NS exam 3 Biology

meaning that it is reduced

5,10-methylene THF gains a hydrogen,

C is correct. In the ETC, carriers travel inside the inner mitochondrial membrane, passing electrons from one to another and pumping protons across the inner mitochondrial membrane. Therefore, they are mobile. In order to travel inside the hydrophobic interior of the membrane, we would expect them to all be hydrophobic. However, cytochrome c is a highly water-soluble protein, unlike other cytochromes. Thus, answer choice C is correct. A: Electrons are passed from carriers with lower reduction potential to those with higher reduction potential. B: An electron from NADH is first accepted by the protein complex NADH-Q reductase, also known as the NADH dehydrogenase complex. H+ ions are pumped through this complex, which spans the membrane, and out of the matrix. D: Each member of the electron transport chain can carry one or two electrons at a time.

A student assisting with the experiment would observe all of the following about the electron transport chain EXCEPT: A. Electrons are passed from carriers with lower reduction potential to those with higher reduction potential. B. Complex I is also a proton pump. C. All electron carriers are mobile and hydrophobic. D. The electron carriers can transport a maximum of 2 electrons.

B is correct. For an enzyme-catalyzed reaction with a very low initial substrate concentration, and where Km >> [S], the Michaelis-Menten equation may be approximated as V = Vmax [S] / Km, where Vmax / Km is a constant of the reaction. Under these conditions, the reaction is approximately first order with respect to S. This can be seen graphically in Figure 1, where at low initial CTP concentrations, the enzyme activity (and the reaction velocity to which it is proportional) increases in a roughly linear fashion with CTP concentration. A generic Michaelis-Menten curve is shown below. Note that at low substrate concentrations, the reaction approximates first-order kinetics, as described above. In contrast, at very high substrate concentrations (where the enzyme is nearly or entirely saturated), the reaction approximates zero-order kinetics, since reaction rate ceases to depend on substrate concentration. A: The reaction would display zero-order kinetics only if increasing CTP concentrations did not lead to an increase in the reaction velocity. As mentioned above, this is generally true under saturating conditions, when substrate concentration is high relative to Km and when V = Vmax. C, D: A second- or third-order relationship at low substrate concentration would show a primarily non-linear relationship between CTP concentration and reaction velocity. problem 27

At very low CTP concentrations, kinetic data fitted to the Michaelis-Menten equation predicts that the initial rate of the CCT-catalyzed reaction is most nearly what order with respect to CTP? A. Zero order B. First order C. Second order D. Third order

Bacteriophages are viruses whose host cells are bacteria. Instead of entering the cell completely, they inject their genetic material into their host through a syringe-like structure known as a tail sheath. Retroviruses are a distinct class of single-stranded RNA viruses, including HIV, that use an enzyme known as reverse transcriptase to synthesize DNA from their RNA genome.

Bacteriophages

During the lytic cycle, the bacteriophage essentially works to replicate at full speed, making full use of the host cell's machinery. Eventually, the host cell is filled with virions to the point that it bursts or lyses, and a tremendous number of new virions spill out into the environment. Alternately, in the lysogenic cycle, bacteriophages can integrate themselves into the host genome, at which point they are referred to as a prophage or a provirus. In response to environmental signals, the prophage can re-emerge from the host genome and resume a lytic cycle. Non-bacteriophage viruses (e.g., those that infect humans) can exhibit a dormant stage that is similar to the lysogenic cycle. Examples of viruses with a prominent dormant stage include HIV and herpesvirus.

Bacteriophages have two distinct life cycles: lytic and lysogenic

Beta-oxidation Beta-oxidation is a process in which fatty acids are broken down into acetyl-CoA, which can be fed into the citric acid cycle (similarly to the effects of glycolysis and the pyruvate dehydrogenation complex). Beta-oxidation also generates the electron carriers NADH and FADH2, which produce energy in the electron transport chain. This process is known as beta-oxidation because the beta carbon of each fatty acid is oxidized to a carbonyl group (C=O). It occurs in the mitochondria in eukaryotic cells. The basic logic of beta-oxidation is to chop up extended fatty acid chains into two-carbon units of acetyl-CoA. Step 1 of beta-oxidation involves forming a C=C double bond between the alpha and beta carbons of the carbonyl group at the head of the acyl-CoA molecule. This is coupled to the formation of FADH2. Then, in step 2, an -OH group is added to the beta carbon. The C-OH bond on the beta carbon is oxidized to C=O in step 3, and NADH is formed. Then, in step 4, the molecule is broken up, yielding an acetyl-CoA group and a shorter acyl-CoA group. This process is easiest to visualize with a saturated fatty acid with an even number of carbons, but special enzymes exist to handle unsaturated fatty acid and odd numbers of carbons. Beta-oxidation can produce very large amounts of ATP, which is connected to the fact that fats are a form of long-term energy storage in the body. For instance, the beta-oxidation of palmitic acid (a saturated fatty acid with 16 carbons) yields approximately 106 ATP.

Beta-oxidation

A is correct. Cell differentiation occurs primarily through different gene expression levels. B: Nearly all cells share the same genome, so differences in the genetic material isn't a realistic explanation. C: Location within the body is not responsible for cell differentiation. A skin cell transplanted to the liver, for example, will not become a liver cell. D: It is DNA methylation, not ethylation, which plays a major role in gene expression and, consequently, cell differentiation

Cell differentiation is mediated primarily by: A. gene expression levels. B. differing genetic material. C. location within an organism. D. DNA ethylation.

Cellular Motility Motility refers to a cell's ability to move. Flagella and cilia are two structures involved in cell motility that are formed from microtubules. Flagella (singular = flagellum) are tail-like appendages that protrude from a cell and allow it to move, although they also can serve as sensory appendages. Flagella are found in both prokaryotes and eukaryotes, but they are structurally distinct. Cilia are relatively small projections that help move substances along the cell surface. A well-known example in the human body is the presence of cilia in the respiratory tract to help move mucus out of the lungs. In eukaryotes, both cilia and flagella are characterized by what is known as a 9+2 structure, in which an outer ring of nine pairs of microtubules surrounds an inner ring of two microtubules. Eukaryotic flagella flap back and forth, and their movement is powered by ATP. In contrast, prokaryotic flagella use a rotary motion, are powered by a proton gradient, and are composed of a protein known as flagellin.

