MCAT Chem/Phys FSQ

In Reaction 1, what is a possible structure for either R1 or R2 of the reactant? CH3 NH2 (CH2)15CH3 (CH2O)10CH3

(CH2)15CH3 A. Methyl is not sufficiently long to be part of a phosphatadylethanolamine. B. If R were NH2 the group would not be a fatty acid acyl chain. C. Because PE is a lipid, the R groups must represent the acyl chains. D. The acyl chains of PE molecules do not contain oxygen atoms.

Smoke detectors fall into two major classes. Ionization detectors, the most common units, contain two parallel electrodes that are typically separated by 3 cm with a 5-V potential difference across them. What is the magnitude of the electric field between the two electrodes in ionization type detectors? 1.5 N/C 1.66 N/C 15 N/C 166 N/C

166 N/C electric field between plates is E = V/d 5/(0.03) = 166.7 V/m = 166.7 N/C

Suppose a defibrillator successfully returns a baby's heart to normal beating. Suppose further that 20 g of blood enters the heart at 25 cm/s and leaves 0.10 s later at 35 cm/s. What is the estimated average force on the 20 g of blood as it moves through the baby's heart? A. 0.020 N B. 0.20 N C. 20 N D. 2000 N

A. 0.020 N According to Newton's second law, the average force is equal to the mass of blood multiplied by the average acceleration of the blood. The average acceleration is (35 cm/s - 25 cm/s)/0.10 s = 100 cm/s2 = 1 m/s2. The average force is 20 g × 1 m/s2 = 0.020 kg × 1 m/s2 = 0.020 N.

Two vectors of magnitudes |A| = 8 units and |B| = 5 units make an angle that can vary from 0° to 180°. The magnitude of the resultant vector A + B CANNOT have the value of: A. 2 units. B. 5 units. C. 8 units. D. 12 units.

A. 2 units. The magnitude of A + B is as small as 3 units (when A and B are anti-parallel and make an angle of 180°) and as large as 13 units (when A and B are parallel and make an angle of 0°). The magnitude of 2 units is smaller than the smallest possible magnitude of vector A + B.

Glycerol 3-phosphate phosphatases (G3PPs) are enzymes that catalyze the hydrolysis of the phosphoester bond in glycerol 3‑phosphate. The active site of G3PP contains Asp14 and Asp16, which act as a nucleophile and general acid, respectively, during catalysis. Based on the information in the passage, which description of an enzyme-substrate covalent intermediate is most likely correct? The substrate is covalently attached to: A. Asp14 through the phosphorus atom of the phosphate group. B. Asp14 through the oxygen atom of the phosphate group. C. Asp16 through the phosphorus atom of the phosphate group. D. Asp16 through the oxygen atom of the phosphate group.

A. Asp14 through the phosphorus atom of the phosphate group. The covalent intermediate will occur through the nucleophilic substitution by the side chain carboxyl of Asp14 at the electrophilic phosphorus atom in the substrate, displacing a leaving group. the phosphorous on G3P wants electrons (to fill shell), so it will take electrons from a nucleophile (donates electrons).

Which single bond present in nitroglycerin is most likely the shortest? A. C-H B. C-O C. C-C D. O-N

A. C-H Since hydrogen has a much smaller atomic radius than second period elements, the covalent bond between C and H is shorter than any of the other bonds listed.

The blood glucose meter uses test strips loaded with an enzyme and additives, and is calibrated to take a fixed volume of solution (usually blood). It measures the amount of Reaction 2, a two-electron oxidation, that takes place. What additional substance is necessary for Reaction 2 to take place? A. FAD B. NADH C. H2O D. Acetyl-CoA

A. FAD an oxidizing agent is required for Reaction 2 to proceed. Of the choices listed, only FAD is a cofactor oxidant.

Which of the following properties of a 2.3 MHz ultrasound wave remains unchanged as it passes into human tissues? A. Frequency B. Wave speed C. Amplitude D. Wavelength

A. Frequency When a wave passes into another medium, its speed changes, but its frequency does not

Researchers studied the kinetics of SidA-catalyzed Reaction 1. They discovered that the reaction proceeds in two steps. In the first step, NADPH reacts with FAD to form NADP+ and FADH-. FADH- quickly reacts with O2 and H+ to form FADH-OOH. This step is pH-independent and occurs at the same rate regardless of which, if any, substance occupies the active site. At this point, FADH-OOH either hydroxylates Compound 1 or decomposes nonproductively to form H2O2 if any other substance (such as L-lysine)occupies the active site. What is the balanced equation for the nonproductive reaction when lysine is the substrate? A. H+ + NADPH + O2 → NADP+ + H2O2 B. 2H+ + O2 → H2O2 C. FAD + NADPH + H+ → FADH2 + NADP+ D. 2H2O + O2 → 2H2O2

A. H+ + NADPH + O2 → NADP+ + H2O2 the passage says that it is a stepwise process, so write all the steps NADPH + FAD ⟶ NADP+ + FADH FADH + H+ + O2 ⟶ FADH-OOH If FADH-OOH decomposes non-productively (in the case of lys): 3. FADH-OOH ⟶ FAD+ + H2O2 Taken together, when you add all the reactions, you will see that FAD, FADH, & FADH-OOH all end up in reactants & products. Therefore, your net balanced reaction is NADPH + H+ + O2 ⟶ NADP+ + H2O2.

In the chromatography of the reaction mixture, water absorbed on cellulose functioned as the stationary phase. What was the principal factor determining the migration of individual components in the sample? A. Hydrogen bonding B. Solute concentration C. Stationary phase concentration D. Thickness of paper

A. Hydrogen bonding Hydrogen bonding affects how pigments travel up the chromatography paper through capillary action by affecting the polarity of the paper and thus the ability of certain pigments to be attracted to the paper and therefore not move as far the relative amount of hydrogen bonding to the stationary phase will determine the relative rate of migration of the various components in the sample. we've separated a homogenous mixture into its individual parts by exploiting differences in their molecular structures.

Radioactive tritium (3H) labeled guanine has been used to measure the rate of biochemical processes that involve its binding or incorporation. Given that water is the solvent for this type of experiment, what is the best site for tritium labeling? A. I B. II C. III D. IV

A. I the best site for tritium labeling would not exchange the tritium ions for protons in water. All of the N−H sites (II−IV) would readily exchange tritium protons due to their lone pair-facilitating protonation and subsequent tritium exchange with water, but the C−H site (I), lacking a lone pair, would retain its tritium label.

The electrolytically generated protons are then transported through the solid SCY electrolyte to react with the nitrogen at the other electrode according to Half-Reaction 3. SCY conductors are favored for this use because their proton conductivities increase substantially with temperature. N2(g) + 6H+ + 6e- → 2NH3(g) Half-Reaction 3 While the Haber process generally operates at about 450°C and 15-30 atm, an electrolytic cell operates most efficiently at about 600°C and atmospheric pressure. In both cases, the upper operating temperatures are limited by the reversible decomposition of ammonia, which is insignificant below 300°C but increases dramatically thereafter. It is possible to design a reactor where the SCY conductor and the nitrogen/ammonia electrode operate at different temperatures. Which combination of temperatures is expected to give the best results? A. SCY temperature higher than electrode temperature B. SCY temperature lower than electrode temperature C. SCY temperature the same as electrode temperature D. The temperature of the components does not make a difference.

A. SCY temperature higher than electrode temperature you want the SCY temperarature to be high so that protons travel faster, which means faster NH3 production. But at the same time, once the protons reach the electrode where they combine with N2 to form NH3, you want the temperature at that electrode to be much lower so that produced NH3 isn't decomposed.

