TPR MCAT CHEM/PHYS Hangups

Saponification

Base mediated hydrolysis of an ester group

Memorize IR spectra

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Study inhibition patterns

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The first daughter nucleus in the 238U decay series is thorium-234. If the next four decays are two β- and two α decays, what is the resulting radionuclide? Question 39 Answer Choices A. Radium-226 Correct Answer B. Polonium-226 C. Radon-228 Your Answer D. Thorium-230

A. Each β- decay increases the number of protons by 1, but leaves the total number of protons and neutrons unchanged. Each α decay decreases the number of protons by 2 and decreases the total number of protons and neutrons by 4. Thus, after two β- and two α decays, thorium-234 becomes radium-226:

Based on the information in the passage, which of the following is true? Question 16 Answer Choices A. Leucine catabolism can lead to synthesis of fatty acids or β-hydroxybutyrate.Correct Answer B. Arginine catabolism can lead to the synthesis of fatty acids or acetone. C. Valine is the only amino acid that is both ketogenic and glucogenic, and does not have an aromatic ring. D. Histidine breakdown can generate acetyl-CoA and α-ketoglutarate, a Krebs cycle intermediate.

A. Figure 1 shows the two ketogenic amino acids, lysine (left) and leucine (right). The passage says ketogenic amino acids are broken down into acetyl-CoA, which can be used in fatty acid biosynthesis or ketogenesis (choice A is correct). Figure 2 shows the four amino acids that are both ketogenic and glucogenic (tryptophan, threonine, tyrosine, isoleucine and phenylalanine, starting at the top left and going around clockwise). Since arginine, valine and histidine are not shown in Figures 1 or 2, they must be solely glucogenic, and cannot be used in the synthesis of acetyl-CoA, and subsequently fatty acids or ketones (choices B, C and D are wrong).

At Pext > 700 atm, which of the following is possible for a sample of one of the real gases given in Figure 1? Question 6 Answer Choices A. Videal is less than the total molecular volume. Correct Answer B. Volume effects are negligible. C. The increased average kinetic energy of the molecules overcomes intermolecular attractions. D. Intermolecular attractions overcome volume effects.

A. In the region Pext > 700 atm in Figure 1, the values of PV/RTfor all the real gases are greater than 1, the ideal. The passage states directly that when PV/RT > 1, volume effects predominate. This eliminates choices B and D. Choice C mentions "the increased average kinetic energy of the molecules." Increased from what? And why and how would it be increased? Choice A is the best choice: At extremely high pressures, the volume of the molecules can become greater than the space between them (Videal).

Trial 2 At 25°C, 50.0 mL of 2.0 M NaOH was combined with 50.0 mL of 2.0 M HCl. The mixture temperature was recorded as 71°C. If Trial 2 were performed with 25.0 mL of NaOH instead of 50.0 mL, the resulting solution would have: Question 26 Answer Choices A. pH < 1 Correct Answer B. T = 71°C C. [H3O+] = [OH-] D. the same ΔH Your Answer

A. Only 0.05 mol of NaOH is added (instead of 0.1 mol). Therefore, only half of the HCl will be neutralized (so [H3O+] > [OH-], eliminate choice C), less heat is liberated (eliminate choice D), and the temperature will not increase as much (eliminate choice B). Although not necessary to answer the question, here's the calculation to show that choice A is indeed correct: 0.05 mol of HCl remains active—in 0.075 L of solution—after the addition of the (reduced amount of) NaOH; since HCl is a strong acid, this means [H+] = (0.05 mol)/(0.075 L) = 2/3 M, which is greater than 0.1 M. As a result, -log[H+] will be less than 1.

1-Methylcyclohexanol reacts with HBr to form 1-bromo-1-methylcyclohexane. The mechanism for this reaction is likely to be an: Question 12 Answer Choices A. SN1 reaction. Correct Answer B. SN2 reaction. C. Nucleophilic addition. D. Addition-elimination.

A. Since no double bonds are formed or broken, this must be a substitution reaction (eliminate choices C and D). Since the hydroxyl in the starting material is on a tertiary carbon atom, the mechanism cannot occur by a bimolecular pathway (eliminate choice B.) The protonated OH group (under acidic conditions) will leave as water to yield a tertiary carbocation, which will be attacked by the bromide ion to give the product.

A sample of gas containing oxygen, nitrogen, argon, and carbon dioxide in equal molar proportions is in a closed container. Which of these gases would escape the fastest if a small hole were punctured in the container? Question 4 Answer Choices A. Nitrogen Correct Answer B. Oxygen C. Argon Your Answer D. Carbon Dioxide

A. Since rate of effusion is inversely proportional to the square root of mass, the gas with the smallest mass will have the highest rate of effusion. Thus, N2 (28 g/mol) will escape fastest.

The standard potential for the reaction K+ + e- K(s) equals -2.93 V, as referenced against 2H+ + 2e- H2(set to 0.0 V by definition). If solid potassium is placed into an aqueous solution of HCl, then: Question 13 Answer Choices A. H2(g) and KCl(aq) are produced. Correct Answer B. Cl2(g) and KCl(aq) are produced. C. Cl2(g), H2(g), and KCl(aq) are produced. D. no reaction occurs.

A. Since the reduction of K+ as given in the question has a potential more negative than the reduction of H+, the oxidation of K(s) to K+ with transfer of electrons to H+ will have a positive potential and be spontaneous. This eliminates choice D. Hydrochloric acid (HCl, a strong acid) will dissociate nearly completely in water into H+ and Cl-. In the presence of a reducing agent (K), H+ ions can accept electrons and be reduced to H2, and K+ and Cl- will remain in solution. Since there is no oxidant which can accept the extra electron from Cl-, Cl2 will not be formed, eliminating choices B and C.

Phase Diagram of water

A. The characteristic that distinguishes the phase diagram for water from the phase diagram for virtually all other substances is that the solid-liquid boundary line in the phase diagram for water has a negative, rather than a positive slope. This is best illustrated by the diagram in choice A.