Cellular Motility

B is correct. This question is asking you to recall the effects of aldosterone and how it achieves those effects. Aldosterone increases H2O and Na+ reabsorption from the kidney while exchanging K+ ions for Na+ ions. The triggers for and results of aldosterone secretion are shown below. A: Hypersecretion of aldosterone will negatively feed back and inhibit renin production. C: Hypersecretion of aldosterone will result in high blood sodium. D: Hypersecretion of aldosterone will result in high blood pressure.

Conn's syndrome, also known as primary hyperaldosteronism, is most likely to cause which symptom? A. High renin concentration B. Low blood potassium C. Low blood sodium D. Hypotension

Effects and Mechanisms of Enzymes Enzymes are biological catalysts. As such, enzymes speed up reaction rate by reducing the activation energy (Ea). These catalysts are highly specific for certain reactions or classes of reactions, and they are sensitive to temperature and pH. Enzymes are not changed or consumed in their associated reactions, do not affect the equilibrium constant (Keq) of these reactions, and do not affect any thermodynamic parameters of the reaction (ΔG, ΔH, and ΔS). For this reason, enzymes cannot make an energetically unfavorable reaction into a favorable one

Effects and Mechanisms of Enzymes

C is correct. The SN2 mechanism is favored by polar aprotic solvents, such as acetone or DMSO. The structure of acetone is shown below; note that it has a dipole moment ("polar"), but does not contain O-H or N-H bonds ("aprotic").

For the reaction below, which solvent will best promote an SN2 mechanism of reaction? A. H2O B. Methanol C. Acetone D. Toluene

Gene Expression The central dogma of molecular biology states that information is passed from DNA to RNA to protein. This means that when a cell needs more of a certain protein, it can increase the degree to which the gene corresponding to that protein is transcribed. Transcribing more or less of a gene in response to the cell's needs is known as gene expression. It plays a major role in the differentiation of organs in multicellular organisms, and can also vary on shorter time scales in response to changing environmental conditions. The details of how gene expression is regulated in eukaryotes are quite intricate, but you should be aware of some key concepts for the MCAT. Promoters are regions of DNA that lie upstream to a given gene and initiate transcription by binding specific transcription factors that contribute to the binding of RNA polymerase. Additionally, expression is upregulated by enhancers, which are DNA sequences that can be located further from the gene of interest, and work by binding transcription factors that twist DNA into a hairpin loop, bringing distant regions into close proximity for transcription to begin. Silencers are the opposite of enhancers in eukaryotic cells; they are regions of DNA to which transcription factors known as repressors bind. Additionally, the methylation of C and A residues can reduce transcription. Methylation is associated with epigenetics, which refers to inheritable phenotypic changes involving mechanisms other than the alteration of the genome itself. Gene expression can also be regulated on the level of nucleosomes (i.e. chromatin and histones). Acetylation promotes transcription by attaching acetyl groups to lysine residues on histones, making them less positively-charged and causing a looser wrapping pattern that allows transcription factors to access the genome more easily. Finally, non-coding RNA plays a role in gene expression. MicroRNA (miRNA) strands are single-nucleotide strands incorporated into an RNA structure with a characteristic hairpin loop, while small interfering RNA (siRNA) molecules are short and double-stranded. Both tend to be approximately 22 nucleotides in length, and silence genes by interrupting expression between transcription and translation.

Gene Expression summary

Glycosidic bonds can be described in terms of the anomers (α or β) and specific carbons involved. Common glycosidic bonds include α(1→4), α(1→6), β(1→4), and β(1→6) bonds. Starch is composed of two main subtypes of polymers: amylose and amylopectin. Amylose is a linear polymer of glucose molecules connected by α(1→4) glycosidic bonds; it comprises about 20%-30% of starch. Amylopectin makes up the remaining 70%−80% of starch; it likewise contains glucose molecules connected by α(1→4) glycosidic bonds, but it has branches due to α(1→6) glycosidic bonds every 24−30 units. It is somewhat easier to break down than amylose. These bonds are also relevant to how humans (as well as most other non-plant forms of life) store energy in the form of glycogen. In humans, glycogen is stored in liver and muscle cells. Structurally, glycogen is similar to amylopectin in that it contains chains of glucose molecules connected by α(1→4) glycosidic bonds, with intervening α(1→6) glycosidic bonds that create branches; the main difference is that glycogen is more heavily branched than amylopectin, with branches occurring every 8−12 units. In contrast, a polysaccharide known as cellulose plays a major structural role in the cell walls of plants. Like starch and glycogen, it is a polymer of glucose, but unlike them, it incorporates the β-anomer of glucose, with glucose subunits connected by β(1→4) glycosidic bonds. This seemingly small structural distinction makes a world of practical difference, as humans lack the necessary enzymes to digest cellulose.

Glycosidic Bonds

B is correct. Triglycerides are composed of 1 glycerol molecule and 3 fatty acid molecules. So, if we start with 3000 triglyceride molecules, we would expect to have 9000 molecules of fatty acids and 3000 molecules of glycerol after hydrolysis. An example of a triglyceride is shown below. A: Triglycerides contain 3 fatty acid components per glycerol component. This answer wrongly assumes a 1:1 ratio of fatty acid: glycerol. C, D: These choices are not correct, since phospholipids are not components of triglycerides. They would not be formed from the hydrolysis described.

