addition of DNA methylase will methylate the promoter region and enhance levels of transcription D) None of the above

D - Methyl groups added to cytosine and adenine can inhibit transcription. As a result, the addition of methylase to methylate these groups will not enhance levels of transcription but rather inhibit it (choices A and C are incorrect). As DNA methylation primarily occurs at cytosine and adenine bases, demethylation of thymine and guanine groups is not expected to affect transcription levels (choice B is incorrect).

A protein denatures as ionic interactions are disrupted by deprotonation of positively charged side chains. This is most likely the result of: A) A decrease in the ionic strength of the solution B) An increase in the ionic strength of the solution C) A decrease in the pH of the solution D) An increase in the pH of the solution

D - Proteins are chains of amino acid residues that perform biological functions and, in most cases, must fold into specific three-dimensional structures to perform their functions. Protein folding is driven by the hydrophobic effect and stabilized by interactions between residues such as hydrogen bonds and ionic bonds. Ionic interactions take place between positively and negatively charged residues known as ionizable residues. Protein folding is sensitive to the ambient pHbecause pH-dependent protonation or deprotonation can alter the charges of ionizable amino acids. By definition, the pH of a solution is equal to −log[H+]. Thus, at low pH, the concentration of protons (H+ ions) is high. Accordingly, ionizable amino acids are more likely to pick up protons from the environment (ie, become protonated) at low pH. In contrast, at high pH, the proton concentration is low and ionizable amino acids tend to lose protons to the environment (ie, become deprotonated). Protonation can neutralize negative charges, and deprotonation can neutralize positive charges. Either case may disrupt ionic bonds. The question states that ionic bonds in a protein were disrupted by deprotonation. Because deprotonation tends to occur in environments with a low concentration of protons (ie, high pH), this result indicates an increase in the pH of the solution. (Choices A and B) Ionic strength is a measure of the concentration of ions in solution. Ions such as Na+ or Cl− can disrupt ionic interactions by competing for ionic bonds. Ionic strength does not alter the protonation state of amino acid side chains.

What kind of bond exists in the sugar phosphate backbone? A) Hydrogen bond B) Peptide bond C) Glycosidic bond D) Phosphodiester bond

D - Right off the bat, we know that hydrogen bonds (A) aren't the answer because hydrogen bonds exist between nucleotide pairs, not the backbone. Peptide bonds (B) exist in proteins, but DNA is a nucleic acid. Glycosidic (C) bonds exist between the pentose sugar and the base of a nucleotide. Phosphodiester bonds (D) is the answer because the sugar phosphate backbone of DNA is connected by phosphate groups bonded to the adjacent pentose sugar.

What is the purpose of the F0 component of ATP synthase? A) It phosphorylates ADP molecules to form ATP B) It regenerates NAD+ C) It catalyzes the reduction of oxygen into water D) It serves as a channel for hydrogen ions to flow into the matrix

D - The F0 component serves as a channel through which hydrogen ions present in the inner mitochondrial space can flow back into the matrix (choice D is correct). This flow is the natural result of the chemiosmotic gradient. The F0 component does not phosphorylate ADP molecules to form ATP; this function is performed by the F1 component (choice A is incorrect). NAD+ is regenerated by lactate dehydrogenase during lactic acid fermentation (choice B is incorrect). Oxygen is reduced into water at complex IV of the electron transport chain (choice C is incorrect).

Palmitic acid synthesis

In the cytoplasm, acetyl-CoA is converted into malonyl-CoA via the addition of a carbon dioxide molecule. This reaction is catalyzed by acetyl-CoA carboxylase, the rate-limiting step of fatty acid synthesis. Finally, fatty acid synthase, a multienzyme complex, catalyzes the polymerization of palmitic acid. This requires NADPH and produces NADP+, carbon dioxide, and water as byproducts. Note that the synthesis of fatty acids does not require the presence of any precursor or template molecules. This is in contrast to the synthesis of DNA or RNA, which requires the presence of a template molecule to form "copies" of itself.