Compound 1 contained five regions: (1) a hydrophobic region, (2) residues that could be cross-linked for additional structural integrity, (3) spacer residues, (4) a residue that was phosphorylated for future attachment of nanocrystals of HA, and (5) residues with affinity for cells. Which amino acid was incorporated into Compound 1 as a future site of covalent attachment to HA prior to mineralization? A. Ser B. Ala C. Tyr D. Thr

A. Ser look at the compound, you can see where the phosphate has been added and see that it's serine which it has been added to Serine and threonine and tyrosine are the three amino acids that can be phosphorylated.

Which forms of guanine and thymine are favored under physiological conditions? A. The keto form of guanine and the keto form of thymine B. The keto form of guanine and the enol form of thymine C. The enol form of guanine and the keto form of thymine D. The enol form of guanine and the enol form of thymine

A. The keto form of guanine and the keto form of thymine Based on the pKa of the protons on the nitrogen atoms in the rings in Figure 2, the keto state is preferred for both guanine and thymine at physiological pH (7.2). The nitrogen atoms are protonated because pH 7.2 is approximately 2 pH units less than the pKa.​ Specifically, the pKa of guanine's hydrogen is 9.2 and physiological pH is 7. Since 7 < 9.2, the hydrogen atom will stay on and the keto form will be preferred (the enol form doesn't have the hydrogen). Same goes for guanine.

In a test, a blood sample with 1/30 dilution ratio was mixed in a transparent cuvette with an oxygen-acceptor solution containing glucose oxidase. The same mixture was also added to five other cuvettes with identical optical properties containing solutions of known glucose concentration. Why was it important that the cuvettes containing the glucose oxidase and the blood sample were identical in terms of optical properties? A. To enable the comparison of the absorption spectra B. To reduce the absorption in the glass walls C. To decrease the uncertainty in the wavelength D. To increase the absorption in the solutions

A. To enable the comparison of the absorption spectra The identical optical properties of the cuvettes ensure that the absorbed radiation is due only to the presence of glucose in the blood and not due to the difference in the absorption features of the walls. While the glass walls have the same optical properties, this does not decrease the uncertainty in the wavelength. Wavelength uncertainty is related only to photon properties. wavelength is unrelated because wavelength is only used for energy not concentration. Think of c=λf and E=hf equations. Properties are independent of the wavelength.

In mammalian cells, iMs have been shown to be subject to two epigenetic modifications: methylation and hydroxymethylation. It is possible that the presence of 5-methylcytosine (5mC) or 5-hydroxymethylcytosine(5hmC) within a C-rich DNA sequence can affect iM stability. Which two classes of enzymes are needed in the two-step conversion of cytosine to 5hmC? A. Transferase and oxidoreductase B. Hydrolase and ligase C. Oxidoreductase and hydrolase D. Transferase and ligase

A. Transferase and oxidoreductase the first step involves transfer of a methyl group to cytosine, and the next step involves the hydroxylation of that methyl group. Therefore, the two classes of enzymes needed are a transferase and an oxidoreductase. Hydroxylation is a chemical process that introduces a hydroxyl group (-OH) into an organic compound.

four variants of RIα were constructed: L203A, I204A, Y229A, and R241A. Based on the data presented in figures 2 and 3, what is the most likely role of Y229 in protein stability and cAMP activation? A. Y229 is important for protein stability but not critical for cAMP activation. B. Y229 is important for cAMP activation but not critical for protein stability. C. Y229 is important for protein stability and critical for cAMP activation. D. Y229 is not important for protein stability and not critical for cAMP activation.

A. Y229 is important for protein stability but not critical for cAMP activation. Removal of Y229 has little effect on protein activation, as the activation curve is similar to WT activation. Y229A mutation doesn't affect cAMP activation. Figure 3 shows a leftward curve shift for the Y229A thermal denaturation profile. This means more of the Y229A mutant protein is unfolded/denatured at lower temperature. So Y229A mutation lowers the stability of the protein, which implies Y229 is necessary for protein stability.

Protein secondary structure is characterized by the pattern of hydrogen bonds between: A. backbone amide protons and carbonyl oxygens. B. backbone amide protons and side chain carbonyl oxygens. C. side chain hydroxyl groups and backbone carbonyl oxygens. D. side chain amide protons and backbone carbonyl oxygens.

A. backbone amide protons and carbonyl oxygens. secondary structure is represented by repeated patterns of hydrogen bonds between the backbone amide protons and carbonyl oxygen atoms. The major force which hold together peptide secondary structure is H-bonding between the peptide backbone

Ca2+(aq) + CO32-(aq) --> CaCO3(s) Equation 1 If the reaction shown in Equation 1 is spontaneous, the value of K must be: A. greater than 1. B. equal to ΔG. C. less than 1. D. equal to zero.

A. greater than 1. a spontaneous reaction is one that exhibits ΔG < 0. Since ΔG = -RTln(Keq), this means that Keq must be > 1.

Four organic compounds: 2-butanone, n-pentane, propanoic acid, and n-butanol, present as a mixture, are separated by column chromatography using silica gel with benzene as the eluent. What is the expected order of elution of these four organic compounds from first to last? A. n-Pentane → 2-butanone → n-butanol → propanoic acid B. n-Pentane → n-butanol → 2-butanone → propanoic acid C. Propanoic acid → n-butanol → 2-butanone → n-pentane D. Propanoic acid → 2-butanone → n-butanol → n-pentane

A. n-Pentane → 2-butanone → n-butanol → propanoic acid order of elution will depend on the polarity of the molecule. Since silica gel (polar) serves as the stationary phase for the experiment, increasing the polarity of the eluting molecule will increase its affinity for the stationary phase and increase the elution time (decreased Rf).

Compound 3 is prepared from Compound 2 (Figure 2) by: A. reduction of the ketone and lactonization of the gamma-hydroxyester. B. hydrolysis of one ester and formation of an acetal from the ketoacid. C. reduction of one ester and formation of an acetal from the gamma-hydroxyketone. D. reduction of one ester and the ketone followed by dehydration to a ketoether.

A. reduction of the ketone and lactonization of the gamma-hydroxyester. First, NaBH4 reduces the ketone to a secondary alcohol, the gamma-hydroxyester intermediate. Secondly, the alcohol group in this intermediate then reacts as a nucleophile with the carbonyl in the ethyl ester in the same molecule, forming a new ester by displacing C2H5OH as a leaving group. This cyclic ester is called a lactone, and the intramolecular transesterification yielding this lactone is called lactonization. This is a HYDRIDE REDUCTION reaction, in which the reagant (NaBH4 or LiAlH4) donates a hydride to the carbonyl carbon. It converts a carbonyl to an alcohol. NaBH4 is a reducing agent, it reduces aldehydes and ketones, not esters.

In a human liver cell, approximately how many glucose molecules would be required to synthesize a 750-amino acid protein? A. 100 B. 250 C. 750 D. 3000

A. 100 To synthesize a 750-amino acid protein would require 3000 ATP molecules (4 ATP per amino acid). A human liver cell typically undergoes aerobic respiration (glycolysis, PDC, Krebs cycle, electron transport, oxidative phosphorylation) and produces 30 ATP per glucose molecule. Thus, 3000/30 = 100 glucose molecules are required for this synthesis

2H2(g) + 2NO(g) → 2H2O(g) + N2(g) (ΔH = -664 kJ/mol) If the heat of formation of H2O(g) is -242 kJ/mol, what is the heat of formation of NO(g)? A. 90 kJ/mol B. 180 kJ/mol C. 211 kJ/mol D. 422 kJ/mol

A. 90 kJ/mol The standard enthalpy of formation for diatomic molecules such as O2 (g) equals 0 because no change occurs in the reaction O2 (g) --> O2 (g). Applying the equation ΔHrxn = ΔHf,products - ΔHf,reactants to the balanced Reaction 1 given in the passage (with ΔHrxn = -664 kJ/mol), we find ΔHrxn = ΔHf, products - ΔHf, reactants =[2(-242 kJ/mol) + 0] - [2(0) + 2x], so -664 = -484 - 2x. From this equation it can be determined that x = 90 kJ/mol.