If 3000 molecules of triglycerides are hydrolyzed into their component molecules, what would the resulting mixture contain? A. 3000 fatty acid molecules and 3000 glycerol molecules B. 9000 fatty acid molecules and 3000 glycerol molecules C. 3000 phospholipid molecules and 3000 glycerol molecules D. 9000 phospholipid molecules and 3000 glycerol molecules

C is correct. The products of the initial round of LCFA β-oxidation are 1) an acyl-CoA group that is two carbons shorter than the parent chain and 2) an acetyl-CoA molecule. According to the reaction pathway illustrated in Figure 1, the first and last reactions of which are pictured here, C2 and C3 carbons in an original LCFA will enter the mitochondria in the C2 and C3 positions of activated fatty acid-CoA derivatives (acyl-CoA). They will eventually become included in the C-2 and C-3 positions of 3-ketoacyl-CoA. Image Image According to the final reaction pictured in the pathway in Figure 1, thiolase cleaves the acyl-sulfur bond of the 3-ketoacyl-CoA thioester, producing an acyl-CoA molecule. This acyl-CoA unit contains the original C3 carbon as its carbonyl carbon. The original C2 carbon is contained in acetyl-CoA as the terminal methyl carbon. A, B: These answers are not sufficient, per the reasoning above. D: Labeled carbons will be found in both of these molecules

If a LCFA is synthesized with isotopically labeled carbons at the C2 and C3 positions, in which product(s) of the initial cycle of its β-oxidation will an isotopically-labeled carbon be found? A. A fatty acyl-CoA molecule only B. An acetyl-CoA molecule only C. Both a fatty acyl-CoA and an acetyl-CoA molecule D. Neither a fatty acyl-CoA nor an acetyl-CoA molecule

Immune System The immune system refers to the complex set of mechanisms that the body uses to protect itself against foreign invaders and malfunctioning cells originating from the body itself. The highest-level distinction in the immune system is between the innate (or non-specific) immune system, which responds generally to threats but does not learn to recognize specific foreign bodies/molecules, and the adaptive immune system, which does. The non-cellular component of the innate immune system includes anatomical barriers and signaling molecules such as cytokines and complement proteins, while the cellular component includes a range of white blood cell types (leukocytes) that play various roles in responding to threats. White blood cells include neutrophils, lymphocytes, monocytes (which differentiate into macrophages or dendritic cells), eosinophils, basophils, and mast cells. The various components of the innate immune system can act independently or be coordinated in the process of inflammation. The adaptive immune system includes B cells and T cells, both of which are lymphocytes that are produced in the bone marrow and mature in the lymphatic system. B cells recognize antigens and secrete large amounts of antibodies in response. The human body utilizes five classes of antibodies: immunoglobulin (Ig) A, IgD, IgE, IgG, and IgM, which differ in the details of their heavy chains. This response is known as humoral immunity. In contrast, T cells correspond to the cell-mediated branch of the adaptive immune system. T cells, which mature in the thymus, recognize cells that were originally self, but have been damaged by viral infections or have malfunctioned in ways likely to turn them into cancer cells. Then, various subgroups of T cells either directly attack compromised/foreign cells or mobilize responses to them based on antigen fragments that are presented by major histocompatibility complex (MHC) class I and II.

Immune System summary

C is correct. For this question, we need to know what the G0 phase (shown below along with the rest of the cell cycle) entails. This is the state that a cell will enter if it does not need to divide. Since epithelial cells are those that divide the most out of the options, choice C is correct. A: Liver cells are more likely to be found in G0 because they do not divide as often as epithelial cells. B: Kidney cells are more likely to be found in G0 because they do not divide as often as epithelial cells. D: Neurons in adults do not divide and are almost always found in G0.

In an adult, which of the following cell types is LEAST likely to enter a programmed G0 phase of the cell cycle? A. Liver cells B. Kidney cells C. Epithelial cells D. Neurons

C is correct. G0 is a non-growing state that accounts for the observed differences in length in the cell cycle. While intestinal cells will divide twice per day, neuronal cells do not divide following initial differentiation and therefore remain permanently in G0. The diagram below depicts the stages of the cell cycle, including G0. A, B: Since gastrointenstinal cells divide more often than neurons, they would be expected to spend more time in S (DNA synthesis) phase and more time in M (mitotic) phase. D: This statement is false, for the reasons given above.

In comparison to a neural cell, a gastrointestinal cell would likely spend: A. less time in S phase. B. less time in M phase. C. less time in G0. D. the same amount of time in all cell cycle phases.

C is correct. [S] is the concentration of the substrate, and Km implies that half of the active sites on the enzymes are filled. Different enzymes have different Km values. When all of the active sites are occupied, a reaction has reached maximum velocity. Therefore, Km is equal to the concentration of the substrate when the rate is half of the maximum velocity. A: Be careful here! This answer is incorrect; [S] = Km, not 2 Km, when the reaction rate is half of its maximum velocity. B: Do not fall for this answer simply because it includes 1/2! At half of the maximum velocity, [S] = Km, not 1/2 Km.

In order for an enzyme obeying the Michaelis-Menten model to reach 1/2 of its maximum velocity: A. [S] must be equal to 2 Km. B. [S] must be equal to 1/2 Km. C. [S] must be equal to 1 Km. D. [S] must be equal to 10 Km.

B is correct. Paragraph 2 states that LCFAs must first enter the cell, where they are converted to their CoA derivatives by thiokinase, a membrane-bound enzyme. Also according to the passage, this conversion occurs prior to their transport into the the mitochondrial matrix by carnitine. This indicates that thiokinase must act after an activated fatty acid enters a cell, but before its transport into the mitochondrial matrix. Thiokinase must then be bound to the cytosolic face of the outer mitochondrial membrane, which makes this choice correct. For reference, the parts and compartments of the mitochondrion are shown below. A: According to the passage, thiokinase acts on LCFAs prior to their transport (or specifically, the transport of their derivatives) into the mitochondrial matrix. Thiokinase is therefore not likely to be located in the matrix or on the inner membrane. C, D: The second paragraph states that thiokinase is a "membrane-bound enzyme." These fractions are not associated with a membrane