The intramolecular cyclization in Reaction 2 is accomplished via which mechanistic pathway? Question 41 Answer Choices A. Nucleophilic addition of the amine on the aldehyde, followed by dehydrationCorrect Answer B. Electrophilic addition of the amine on the aldehyde, followed by dehydration C. Nucleophilic substitution of the nitrogen by the oxygen D. Nucleophilic substitution of the oxygen by the nitrogen

A. The difference between the reactant and product is the formation of a C—N bond and loss of the oxygen atom to form a C—C double bond. If oxygen acted as a nucleophile to substitute for nitrogen, the O would still be in the product (eliminate choice C). Likewise, if nitrogen played the role of nucleophile to substitute for the oxygen, no double bond could form between carbons (eliminate choice D). Nitrogen is a nucleophile, or Lewis base (eliminate choice B), and undergoes addition to the electrophilic carbonyl carbon, followed by dehydration of water to form the C—C double bond (choice A is correct).

In covalent bonding between a metal and a non-metal where the non-metal acts as a ligand, the non-metal acts as a: a) lewis acid b) lewis base c) oxidant d) reductant

B) In almost all cases, when a nonmetal atom forms a covalent bond with a metal ion (more appropriately called a coordinate covalent bond), the nonmetal is acting as a ligand and a Lewis base by donating a pair of electrons to form the bond.

Which of the following would be increased the most in an individual with uncontrolled diabetes? Question 56 Answer Choices A. Acetyl-CoA Your Answer B. Ketone bodies Correct Answer C. Lactic acid D. NADH

B. An individual with uncontrolled diabetes would have hyperglycemia and a relative (or complete) deficiency of insulin. This means that glucose levels would be elevated in the blood and unable to get inside of insulin-responsive tissues. Thus, any molecule that is produced mainly through glucose catabolism and cannot be made from alternative pathways (such as lactic acid), would be decreased, not increased (choice C is incorrect). In addition to being produced during cellular respiration, NADH can be made in the Krebs cycle and acetyl-CoA can be made from fatty acid oxidation would enter the Krebs cycle; however the greatest increase would be in the production of ketone bodies (choices A and D are eliminated, choice B is correct). Ketone bodies represent an alternate energy molecule in the absence of glucose.

Which of the following inhibitors would have no effect on the slope of a Lineweaver-Burk plot? Question 28 Answer Choices A. Competitive Inhibitor B. Uncompetitive Inhibitor Correct Answer C. Noncompetitive Inhibitor D. Mixed Inhibitor

B. As the slope of the Lineweaver-Burk plot is Km/Vmax, the correct answer is an inhibitor that has the same effect on both Km and Vmax. An uncompetitive inhibitor leads to an equal decrease in the both the Km and Vmax, leading to a series of parallel lines on a Lineweaver-Burk plot (choice B is correct). Competitive, noncompetitive, and mixed inhibitions would all affect the slope of a Lineweaver-Burk plot (choices A, C, and D are wrong).

A converging glass lens forms a real image of a red object at a distance equal to twice its focal length. If a blue object is placed adjacent to the red object, will its image form closer to the lens or farther from the lens than the image of the red object? Question 27 Answer Choices A. Farther, because blue light is refracted more due to its shorter wavelength B. Closer, because blue light is refracted more due to its shorter wavelength Correct Answer C. Farther, because blue light is refracted less due to its longer wavelength D. Closer, because blue light is refracted less due to its longer wavelength

B. Choices C and D can be eliminated immediately since blue light has a shorter wavelength than red light. The refractive index of a transparent medium increases with increasing frequency of the transmitted light. Thus, the refractive index of the lens for blue light would be slightly higher than for red light. This would imply that blue light experiences more refraction than red light. Since the lens is a converging one (because diverging lenses cannot form real images), the blue light would be bent more sharply toward the axis than the red light, so the image of the blue object would be formed closer to the lens.

Trial 2 At 25°C, 50.0 mL of 2.0 M NaOH was combined with 50.0 mL of 2.0 M HCl. The mixture temperature was recorded as 71°C. At 25°C, the pH of the final solution obtained in Trial 2: Question 24 Answer Choices A. should be lower than 7. B. should be equal to 7. Correct Answer C. should be higher than 7. D. cannot be determined from the information given.

B. Exactly the same amount of acid and base are combined in Trial 2, so only their conjugates remain in solution. Recall the first conjugate rule: the conjugate of a strong acid or strong base is pH neutral. Since HCl and NaOH are both strong, this rule tells us that their conjugates (Cl- and Na+) must be pH neutral. Choice B is the correct answer.

If the front mirror in Figure 1, which is made of glass of refractive index 3/2, were repositioned so that the laser beam strikes at an angle of 30° to its normal, what would be the angle of reflection? Question 17 Answer Choices A. sin-1(1/3) B. 30° Correct Answer C. sin-1(1/31/2) D. 60°

B. If the laser beam strikes the mirror at an angle of 30° relative to the normal, then this is the angle of incidence, and, by the Law of Reflection, it is also the angle of reflection. The refractive index of the mirror is irrelevant.

Would an individual without a parathyroid gland be expected to have difficulty breathing? Question 3 Answer Choices A. Yes; skeletal muscle contraction requires calcium influx through calcium channels on the plasma membrane during each action potential. B. Yes; decreased calcium induces convulsions and tetany. Correct Answer C. No; the individual would probably have enhanced kidney function. D. No; respiration is influenced primarily by blood concentrations of carbon dioxide and oxygen.

B. In the absence of parathyroid hormone, plasma calcium drops and hypocalcemia results. The hypocalcemia in a person with complete absence of PTH would probably be severe, resulting in convulsions and tetany of skeletal muscle as described in the first paragraph of the passage. Respiration requires properly functioning nervous stimulation, and the diaphragm is skeletal muscle. Tetany of the diaphragm would make breathing very difficult (choice B is correct). It is true that contraction of skeletal muscle requires calcium, but this calcium comes from the sarcoplasmic reticulum where it is stored, not from the extracellular environment (choice A is wrong). There is no reason to assume that the person would have enhanced kidney function, nor do the kidneys have anything to do with respiration (choice C is wrong). The rate of respiration is influenced by carbon dioxide and oxygen levels, not the mechanism of respiration (choice D is wrong).

