Physics/Ochem/GenChem 30 Question practice

For these reactions, the carbon adjacent to the carbonyl group (carbon 2) is the α-carbon. The α-carbons in enols are susceptible to substitution because the carbocation intermediate formed during the substitution reaction is stabilized by resonance with the enol hydroxyl group. In this case, the α-carbon becomes brominated after the addition of Br2. Note that a halonium/bromonium intermediate is not formed from the reaction of the alkene C=C bond with bromine because the electron-withdrawing O and Br groups destabilize any positive charge localization on the carbonyl carbon. The acyl bromide bromine on the carbonyl carbon (carbon 3) is eliminated during hydrolysis. Therefore, carbon 2 is the only carbon to remain brominated after hydrolysis, yielding the final product, an α-bromoacid.

2. Researchers want to synthesize valine from isovaleric acid. The enol shown below, with certain carbon atoms labeled, is formed from the reaction of isovaleric acid with PBr3. If Br2 and then water are added to the enol, which of the following carbon atoms will be brominated in the final product? A.Carbons 1 and 2 only B.Carbon 2 only C.Carbons 1 and 3 only D.Carbon 3 only Reactions at the α-position (ie, 2-position) of carboxylic acids are an extremely important class of reactions in organic chemistry and biochemistry. Substitution of bromine on the α-carbon of a carboxylic acid, known as the Hell-Volhard-Zelinsky (HVZ) reaction, begins with the addition of PBr3. This step converts the carboxylic acid to an acyl bromide, which can then tautomerize from the keto form to the enol form. When Br2 is added to the reaction, the α-carbon of the enol acts as a nucleophile and attacks bromine. One of the bromine atoms in Br2 withdraws electrons away from the other, giving it a partial positive charge and making it a good electrophile. The electrophilic bromine is added to the α-carbon, forming an α-bromo acyl bromide and HBr. When water is added, it hydrolyzes the acyl bromide by attacking the electrophilic carbonyl carbon while the bromine on the carbonyl carbon acts as the leaving group. This yields an α-bromoacid, which is a useful intermediate in the synthesis of α-amino acids such as valine.

Alcohols are weak electrophiles and poor leaving groups; therefore, a nucleophile is unlikely to displace the hydroxyl group (-OH) of an alcohol. To improve its leaving group ability, an alcohol can be converted into a tosylate (-SO3C6H4CH3) or a mesylate (-SO3CH3). Tosylates are produced by the reaction of an alcohol with p-toluenesulfonyl chloride (TsCl) and a base, such as pyridine. The hydroxyl O acts as the nucleophile to attack the S of TsCl, and Cl acts as the leaving group. Then pyridine deprotonates the O atom, giving the tosylate product. Therefore, a bond between S and O is formed during the synthesis of a tosylate.

During the synthesis of a tosylate, which of the following pairs of atoms form a covalent bond? A. C and Cl B. S and O C. S and Cl D. C and O Substitution reactions require strong nucleophiles and electrophiles, as well as good leaving groups. The nucleophile attacks the electrophile and displaces the leaving group. A bond is formed between the nucleophile and electrophile, whereas the bond between the electrophile and the leaving group is broken. Alcohols are weak electrophiles and poor leaving groups; therefore, a nucleophile is unlikely to displace the hydroxyl group (-OH) of an alcohol. To improve its leaving group ability, an alcohol can be converted into a tosylate (-SO3C6H4CH3) or a mesylate (-SO3CH3). Tosylates are produced by the reaction of an alcohol with p-toluenesulfonyl chloride (TsCl) and a base, such as pyridine. The hydroxyl O acts as the nucleophile to attack the S of TsCl, and Cl acts as the leaving group. Then pyridine deprotonates the O atom, giving the tosylate product. Therefore, a bond between S and O is formed during the synthesis of a tosylate.