In order to be β-oxidized, LCFAs must first enter the cell, where they are converted to their CoA derivatives by thiokinase, a membrane-bound enzyme. Because activated fatty acids, including LCFA-CoA derivatives, are impermeable to the inner mitochondrial membrane, they must be transported into the mitochondrial matrix by carnitine, a specialized carrier protein. First, the fatty acid acyl group is transferred from CoA to carnitine by the enzyme CPT-I, forming acylcarnitine. Acylcarnitine is then transported across the inner mitochondrial matrix in exchange for free carnitine in a transfer reaction catalyzed by CPT-II. Once reformed, carnitine is returned to the intermembrane space for re-use by facilitated diffusion. Fatty acids shorter than 12 carbons are able to cross the inner mitochondrial matrix without the aid of the CPT system. Researchers studying the energy yield of fatty acid metabolism estimated the amount of ATP produced in the mitochondria of cultured human cells by the β-oxidation of saturated, straight-chain fatty acids of various lengths in both the presence and the absence of oligomycin, an inhibitor of ATP synthase Four fractions were obtained from the differential centrifugation of a homogenized human liver cell sample: F1: Inner mitochondrial membrane plus matrix F2: Outer mitochondrial membrane F3: Enzymes localized to the intermembrane space F4: Cytosolic enzymes In which of the following fractions was thiokinase most likely isolated? A. F1 B. F2 C. F3 D. F4

A is correct. Malonyl-CoA, an indicator of ongoing fatty acid synthesis, inhibits β-oxidation by preventing the movement of long-chain acyl groups into the mitochondrial matrix, thereby preventing a futile cycle of fatty acid synthesis followed by immediate β-oxidative catabolism of those newly synthesized fatty acids. B, C: Inhibition of β-oxidation reduces the available pool of reducing equivalents that would otherwise be produced directly from the reactions of β-oxidation, as well as by the oxidation of acetyl-CoA, the production of which slows as fewer fatty acids flow through the cycle. D: The passage provides no direct or indirect evidence to suggest that the allosteric effect of malonyl-CoA is modulated by insulin. Additionally, decreased insulin secretion tends to globally increase the rate of catabolic processes, such as β-oxidation, and decrease the rate of anabolic ones, including fatty acid synthesis.

Malonyl-CoA is an intermediate in cytosolic fatty acid biosynthesis. Its inhibition of the β-oxidation of long-chain fatty acids: A. prevents the degradation of newly synthesized fatty acids. B. ensures that ample reducing equivalents are available for fatty acid synthetic activity. C. increases the cellular pool of acetyl-CoA available to condense with oxaloacetate. D. is up-regulated in the presence of decreased insulin secretion.

D is correct. The final paragraph of the passage mentions that p53 is so named for its estimated molecular weight using SDS-PAGE. This question is asking us to determine a plausible reason for the 9.3-kDa discrepancy. SDS-PAGE is used to grant a uniform negative charge to all proteins in an assay. However, if the protein has enough charged residues on its own, it causes the measurements to be less accurate. If there are enough positive charges, the negative charge of the protein will not be as great as anticipated and the molecule will travel a smaller distance, which, to the individual running the assay, usually means the mass of the protein is greater and would lead to the discrepancy described. A, C: The SDS-PAGE measurement is artificially large (53 kDa rather than the actual weight of 43.7 kDa). This means that the p53 protein migrated a smaller distance, not a greater distance. (Remember, the smaller the distance traveled in SDS-PAGE, the larger we would expect the protein to be.) B: In SDS-PAGE, proteins migrate toward a positively-charged pole. As such, containing many negatively-charged residues would make a protein migrate a larger distance, not a smaller one

Methylation by SAM has the regulatory effect of reducing gene expression of a methylated gene. The body has many processes in place to repair these DNA problems. Perhaps the most important protein to have been discovered with regards to regulation of DNA repair is called p53, named so for the weight of the protein as estimated by SDS-PAGE. This protein has been shown to be involved in countless processes, resulting in death of damaged cells or the repair of their DNA. P53 is activated by a wide variety of stimuli including DNA damage and increased free radicals. P53 levels are maintained at a normal level by their continued degradation. The protein HDM2 is an integral protein involved in the degradation of p53. The true weight of the p53 protein, as calculated by summing the weight of the amino acids encoded by its gene, is 43.7 kDa. Which of the following possible mechanisms would best explain this discrepancy from its estimated weight, as mentioned in the passage? A. The protein migrated a greater distance because it contains many positively charged residues. B. The protein migrated a smaller distance because it contains many negatively charged residues. C. The protein migrated a greater distance because it contains many negatively charged residues. D. The protein migrated a smaller distance because it contains many positively charged residues.

Oxidative Phosphorylation The final major process of aerobic respiration is oxidative phosphorylation, through which the majority of aerobically-derived ATP is synthesized. This process begins by passing electrons through a series of chemical reactions, known as the electron transport chain (ETC), to a final electron acceptor, oxygen. This is the only time in eukaryotic aerobic respiration where oxygen is directly required. These reactions take place in specialized proteins where the energy from NADH and FADH2 is used up, molecular oxygen is reduced into water, and approximately 30-36 ATP are created from ADP and inorganic phosphate. The electron transport chain (ETC) uses free oxygen as the final electron acceptor of the electrons removed from NADH and FADH2 formed in glycolysis and the Krebs cycle (also known as the citric acid cycle). The ETC is composed of four large protein complexes (Complexes I-IV) embedded in the inner mitochondrial membrane and two small electron carriers shuttling electrons between them. Complex I is known as NADH dehydrogenase, II is known as succinate dehydrogenase, III is known as cytochrome bc or c, and IV is known as cytochrome c oxidase. Electrons are released from NADH and FADH2 through a series of reactions. In the ETC, the energy released from the series of electron transfers is used to pump H+ across the membrane. The unequal concentrations of H+ ions across the membrane establishes an electrochemical gradient, leading to chemiosmosis, or the passive diffusion of the protons down their concentration gradient, which is coupled to ATP synthase. This proton movement generates 90% of the ATP synthesized during oxidative phosphorylation. The electrons passing through the electron transport chain gradually lose energy until eventually they are donated to O2, which accepts two H+ ions and is transformed into water. If the proton gradient is disrupted or destroyed, chemiosmosis can become uncoupled from the ETC, resulting in little to no ATP generation despite the transfer of electrons carrying on. Many poisons and toxins act by uncoupling the proton gradient from ATP synthase