Increased intermolecular attractions cause the ratio PV/(RT) to decline because individual molecules: Question 8 Answer Choices A. eventually combine with other molecules, decreasing the number of particles in the container. B. lose kinetic energy to potential energy and strike the side of the container with less force. Correct Answer C. increase in speed due to electron repulsion and strike the side of the container with increased force. D. transfer electrons during collisions with other molecules in the container.

B. Intermolecular attractions increase the potential energy between molecules, decreasing their kinetic energy. Note that the molecules do not undergo any reactions, eliminating choices A and D. Choice C is incorrect because intermolecular attractions do not result in electron repulsion.

Which of the following methods should be used to purify the mesylate mixture produced from Reaction 1 from any remaining starting material? Question 43 Answer Choices A. NMR spectroscopy B. HPLC Correct Answer C. Thin-layer chromatography D. Size-exclusion chromatography

B. NMR spectroscopy is not a separation technique (eliminate choice A). Although size-exclusion chromatography is a separation technique that will allow for quantitative purification of mixtures, the differences in size of the compounds in the mesylate mixture are not significant enough for the technique to be efficient (eliminate choice D). Thin-layer chromatography (TLC) is also a separation technique, but it uses very small amounts of material to determine polarity of compounds and monitor reactions. It is not a technique that will allow for quantitative purification of a reaction mixture (eliminate choice C). HPLC (choice B) is the best choice as this form of chromatography would easily quantitatively purify a mesylate from its parent alcohol based on the difference in polarities of these functional groups.

A Q-switched laser can be used to treat skin blemishes and to remove tattoos. The Q switch momentarily interrupts the inducing light creating a build-up of energy within the crystal. This does not increase the overall energy of the laser, but concentrates it into shorter time periods or pulses. A longer interruption with the Q-switch most likely would increase the: Question 14 Answer Choices A. total amount of work done by the laser. B. power of each laser pulse. Correct Answer C. wavelength of the laser light. D. frequency of the laser light.

B. Since the overall energy of the laser does not change, neither will the frequency, wavelength, nor work done by the laser. This leaves choice B: Concentrating the energy into a shorter time period increases the power of each pulse (since power equals energy delivered per unit time, by definition).

Which of the following is the correct rate law expression for the fast step in Reaction 2? Question 37 Answer Choices A. Rate = k [HSO3-]2 B. Rate = k [HSO3-]2 [O2] Correct Answer C. Rate = k [2HSO3-] [O2] D. The rate law cannot be determined from the information given. Your Answer

B. Since this is an elementary reaction (all reactions in a mechanism are elementary), we can determine the rate law using the stoichiometric coefficients on the reactants (eliminate choice D). To write the rate law from an elementary reaction, we simply raise each reactant concentration to the exponent equal to the coefficient of that species in the elementary reaction. Choice A is wrong since it does not include the dissolved O2, and choice C is wrong because the concentration of HSO3− should be squared, not doubled.

Stearic acid (or octadecanoic acid) is a saturated fatty acid with the formula CH3(CH2)16COOH After undergoing β-oxidation, it will generate: Question 58 Answer Choices A. 9 acetyl-CoA, 9 NADH and 9 FADH2. B. 9 acetyl-CoA, 8 NADH and 8 FADH2. Correct Answer C. 8 acetyl-CoA, 8 NADH and 8 FADH2. D. 8 acetyl-CoA, 7 NADH and 7 FADH2.

B. Stearic acid has 18 carbons and so will undergo 8 rounds of β-oxidation. This will generate 9 molecules of acetyl-CoA (eliminate choices C and D), 8 molecules of NADH and 8 molecules of FADH2(choice B is correct, eliminate choice A

AgI(s) --> Ag+(aq) + I-(aq) Reaction 1 Why does the solubility of AgI decrease in an aqueous solution when NaI is added? Question 20 Answer Choices A. A lower [Ag+] shifts Reaction 1 to the right. B. A higher [I-] shifts Reaction 1 to the left. Correct Answer C. A higher [Na+] shifts Reaction 1 to the right. Your Answer D. A higher [Ag+] shifts Reaction 1 to the left.

B. The equilibrium described by Reaction 1 will be shifted to the left if more I- ions are present in solution (which would occur if NaI were added), thus precipitating more AgI(s) and decreasing its solubility compared to that in pure water. This phenomenon is known as the common ion effect, and is a consequence of Le Châtelier's Principle. While both choices A and D are true statements, only a shift to the left indicates a decrease in solubility of AgI (eliminate choice A); the addition of NaI also has no effect on the concentration of Ag+ ions in order to cause a shift (eliminate choice D). Since Na+ is not a part of Reaction 1 it cannot affect the equilibrium position of the reaction (eliminate choice C)

A possible contributor to Alzheimer's disease is the clumping of amyloid proteins formed from the degradation of larger proteins in the fatty tissues surrounding nerve cells. This degradation could be a result of: Question 44 Answer Choices A. formation of more peptide linkages via the condensation of amino and carboxylic acid functional groups attached at the ends of the larger protein unit. B. hydrolysis of some of the amide peptide linkages, creating smaller polypeptide units.Correct Answer C. the destabilizing effects of 2° structures in the proteins. D. oxidative cleavage of the disulfide linkages between cysteine units of the proteins.

B. The question states that the degradation of a larger protein is responsible for generating smaller amyloid protein segments. Proteins are polypeptides with an amide linkage holding adjacent amino acids together. These peptide bonds will break when hydrolyzed and form smaller chains (choice B is correct). Condensation reactions result from the combination of two molecules with the removal of water (the opposite of hydrolysis) and will only lead to larger molecules not smaller ones (eliminate choice A). The 2° structures of proteins are α-helices and β-sheets. These structures are energetically favorable, thus stabilizing, as they are held together by hydrogen bonds (eliminate choice C). Cleavage of the disulfide bond in the 4° structure of a protein can lead to smaller subunits. However, that process is a reduction, not an oxidation (eliminate choice D).