(Choice A) The solution has a 32P concentration of 1.2 M after 2.0 half-lives have passed. (Choice C) The solution has a 32P concentration of 2.3 M after 1.0 half-life has passed. (Choice D) The solution has a 32P concentration of 3.2 M after 0.5 half-life has passed. A value of 3.2 mmol (with units of quantity, not concentration) is also obtained by failing to divide the number of millimoles by the sample volume in the last step of the calculation. Educational objective:A half-life is the time required for an amount of a given radioactive isotope to decrease by half. With each subsequent half-life interval that passes, the amount that remains decreases by half again, becoming exponentially smaller and eventually approaching zero. The molar concentration of a solution at a given time is determined by dividing the number of moles of analyte by the sample volume (in liters).

Question 10 Phosphorus-32 is a radioactive beta-emitter used in the treatment of polycythemia vera (excess red blood cells). The radioactive decay profile for the number of 32P atoms (millimoles) in a 2.0-mL sample was plotted, as shown below. Based on the graph, what is the concentration of 32P in the 2.0-mL sample after 1.5 half-lives have passed? A. 1.2 M B. 1.6 M C. 2.3 M D. 3.2 M In a radioactive decay, the half-life is the time required for an amount of a given isotope to decrease by half. With each subsequent half-life interval that passes, the amount remaining again decreases by half and the fraction of the initial amount that remains becomes exponentially smaller, eventually approaching zero. Accordingly, the fraction remaining after n half-lives can be calculated: Fraction remaining=12nFraction remaining=12n Following a second half-life interval (n = 2), the fraction remaining is again cut in half, making the resulting amount equal to one fourth the initial amount. Three half-lives (n = 3) would leave an amount equal to one eighth the initial amount. In the graph of the radioactive decay of 32P, the sample initially (0 days) has 9.0 mmol of 32P atoms. The time required for this amount to decrease by 50% to 4.5 mmol is 14 days; therefore, the half-life of 32P is 14 days. As such, a time interval of 1.5 half-lives is equal to 21 days. 1.5 half-lives×14 days1 half-life=21 days1.5 half-lives×14 days1 half-life=21 days At 21 days (1.5 half-lives), the graph shows that there are 3.2 mmol of 32P atoms in the sample. To determine the molar concentration (mol/L) of 32P in the sample, the number of moles should be divided by the 2.0-mL sample volume: Once the 1.5 con. mmole of half life is found; use molarity to solve for concentration of sample.

Educational objective:Variables either have a dimension and are expressed in terms of a system of units, or they are dimensionless, pure numbers. Dimensional analysis is used to determine the possible units of a variable in an equation.

Question 11 Which of the following units can be used to express Reynolds number? Re = pvd/n (viscosity = kg.m-1s-1 A. g⋅cm/s B. kg2⋅m2⋅s2 C. kg⋅m/s D. kg⋅m−1⋅s−1/kg⋅m−1⋅s−1

Therefore, the continuity equation implies v is inversely proportional to the square of conduit diameter and increases by a factor of four as conduit diameter is halved. (Choices A and B) v in a laminar flow is inversely proportional to the square of flow diameter. Therefore, velocity must increase if the diameter of the flow is reduced. (Choice C) v is inversely proportional to the square of diameter, not the diameter. Consequently, v must increase by a factor of four. Educational objective:The continuity equation expresses the conservation of mass observed as fluids flow through a conduit. The continuity equation indicates that velocity is inversely proportional to the square of diameter.

Question 13 The diameter of a segment of an artery is reduced by a factor of two due to an obstruction. Assume that the flow is incompressible and laminar, and therefore follows the continuity equation. Compared to an unobstructed segment of the artery, the velocity of blood in the obstructed segment of the artery is: A. 1/4 as large. B. 1/2 as large. C. 2 times as large. D. 4 times as large. The flow of a fluid refers to the motion of many individual particles that make up the fluid. If the flow lines used to conceptualize the flow of the particles remain parallel to each other and do not change over time (ie, each of the fluid layers move past each other with little friction), the flow is referred to as laminar. For incompressible laminar flows traveling at velocity v between two regions of a conduit (eg, pipe) with cross-sectional area A, the quantity of mass entering one portion of a pipe over a period of time must equal the quantity of mass exiting that region simultaneously during that period. This relationship is stated by the continuity equation:

Consequently, the kinetic energy of the flow is dissipated by frictional shearing forces such that the velocity decreases and the fluid gradually returns to a steady state. (Choice A) The incompressibility (or compressibility) of a fluid does not affect the transition from turbulent to laminar flow. (Choice B) In many fluid flows, the motion of fluid particles is essentially horizontal such that gravitational potential energy does not play a role in the transition from turbulent to static flow. (Choice C) Convection refers to the transfer of heat due to the large-scale motion of molecules. There is no transfer of heat by convection as the turbulent flow settles down to a resting state. Educational objective:Viscosity is a measure of internal friction in a fluid. Kinetic energy in fluid flows is dissipated by the viscous shear force acting between different layers of the fluid flow.

Question 14 A fluid is stirred until it becomes turbulent. After the stirring stops, the fluid gradually returns to its original static state. The turbulent flow does not continue indefinitely because: A. the fluid is incompressible. B. the kinetic energy in the flow is transferred to gravitational potential energy. C. the energy put into the flow by stirring is transferred by convection. D. the energy put into the flow by stirring is dissipated by the fluid's viscosity. Viscosity is an intrinsic property of a fluid that characterizes the amount of friction resisting motion inside the fluid itself. The friction due to viscosity arises from interactions between the molecules in a liquid or gas. The viscous friction force can be conceptualized as a shearing force between fluid layers that are flowing past each other with different velocities ( v→1v→1 and v→2v→2). The shearing force F→shearF→shear is proportional to gradient of the velocity Δv→Δv→ between each layer, which is a value that characterizes how the velocity changes within the fluid between different locations separated by some distance Δy: where the coefficient η is the dynamic viscosity. The more viscous the fluid, the larger the coefficient η and the larger the shear force for given velocity gradients. As a static fluid is stirred up, energy is transferred into the kinetic energy associated with the motion of the fluid. At first, the flow is laminar but it eventually becomes turbulent as the velocity of the fluid increases. In a turbulent flow, the velocity varies dramatically from point to point. Once the stirring stops, the only force acting on the flow is the viscous shearing force resulting from the velocity gradients present within the fluid.

A. Therefore, the blood experiences an acceleration of −67.0 cm/s2 over 600 ms as they travel from the aorta to the brain Educational objective:Average acceleration is the change of velocity over a period of time. The average acceleration can be positive or negative, depending on whether the final velocity is larger or smaller than the initial velocity.

Question 15 The blood leaving the aorta reaches the circle of Willis, a collection of arteries that supplies blood to the brain. Assume that it takes 600 ms for blood to reach the circle of Willis and that its average velocity is reduced to 61 cm/s. What is the average acceleration experienced by the blood? A. −67.0 cm/s2 B. −0.670 cm/s2 C. 0.670 cm/s2 D. 67.0 cm/s2 Blood Velocity within the Aorta of an Experimental Subject (mean velocity v̅ = 101 cm/s) In this expression, Δv represents the difference between final velocity vf and initial velocity vi, and Δt represents the difference between final time tf and initial time ti. If vf is greater than vi (vf > vi), then the average acceleration is positive and the object is accelerating. Conversely, average acceleration is negative when vi is greater than vf (vi > vf). When blood flows from the aorta to the brain, it ascends the neck structures with negligible horizontal movement; therefore, the blood fluid flow and any associated acceleration occurs essentially along one dimension. Specifically, the viof blood is the average velocity of blood within the aorta featured in Table 1: vi=101 cm/svi=101 cm/s The blood flow between aorta and brain takes place over 600 ms (0.6 s) and the vf of blood is 61.0 cm/s. Consequently, the average acceleration of the blood is calculated as:

Educational objective:Standard deviation is a measure of how much the points in a data set tend to deviate from the mean value. Standard deviation is indicated by the average magnitude of the fluctuations about the mean in a graph.