Oxidative Phosphorylation

A is correct. According to paragraph 2, "Arg681 engages in two direct interactions: a salt bridge with Asp145 of chain A and an intra-subunit interaction with Cys683." Since the passage describes the interaction between Arg681 and Cys683 as "intra-subunit," the two residues must be located on the same chain (eliminate choices B and C). In order for arginine to participate in a salt bridge, it must be ionized, indicating that a nitrogen of the side chain guanidinium group must be protonated. Cysteine is not ionized at physiological pH (~7.4), but it is polar. Thus, its interactions with the other residue are most likely to involve ion-dipole interactions. This is sufficient information to eliminate all answer choices other than A. B, C: Arg681 and Cys683 are both located on chain B. D: Two sulfhydryl groups present in individual cysteine side chains are required for the formation of a disulfide bond

Plasmodium CCT functions as a dimer containing two functionally equivalent catalytic domains, PfCCT and MK, located on the subunit chains A and B, respectively. Molecular dynamics (MD) simulations were used to generate thermodynamic data for the wild-type enzyme (MKWT) and for its point mutant (MKR681H) resulting from the substitution of of arginine 681 with histidine. Data revealed that Arg681 engages in two direct interactions: a salt bridge with Asp145 of chain A and an intra-subunit interaction with Cys683. Interaction energies between chains A and B of both enzymes were calculated in the equilibrated phase according to the equation: At physiological pH, what will be the principal interactions between Arg681 and Cys683? A. Ion-dipole interactions between two chain B residues B. Dipole-dipole interactions between chain A and chain B residues C. Hydrogen bonds between a chain A and a chain B residue D. Disulfide bonds between two chain B residues

Reaction Rates The rate of a reaction—or, simply put, how fast it happens—is expressed as a function of the rate constant k and the concentration of some or all of the reactants. The rate law of a general reaction of the form aA + bB → cC + dD is: rate = k[A]x[B]y. Reaction rate itself has units of M/s or mol/L∙s. [A] and [B] are concentrations of the reactants in units of M or mol/L. The exponents x and y must be experimentally determined; they do not correspond to the stoichiometric coefficients a and b in the reaction formula. The order of a reaction is defined by the sum of the exponents (x + y) in the rate law. If the exponents sum to zero, the reaction is zero-order. If the reactants sum to 1, the reaction is first-order. If they sum to 2, it is second-order, and so on. Rate laws can be determined experimentally using the method of initial rates. In this method, multiple trials are run with variations in the concentration of individual reactants. The initial rate is then measured, with the goal of identifying how changes in the concentration of a reactant affect the rate. The rate constant k can be calculated from any experimental trial after the exponents of the rate law have been established, and the units of k can be derived once those exponents are known through simple algebra. It should be noted, though, that while the units of M (alternatively expressible as mol/L) vary according to the overall order of the reaction, the units of k must always involve inverse seconds. Although the rate constant must be experimentally determined, it varies according to the activation energy (Ea) and temperature. Decreasing Ea through a catalyst/enzyme and increasing the temperature will increase the reaction rate. Physiologically, zero-order reactions are exemplified by enzyme-catalyzed reactions in which the enzyme is saturated—that is, when concentrations of the reactant far exceed the available active sites on enzymes. In such a situation, the catalysis is the rate-limiting step and the concentration of the reactant is irrelevant. First-order reactions are exemplified by radioactive decay and SN1 reactions that are dependent on carbocation formation. Second-order reactions physically involve collisions between two reactant molecules, as in SN2 reactions. In multi-step reactions, the step with the slowest rate determines the overall rate of the reaction, so it is known as the rate-limiting step.

Reaction Rates

Regulation of Fluid Balance Fluid regulation is crucial for the maintenance of life. There are two basic scenarios that the body can encounter: too little fluid and too much fluid. Having too little fluid in one's system manifests in three important ways: reduced blood volume (because relatively little water is present in the blood plasma), reduced blood pressure (a consequence of reduced blood volume—less liquid is present to exert pressure against the walls of the blood vessels), and increased blood osmolarity (the same solutes are present, but less solvent is available). Correspondingly, increases in the amount of fluid present in the system manifest as increased blood volume (because more water is present in the blood plasma), increased blood pressure (more blood volume means more pressure against the walls of the blood vessels), and decreased blood osmolarity (the same solutes are present, but more solvent is available). Two major hormones respond to low fluid levels by increasing fluid retention: aldosterone (the main example of a class of steroid hormones known as mineralocorticoids) and anti-diuretic hormone (ADH), a peptide hormone that is also known as vasopressin. However, these two hormones have different mechanisms. Aldosterone works by increasing sodium absorption in the distal convoluted tubule and collecting duct of the nephron, which drives water absorption. Aldosterone also increases excretion of potassium and hydrogen ions in the urine. In contrast, ADH increases the permeability of the collecting duct to water, thereby increasing water absorption. Thus, ADH acts to reduce the osmolarity of blood by increasing the amount of water present without changing the solute levels, whereas aldosterone does not affect osmolarity because sodium reabsorption drives water absorption. Atrial natriuretic peptide (ANP; also known as atrial natriuretic factor [ANF]) is a hormone that the endocrine system uses to deal with the problem of excess blood volume. Essentially, it is the opposite of aldosterone. It is released in response to high blood volume and decreases sodium reabsorption in the distal convoluted tubule and the collecting duct, as well as increasing the glomerular filtration rate and inhibiting aldosterone release.

Regulation of Fluid Balance

B is correct. This question is asking us to determine the weight of a strand of genetic material that contains uracil instead of thymine. We are told in the passage that uracil can be formed from the direct deamination of cytosine, while thymine is formed from a deamination and a methylation. Therefore, thymine weighs more than uracil by one methyl group. The ribose ring of RNA will also be heavier than the deoxyribose ring of DNA because the DNA is "de-oxy," meaning it lacks the 2' hydroxyl group that is present on the ribose of RN

Researchers compare a DNA sequence and an identical sequence of RNA where all thymines are replaced with uracil. Which of the following describes changes that they can expect? I. The uracil base will be heavier than the thymine base because of an extra methyl group. II. The thymine base will be heavier than the uracil base because of an extra methyl group. III. The carbohydrate ring in RNA is heavier than the carbohydrate ring in DNA. IV. The carbohydrate ring in DNA is heavier than the carbohydrate ring in RNA. A. I and III only B. II and III only C. I and IV only D. II and IV only

Saponification Saponification is a special case of acid-base chemistry involving carboxylic acids (Latin sapo means "soap," so this literally means something like "soapification"). Under basic conditions (e.g., when mixed with NaOH or KOH), carboxylic acids are deprotonated and their conjugate bases form salts, according to the following template: RCOOH + Na+ + OH− → RCOO−Na+ + H2O. The acid-base chemistry of this process is straightforward, although it is worth noting that the process can be combined with the base-catalyzed hydrolysis of a triglyceride containing three fatty acid chains. As such, saponification is mostly of note for historical interest and because its wide range of applications opens the door for it to be introduced as a way of testing organic chemistry and general chemistry principles in a passage on Test Day.