CaSO4(s) --> Ca2+(aq) + SO42-(aq) Calcium sulfate is added to 1 L of distilled water at 350 K until no more salt dissolves. If a few drops of 2 M calcium chloride is added to the solution, which of the following would most likely occur? A. Calcium sulfate would precipitate out of solution. B. There would be no effect. C. The Ksp of CaSO4 would decrease. D. More sulfate ion would be present in solution.

A. Calcium sulfate would precipitate out of solution. Adding calcium chloride to the solution will increase the concentration of aqueous Ca2+. Le Châtelier's principle states that increasing the concentration of products will cause a shift in the equilibrium of the reaction to offset the imbalance, resulting in an increase in reactants. Since the reactant in this case is solid CaSO4, calcium sulfate will precipitate out of the solution.

The 1H NMR spectrum of pyruvate would contain one resonance composed of a: A. singlet. B. doublet. C. triplet. D. quartet.

A. singlet. Proton spin-spin splitting occurs when nonequivalent Hs are bonded to neighboring atoms. Since this molecule has only one kind of equivalent H, it will not experience any splitting, so its 1H NMR spectrum presents only one singlet

A 60 kg person stands inside a 20,000 kg railcar that is attached to a locomotive. For experimental purposes, the person has a ball, a string, a 600 nm wavelength laser, a concave lens, a 100 mF capacitor, and some 12 V, 100 W lightbulbs. The ball and string can be joined to make a pendulum of mass M and length L. What is the total amount of charge and energy, respectively, that the capacitor will store if it is connected to the battery on the railcar? A. 0.12 C and 14.4 J B. 1.20 C and 7.20 J C. 1.44 C and 0.12 J D. 7.20 C and 1.20 J

B. 1.20 C and 7.20 J charge is given by the relation 100 mF × 12 V = 1.2 C and the energy is equal to 0.5 × 100 mF × (12 V)2 = 7.2 J. use capacitance equation and the E=(1/2)CV^2

In the above figure, an object O is at a distance of three focal lengths from the center of a convex lens. What is the ratio of the height of the image to the height of the object? A. 1/3 B. 1/2 C. 2/3 D. 3/2

B. 1/2

When two amino acids are joined via a peptide bond, what is the mass of the byproduct of this reaction? (Note: Assume that the amino acids were not modified by protecting groups.) A. 17 amu B. 18 amu C. 32 amu D. 44 amu

B. 18 amu the formation of a peptide bond is accompanied by the formation of water as a by-product, and the mass of water is 18 amu. water is released as a by-product of peptide bond formation from the OH of the carboxyl group and the NH of the amino group

If no braking occurs, a total of how much power would be required to keep the railcar moving at 40 m/s? The rolling friction of the wheels and the internal friction between the wheels and axles contribute a continuous 1000 N decelerating force any time that the railcar is in motion. A. 16 kW B. 40 kW C. 600 kW D. 800 kW

B. 40 kW the power required must match the work done by the friction force that tends to slow down the railcar, which is equal to the decelerating force multiplied by the constant speed, so 1000 N × 40 m/s = 40 kW.

What is the pH of a buffer solution that is 0.2 M in HCO3- and 2 M in H2CO3? (Note: The first pKa of carbonic acid is 6.37.) A. 4.37 B. 5.37 C. 6.37 D. 7.37

B. 5.37 The pH of the solution can be calculated using the Henderson-Hasselbach equation: pH = pKa + log([base]/[acid]). Plugging in the values provided in the question gives pH = 6.37 + log(0.2/2) = 5.37.

The blood glucose meter uses test strips loaded with an enzyme and additives, and is calibrated to take a fixed volume of solution (usually blood). It measures the amount of Reaction 2, a two-electron oxidation, that takes place. The glucose meter measures the current produced during Reaction 2. If 0.67 μmol of electrons were measured, what mass of glucose was present in the sample? (Note: The molar mass of glucose is 180 g/mol = 180 μg/μmol.) A. 20 μg B. 60 μg C. 90 μg D. 270 μg

B. 60 μg stoichiometry of the reaction is 2 mol e- per mole of glucose consumed. The device measured 0.67 (2/3) μmol of electrons, indicating that 0.33 (1/3) μmol of glucose was consumed. This weighs 60 μg, based on its molar mass of 180 g/mol.

A study was done on the tissues of rats treated with microbubbles burst by 2.3 MHz ultrasound. It was observed that the burst microbubbles made openings of ~2.5 × 104 μm2 through the capillary walls. Assume that in the study with the rat tissues, fluid flows at a speed of 0.30 mm/s through a typical capillary opening caused by a burst microbubble. Given this, which of the following is closest to the volume flow rate of fluid passing through the opening? A. 4.5 × 106 μm3/s B. 7.5 × 106 μm3/s C. 1.2 × 107 μm3/s D. 4.5 × 107 μm3/s

B. 7.5 × 106 μm3/s the volume flow rate is given by 2.5 × 104 µm2 × 0.30 mm/s = 7.5 × 106 µm3/s. f=Av

The imidazoline family of compounds inhibits the NF-κB signaling pathway. Based on this observation, researchers were interested in determining if a member of this family of compounds, Compound 1 (Figure 1), could selectively sensitize cancer cells to the potent chemotherapeutic agent camptothecin (Compound 2, Figure 1). Assume that cellular uptake rates and drug delivery rates of compounds 1 and 2 are identical. If the proposed mechanism of sensitization by Compound 1 is correct, what cancer cell treatment protocol is most likely to produce the most apoptosis 20 h after treatment? A. Administration of both Compound 1 and Compound 2 simultaneously B. Administration of Compound 1 followed by Compound 2 after 0.5 h C. Administration of Compound 2 followed by Compound 1 after 0.5 h D. Administration of Compound 2 followed by Compound 1 after 1 d

B. Administration of Compound 1 followed by Compound 2 after 0.5 h compound 1 makes compound 2 more effective. We want compound 1 to act on all the cells before being introduced to compound 2. If the chemotherapeutic agent is administered at the same time as the sensitization agent, there is good reason to believe that some cells will be exposed to Compound 2 prior to Compound 1 as a result of random probability.

The graph below shows the relationship between the predominant form of iron as a function of solution pH and applied potential. Based on the graph, which of the following statements is true? A. At a potential of -0.4 V, as pH increases, Fe2+ is reduced and precipitates as Fe(OH)3. B. At a potential of -0.44 V, the equilibrium between Fe and Fe2+ is independent of solution pH below pH 6. C. At pH = 1, as the potential is changed from -0.2 to +0.8, Fe3+ is reduced to Fe2+. D. At pH = 8 and V = -0.1 V, Fe(OH)2 is the predominant form of iron.

B. At a potential of -0.44 V, the equilibrium between Fe and Fe2+ is independent of solution pH below pH 6. At an applied potential of −0.44V and pH below 6, there is an equilibrium between Fe(s) and Fe2+(aq), as shown by the line of demarcation. The fact that this line moves horizontally with increasing pH up until pH = 6 means that this equilibrium is unaffected by changing the pH.

An inflatable cuff was used to temporarily stop blood flow in an upper arm artery. While releasing the pressure to deflate the cuff, a stethoscope was used to listen to blood flow in the forearm. The blood pressure reading was 130/85. Given this information, which of the following statements is LEAST likely to be true? A. 85 mmHg was the diastolic pressure. B. Blood flow was heard when the pressure of the cuff was greater than 130 mmHg. C. 130 mmHg was the systolic pressure. D. Blood flow was heard when the pressure of the cuff was 90 mmHg.