Can glucogenic amino acids be converted into glucose? Question 15 Answer Choices A. Yes: pyruvate and oxaloacetate can be converted directly into glyceraldehyde-3-P, which is a major intermediate in both gluconeogenesis and glycolysis. Your Answer B. Yes: pyruvate and Krebs cycle intermediates can be converted into oxaloacetate, then phosphoenolpyruvate, which can enter gluconeogenesis. Correct Answer C. No: pyruvate and Krebs cycle intermediates are formed as part of glucose breakdown and this process is important to generate ATP for the cell. D. No: glucose is obtained from the diet and stored in the liver; it cannot be made as a new molecule because cellular respiration has several steps with a -ΔG.

B. This is a typical two by two question. The passage says that glucogenic amino acids are broken down into citric acid cycle intermediates or pyruvate. The Krebs cycle regenerates oxaloacetate (OAA), and in the first step of gluconeogenesis, pyruvate is also converted into OAA (by the enzyme pyruvate carboxylase). OAA is converted into phosphoenolpyruvate (PEP) by the enzyme phosphoenolpyruvate carboxykinase (or PEPCK). Gluconeogenesis can then continue to run, and will generate glucose from these non-carbohydrate precursor molecules (choice B is correct). Although glyceraldehyde-3-P is a major intermediate in both gluconeogenesis and glycolysis, pyruvate and oxaloacetate cannot be converted directly into this molecule (choice A is incorrect). Both choices C and D are incorrect because they start with "No"; as discussed above, glucogenic amino acids can be converted into glucose (choices C and D are wrong).

In a facultative anaerobe, which of the following processes occurs under both aerobic and anaerobic conditions? Question 11 Answer Choices A. Fermentation B. Krebs cycle C. Glycolysis Correct Answer D. Oxidative phosphorylation

C. A facultative anaerobe can survive through fermentation when oxygen is not available but will use oxidative respiration when oxygen is available. Glycolysis will occur under both aerobic conditions (in which case the pyruvate will go on to enter the Krebs cycle as acetyl-CoA) and under anaerobic conditions (fermentation reduces pyruvate to alcohol or lactate), so choice C is correct. Fermentation occurs only in anaerobic conditions (choice A is wrong), while the Krebs cycle and oxidative phosphorylation can occur only in aerobic conditions (choices B and D are wrong).

An object is floating in a fluid of 1.5 specific gravity. If the volume of the fluid displaced by the floating object is 5 × 10-3 m3, what is the object's mass? Question 14 Answer Choices A. 2.5 kg B. 5.0 kg C. 7.5 kg Correct Answer D. Cannot be determined from the information given

C. Because the object is floating, the object's weight is balanced by the buoyant force; that is, mg= ρfluidVsubg, or, after canceling the g's, m = ρfluidVsub. With ρfluid = 1.5ρH2O = 1500 kg/m3 and Vsub = 5 × 10-3 m3, we find that m = ρfluidVsub = (1500 kg/m3)(5 × 10-3 m3) = 7.5 kg

Eukaryotic cells use reciprocal regulation to make sure β-oxidation and fatty acid biosynthesis do not occur at the same time. This is accomplished by malonyl CoA, which inhibits the carnitine shuttle required to transfer activated fatty acids from the cytoplasm to the mitochondrial matrix. Each of the following is a true statement EXCEPT: Question 34 Answer Choices A. both β-oxidation and fatty acid biosynthesis occur in four steps; fatty acid biosynthesis involves elongation, two redox reactions and a dehydration. B. β-oxidation involves the reduction of both FAD and NAD+; fatty acid biosynthesis oxidizes two NADPH (generated by the pentose phosphate pathway) to two NADP+. C. because of this reciprocal regulation, both β-oxidation and fatty acid biosynthesis occur in the mitochondrial matrix. Correct Answer D. β-oxidation generates acetyl CoA, while fatty acid biosynthesis uses malonyl CoA, which is made from acetyl CoA by acetyl CoA carboxylase.

C. Choices A, B and D are true statements. Choice C is false (and the correct answer); β-oxidation occurs in the mitochondrial matrix and fatty acid biosynthesis occurs in the cytoplasm.

Which of the following statements best supports why two equivalents of hydroxide are used in the basic hydrolysis of dichorine monoxide in Reaction 1? Question 35 Answer Choices A. The second hydroxide is necessary to neutralize the HCl produced in situ during the course of the reaction. B. Adding a second equivalent of hydroxide will double the rate of reaction. C. Adding only one equivalent will lead to the formation of a buffer solution of hypochlorous acid and its conjugate base, and therefore an incomplete reaction. Correct Answer D. Chlorine in the +1 oxidation state is highly reactive, and only stable in highly basic environments.

C. In Reaction 1, chlorine is in the +1 oxidation state throughout, and there is no redox reaction available to take it to the -1 state it would require to form HCl (eliminate choice A). Choice B is a true statement only if the hydroxide ion is part of the overall rate law for the reaction and is first order in hydroxide. Regardless, this statement doesn't address the exact stoichiometric amount in question (eliminate choice B). Choice D is false because Cl+ should find no problem with an acidic environment where the most active species is H+. One equivalent of hydroxide will hydrolyze the dichlorine monoxide as follows: Cl2O + OH- HOCl + OCl-. Since this reaction forms a weak acid and its conjugate base in equal amounts, a buffer solution is produced. When the second equivalent of OH- is added, it will undergo another acid/base reaction to deprotonate the hypochlorous acid to the hypochlorite ion, giving the products in Reaction 1 of the passage.