Saponification

A is correct. Based on the passage, DNA coding for the antibody cetuximab was spliced into the viral vector, resulting in AAVrh.10CetMab. The only way that this splicing process could take place is if the viral genetic material were made of DNA. B, C: The genomic material of the viral vector must have been DNA for the splicing process to have worked effectively. D: Retroviruses have positive-sense RNA genomes, not DNA genomes. There is also no evidence in the passage that the mechanism of antibody delivery involved reverse transcriptase.

The AAVrh.10 viral vector was most likely which kind of virus? A. A DNA virus B. A positive-sense RNA virus C. A negative-sense RNA virus D. A retrovirus

The following features are associated with reduction in organic chemistry: (1) gain of an electron, (2) decreased oxidation state, (3) formation of a C-H bond (e.g. alkene → alkane), and (4) loss of a C-O or C-N bond (or any bond between carbon and an electronegative atom). Conversely, oxidation is associated with (1) loss of an electron, (2) increased oxidation state, (3) loss of a C-H bond (e.g. alkane → alkene), and (4) gain of a C-O or C-N bond (or any bond between carbon and a highly electronegative atom). Oxygen-containing organic compounds exist on a spectrum of oxidation from alcohols (most reduced/least oxidized) to aldehydes/ketones (intermediate reduction/oxidation) to carboxylic acids (least reduced/most oxidized). A primary alcohol can be oxidized to an aldehyde by a mild oxidizing agent (such as PCC) or to a carboxylic acid by a strong oxidizing agent like NaCr2O7. A secondary alcohol will be oxidized to a ketone by either a mild or a strong oxidizing agent. A strong oxidizing agent will likewise oxidize an aldehyde to a carboxylic acid. A strong reducing agent, such as LiAlH4, can reduce a carboxylic acid directly to an alcohol, while weak reducing agents such as NaBH4 will not reduce carboxylic acids at all. A special agent, DIBAL, can reduce a carboxylic acid to an aldehyde when applied at a precise 1:1 ratio. Both mild and strong reducing agents can reduce aldehydes and ketones to primary and secondary alcohols, respectively.

The following features are associated with reduction in organic chemistry

A is correct. This question asks us to find the probability of a mother giving birth to have a phenotypically normal child. The chances of nondisjunction are 1/130 (0.008), or about 0.8%. If nondisjunction occurs, there is a 100% chance that she will pass on either no copies of her X chromosome and have a child with Turner syndrome (XO) or pass on two copies of the X chromosomes and have a baby with Trisomy X (XXX) or Klinefelter syndrome (see the Punnett square below). The only viable monosomy is the XO genotype. A YO genotype is not viable, meaning we cannot include it in our calculations, since the question asks us to assume a viable birth. Therefore, we simply take 1 - (probability of nondisjunction) = 1 - 0.008 = 0.992 = 99.2%. Image Note that nondisjunction refers to the failure of chromosomes or chromatids to properly separate during anaphase. Nondisjunction can occur during meiosis I or meiosis II. In the diagram below, the left image shows nondisjunction during meiosis II, while the right-hand image shows the same event occurring during meiosis I. B: This answer is the result of keeping the YO as a possible outcome when calculating probability. C, D: These choices more closely resemble the probability of having an non-phenotypically-normal baby.

The instance of nondisjunction for the X chromosome in females over the age of 30 is about one out of every 130 live births. If a woman over 30 gives birth to a viable baby, assuming the risk of nondisjunction from the father is negligible, what is the likelihood that it will have a normal phenotype? A. 99.2% B. 99.6% C. 0.6% D. 0.4%

C is correct. Passive immunity is the transfer of active humoral immunity in the form of ready-made antibodies, from one individual to another. A: Natural immunity is immunity that is present in the individual at birth, prior to exposure to a pathogen or antigen, and that includes intact skin, salivary enzymes, neutrophils, natural killer cells, and complement. B: Cell-mediated immunity is an immune response that does not involve antibodies, but rather involves the activation of phagocytes, antigen-specific cytotoxic T-lymphocytes, and the release of various cytokines in response to an antigen. D: The innate (non-specific) immune system includes anatomical barriers, secretory molecules, and cellular components. Among the mechanical anatomical barriers are the skin and internal epithelial layers, the movement of the intestines, and the oscillation of bronchopulmonary cilia.

The passage of IgG antibodies from mother to fetus illustrates: A. natural immunity. B. cell-mediated immunity. C. passive immunity. D. nonspecific immunity.

B is correct. Translation, the production of gene products, aka proteins, occurs at the ribosomes. A: The nucleolus is the nuclear subdomain that assembles ribosomal subunits in eukaryotic cells. C: The lysosome is the organelle in the cytoplasm of eukaryotic cells containing degradative enzymes enclosed in a membrane. D: A centriole is composed of short lengths of microtubules arranged in the form of an open-ended cylinder, designed to help chromosomes separate during cell division.

The production of the identified immunoglobulin would most directly involve which eukaryotic structure? A. Nucleolus B. Ribosome C. Lysosome D. Centriole

C is correct. Transduction is a form of horizontal gene transfer in bacteria in which bacteriophages (viruses that infect bacteria) transmit genomic material. A: Transformation involves direct uptake of genetic material from the environment and is not mediated by viruses. B: Conjugation is a horizontal gene transfer process in bacteria in which plasmid DNA is transferred from one bacterium to another through a pilus. Viruses are not involved. D: Binary fission is how bacteria reproduce asexually.