B. Blood flow was heard when the pressure of the cuff was greater than 130 mmHg. the cuff was inflated to temporarily stop blood flow in the artery. The systolic pressure is determined from the first sound of blood flow that can be heard once the pressure exerted by the inflatable cuff falls below the pressure in the artery. The blood pressure reading was 130/85, which indicates that blood flow started again when the pressure was 130 mmHg. Therefore, blood flow was not heard when the pressure of the cuff was greater than 130 mmHg. D is not the correct response because blood flow would be heard when the pressure of the cuff is 90 mmHg, as this pressure is higher than the diastolic pressure.

Which variant of DNA polymerase will most likely retain catalytic activity? A. D429A B. D429E C. D429K D. D429F

B. D429E Since the side chain carboxylate groups bind the metal ions, the only variant that would retain this function is D429E.

Absorption of ultraviolet light by organic molecules always results in what process? A. Bond breaking B. Excitation of bound electrons C. Vibration of atoms in polar bonds D. Ejection of bound electrons

B. Excitation of bound electrons The absorption of ultraviolet light by organic molecules always results in electronic excitation. Bond breaking can subsequently result, as can ionization or bond vibration, but none of these processes are guaranteed to result from the absorption of ultraviolet light.

The electric field inside each of the conductors that forms the capacitor in the defibrillator is zero. Which of the following reasons best explains why this is true? A. All of the electrons in the conductor are bound to atoms, and thus there is no way for an external electric field to penetrate atoms with no net charge. B. Free electrons in the conductor arrange themselves on the surface so that the electric field they produce inside the conductor exactly cancels any external electric field. C. Free electrons in the conductor arrange themselves on the surface and throughout the interior so that the electric field they produce inside the conductor exactly cancels any external electric field. D. All electrons in the conductor, both free and bound, arrange themselves on the surface so that the electric field they produce inside the conductor exactly cancels any external electric field.

B. Free electrons in the conductor arrange themselves on the surface so that the electric field they produce inside the conductor exactly cancels any external electric field. Conductors contain both atom-bound electrons and free electrons. Free electrons arrange themselves on the surface of conductors, and their collective electric field produced inside the conductor cancels any external electric field. The resulting electric field inside the conductor is zero. FREE ELECTRONS MOVE ON THE SURFACE

[Cu(H2O)4]2+(aq) + 4NH3(aq) ---> [Cu(NH3)4]2+(aq) + 4H2O(l) Would the concentration of [Cu(NH3)4]2+ increase if the equilibrium were disturbed by adding hydrochloric acid? A. Yes, because the equilibrium in Equation 1 would shift to the left B. No, because the equilibrium in Equation 1 would shift to the left C. Yes, because the equilibrium in Equation 1 would shift to the right D. No, because the equilibrium in Equation 1 would shift to the right

B. No, because the equilibrium in Equation 1 would shift to the left Hydrochloric acid will protonate ammonia in a Brönsted acid-base reaction and reduce the amount of ammonia present. The disturbed equilibrium responds in a way to restore ammonia, but this causes the amount of [Cu(H2O)2(NH3)2]2+ to decrease.

A carbonyl group contains what type of bonding interaction(s) between the C and O atoms? A. One σ only B. One σ and one π only C. One π only D. One σ and two π only

B. One σ and one π only

Based on the reaction scheme in Figure 1, what is the mechanism of substrate binding to RT? A. Random order B. Ordered C. Ping-pong D. Double-displacement

B. Ordered the TP substrate binds first without any catalysis occurring and then the dNTP substrate binds. This is an ordered mechanism.

In [Cu(NH3)4]2+, the subscript 4 indicates which of the following? A. The oxidation number of Cu only B. The coordination number of Cu2+ only C. Both the oxidation number of Cu and the coordination number of Cu2+ D. Neither the oxidation number of Cu nor the coordination number of Cu2+

B. The coordination number of Cu2+ only For molecules and polyatomic ions the coordination number of an atom is determined by simply counting the other atoms to which it is bonded (by either single or multiple bonds). number 4 reflects only the number of ammonia molecules that bind to the central Cu2+ cation and does not indicate anything about its oxidation number.

Which statement about the unfolding cooperativity and pK of the oligonucleotides is consistent with the data in Figure 1? A. The oligonucleotide with the highest pK displays the highest unfolding cooperativity. B. The oligonucleotide with the lowest pK displays the highest unfolding cooperativity. C. The oligonucleotide with the second highest pK displays the highest unfolding cooperativity. D. The oligonucleotide with the second highest pK displays the lowest unfolding cooperativity.

B. The oligonucleotide with the lowest pK displays the highest unfolding cooperativity. the pK is the pH at which the fraction of folded DNA is 0.5. This occurs at the lowest value in 5hmC-WT. Cooperativity is measured as the slope of the unfolding transition. This is also highest in 5hmC-WT. the pK in the graph would be the pH at which you have 50% of the signal. And we know a cooperative graph would have the sigmoidal shape you see with the 5hmc-WT curve, so this has the highest cooperativity for unfolding. (think about Michaelis-Menten graphs)

A patient puts on a mask with lateral openings and inhales oxygen from a tank as shown. What phenomenon causes static air to be drawn into the mask when oxygen flows? A. Doppler effect B. Venturi effect C. Diffusion D. Dispersion

B. Venturi effect Air enters the mask because the static pressure of the air is larger than the static pressure of the oxygen in flow. This is the Venturi effect, and the mask is called the Venturi mask.

When a strip of Zn is placed in a beaker containing 0.1 M HCl, H2(g) evolves. If a strip of Al is placed in a beaker containing 0.1 M HCl, does H2(g) evolve? A. Yes; Al is reduced and H+(aq) is oxidized. B. Yes; Al is oxidized and H+(aq) is reduced. C. No; Al is reduced and Cl-(aq) is oxidized D. No; Al is oxidized and H2O(l) is produced.

B. Yes; Al is oxidized and H+(aq) is reduced. Since a new solid forms when Al(s) is mixed with Zn2+(aq), it is reasonable to assume that Al(s) is more susceptible to oxidation than Zn(s). Al(s) + Zn2+(aq) forms a solid, meaning that Zn2+ is reduced. Zn(s) + Al3+(aq) does not, meaning that Al3+ is not reduced. The conclusions we can draw from this are 1) that Zn2+ has a higher reduction potential than Al3+ and 2) that Al(s) has a higher oxidation potential than Zn(s).

Additional opsin based receptors are expressed in specialized cone cells and have maximum absorptions at 420 nm, 530 nm, and 560 nm when reconstituted with 11-cis-retinal. The production of a variety of opsins functions to: A. increase sensitivity to low light. B. enable the detection of different colors. C. ensure fast recovery of 11-cis-retinal after exposure. D. increase refractive index of the eye lens.

B. enable the detection of different colors. The wavelength of light absorbed by a molecule depends on its structure, and so the production of a variety of structurally related opsins functions to enable the detection of different colors.

N2(g) + H2(g) --> NH3(g) In the overall electrochemical reaction: A. nitrogen is oxidized at the anode, and hydrogen is reduced at the cathode. B. nitrogen is reduced at the cathode, and hydrogen is oxidized at the anode. C. nitrogen is reduced at the anode, and hydrogen is oxidized at the cathode. D. nitrogen is oxidized at the cathode, and hydrogen is reduced at the anode.

B. nitrogen is reduced at the cathode, and hydrogen is oxidized at the anode. oxidation always occurs at the anode and reduction at the cathode of an electrochemical cell. Since nitrogen decreases in oxidation state during the reaction, it is reduced. Hydrogen, on the other hand, increases in oxidation state and is, therefore, oxidized.