Brown fat will increase energy expenditure in the form of heat better than any other adipose tissue in the body. In brown fat, leptin indirectly activates uncoupling protein 1 (UCP-1) located in the inner mitochondrial membrane. UCP-1 activation will dissipate the proton gradient across the inner mitochondrial membrane and STOP which of the following reactions in mitochondria? Question 8 Answer Choices A. Electron transport chain Your Answer B. Krebs cycle C. Oxidative phosphorylation Correct Answer D. NAD+ reduction

C. In the absence of the proton gradient across the inner mitochondrial membrane no electrochemical gradient has been built and oxidative phosphorylation, i.e. the formation of ATP by ATP synthase, will not proceed (C is correct). The electron transport chain will continue and pass electrons along the transport chain, but the pumped protons will return to the matrix of the mitochondria with the help of UCP-1 activation (choice A will continue and can be eliminated). The Krebs cycle provides NADH and FADH2 as substrates to the electron transport chain and receives NAD+ and FAD for continuation of its chemical cycle. This action can proceed unhindered during UCP-1 activation and choice B can be eliminated. Reduction of NAD+ in the mitochondria takes place in the Krebs cycle and the pyruvate dehydrogenase complex reaction and occurs prior to the action of events induced by UCP-1 (choice D will continue and can be eliminated).

Which of the following accurately lists the composition of the endoplasmic reticulum, phosphofructokinase and alanine, respectively? Question 46 Answer Choices A. Double membrane surrounding a lumen, amino acids linked together by covalent proteolytic bonds, four different groups on a chiral carbon. B. Single membrane surrounding a lumen, amino acids linked together by hydrogen bonds, four different groups on a chiral carbon. C. Single membrane surrounding a lumen, amino acids linked together by covalent peptide bonds, four different groups on a chiral carbon. Correct Answer D. Single membrane surrounding a lumen, amino acids linked together by covalent peptide bonds, three different groups on a chiral carbon.

C. The endoplasmic reticulum is an organelle with a single membrane surrounding a lumen (eliminate choice A). Phosphofructokinase is an enzyme of glycolysis, and so is a protein made of amino acids linked together by covalent peptide bonds (choice B is wrong). Alanine is a chiral amino acid, and so has four different groups on a chiral carbon (another reason to eliminate choice D). Choice C provides an accurate description and is correct.

The passage describes the simplest interpretation of the basic properties of a laser, but some modifications are necessary to build a workable laser. Which of the following, if true, are possible problems with the simple model described in the passage? It is difficult to keep a collection of atoms in their excited state long enough to be stimulated to emit the induced photon. Atomic energy levels are quantized. Atoms in the crystal that happen to be in their ground state will undergo absorption, thereby removing photons from the beam as it builds up. Question 15 Answer Choices A. I only B. III only C. I and III only Correct Answer D. II and III only

C. The laser acquires its intense energy by induced emission, so if Statement I were true (and it is), then too many atoms could emit energy by spontaneous emission, thereby decreasing the efficiency of the laser. This eliminates Choices B and D. If Statement III were true (and it is), this would also represent a mechanism by which the laser's intensity would be reduced (eliminating choice A). Statement II, although true, does not represent a problem; indeed it is a basic quantum mechanical principle that permits lasers to exist, because only certain photon energies are allowed to be absorbed or emitted by atoms and molecules.

Barth syndrome is a rare X-linked recessive genetic disorder and is caused by a mutation in the gene coding for tafazzin, an enzyme involved in remodeling of cardiolipin acyl chains. Which of the following is true? Question 32 Answer Choices A. Patients with Barth syndrome are unable to biosynthesize premature cardiolipin. B. Two normal parents could have both a son and a daughter with Barth syndrome. C. Patients with Barth syndrome commonly suffer from cardiomyopathy and general weakness.Correct Answer D. If a carrier female mates with a normal man, ¾ of their children will be either affected or a carrier.

C. The passage says that immature cardiolipin is generated and then remodeled to form mature cardiolipin. If tafazzin functions in this reaction, patients with Barth syndrome will be unable to make mature cardiolipin but will have no problem generating premature cardiolipin (choice A is wrong). If Barth syndrome is an X-linked recessive disorder, the disease causing allele will be recessive; let's call it t. The normal allele will be dominant (T). Two normal parents could be XTXt and XTY, or XTXT and XTY. Since the father has the dominant normal allele, there is no possibility of these people having a daughter with Barth syndrome, no matter the genotype of the mother (choice B is wrong). The passage says cardiolipin is found in heart and liver tissue. It is therefore possible that patients with Barth syndrome (and thus defective cardiolipin synthesis) could suffer from cardiomyopathy and general weakness (choice C is correct). If a carrier female (XTXt) mates with a normal man (XTY), their offspring could be XTXT (normal female, 25% probability), XTXt (carrier female, 25% probability), XTY (normal male, 25% probability) or XtY (male with Barth syndrome, 25% probability). ½ of the offspring will be normal and ½ will be either affected or a carrier (choice D is wrong).

At physiological pH, the amino acid with the formula C4H7NO4 has: Question 17 Answer Choices A. two deprotonated carboxylic acid groups and is ketogenic. B. one deprotonated carboxylic acid group and is ketogenic. C. two deprotonated carboxylic acid groups and is glucogenic. Correct Answer D. one deprotonated carboxylic acid group and is glucogenic.

C. This is a typical two-by-two question. Since none of the pictured amino acids in Figures 1 or 2 have four oxygen atoms, the amino acid with the formula C4H7NO4 must be glucogenic (choices A and B can be eliminated). Only aspartic acid and glutamic acid have four oxygen atoms, two of them found in the carboxylic acid end of the backbone, and two of them found in the carboxylic acid end of the side chain, so this question must be asking about one of these amino acids. In actual fact, aspartic acid has the formula C4H7NO4, but you don't need to figure this out to answer the question. At physiological pH (about 7.4), both the carboxylic acid end of the amino acid and the side chain COOH will be deprotonated. This is because physiological pH is above the pKa values for these acidic protons (carboxyl pKa is approximately 2, and the R-group pKa is approximately 4). Thus, aspartic acid will actually be in its aspartate form, and will have two deprotonated carboxylic acid groups at physiological pH (choice C is correct and choice D is wrong).