Viruses are directly involved in which of the following processes in bacteria? A. Transformation B. Conjugation C. Transduction D. Binary fission

D is correct. Viruses are unique in that they occupy a gray area between living and non-living. They have been described as non-living. Prions, or pathogenic misfolded proteins, are also non-living biological substances that are pathogens, but they are not considered organisms. A: This choice is not correct, since other types of organisms (such as bacteria and fungi) are able to reproduce inside other organisms, as with a fungal infection. B: This choice is not correct, since some viruses use DNA as genetic material. C: Other types of infectious agents are obligate parasites (e.g., tapeworms).

Viruses are distinct from most biological organisms in that they: A. are able to reproduce inside a host. B. only use RNA as genetic material. C. are obligate parasites. D. are pathogenic entities which have been described as not being "living."

B is correct. This question is asking us to simply recall a fact about the S-phase of a cell, aka DNA replication. You should know that DNA ligase is the enzyme that connects Okazaki fragments on the lagging strand of DNA being replicated. An overview of some of the key structures and enzymes of DNA replication is shown below. A: This is the function of primase. C: This is the function of DNA polymerases, not ligase. D: This is the function of helicase.

What is the function of ligase during S-phase? A. Begins synthesis of a new strand of copied DNA B. Binds together pieces of the lagging strand C. Repairs pieces of mis-replicated DNA D. Opens the double helical strand of DNA to begin replication

A is correct. The passage states that DMD is X-linked recessive, which means that a woman must have a mutated copy of the dystrophin gene on both of her X chromosomes in order to express the trait. Because the father is unaffected and will pass his X chromosome onto his daughter, there is a 0% chance that she will be affected. The father cannot have a copy of the DMD mutation, or he would be affected himself. X-linked recessive diseases are extremely rare in women. Image B, C, D: Again, since the father is unaffected, he must not carry the DMD allele. As such, there is a 0% chance that his daughter will have this recessive disease.

What is the probability that the daughter of an unaffected father and a mother who is a carrier for the DMD gene is affected by the disease? A. 0% B. 25% C. 50% D. 100%

C is correct. It is stated in the first paragraph that the polymorphism involves the conversion of a cytosine to a thymine. For the MCAT, you should know that adenine and guanine are purines (2-ring stuctures) and that thymine, cytosine, and uracil are pyrimidines (single-ring structures). As such, this is a conversion from a pyrimidine (single-ring structure) to another pyrimidine (single-ring structure). https://www.atdbio.com/img/articles/purine-pyrimidine-large.png A, B, D: These choices all involve at least one 2-ring structure, or purine (adenine or guanine). Neither purine is involved in this polymorphism.

When cytosine is replaced by thymine at the 677th position The polymorphism located at the 677th location involves the conversion of a: A. 2-ring structure to a single-ring structure. B. single-ring structure to a 2-ring structure. C. single-ring structure to a single-ring structure. D. 2-ring structure to a 2-ring structure.

A is correct. Certain tissues in the body are able to regenerate due to the ability of their cells to undergo mitosis when necessary. Skin grows by mitosis when injured; however, if control of this growth is lost, then conditions such as keloids of skin cancer (melanoma) may result. B: Meiosis is used to generate sex cells (gametes), not somatic cells. C: Excessive proliferation would not clearly be linked to defects in aerobic metabolism. D: Phagocytosis is not involved in prevention of skin overgrowth.

When human skin suffers a cut, the process of healing rapidly begins allowing for wound closure and healing within a few days. Keloids occur when skin around wounds continues to grow after the skin has healed. A disruption in the regulation of which cellular process is probably responsible for this condition? A. Mitosis B. Meiosis C. Aerobic respiration D. Phagocytosis

A is correct. The membrane-spanning domain of EGFR is in contact with the nonpolar tails of the phospholipid molecules that comprise the bilayer membrane. This region would be expected to be abundant in nonpolar amino acid residues, such as Val (valine). B: Tyr (tyrosine) has a polar hydroxyl group. C: Glu (glutamate) has a negatively-charged COO- group at physiological pH. Charged groups are polar and would be less likely than nonpolar residues to be found in the transmembrane domain. D: Lys (lysine) has a positively-charged NH3+ group at physiological pH.

Which of the following amino acid residues is most likely to be found in the transmembrane domain of EGFR? A. Val B. Tyr C. Glu D. Lys

B is correct. This question is asking us to determine the best support for a claim made in the passage. Choice B directly relates p53 to the incidence of tumors in humans. More importantly, it supports the claim that p53 is the "most important" protein that may have a role in DNA repair, since p53 mutations are implicated in more tumors than are mutations in any other gene. A: This could be true even if p53 played only a very minor role in DNA repair. C: The simple fact that cancerous cells lack a protein do not mean that protein has anything to do with the incidence of that cancer (or with DNA repair). D: This could also be true even if p53 did not function in DNA repair.

Which of the following discoveries would most support the statement that p53 is the most important protein regulator of DNA repair to have been discovered? A. When mutated, p53 continues to function to repair damaged DNA. B. More human tumors can be traced to a mutation in the p53 protein gene than in any other protein. C. Most cancerous cells lack the p53 protein. D. p53 levels are always elevated in patients with cancer.

B is correct. An organic compound must contain carbon and hydrogen in its formula. Furthermore, there must be a covalent bond between a carbon and hydrogen atom in the molecular structure. Organic acids are weak acids, generally having formulas of R-CO2H, with the acidic hydrogen bonded to an oxygen atom. Even knowing this information, you may be intimidated, as you may not be familiar with the structures of all of these choices. However, you can begin with the most familiar compound, carbonic acid. The Lewis structure of this compound is shown below. Image From this structure alone, we can see that neither hydrogen atom is directly bound to carbon, so carbonic acid is not an organic acid and must be the correct answer here. A, C, D: The structures of these choices are shown below; all fit the criteria required for organic acids.