[Cu(H2O)4]2+(aq) + 4NH3(aq) [Cu(NH3)4]2+(aq) + 4H2O(l) Equation 1 The value of the formation constant of [Cu(NH3)4]2+ is 5.6 10^11 at 25°C. At 25°C, the formation of [Cu(NH3)4]2+ according to Equation 1 is most likely a: A. spontaneous process with positive ΔG°. B. spontaneous process with negative ΔG°. C. nonspontaneous process with positive ΔG°. D. nonspontaneous process with negative ΔG°.

B. spontaneous process with negative ΔG°. the equilibrium constant for the reaction is very large (much greater than 1). This necessarily means that ΔG° is negative and the reaction is spontaneous. use equation deltaG= -RTln(Keq) when Keq is greater than 1, ln(Keq) is positive when Keq is 0<x<1, ln(Keq) is negative

How far from a converging lens with a focal length of 6 cm must an object be placed in order to form a virtual upright image at a distance of 12 cm from the lens? A. 3 cm B. 4 cm C. 6 cm D. 12 cm

B. 4 cm Substitute f = 6 cm and i = -12 cm (i is negative since the image is virtual) into the lens equation and solve for the object distance, o:

Which organ is involved in the regulation of all of the following: gluconeogenesis, urea production, and drug detoxification? A. Spleen B. Liver C. Kidney D. Small intestine

B. Liver spleen: filters blood, removes old red blood cells, provides a site for B cells to mature, and stores blood kidney: filters blood, maintains blood pressure, and regulates body water balance, electrolyte balance, and pH small intestine: is the site of digestion and absorption of food molecules

The solubility of a salt changes with the temperature of solution. An aqueous solution is saturated with NaCl, with excess salt present. If the system's temperature is increased, all of the following would occur EXCEPT: A. the Ksp of NaCl will increase. B. the amount of NaCl precipitate will increase. C. the amount of dissolved NaCl will increase. D. the entropy of the system will increase.

B. the amount of NaCl precipitate will increase. More precipitate forms when a substance becomes less, not more soluble. Substances with higher values of Ksp are more soluble.

Which of the following relations best fits the data in table 1? A. v ∝ V B. v ∝ (ρ - ρo) C. v ∝ 1/ρ D. v ∝ ρo

B. v ∝ (ρ - ρo) The nucleolus and chromosome have the same values for v, so the best fit for an independent variable will also have the same values for those two quantities, eliminating choice A. In general the values for v increase from row to row, eliminating choice C, which has the opposite pattern (because of the inverse proportion). Choice D makes no sense because the solvent density is a constant for the experiment. Choice B is in fact correct: the constant speed at which particles settle into their respective density layers is directly proportional to the difference between their density and the density of the solvent (as well as to the area of the particles, and inversely to the viscosity of the fluid).

What is the net charge of sT-loop (amino acid sequence KTFCGPEYLA) at pH 7.2? A. -2 B. -1 C. 0 D. +1

C. 0 at pH 7.2, the N-terminus will be positively charged and the C-terminus will be negatively charged. In addition, the lysine side chain will carry one positive charge and the glutamic acid side chain will carry one negative charge.

A tall tube is evacuated, and its stopcock closed. The open end of the tube is immersed into a container of water (density 103 kg/m3) that is open to the atmosphere (pressure 105 N/m2). When the stopcock is opened, how far up the tube will the water rise? A. 1 m B. 5 m C. 10 m D. 20 m

C. 10 m use equation F=pVg (archimede's principle: the magnitude of the bouyant force equals the weight of the fluid displaced by the object. set F equal to P*A, because pressure = F/A the water will rise to a height such that the weight (mass multiplied by gravitational acceleration) of the water column equals the atmospheric pressure multiplied by the tube cross-sectional area A. Because mass is density times volume, it follows that 103 kg/m3 × h × A × 10 m/s2 = 105 N/m2 × A, where h is the height sought. Solving for hyields h = 105 N/m2/(104 N/m3) = 10 m.

A 60-Ω resistor is connected in parallel with a 20-Ω resistor. What is the equivalent resistance of the combination? A. 80 Ω B. 40 Ω C. 15 Ω D. 3 Ω

C. 15 Ω equivalent resistance is given by the expression (1/(60) + 1/(20))-1 = 15 .

One company sells a defibrillator for home use that uses a 9-volt DC battery. The battery is rated at 4.2 A•hr (amp•hour). Roughly how much charge can the battery deliver? A. 4.2 C B. 38 C C. 15,000 C D. 136,000 C

C. 15,000 C definition of current is flow of charge per unit time. Thus, charge equals current multiplied by time, hence 4.2 A × 1 hr = 4.2 A × 3600 s = 15,120 C ≈ 15,000 C.

The best-performing PANI had a maximum conductivity of 5.0 × 10-3 (Ω∙cm)-1. What is the resistivity of the best-performing PANI described in the passage? A. 0.002 Ω•cm B. 50 Ω•cm C. 200 Ω•cm D. 500 Ω•cm

C. 200 Ω•cm resistivity is the inverse of the conductivity, which is 1/5.0 × 10-3 (Ω∙cm)-1 = 200 Ω∙cm. Resistivity is a measure of the resistance of a given size of a specific material to electrical conduction. High resistivity designates poor conductors.

A person, whose eye has a lens-to-retina distance of 2.0 cm, can only clearly see objects that are closer than 1.0 m away. What is the strength S of the person's eye lens? (Note: Use the thin lens formula A. -50 D B. -10 D C. 51 D D. 55 D

C. 51 D strength of the eye lens is equal to the inverse of the focal length of the eye lens. Its numerical value is given by ​​-1+(0.02 m)-1=1 D+50 D=51 D. Lens power (P)= 1/f

What is the concentration of Ca2+(aq) in a saturated solution of CaCO3? (Note: The solubility product constant Ksp for CaCO3 is 4.9 × 10-9.) A. 2.4 × 10-4 M B. 4.9 × 10-5 M C. 7.0 × 10-5 M D. 4.9 × 10-9 M

C. 7.0 × 10-5 M The solubility product constant, Ksp​, is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in solution. The more soluble a substance is, the higher the Ksp value it has. Since equal quantities of Ca2+(aq) and CO32-(aq) are produced when CaCO3 dissolves, take the square root of 4.9 x 10^-9.

The pH of a 1 L phosphate buffer solution was measured as 7.6, but the experimental procedure calls for a pH 7.2 buffer. Which method will adjust the solution to the proper pH? (Note: The pKa values for phosphoric acid are 2.2, 7.2, and 12.3.) A. Add enough 1 M Na2HPO3 to increase the phosphate anion concentration ten-fold. B. Add 1 M NaOH to neutralize a portion of the hydronium ions found in the solution. C. Alter the ratio of monosodium/disodium phosphate added to favor the monosodium species. D. Add 100 mL distilled, deionized water to dilute the basicity of the buffer.

C. Alter the ratio of monosodium/disodium phosphate added to favor the monosodium species. in order to lower the pH of a buffer, the proportion of acidic buffer component must be increased. Adding strong base, diluting with water, or adding a different basic salt will not lower the pH.

What type of functional group is formed when aspartic acid reacts with another amino acid to form a peptide bond? A. An amine group B. An aldehyde group C. An amide group D. A carboxyl group

C. An amide group

What can be inferred from the results of Scheme A about the relative thermodynamic stabilities of Compounds 4a and 4b? A. They are equally stable. B. Compound 4a is less stable than Compound 4b. C. Compound 4a is more stable than Compound 4b. D. Nothing can be inferred about relative thermodynamic stabilities.

C. Compound 4a is more stable than Compound 4b. there is a higher percentage of Compound 4a than Compound 4b at equilibrium (a thermodynamic mixture). For a thermodynamic mixture of isomeric products, the relative mole ratio of products is directly related to the relative stability of these products. stability of isomers must be correlated to percent ratio of products.