What is the correct relationship between the first ionization energies of the elements mercury, thallium, and lead? Question 57 Answer Choices A. Hg > Tl > Pb B. Tl > Pb > Hg C. Pb > Tl > Hg Your Answer D. Hg > Pb > Tl Correct Answer

D. According to periodic trends, as you move from left to right within a row, ionization energy should increase due to increased effective nuclear charge. However, these three elements (Hg, Tl, and Pb—atomic numbers 80, 81, and 82, respectively) provide an example of an exception to that rule. The electron configurations of these atoms are as follows: Hg: [Xe] 4f14 5d10 6s2Tl: [Xe] 4f14 5d10 6s2 6p1Pb: [Xe] 4f14 5d10 6s2 6p2 In Hg, all orbitals are full, which is relatively stable. It is the least likely atom to give up any electrons, and therefore it has the highest ionization energy (eliminate choices B and C). Tl is the opposite. If Tl gives up a single electron, it would result in the electron configuration of Hg, which is more stable. Therefore, Tl has the lowest ionization energy of the three, making the answer choice D.

The Ka for the first ionization of sulfurous acid (1.7 × 10-2) is significantly larger than the Ka for the second ionization (6.4 × 10-8). A likely explanation for this is that: Question 48 Answer Choices A. the electronegativity of the remaining hydrogen atom increases after the first hydrogen ion has been removed. B. the second ionization can only take place if the first ionization proceeds to completion. C. neutral hydrogen is difficult to ionize in aqueous solution. D. the remaining hydrogen atom experiences greater electrostatic attraction after the loss of the first hydrogen ion. Correct Answer

D. After loss of the first proton, the remaining hydrogen is bound to a negatively-charged molecule. Electrostatic attraction between this remaining hydrogen and the negatively-charged molecule would disfavor loss of the second proton, resulting in a smaller K. Therefore, K2 << K1.

When carbon dioxide in a closed container is subjected to external pressures less than 650 atm, the deviation from ideality is primarily due to the fact that: Question 7 Answer Choices A. calculated gas pressure is less than actual gas pressure. B. calculated volume is less than actual volume. C. the number of collisions between gas particles decreases with increasing pressure. D. actual gas pressure is less than calculated gas pressure. Correct Answer

D. At external pressures less than 650 atm, Figure 1 shows that the PV/(RT) ratio for CO2 is less than 1. Therefore, the actual gas pressure is less than the calculated gas pressure, which corresponds to a ratio of 1 (eliminate choices A and B). The number of collisions is directly proportional to pressure. Additionally, only the collisions of the molecules with the wall of the container can be measured making choice C false (eliminate choice C).

Ghrelin, the "hunger hormone," is released when the stomach is empty. It increases hunger and promotes secretion and motility in the digestive tract.The receptor for ghrelin is found on the same cells in the brain as the receptor for leptin. Which of the following is a true statement? Question 7 Answer Choices A. The interaction of leptin and ghrelin can be considered competitive inhibition because they produce antagonistic effects.Your Answer B. The interaction of leptin and ghrelin is considered to be allosteric inhibition. C. The interaction of leptin and ghrelin is considered uncompetitive because ghrelin only binds when leptin is already bound. D. There is no interaction between leptin and ghrelin because they operate on separate receptors.

D. Even though they bind to the same cells in the brain, the fact that leptin and ghrelin bind to separate receptors means that there cannot be any kind of inhibition or interaction between them (choice D is correct). For competitive inhibition to occur, two things must be trying to bind to one site; for example, if both leptin and ghrelin bound to the same site on the same receptor. The fact that they have separate receptors prevents this (choice A is wrong). Likewise, allosteric inhibition requires that the inhibitor bind to a separate site on the same enzyme or receptor (choice B is wrong), and there is nothing in the question stem or passage to suggest that ghrelin can only bind when leptin is bound (choice C is wrong). Question states they act on the same CELL, not receptor.

Based on the information in the passage, D-2HG is most likely a(n): Question 54 Answer Choices A. tumor suppressor because it decreases the activity of histone demethylase. B. tumor suppressor because it may affect cell mobility at low concentrations. C. oncogene due to its ability to stimulate cell growth at moderate intracellular concentrations.Your Answer D. oncometabolite because small concentrations of this molecule enhanced cell growth.Correct Answer

D. Figure 3 shows that when cells were treated with D-2HG in DMSO for 24 hours and allowed to grow for three weeks, more colonies were present than when cells were treated with DMSO only. This means D-2HG is likely promoting cell growth, and is thus oncogenic instead of tumor suppressive (choices A and B are wrong). Also note that the data in Figure 2 is not useful in answering this question (another reason to eliminate choice A). D-2HG is a small molecule and metabolite, not a gene or protein (choice D is correct; choice C is wrong). Also note that the passage does not give information on the mechanism of action of D-2HG in the soft agar colony formation assay; you cannot be sure this small molecule is acting intracellularly. It could be binding a receptor at the plasma membrane (another reason to eliminate choice C).

Which of the following can be inferred from information in the passage? Cardiomyocytes and hepatocytes are mitochondrial rich cells. Slow twitch fibers have higher cardiolipin content than fast twitch fibers. Prokaryotes pump H+ across their plasma membranes, while eukaryotes do not Question 33 Answer Choices A. III only B. I and II only Your Answer C. II and III only D. I, II, and III Correct Answer

D. Item I is true: the passage states that cardiolipin constitutes 20% of the total lipid composition of the inner membranes of some eukaryotic organelles, and the table suggests mitochondria are affected in cardiolipin synthesis mutants. This means cardiolipin is likely a component of the inner mitochondrial membrane. The passage also says cardiolipin can be isolated from heart and liver tissue. This means these two tissues must be rich in mitochondria (choices A and C can be eliminated). Note that both remaining choices include Item II, so it must be true: slow twitch fibers have many mitochondria and therefore will have more cardiolipin. Item III is true: based on the endosymbiotic theory, eukaryotic cells acquired mitochondria via endocytosis of a prokaryotic cell. The bacterial plasma membrane contained the electron transport chain machinery and became the inner mitochondrial membrane. The endosome membrane became the outer mitochondrial membrane. As a result, while prokaryotes pump H+ across their plasma membrane, eukaryotes pump protons across the inner mitochondrial membrane. (choice B can be eliminated and choice D is correct).