Which of the following is NOT considered an organic acid? A. Folic acid B. Carbonic acid C. Ascorbic acid D. Citric acid

C is correct. This question is asking you to recall the function of glucagon and where it is likely to induce this function. Glucagon is released by the pancreas as a response to low blood glucose levels. Its main purpose is to increase glycogenolysis to increase blood glucose. The glycogen stored in the liver is broken down to create glucose that is released into the bloodstream; therefore, most of glucagon's action occurs in the liver. A: The kidney has glucagon receptors, but not as many as the liver. B: There is no reason to think that the heart would have the greatest need to be acted on by glucagon. D: Although skeletal muscle cells do store glycogen, they use their glycogen to power rapid contractions when needed, not to create glucose to be released into the blood. Therefore, they would not have a great need for glucagon receptors.

Which of the following locations is expected to have the highest number of glucagon receptors? A. Kidney B. Heart C. Liver D. Skeletal muscle

B is correct. This question is asking us to determine whether the molecules given are chiral. We should know that glycine (shown below) is the only amino acid that is not chiral, since it contains two H atoms bound to its alpha carbon. A: Lysine is chiral and includes one stereocenter, so it can exist as two enantiomers. C: Tyrosine does contain one stereocenter. D: Arginine is chiral.

Which of the following statements does NOT accurately describe amino acid stereochemistry? A. Lysine can exist as two enantiomers. B. Glycine is a chiral molecule. C. Tyrosine contains one stereocenter. D. Arginine is a chiral molecule.

C is correct. This question is asking us to determine the least likely mutation. Methylation and deamination cannot convert a purine to a pyrimidine or vice versa, because purines and pyrimidines differ based on whether they have a 1-ring structure (pyrimidines) or a 2-ring structure (purines), and the loss or addition of a single functional group (as in (de)methylation or (de)amination) is insufficient to convert between these very different structures. Furthermore, in addition to being able to predict this through general knowledge, the passage confirms this intuition by stating that purines are likely to mutate into other purines. Thus, choice C is correct as it depicts a mutation from cytosine (a pyrimidine) to adenine (a purine).

Which of the following substitutions is LEAST likely to occur via methylation or deamination? A. AGTCATA → AATCATA B. AGTCATA → AGTCGTA C. AGTCATA → AGTAATA D. AGTCATA → AGTTATA

C is correct. In order for cells to travel to the site of injury, they need to migrate. Microtubule de-polymerization is responsible for separating chromosomes during anaphase of mitosis or meiosis I or II. It does not contribute to overall cell migration. A, D: Flagella and cilia are structures that allow simple eukaryotic and prokaryotic cells to propel themselves or nutrients in their environment. These structures are shown below; however, since both are involved in cellular movement, neither is the answer to this LEAST question. B: Rapid actin polymerization near the edge of the cellular membrane is responsible for cellular motility in complex eukaryotic cells.

Which of the following would be LEAST useful in cellular movement? A. Flagella B. Actin polymerization C. Microtubule depolymerization D. Cilia

A is correct. The blood-brain barrier is primarily composed of endothelial cells with tight junctions (shown below) that prevent the movement of most solutes. Astrocytes also contribute to the function of the blood-brain barrier; however, these glial cells do not have dendrites or synapses, which are characteristic of neurons and their interconnections, respectively. B, C, D: Synapses and dendrites are characteristic of neurons and are therefore not present in the endothelial cells of the blood-brain barrier.

Which of the structures listed below is found in the blood-brain barrier? A. Tight junctions B. Synapses C. Dendrites D. More than one of the above

The four levels of protein structure contribute to the multiple functions that proteins play inside the body. The first level, referred to as the primary structure, is the amino acid sequence that codes the protein. Each amino acid is coded for by a three-nucleotide sequence known as a codon. The second level of protein structure refers to its two-dimensional arrangement (secondary structure). This folding is due to hydrogen bonds between groups along the peptide backbone that link different segments of the polypeptide chain to each other. These segments do not necessarily follow each other in the primary sequence and may be located in very distant regions of the polypeptide chain. There are two important types of secondary structure: the alpha-helix and the beta-sheet. In the alpha-helix, there are 3.6 residues per turn, which means that there is one residue every 100 degrees of rotation of the molecule. In beta-sheets, the protein is arranged into several β-strands, which are stretched segments of the polypeptide chain that are also kept together by hydrogen bonds. Beta-sheets may be parallel (the strands point in the same direction) or anti-parallel (the strands point in opposite directions). The third level of protein structure, known as tertiary structure, refers to the actual three-dimensional shape of the native protein. These folds are caused and maintained primarily by hydrophobic/hydrophilic interactions and hydrogen bonds between the side chains of amino acids and between the amino acids and the protein's environment. Disulfide linkages between cysteine residues also contribute to tertiary structure by stabilizing the three-dimensional shape of the protein. Independent folding units of the protein structure are called domains. Some proteins contain a single domain, while others may contain several domains. The third level of structure is what gives proteins their biological activity and is the highest level of structure shared by all proteins. The fourth level of protein structure, which is not present in all proteins, is quaternary structure. Quaternary structure exists when a single protein consists of two or more polypeptide chains (called subunits). For example, hemoglobin is made up of four identical heme subunits joined together. The number of subunits will be reflected in the name given to its structure (dimer, trimer, pentamer). If the subunits are identical, the prefix "homo" is used, and if they differ, the term "hetero" is used. Thus, a protein with two identical subunits would be a homodimer, while a heterotetramer would be a protein with four subunits, at least some of which are different. These subunits can interact with each other, contribute to an active site or to the dynamics of the complex, or interact with some target molecules. The most common way the MCAT hints at the quaternary structure of a protein is to show that disulfide bond cleavage splits the protein into two or more pieces on a gel.

know the The four levels of protein structure

post-transcriptional modifications. s, this most likely includes polyadenylation, which is needed to export mRNA from the nucleus to the cytoplasm (RN II). Glycosylation is described in the passage, but it is a post-translational modification

polyadenylation and Glycosylation

solvents have a notable effect on the rates of these reactions. Polar protic solvents (such as water and ethanol) tend to stabilize ions in solution. Since they can stabilize the carbocation, these protic solvents are best used for SN1 reactions. However, SN2 reactions rely not on a carbocation, but on a strong nucleophile displacing the leaving group. Protic solvents tend to stabilize (weaken) this nucleophile, so they should not be used for SN2 procedures. Instead, polar aprotic solvents, such as acetone, are a better choice.

solvents for sn1 and sn2 reactions