What causes duplex DNA with a certain (A + T):(G + C) ratio to melt at a higher temperature than comparable length duplex DNA with a greater (A + T):(G + C) ratio? A. Stronger van der Waals forces of pyrimidines B. Stronger van der Waals forces of purines C. Increased π- stacking strength D. Reduced electrostatic repulsion of phosphates

C. Increased π- stacking strength G-C base pairs form stronger π-stacking interactions than A-T base pairs, thereby creating the most thermal stability. This disparity has often been used to explain the increased melting temperature of DNA rich in GC content. G-C has 3 H-bonds while A-T has 2 H-bonds. More bonds generally = more stronger. pyrimidines and purines don't pair with the same in each pair.

The term "ideal gas" refers to a gas for which certain assumptions have been made. Which of the following is such an assumption? A. The law PV = nRT2 is strictly obeyed. B. Intermolecular molecular forces are infinitely large. C. Individual molecular volume and intermolecular forces are negligible. D. One gram-mole occupies a volume of 22.4 L at 25°C and one atmosphere pressure.

C. Individual molecular volume and intermolecular forces are negligible. STP (when 1 mol occupies a volume =22.4 L) for gasses is at T=0degrees C=273K and P=1 atm. don't confuse with standard conditions in thermodynamics (not this!) T = 25decrees C = 297K, P = 1 atm, molar concentration = 1M.

Given that Tris has a pKa of 8.07, for how many of the experiments would Tris have been an acceptable buffer? A. None of the experiments B. Only 1 of the experiments C. Only 2 of the experiments D. All three of the experiments

C. Only 2 of the experiments A buffer has a buffering capacity that is ±1 pH unit away from the pKa, which means that only two of the experiments would have used Tris. The buffering region is about 1 pH unit on either side of the pKa of the conjugate acid.

In mammalian cells, iMs have been shown to be subject to two epigenetic modifications: methylation and hydroxymethylation. It is possible that the presence of 5-methylcytosine (5mC) or 5-hydroxymethylcytosine(5hmC) within a C-rich DNA sequence can affect iM stability. Based on the information in the passage and in Figure 1, what effect does epigenetic modification have on iM pH-dependent denaturation? A. Both methylation and hydroxymethylation result in significantly decreased stability because cytosine is more readily deprotonated. B. Both methylation and hydroxymethylation result in significantly decreased stability because cytosine is more readily protonated. C. Only hydroxymethylation results in significantly decreased stability because cytosine is more readily deprotonated. D. Only hydroxymethylation results in significantly decreased stability because cytosine is more readily protonated.

C. Only hydroxymethylation results in significantly decreased stability because cytosine is more readily deprotonated. the drop in pK of the transition denotes a decrease in stability. Because DNA unfolding occurs as the pH increases, it can be inferred that this is due to cytosine deprotonation. Since 5hmC-WT has the lowest pK, hydroxymethylation decreases the stability by increasing the acidity of cytosine. more of the 5hmC-WT became basic (deprotonated) sooner compared to the other strains. Because the curve is decreasing with decreasing acidity, that means that something got deprotonated earlier compared to WT (decreasing acidity = more basic). So that means that cytosine essentially got deprotonated earlier.

Which statement correctly describes the structure of the DNA double helix? A. Nitrogenous bases pair with other bases in the same purine or pyrimidine groups. B. The two DNA strands of the double helix are oriented in the same direction. C. The amount of guanine will equal the amount of cytosine in a DNA sequence. D. Sugar-phosphate backbones form the interior of the double helix.

C. The amount of guanine will equal the amount of cytosine in a DNA sequence. as guanine and cytosine form base pairs on opposite DNA strands, they will occur in equal amounts within a specific DNA sequence.

The blood glucose meter uses test strips loaded with an enzyme and additives, and is calibrated to take a fixed volume of solution (usually blood). It measures the amount of Reaction 2, a two-electron oxidation, that takes place. The enzyme used in the blood glucose meter described in the passage is classified as: A. a transferase. B. an isomerase. C. an oxidoreductase. D. a hydrolase.

C. an oxidoreductase. Reaction 2 is an oxidation of glucose, and thus the enzyme that catalyzes this reaction is classified as an oxidoreductase.

The Na2CO3 used in Step 5 of Figure 3 is necessary to: A. dissolve the bromoester. B. remove an α-hydrogen from the bromoester. C. convert the ammonium salt in Compound 4 to the amine. D. replace the chloride ion in Compound 4 with the carbonate ion.

C. convert the ammonium salt in Compound 4 to the amine. Na2CO3 acts as a Brønsted base. Until a proton is removed from the alkylammonium salt in Compound 4, it does not have the lone pair needed to be a nucleophile. As a nucleophile, Compound 4 performs the SN2 reaction that furnishes Compound 5.

This synthetic T-loop (sT-loop) was incubated with 32P-labeled ATP in the presence of PDK1 for different time periods at 37 ° C and pH 7.2, and the amount of radioactivity incorporated into sT-loop was measured by detection of β- decay. In designing the experiment, the researchers used which type of 32P labeled ATP? A. α32P-ATP B. β32P-ATP C. γ32P-ATP D. δ32P-ATP

C. γ32P-ATP Because the kinase (I think that's the enzyme) transfers a phosphate group to the substrate, it makes sense that it would transfer the one at the very end (with the free end). The three phosphate groups, in order of closest to furthest from the ribose sugar, are labeled alpha, beta, and gamma.

If a rotor with a radius of 10 cm spins at a rate of 60,000 rpm, how many gees (that is, how many times the acceleration of gravity) of centrifugal acceleration are experienced within test tubes at the outer rim? A. 4,000g B. 40,200g C. 402,000g D. (4 × 107)g

C. 402,000g The question asks about acceleration: using Newton's second law, one can adjust the centrifugal pseudo-force equation to find the acceleration by dividing out the mass. Thus a = ω2r, where ω = 60,000 rpm × 1 min / 60 s × 2π rad/rev ≈ 6,000 radians per second. This yields a = (6 × 103 rad/s)2(0.10 m) = 3.6 × 106 m/s2, which is about (3.6 × 105)g, or 360,000g (the actual value is higher because we underestimated pi and overestimated g as 10 m/s2).

Lysine is an amino acid with a basic side chain. Its isoelectric point would be closest to: A. 6. B. 7. C. 8. D. 12.

C. 8. All amino acids have one low pKa ~2 (for the acidic carboxyl group) and one high pKa ~10 (for the basic ammonium group), which are averaged to calculate the pI. When an acidic or basic side chain contributes a third pKa value, the pI is determined by averaging the two lowest or the two highest pKa values of the three. Thus, a basic side chain, such as Lys, will contribute a high (~10) third pKa value and is expected to increase the pI to a point above neutral pH (eliminate choices A and B; choice C is correct). A pI of 12 implies an average of two very high pKa values, which are not found in amino acids (except arginine)

In modern radiotherapy, superconducting wire coils are used to create magnetic fields that focus particle beams in order to deliver the maximum dose of ionizing radiation to a tumor site while minimizing the destruction of healthy tissue. For which of the following types of radiation would this technique NOT be useful? A. Alpha particles B. Proton beams C. Gamma photons D. Beta particles

C. Gamma photons Magnetic fields exert forces on moving charged particles. Since photons have no charge, they are unaffected by magnetic fields (choice C is correct). All of the other particles named are positively or negatively charged.