Question 16. Which graph depicts the flow with the largest turbulent kinetic energy? Therefore, the flow with the largest turbulent kinetic energy has the largest average spread in the fluctuations about its mean value. This behavior is best represented by the graph in Choice D. (Choice A) This graph depicts a flow with the highest mean velocity. However, TKE is small because it is proportional only to the average magnitude of the fluctuations about the mean. (Choice B) The flow depicted in this graph has both a low mean velocity and a low TKE because the position of the mean is low on the y-axis and the fluctuations about the mean are small. (Choice C) This graph does not depict maximal TKE because the fluctuations about the mean are smaller than those featured in the graph in Choice D.

(Choice A) Enantiomers are nonsuperimposable mirror images in which all stereocenters are inverted. The stereocenter configurations in Compounds 1 and 2 are identical, and therefore cannot be enantiomers. (Choice B) Diastereomers are stereoisomers in which at least one, but not all, stereocenters are inverted. Because none of the stereocenters in Compounds 1 and 2 are inverted, they cannot be diastereomers. (Choice D) Compounds 1 and 2 have the same molecular formula and the same connectivity, so they are not constitutional isomers. Educational objective:Conformational isomers are structures that have the same formula and connectivity, and can be interconverted by the rotation of σ bonds. Because conformational isomers are identical except for structural bond rotations, they are the same compound.

Question 17 Compounds 1 and 2 can be described as: A. enantiomers. B. diastereomers. C. conformational isomers. D. constitutional isomers. Isomers are compounds that have the same molecular formula but a different arrangement of atoms. Conformational isomers are structures that have the same connectivity and can be interconverted by the rotation of σ bonds. Because conformational isomers are identical except for bond rotations, they are the same compound. Six-membered rings can be depicted in two different ways: as a wedge-dash projection or a chair conformation. In wedge-dash projections, wedges indicate the substituent is up and above the ring, and dashes indicate the substituent is down and below the ring. Substituents can be either axial or equatorial in a chair conformation. However, axial and equatorial do not dictate whether the substituent is up and above the ring or down and below the ring. The wedge-dash projection of Compound 1 shows the -OH and the -CH3 both on a dash (down). Likewise, the chair conformation of Compound 2 shows the -OH and -CH3 as equatorial down. To represent this wedge-dash projection as a chair conformation, both substituents can be shown as axial down. A chair flip puts the substituents in an equatorial down position (still on the same side of the ring), yielding an identical structure to Compound 2. Because these compounds have the same connectivity and can be interconverted from one to the other through bond rotations (via chair flip), they are conformational isomers.

Educational objective:Chemical reactions can be categorized broadly into five main types: combination, decomposition, single replacement, double replacement, and combustion. Double replacement reactions involve the exchange of the bonding partners of two reactant compounds, forming two new product compounds.

Question 18 Antacids act by functioning chemically as weak Brønsted-Lowry bases that neutralize excess stomach acid. Equations 1 and 2 describe the chemical behavior of the antacid sodium bicarbonate with stomach acid (HCl). NaHCO3(aq) + HCl(aq) → H2CO3(aq) + NaCl(aq) Equation 1 H2CO3(aq) → CO2(g) + H2O(l) Equation 2 In Equation 1, carbonic acid (H2CO3) and sodium chloride form during a: A. single replacement reaction. B. double replacement reaction. C. combination reaction. D. decomposition reaction.

(Choice B) This pair of structures represents a dipole-induced dipole interaction in which the permanent dipole from the polar O-H bond distorts the electron cloud of a neighboring nonpolar C-H bond and induces a weak temporary dipole. Thus, the O-H dipole and the induced C-H dipole form a weak attraction. (Choice D) This pair of structures represents London forces that form when two nonpolar bonds induce momentary dipoles in each other by distortions in the electron cloud around the atoms. These short-lived instantaneous dipoles invoke weak mutual attractions. Educational objective:Ion-dipole interactions occur between the partial charge of a permanent dipole and the full charge of an ion. Dipole-induced dipole interactions occur when a permanent dipole induces a weak temporary dipole in a nonpolar bond or atom. Dipole-dipole interactions occur between the partial charges of two permanent dipoles.