The serotonin transporter (or SERT) removes serotonin from the synaptic cleft and recycles it back into the presynaptic cell. It thus terminates the effects of serotonin and this mechanism has been targeted in treatments for alcoholism, clinical depression, obsessive-compulsive disorder, and hypertension. SERT spans the plasma membrane 12 times and is also a: Question 27 Answer Choices A. glycosylated phospholipid, with both hydrophobic and hydrophilic regions. B. peptide chain with at least four levels of protein structure, held together by disulphide and peptide bonds. C. protein with twelve hydrophobic domains, none of which contain the amino acid proline and at least some of which contain -sheets stabilized by covalent bonds. D. protein with twelve transmembrane domains, each of which is an -helix with no proline and external hydrophobic residues, stabilized by hydrogen bonds. Correct Answer

D. Membrane transport is mediated by proteins, not phospholipids (eliminate choice A). All proteins have at least three levels of protein structure, but only some have quaternary structure. There is no information in the question stem to support the fact that SERT contains more than one peptide chain (eliminate choice B). Transmembrane domains are -helices with external hydrophobic residues. They cannot contain proline because of its secondary amine structure. Both -helices and -sheets are stabilized by hydrogen bonds, not covalent bonds (eliminate choice C, choice D is correct).

A researcher obtains a sample of recombinant mutant isocitrate dehydrogenase and wants to confirm its activity in vitro. To do this, she could: Question 56 Answer Choices A. Add isocitrate as the substrate and analyze reaction completion via size exclusion chromatography. B. Add isocitrate as the substrate and analyze reaction completion via 1H NMR. C. Add a-ketoglutarate as the substrate and analyze reaction completion via size exclusion chromatography. D. Add a-ketoglutarate as the substrate and analyze reaction completion via 1H NMR.Correct Answer

D. Notice this is a two-by-two question setup. Based on information in the passage, mutant isocitrate dehydrogenase converts α-ketoglutarate into D-2-hydroxyglutarate, which is shown in reaction 2 of Figure 1. Wildtype (not mutant) isocitrate dehydrogenase uses isocitrate as a substrate (eliminate choices A and B). α-ketoglutarate and D-2-hydroxyglutarate have different different 1H NMR signals (because they contain different hydrogen atoms) which could be used to distinguish between α -ketoglutarate and D-2-hydroxyglutarate, and thus would give information on whether the mutant isocitrate dehydrogenase sample obtained was functional. Size exclusion chromatography is used to separate large polymers from small oligomeric fragments, and would not be useful here (eliminate choice C, option D is correct).

A patient with elevated parietal cell secretions but restricted mucus secretions will most likely exhibit: Question 18 Answer Choices A. normal protein catabolism, a gastric ulcer, and a gastric pH that is lower than normal. B. reduced protein catabolism, a gastric ulcer and a gastric pH that is lower than normal. C. increased protein catabolism and a gastric ulcer, but normal gastric pH. D. increased protein catabolism, a gastric ulcer and a gastric pH that is lower than normal. Correct Answer

D. Parietal cells are found in the stomach and are responsible for producing HCl, which causes the stomach contents to have a low pH. HCl also non-specifically digests proteins by cleaving peptide bonds, and activates pepsinogen. Elevated parietal cell secretions would mean more acid, and a lower gastric pH (choice C is wrong). More acid would also lead to slightly more protein break down (choices A and B are wrong; choice D is correct). The mucus-secreting cells in the stomach protects gastric epithelial cells from the low pH in the stomach. Reducing mucus secretions can increase the probability of a gastric ulcer, where there is a break in the lining of the stomach.

Which of the following statements regarding RNA molecules is NOT true? Question 12 Answer Choices A. RNAs can act as enzymes to catalyze reactions. B. Some RNAs have more than four different types of bases. C. Some RNAs are synthesized in the nucleolus. D. RNAs are insusceptible to alkaline hydrolysis. Correct Answer

D. RNA molecules have decreased stability compared to DNA in part because of their susceptibility to alkaline hydrolysis due to the presence of hydroxyl group at 2'-C position (choice D is not true of RNA and is the correct answer choice). Some RNAs have enzymatic function (such as in telomerase) and they are termed ribozymes (choice A is true and can be eliminated). tRNA has unique and modified bases apart from the traditional four bases A,U,C, and G (such as inosine, choice B is true and can be eliminated). rRNA is synthesized in the nucleolus (choice C is true and can be eliminated).

Tautomers

Easy way to think about it: Resonance forms that shift around atoms, not only electrons.

If the spring in the gun is compressed a distance x from rest before the gun is fired, which of the following expressions will give the value of the gun's spring constant in terms of the measured and calculated values of the experiment? Question 5 Answer Choices A. B. C. D. Correct Answer

D. The elastic potential energy of the compressed spring, (1/2)kx2, is converted to the kinetic energy, (1/2)mv2, of the bullet. Setting (1/2)kx2 equal to (1/2)mv2, we find that k = mv2/x2.