In non-cancerous cells, is the production of lactic acid expected under highly aerobic conditions? A. Yes; under aerobic conditions lactic acid production is regularly used to form NADH B. Yes; under aerobic conditions lactic acid production is regularly used to form NAD+ C. No; under aerobic conditions NAD+ is regenerated by the electron transport chain D. No; lactic acid is not regularly produced under any conditions in the human body

C. No; under aerobic conditions NAD+ is regenerated by the electron transport chain The purpose of lactic acid production (from pyruvate) is to regenerate NAD+ needed for glycolysis under anaerobic conditions. Under aerobic conditions, NADH is converted back to NAD+ by the electron transport chain, using oxygen as the final electron acceptor

2H2(g) + 2NO(g) → 2H2O(g) + N2(g) Reaction 1 What is the rate law for Reaction 1? A. Rate = k [H2] [NO] B. Rate = k [H2]2 [NO] C. Rate = k [H2] [NO]2 D. Rate = k [H2]2 [NO]2

C. Rate = k [H2] [NO]2 Do not confuse with equilibrium expression (Keq), where the products are over reactants, and the coefficient of each species becomes an exponent on its concentration. In rate law expression, only reactants are included, and rate law can only be determined experimentally.

Sterilization is also caused by the UV radiation (200-300 nm) generated by the electron transitions within the plasma plume. What is the maximum energy of the UV photons generated by this plasma pencil? (Note: Speed of light is c = 3.0 × 108 m/s; Planck's constant is h = 6.63 × 10-34 J•s.) A. 7 × 10-16 J B. 5 × 10-17 J C. 3 × 10-17 J D. 1 × 10-18 J

D. 1 × 10-18 J the energy of a photon is given by E = hf. The passage gives the wavelength of the UV radiation. Inserting this into the formula for photon energy yields E = hc/λ = 6.6 × 10-34 J × (3 × 108 m/s)/(200 × 10-9 m) ≅ 1 × 10-18 J.

What is the ratio of the minimum sound intensities heard by a 64-year-old male (20 dB) and a 74-year-old female (40 dB)? A. 20 B. 40 C. 50 D. 100

D. 100 the relative intensities of the two sound waves are 20 dB and 40 dB, respectively. The difference is 20 dB, meaning that the decimal log of the ratio of their intensities is 2, which means that the ratio of their intensities is 100.

What is the work generated by a healthy adult who circulates 9 L of blood through the brachial artery in 10 min? A. 2 kJ B. 12 kJ C. 20 kJ D. 120 kJ

D. 120 kJ Formulas needed: rate = d/t P = W/t, or W= Pt flow of 9 liters in 10 minutes means a flow rate of 900 mL/min, and according to the graph, it corresponds to a power of 200 W. The work is then 200 W x 600 s = 120 kJ.

4C3H5N3O9(l) → 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g) At STP, the volume of N2(g) produced by the complete decomposition of 1 mole of nitroglycerin would be closest to which of the following? A. 5 L B. 10 L C. 20 L D. 30 L

D. 30 L Based on the balanced equation provided, 4 moles of nitroglycerin produce 6 moles of N2(g). Therefore, 1 mole of nitroglycerin will produce 1.5 moles of N2(g). At STP 1.5 moles of N2(g) will occupy 33.6 L since the molar volume of an ideal gas at STP is 22.4 L/mol.

4C3H5N3O9(l) → 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g) Based on the passage, the magnitude of ΔH° (in kJ) for the decomposition of 2 moles of nitroglycerin at 25°C is closest to which of the following? A. 500 B. 1000 C. 2000 D. 3000

D. 3000 Two moles of nitroglycerin produce 6 moles of CO2(g) and 5 moles of H2O(g). apply Hess's Law

What volume of a 0.120 M CaI2 solution would contain 0.078 mol of the solute? A. 35.0 mL B. 65.0 mL C. 350 mL D. 650 mL

D. 650 mL obtain the volume of solution necessary to provide a given amount of solute in moles

Compound 1 contained five regions: (1) a hydrophobic region, (2) residues that could be cross-linked for additional structural integrity, (3) spacer residues, (4) a residue that was phosphorylated for future attachment of nanocrystals of HA, and (5) residues with affinity for cells. Which amino acid residues were incorporated into Compound 1 to promote the adhesion of cells on the scaffold surfaces? A. Arg and Gly B. Cys and Gly C. Cys and Asp D. Asp and Arg

D. Asp and Arg The 5th region is stated as being involved in adhesion to other cells. If you look at the figure, the two rightmost side chains are Arg & Asp

Which of the following energy conversions best describes what takes place in a battery-powered resistive circuit when the current is flowing? A. Electric to thermal to chemical B. Chemical to thermal to electric C. Electric to chemical to thermal D. Chemical to electric to thermal

D. Chemical to electric to thermal chemical energy of the battery elements is used as electrical energy to set the charge carriers in motion through the resistor, where they experience drag from the crystal lattice of the resistive conductor and dissipate their energy as heat from the resistor.

This experiment was repeated in the presence of a synthetic peptide that mimics the HM domain (sHM) of Ser/Thr kinases with the amino acid sequence FLGFTY. Phosphorylated sHM (spHM) was also used in place of sHM. When used in place of spHM, which peptide would be most likely to achieve the same experimental results? A. FLGFAY B. FLGFQY C. FLGFGY D. FLGFEY

D. FLGFEY the phosphorylated threonine would most likely be mimicked by glutamic acid in terms of size and charge. amino acids that can be phosphorylated: serine, threonine, tyrosine, histidine, aspartic acid and glutamic acid

A common column material used in size-exclusion chromatography is dextran, a polysaccharide of glucose. Which type of interaction most likely occurs between proteins and the dextran column material? A. Aromatic B. Hydrophobic C. Salt bridge D. Hydrogen bonding

D. Hydrogen bonding a polysaccharide of glucose has numerous hydroxyl groups that can hydrogen bond to the polar side chains that are typically exposed on a protein surface.

Horseradish peroxidase (HRP) is a 44 kDa monomeric protein that contains a single iron(III) heme cofactor and two calcium ions. The heme iron is coordinated by a single histidine residue, and the calcium ions are coordinated by various Asp, Ser, Thr, Val, Gly, and Ile residues. Which atom is most likely involved in the coordination of calcium ions found in HRP? A. Hydrogen B. Carbon C. Nitrogen D. Oxygen

D. Oxygen an atom must be a Lewis base to coordinate to calcium ions, and oxygen is the only Lewis basic atom present in the side chains or backbones of the listed amino acids that has either a partially negative charge (in the peptide backbone or Ser side chain) or a negative charge (in the Asp side chain). The only other Lewis basic atom present is nitrogen (distractor C), which has a partially positive chain in the peptide backbone due to resonance, and is thus less likely to coordinate calcium ions.

Which of the following reasons best explains why it is possible to separate a 1:1 mixture of 1-chlorobutane and 1-butanol by fractional distillation? A. Both 1-chlorobutane and 1-butanol are polar. B. Both 1-chlorobutane and 1-butanol are nonpolar. C. The boiling point of 1-chlorobutane is substantially higher than that of 1-butanol. D. The boiling point of 1-chlorobutane is substantially lower than that of 1-butanol.

D. The boiling point of 1-chlorobutane is substantially lower than that of 1-butanol. The molecules have similar molecular weights, but 1-butanol has a hydroxyl functional group that can participate in hydrogen bonding. Hydrogen bonding is a particularly strong force of intermolecular attraction.

Which experimental condition is NOT necessary to achieve reliable data for Michaelis-Menten enzyme kinetics? A. Initial velocity is measured under steady state conditions. B. Solution pH remains constant at all substrate concentrations. C. The concentration of enzyme is lower than that of substrate. D. The reaction is allowed to reach equilibrium before measurements are taken.

D. The reaction is allowed to reach equilibrium before measurements are taken. once the reaction reaches equilibrium, measurement of Vo will be impossible and the kinetic data will look the same regardless of substrate concentration. Hence, response D is not necessary (nor desirable) to achieve reliable data for Michaelis−Menten enzyme kinetics. In contrast, Distractors A−C are essential to obtain reliable Vo versus substrate concentration data to calculate KM and Vmax using Michaelis−Menten enzyme kinetics.