Citric Acid Cycle (Krebs Cycle)

In the mitochondrial matrix, acetyl-CoA will enter the citric acid cycle, also known as the Krebs cycle. Although this process does not use oxygen, it is considered aerobic. Why is this? The citric acid cycle needs both electron carriers--NAD+ and FAD--to continue producing NADH and FADH2. NAD+ and FAD are regenerated at a later point in the pathway--during the electron transport chain--by donating hydride ions to an oxygen atom, ultimately forming water. Without the presence of oxygen, the citric acid cycle cannot receive NAD+ and FAD, and will ultimately grind to a halt. For the MCAT, you should be able to recognize the names and functions of the enzymes that carry out this process. The mnemonic "Can I Keep Selling Seashells For Money, Officer" is often used to memorize the TCA cycle's reactants. One enzyme that is particularly important is isocitrate dehydrogenase, as it catalyzes the rate-limiting step of the TCA cycle. Instead of memorizing the exact structure of every intermediate molecule in this pathway, simply memorizing the number of carbons can help you recognize structures if presented on your exam. A helpful trick is to remember that the number of carbons present on a molecule can help you recall their name. Citrate and isocitrate have 6 carbons each. Alpha-ketoglutarate has 5 carbons. Succinyl-CoA, succinate, fumarate, malate, and oxaloacetate each have 4 carbons. Acetyl-CoA has 2 carbons.

Glucagon

Is secreted by α-cells when blood glucose levels are low. This hormone promotes both anabolic and catabolic pathways that increase glucose concentration in the blood, such as glycogenolysis and gluconeogenesis.

The Electrochemical Gradient

It might be worth memorizing the number of protons that are pumped into the intermembrane space for each molecule of NADH or FADH2. NADH either directly or indirectly donates electrons to complexes I, III, and IV. Thus, for each NADH molecule, 10 protons are pumped into the intermembrane space (four protons at complex I, four protons at complex III, and two protons at complex IV). FADH2 either directly or indirectly donates electrons to complexes II, III, and IV. Thus, only six protons are pumped into the intermembrane space for each FADH2 molecule (none at complex II, four at complex III, and two at complex IV). The electrochemical gradient is essential to produce ATP through ATP synthase. This is why it is so crucial for both the outer and inner membranes to be impermeable to ion exchange: if H+ ions were able to leak from either membrane, the gradient would rapidly dissipate.

Size Exclusion Chromatography

Like gel electrophoresis, size exclusion chromatography separates molecules by size. Unlike gel electrophoresis, small molecules move more slowly through size exclusion columns than do large molecules. The column is filled with tiny beads that have pores of varying sizes. Molecules that are too large to fit into any of the pores are "excluded" from some of the column volume and travel only through the space surrounding the beads, called the "void volume." Molecules that can fit into the large pores travel through the additional volume that they provide, and those that are small enough to fit inside all the pores travel through the entire available volume. Because they have a longer effective distance to travel before reaching the end of the column, small molecules take longer to elute. Peptide 3 is composed of only 9 amino acids and is substantially smaller than the other peptides, so it will elute last.

Structural Lipids

Lipids are a broad class of hydrophobic biomolecules that are categorized at multiple levels. At a general level, they are classified as hydrolyzable or nonhydrolyzable based on whether they contain hydrolyzable linkages such as esters or amides. Hydrolyzable lipids are further classified based on their backbone structure and modifications. For example, lipids built on a glycerol backbone are glycerolipids, whereas those built on sphingosine backbones are sphingolipids. A lipid with a phosphate group modification is called a phospholipid, and a lipid with a carbohydrate attached is called a glycolipid. Multiple characteristics may be present at once, yielding more complex names such as glycerophospholipids, phosophosphingolipids, glycosphingolipids, or phosphoglycolipids. The passage states that GC is a hydrolyzable lipid. The structure in Figure 1 shows that it is built on a sphingosine backbone, making it a sphingolipid. Figure 1 and the passage also show that GC includes the carbohydrate galactose, attached to the lipid by a glycosidic bond (glycosylation). Because GC is a sphingolipid that has been glycosylated, it can be classified as a glycosphingolipid.

Amino Acid Bonds

Peptide bonds are amide bonds that form between the N-terminus of one amino acid or peptide and the C-terminus of another. At physiological pH, the N- and C-termini consist of an NH3+ group and a COO− group, respectively. When these groups react with each other, they condense to form an amide group with the formula NHCO. Therefore, 1 O atom and 2 H atoms are lost from the reacting species. These atoms combine to form water.

Step 7: Phosphoglycerate kinase

Phosphoglycerate kinase catalyzes the reversible conversion of 1,3-bisphosphoglycerate into 3-phosphoglycerate, or the removal of a phosphate group from 1,3-bisphosphoglycerate. This generates one ATP per molecule of phosphoglycerate (or 2 ATP per glucose molecule).