Which of the following statements concerning Reaction 1 is FALSE? Question 36 Answer Choices A. The chlorine atom in the product has a +1 oxidation state. B. The dichlorine oxide starting material has a bent molecular geometry. C. Addition of acid to the reaction mixture would decrease the potency of the bleach solution formed. Your Answer D. Increasing the partial pressure of Cl2O(g) would increase the equilibrium constant, Keq, of the reaction. Correct Answer

D. The equilibrium constant, K, depends on the temperature only. That is, the ratio of products to reactants (with the appropriate stoichiometric coefficients as exponents on the concentrations) at equilibrium will remain the same at a given temperature. Therefore, choice D is false. Choice A is true since the rules for assigning oxidation states gives the oxygen atom a -2 oxidation state first, leaving chlorine with a +1 oxidation state in the ion OCl-. Choice B is true, since by VSEPR theory we'd predict the two lone pairs on the oxygen create a roughly 109° bond angle between the Cl—O bonds. Choice C is true, since Le Châtelier's Principle tells us that this reaction will go backward if we lower [OH−]; if the reverse reaction is favored, the active bleaching ingredient will be depleted.

How much heat was evolved when 4.0 g of NaOH was added to 100 mL of water? Question 23 Answer Choices A. 10/13 × 4.5 J B. 13/10 × 4.5 J C. 10/13 × 4.5 kJ D. 13/10 × 4.5 kJ Correct Answer

D. The passage tells us (in the Calibration section) that Q is 4.5 kJ (since we used 100 mL), and ΔTcalibration is 35°C - 25°C = 10°C. Therefore, the heat capacity is CP = Q/ΔTcalibration = (4.5 kJ)/(10°C), eliminating choices A and B. The value of ΔT in Trial 1 was 38°C - 25°C = 13°C, so the heat evolved was Q = CPΔT = (4.5 kJ)/(10°C) × 13°C = (13/10) × 4.5 kJ (eliminate choice C). Thermodynamics: Calorimetry

Electric force

F = qE

A CO2 laser used as a laser scalpel produces a beam of laser light with a wavelength of 10.6 µm. Compared to the excited electrons in the laser shown in Figure 1 (wavelength = 1.06um), the excited electrons in the CO2 laser most likely have: Question 16 Answer Choices A. greater mass. B. less mass. C. a greater energy difference between their normal state and their excited state. Your Answer D. a smaller energy difference between their normal state and their excited state. Correct Answer

D. The wavelength of the CO2 laser is 10 times greater than the wavelength of the laser shown in Figure 1. Therefore, the frequency and the energy of the CO2 laser are 10 times lower. Since the laser light is the energy released by transitions of electrons dropping to a lower energy level, less emitted energy implies a smaller difference between the energy levels. The mass of an electron is independent of its atomic or molecular energy state, eliminating choices A and B (the mass of the atom or molecule as a system changes with electron excitation according to relativity theory, but not the mass of the electron itself, and anyway that is well beyond the scope of the MCAT).

Trial 2 At 25°C, 50.0 mL of 2.0 M NaOH was combined with 50.0 mL of 2.0 M HCl. The mixture temperature was recorded as 71°C. If a neutralization similar to Trial 2 were performed with the same molar values of NaOH and acetic acid, which of the following results is the most likely to occur? Question 25 Answer Choices A. A higher temperature due to a larger |ΔH| B. A lower temperature due to a larger |ΔH| C. A higher temperature due to a smaller |ΔH| D. A lower temperature due to a smaller |ΔH| Correct Answer

D. This is a 2 × 2 style question, and in this case two choices can be eliminated for being internally inconsistent. If the ΔHis larger for an exothermic reaction, more energy will be given off, resulting in a larger increase in temperature (eliminate choice B). Similarly, if less energy is given off, the temperature change should be smaller (eliminate choice C). The heat of neutralization is defined as the enthalpy change resulting from 1 mole of water being produced from the reaction H+(aq) + OH-(aq) → H2O(l). Since weak acids (like acetic acid) are only partially ionized, their neutralization by a strong base will not result in an equal production of water molecules compared with a strong acid/strong base neutralization (eliminate choice A).

Although zinc metal is attacked by hydrochloric acid according to the reaction Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g) copper metal is unaffected by HCl. Why? Question 6 Answer Choices A. Cl- is a better oxidizing agent than zinc, but is not a better oxidizing agent than copper. B. Zinc is a better oxidant than copper. Your Answer C. Zinc is a better oxidant than H+. D. Cu+ is a better oxidant than H+. Correct Answer

D. This is a classic redox tug-of-war question. The first problem is to decide who's pulling on whose electrons. You should realize that HCl(aq) and ZnCl2(aq) are really H+, Cl-, and Zn2+ in solution. With this in mind, it is easier to figure out that chloride is a spectator ion; it doesn't affect anything (eliminating choice A). Next, it is apparent that the H+ ion is pulling away (oxidizing) the zinc's electrons, but it cannot take away the copper's electrons. This means that H+ is a better oxidizing agent than Zn+ or Zn2+ but is not as strong as the Cu+ ion. Therefore, choices B and C are incorrect and choice D is the best answer.

If a fully saturated solution of AgI, with precipitate present, were treated with NaCl instead of NaI, which of the following observations is likely? Question 21 Answer Choices A. As NaCl is added, all precipitates are dissolved into the aqueous solution. B. The decrease in [AgI] is even more drastic than with the addition of NaI in Figure 1. C. There is no change in the amount of undissolved AgI. D. The concentration of [I-] increases. Correct Answer

D. Unlike NaI, NaCl does not have a common ion with AgI and will therefore NOT cause a decrease in the solubility for AgI with increasing concentration (eliminate choice B). The following will act as a competing reaction when [Cl-] concentrations become sufficiently large: Ag+ (aq) + Cl- (aq) → AgCl (s) With this in mind, there will be no situations wherein the solution is free of precipitate (eliminate choice A). As the dissolved [NaCl] concentration increases, AgCl will be precipitated from solution, which will enable additional AgI to dissolve (eliminate choice C). The increased dissolution of AgI will cause the increase in [I-], even as [Ag+] levels remain low.

What is oxidative phosphorylation?

The formation of ATP by ATP synthase using the proton gradient established by the ETC

(T/F) All proteins destined for the plasma membrane pass through the secretory pathway meaning postrpanlastional modification occurs in the ER before transportation to the plasma membrane.

True

Electric field voltage

V = Ed (d = distance between plates)

Study lenses and refraction and optics and stuff

do it