Question 19 Which of the following pairs of Lewis structures best represents a dipole-dipole interaction between permanent dipoles in two lactate molecules in a Plasma-Lyte A solution? A dipole consists of opposite electric charges separated across a distance (eg, a chemical bond). In molecules, permanent dipoles form across some covalent bonds due to unequal sharing of electrons between two atoms caused by large differences in electronegativity. In a dipole, the region with greater electron density gains a partial negative charge (δ−) and the region with less electron density acquires a partial positive charge (δ+). Many intermolecular forces (ie, noncovalent interactions between atoms and molecules) involve the partial charges of bond dipoles. A dipole-dipole interaction occurs when the opposite partial charges of the permanent dipoles of the polar bonds in neighboring molecules form a mutual attraction. In the structure of lactate, the C=O bond is polar and exhibits a permanent dipole. As such, the opposite partial charges of two C=O bonds can align and participate in dipole-dipole interactions, as shown in choice C. (Choice A) This pair of structures represents an ion-dipole interaction in which the full charge of an ion is attracted to the opposite partial charge of a polar bond dipole. In addition, the incorrect oxygen atom is deprotonated in one of the lactate structures.

Dextrose is not an electrolyte and does not conduct electricity in solution because it is a molecular compound made of covalently bonded nonmetal atoms that do not form ions in solution. However, sodium chloride is an electrolytebecause it is an ionic compound made of metal and nonmetal ions that dissociate in solution. Therefore, the presence of the sodium chloride electrolyte causes the dextrose saline solution to be electrically conductive. (Choices A, C, and D) The dextrose saline solution is a mixture of an electrolyte (sodium chloride) and a nonelectrolyte (dextrose). Although dextrose does not support conductivity, the presence of sodium chloride makes the solution electrically conductive. Educational objective:Soluble ionic compounds act as electrically conductive electrolytes in solution, but covalent molecular compounds generally do not conduct electricity in solution.

Question 20 Electrolytes are solutes that enable the conduction of electricity within a solvent. Based on its composition, the dextrose saline solution from Table 1 is: A. conductive, because dextrose is an electrolyte. B. conductive, because sodium chloride is an electrolyte. C. conductive, because both dextrose and sodium chloride are electrolytes. D. nonconductive, because neither dextrose nor sodium chloride are electrolytes. Chemical compounds can be classified by their bonding characteristics and physical properties: Ionic compounds consist of oppositely charged metal and nonmetal ions bound by strong electrostatic attractions (ie, ionic bonds) that result in solids with very high melting points. Ionic compounds do not conduct electricity as solids because the ions are held in place, but as liquids (ie, when melted or dissolved in solution) ionic compounds are conductive because the ions can move freely and act as charge-carrying electrolytes. Metallic compounds consist of metal nuclei surrounded by a delocalized "sea" of electrons that can move freely. These strong metallic bonds give metals a high melting point and make them electrically conductive in both solid and liquid states. Molecular compounds consist of covalently bonded nonmetals that have weak, noncovalent, intermolecular interactions between neighboring molecules that result in relatively low melting points (less than ~300 °C). Because the bonding electrons are localized (ie, confined within the covalent bonds), molecular compounds are generally nonconductive as both solids and liquids.

(Choice A) Interaction I cannot be a typical ion-dipole interaction because C-H bonds are nonpolar (ie no significant permanent dipole). (Choice B) Interaction II is a repulsive ion-dipole interaction because the partial negative charge on the oxygen atom is oriented toward a negatively charged chloride ion (ie, like charges repel). (Choice D) Interaction IV is attractive but results from a special type of dipole-dipole interaction called hydrogen bonding. Interaction IV cannot be an ion-dipole interaction because it does not involve an ion (ie, no fully charged species). Educational objective:Ion-dipole interactions occur between the partial charge of a permanent dipole and the full charge of an ion. Opposite charges attract whereas like charges repel.