Pyruvate oxidation

Recall that under aerobic conditions, the pyruvate molecules generated by glycolysis will enter the mitochondria. In the mitochondrial matrix, an enzyme known as pyruvate dehydrogenase catalyzes the irreversible conversion of pyruvate into acetyl-CoA and generates NADH and carbon dioxide as byproducts. This enzyme is subject to negative feedback as its product, acetyl-CoA, will inhibit its activity. The pyruvate dehydrogenase enzyme is not classified as part of glycolysis, nor as part of the citric acid cycle. Rather, it is an intermediate enzyme that catalyzes a transitionary step between the two pathways.

Most important steps in glycolysis:

Step 1: Hexokinase/Glucokinase Step 3: Phosphofructokinase 1 (PFK-1) Step 6: G3P dehydrogenase Step 7: Phosphoglycerate kinase Step 10: Pyruvate kinase

Amino Acid Structure

Steric effects occur within a molecule when electron clouds in close proximity interact and repel each other. Molecules (or substituent branches) that contain many atoms or atoms with large electron clouds are more prone to experiencing steric effects than molecules with few atoms or atoms that have small electron clouds. Molecular structures that are prone to steric effects (ie, large molecules) are constrained in the conformations they can adopt because many conformations result in repulsive interactions and are not stable. The twenty standard amino acids found in natural peptides and proteins vary greatly in the sizes of their side chains, and therefore they vary in the sizes of their electron clouds. The amino acids phenylalanine (F), isoleucine (I), leucine (L), lysine (K), methionine (M), arginine (R), tryptophan (W), and tyrosine (Y) all contain large side chains that make substantial contributions to steric constraints. Peptides that contain these amino acids experience greater steric effects than peptides of the same length that contain smaller amino acids. Each of the peptides in this question is made from six amino acids. However, the peptide FYLWIR consists entirely of amino acids with large side chains and will exhibit the greatest steric constraints of the choices given.

Mitochondrial structure

The outermost layer of the mitochondrion, the outer membrane, acts as a barrier between the cytosol and the organelle's interior. To serve this function, it is highly impermeable to the passage of ions. The inner layer, or the inner membrane, is the site of the electron transport chain (we'll discuss this more later in this guide). The inner membrane has many folds called cristae, which increase surface area. The inner mitochondrial membrane is also highly impermeable to ion exchange. The space enveloped by the inner mitochondrial membrane is known as the mitochondrial matrix. Finally, in between the inner membrane and the outer membrane is the intermembrane space, another important site for the electron transport chain. These regions and their acidity, or hydrogen ion concentration, will become exceedingly important for understanding how the electron transport chain functions.

The oxaloacetate shuttle

The oxaloacetate that is now present in the cytoplasm then re-enters the mitochondrial matrix in a series of steps: 1. Oxaloacetate is converted to malate. 2. Malic enzyme catalyzes the conversion of malate into pyruvate and produces NADPH as a byproduct. This NADPH will be critical in later steps of synthesis. 3. Pyruvate enters the mitochondrion and is converted into oxaloacetate by pyruvate carboxylase. Here, oxaloacetate can again be paired with acetyl-CoA to form citrate via citrate synthase. Why go through the trouble of shuttling citrate and pyruvate back and forth? Note that the oxaloacetate shuttle results in the production of NADPH. This NADPH is a crucial electron carrier that will be needed later in the synthesis. Without the oxaloacetate shuttle, these electrons would not be able to move from the inner mitochondrial membrane into the cytoplasm.

Oxidative phosphorylation

The protons making up the electrochemical gradient are used to make ATP through a process known as oxidative phosphorylation. As the name implies, this process transfers energy that was released through the oxidation of NADH and FADH2 into a new bond: one between ADP and an inorganic phosphate molecule. This results in the production of ATP. Spanning the inner mitochondrial membrane is the enzyme ATP synthase, sometimes referred to as complex V. The enzyme itself is composed of two components. The F0 component serves as a channel through which hydrogen ions in the inner mitochondrial space flow into the mitochondrial matrix. This flow of protons is also referred to as chemiosmosis. The F1 component uses the energy released by the gradient to create ATP via the phosphorylation of ADP. This phosphorylation is accomplished through conformational changes induced by the spinning of the F1 component—leading to the F1 component being known as a "molecular motor." Decoupling agents, or uncouplers, are molecules that inhibit the synthesis of ATP by destroying the proton gradient. Defects in proteins known as cytochromes commonly decouple this gradient.