A person is sitting on a chair as shown. Why must the person either lean forward or slide their feet under the chair in order to stand up? A. To increase the force required to stand up B. To use the friction with the ground C. To reduce the energy required to stand up D. To keep the body in equilibrium while rising

D. To keep the body in equilibrium while rising as the person is attempting to stand, the only support comes from the feet on the ground. The person is in equilibrium only when the center of mass is directly above their feet. Otherwise, if the person did not lean forward or slide the feet under the chair, the person would fall backward due to the large torque created by the combination of the weight of the body (applied at the person's center of mass) and the distance along the horizontal between the center of mass and the support point.

What kind of image is formed by the lenses of the glasses worn by a 68-year-old male who sees an object 2 m away? (focal length is negative) A. Real and enlarged B. Real and reduced C. Virtual and enlarged D. Virtual and reduced

D. Virtual and reduced lenses have a negative focal length which means they are diverging lenses. Such lenses form virtual and reduced images of objects situated at distances larger than the focal length.

The intensity of the radiation emitted by the oxygen sensor is directly proportional to the: A. propagation speed of the radiation. B. wavelength of the radiation. C. polarization of photons emitted. D. number of photons emitted.

D. number of photons emitted. energy of electromagnetic radiation is directly proportional to the number of photons, and the intensity of electromagnetic radiation is defined as energy emitted per unit time. Thus, intensity is directly proportional to the number of photons emitted.

In order to examine the effects of epigenetic modification on iM stability, researchers subjected three 22-nucleotide sequences to both pH and thermal denaturation. The pH-dependent unfolding of each oligonucleotide solution was monitored using circular dichroism (CD) spectroscopy. What is the correct expression for the ΔG′° for the transition observed in the experiments described in the passage? A. ΔG′° = -RTe([native]/[unfolded]) B. ΔG′° = -RTe([unfolded]/[native]) C. ΔG′° = -RTln([native]/[unfolded]) D. ΔG′° = -RTln([unfolded]/[native])

D. ΔG′° = -RTln([unfolded]/[native]) the equilibrium constant for DNA unfolding is Keq = [unfolded]/[native] as the unfolded DNA is considered to be a product. Thus, ΔG′° = −RTlnKeq = −RTln([unfolded]/[native]).

Which of the following ranks the redox-active species of the electron transport chain in order of decreasing electron affinity? A. O2 > FAD > CoQ > NAD+ B. NAD+ > FAD > CoQ > O2 C. O2 > CoQ > NAD+ > FAD D. O2 > CoQ > FAD > NAD+

D. O2 > CoQ > FAD > NAD+ D. Because O2 is the terminal species in the electron transport chain (ETC), it must have the highest electron affinity and thus be first in the list (eliminating choice B). Since NADH donates its electrons to the first electron carrier in the ETC, the product of its oxidation, NAD+, would have the lowest electron affinity and would be last in the list (eliminating choice C). FADH2 enters the ETC by donating its electrons directly to ubiquinone (CoQ) so the oxidized FAD will have a lower electron affinity than CoQ (eliminating choice A).

A biochemist determines reaction rate using varying substrate concentrations and one unit of Taq polymerase at 72oC. The following data are obtained. Which of the following is true when the reaction rate is 5 fmol/min? A. There are significantly more substrate molecules than free enzymes. B. There is twice the number of substrate molecules compared to free enzymes. C. There are approximately equal numbers of substrate molecules and free enzymes. D. There is an excess of free enzymes, compared to substrate molecules.

D. There is an excess of free enzymes, compared to substrate molecules. When the reaction rate is 5 fmol/min, the reaction rate is increasing linearly with more substrate. This means free enzymes are in excess, compared to the concentration of substrate molecules If there is only a little substrate, then the rate V is directly proportional to the amount of substrate added. But eventually there is so much substrate that the active sites of the enzymes are occupied much of the time, and adding more substrate doesn't increase the rxn rate as much (slope decreases).

A galvanometer is to be used as an ammeter to measure the current passing through a resistor. The galvanometer should be inserted into the circuit: A. in parallel with the resistor, so it experiences the same voltage drop. B. in series with the resistor, so it experiences the same voltage drop. C. in parallel with the resistor, so it experiences the same current. D. in series with the resistor, so it experiences the same current.

D. in series with the resistor, so it experiences the same current. Circuit elements in parallel experience the same voltage drop, and those in series experience the same current; this eliminates choices B and C. Since the goal is to measure the current through a resistor, the galvanometer (with as little internal resistance as possible) should be placed in series with the resistor in order to measure that current.

The Gram staining in Gram-positive bacteria is due to the presence of: A. a cell membrane. B. a capsule. C. a nucleus. D. peptidoglycan.

D. peptidoglycan. Gram staining is a common technique used to differentiate two large groups of bacteria based on their different cell wall constituents. Gram positive bacteria stain violet due to the presence of a thick layer of peptidoglycan in their cell walls, which retains the crystal violet these cells are stained with. Peptidoglycan serves a structural role in the bacterial cell wall, giving structural strength

Fructose-6-phosphate and glucose-6-phosphate are best characterized as: A. epimers. B. enantiomers. C. stereoisomers. D. structural isomers.

D. structural isomers. Structural isomers have the same molecular formula, but different connectivities. Fructose: five membered ring Glucose: six membered ring

Matrix-Assisted-Laser-Desorption/Ionization (MALDI) is a soft ionization technique used in conjunction with mass spectrometry (MS) to analyze proteins, protein fragments, and peptides. Which laser is suitable for the MALDI technique after its frequency is doubled? Laser A: wavelength 826 nm, power 1.2 mW Laser B: wavelength 714 nm, power 1.2 mW Laser C: wavelength 650 nm, power 1.5 mW Laser D: wavelength 532 nm, power 1.5 mW

Laser D: wavelength 532 nm, power 1.5 mW the wavelength must be either 266 nm or 325 nm, and by doubling the frequency of the laser whose wavelength is 532 nm, the resulting wavelength is 532 nm/2 = 266 nm, because electromagnetic radiation wavelength and frequency are inversely proportional to each other. Also, the power of the radiation must be 1.5 mW. It is giving you a set of hypothetical lasers in the answer choices and asking, "among these hypothetical lasers, doubling the frequency of which one would make it a suitable MALDI laser?" The two options for potential MALDI lasers are listed in the passage: wavelength 266 and power 1.5/ wavelength 325 and power 2.2 among the answer choices, if you double the frequency of Laser D, the wavelength will drop by half, from 532 nm to 266, and since laser D also has a listed power of 1.5, it corresponds exactly to the qualities of the first MALDI laser listed in the passage.

τ = NIAB sinθ Which one of the following graphs best illustrates how τ0, the initial torque on the coil, varies as I, the current flowing through the coil, changes?

Since τ0 = NIAB sinθ0, the initial torque τ0 is proportional to I. Therefore, the graph of τ0 vs. I must be a straight line through the origin.

An individual protein molecule in its native conformation can be mechanically unfolded using an applied external force. In the absence of applied force, the rate constant for unfolding of the protein was k°u (Equation 1). What are the units for the rate constant ku discussed in the passage? M-1•s-1 M•s M•s-1 s-1

s-1 This problem is dimension analysis. You need to know that the rate has units M/s no matter what. So, depending on whether the reaction is first or second order, that would change the units of the constant so that everything cancels out to give M/s. Example: Rate = k [A][B] This is a second order reaction. [A]*[B] equals M^2 . And so, the rate constant has units 1/(M•s) . This allows the cancellations to happen, giving the rate the needed units of M/s