Question 21 Which labeled intermolecular interaction in Figure 1 represents an attractive ion-dipole interaction? A. I B. II C. III D. IV Permanent dipoles (ie, opposite electric charges separated across a distance) form across covalent bonds that have an unequal sharing of electrons between two atoms due to a significant difference in electronegativity. The more electronegative atom has more electron density and a partial negative charge (δ−), and the less electronegative atom has less electron density and a partial positive charge (δ+). Many intermolecular forces (ie, noncovalent interactions between atoms and molecules) involve attractions between the partial charges of bond dipoles. Opposite charges attract whereas like charges repel; as such, charged ions can form ion-dipole interactions in which the full charge of the ion is attracted to the opposite partial charge (and repelled by the like partial charge) of a polar bond dipole. In the structure of dextrose, the C-O and O-H bonds are polar (ie have a permanent dipole) due to the large difference in electronegativity between the atoms. As such, only interactions II and III are ion-dipole interactions. Hydrogen is less electronegative than oxygen, so hydrogen has less electron density and carries a partial positive charge. Therefore, interaction III is attractive because the partial positive charge of the dipole (on the hydrogen atom) is oriented toward a negatively charged chloride ion (ie, opposite charges attract).

Educational objective:The molar mass permits conversions between a given amount of mass and the number of moles contained in that mass. Mole ratios from a balanced reaction equation can be applied to determine the number of moles of one chemical species relative to the moles of another chemical species in a reaction.

Question 22 During an investigation of metabolic rate, isotopically labeled dextrose (shown below) was prepared with a carbon-13 atom in the substituent (indicated by *) to function as a radioactive tracer. If 90.5 g of isotopically labeled dextrose is fully metabolized into carbon dioxide and water, how many moles of unlabeled carbon dioxide are produced? (Note: The molar mass of the 13C-labeled dextrose is 181 g/mol.) A. 0.42 mol B. 0.50 mol C. 2.50 mol D. 3.00 mol waste too much time.

(Choice A) The reaction is not a chelate formation reaction because nickel(II) bromide is an ionic compound and does not have a ringed structure formed from pincer-like coordinate bonds between the metal and a ligand. (Choice C) The reaction yields a blue-green solution of nickel(II) bromide rather than an insoluble solid precipitate. (Choice D) Neither Ni nor Br2 are a Brønsted-Lowry acid or base. Educational objective:Many reactions can be classified as one of five common types, but they can also be further classified based on the results of the reaction. Classifications include precipitate formation, chelate formation, and oxidation-reduction reactions.

Question 23 Shiny nickel metal granules added to orange liquid bromine mixed in alcohol produced a combination reaction yielding a blue-green solution of nickel(II) bromide. In the reaction between Ni and Br2: A. bromine forms a chelate with nickel. B. nickel undergoes an oxidation-reduction reaction with bromine. C. the reaction forms a precipitate of nickel(II) bromide. D. nickel and bromine participate in a Brønsted-Lowry acid-base neutralization. Many reactions can be classified as one of five common types, but they can also be further classified based on the results of the reaction. For example, reactions producing an insoluble solid (precipitate) crashing from the solution are said to be precipitate formation reactions whereas those involving Brønsted-Lowry acids and bases that react to produce an ionic salt and water are classified as acid-base neutralization reactions. Two other important reaction classifications are chelate formation and redox reactions. In chelate formation reactions, a metal cation and a ligand react to form one or more rings via a pincer-like coordinate bonding arrangement. In an oxidation-reduction (redox) reaction, the oxidation states of some atoms change during the conversion of the reactants to the products. The given reaction between Ni and Br2 is a combination reaction in which two reactants form a single ionic product, NiBr2. The assigned oxidation states for the elements in this reaction show that nickel goes from an oxidation state of zero in its elemental metal form to a state of +2 in the product (an oxidation), and bromine goes from an oxidation state of zero to a state of −1 (a reduction). Therefore, in addition to being a combination reaction, the reaction between Ni and Br2 is also an oxidation-reduction reaction.

(Choice A) This structure is a tautomer of the purine adenine, which contains an amine instead of a ketone. Therefore, its tautomer contains an imine (C=N) rather than an enol. (Choice C) This structure is the tautomer of cytosine, which is a pyrimidine and contains only one nitrogenous aromatic ring. Like adenine, cytosine's tautomer is an imine. (Choice D) This structure is a tautomer of thymine, which is a pyrimidine and contains a lactam. Like guanine, thymine contains a ketone, and its tautomer contains an enol. Educational objective:Tautomerization is a type of isomerization involving the transfer of a hydrogen atom from one position to another in a molecule and the movement of a double bond. The nucleotides can tautomerize, with guanine and thymine shifting between keto (major) and enol (minor) forms whereas adenine and cytosine shift between amines and imines.

Question 24 Which structure is a tautomer of guanine? Tautomerization is a type of isomerization that involves the transfer of hydrogen from one position to another within a molecule and the movement of a double bond to an adjacent atom. These two forms, called tautomers, are in equilibrium. Tautomerization frequently involves a keto form (compound with C=O) as the major tautomer whereas the enol form is the minor tautomer. Tautomers are classified as constitutional isomers rather than resonance structures because a bond is broken during the hydrogen transfer to yield a different compound. On the other hand, resonance structures are forms of the same compound in which electrons move from one atom to another without breaking any sigma bonds. The nitrogenous base guanine is a component of nucleotides and one of the four bases that are found in DNA. Guanine contains a δ-lactam and can be classified as a purine. It is a fused ring system with a ketone in the pyrimidine ring. Therefore, its tautomer must contain an enol.

(Choices A and D) The mass of the drug is not needed to calculate velocity because it is not included in any of the energy terms for fluid flow. (Choices C and D) The volume of the drug is not needed to calculate velocity because it is not included in any of the energy terms for fluid flow. Educational objective:The Venturi effect describes the decreased pressure associated with increased velocity of a fluid flowing in a pipe. This effect derives from a special case of Bernoulli's equation, in which the height of the fluid remains constant.

Question 25 If the density of the drug is known, what additional information is required to calculate the velocity of the drug as it enters the blood vessel? (Note: Assume ideal fluid flow.) A. Pressure in syringe, pressure in blood vessel, and mass of drug B. Pressure in syringe, velocity in syringe, and pressure in blood vessel C. Pressure in syringe, velocity in syringe, and volume of drug D. Velocity in syringe, mass of drug, and volume of drug Bernoulli's equation describes the flow of an ideal fluid within a pipe from point A to point B. At each of these two points, the total energy of the fluid remains constant. Bernoulli's equation includes three energy terms, pressure P, potential energy, and kinetic energy. Bernoulli's principle is derived from Bernoulli's equation under the special condition of constant height for the fluid. Hence, the potential energy terms cancel, and the equation depends only on P, fluid density ρ, and velocity v: PA+12ρv2A=PB+12ρv2BPA+12ρvA2=PB+12ρvB2 This equation describes the Venturi effect, in which decreased P is associated with increased v (and vice versa) to maintain a constant sum of the energy terms. Furthermore, solving for the velocity at point B vB yields: vB=Sqrt(2(PA−PB)/ρ)+(v^2A) In this question, the ρ of the drug is known and the researchers want to calculate vB. Therefore, the pressure in the syringe PA, velocity in the syringe vA, and pressure in the blood vessel PB are required to calculate vB.

(Choices B, C, and D) The frequency, velocity, and energy of the gamma rays remain constant. Only the gamma ray intensity increases when the dose increases. Educational objective:The intensity of electromagnetic radiation is the amount of power, or energy per unit time, delivered per unit area. The gamma ray intensity increases with increased drug dose because more gamma ray photons are emitted.

Question 26 Which of the following statements best explains why the higher dose of gamma-emitting drug has a greater therapeutic response? A. The intensity of the gamma rays is greater. B. The frequency of the gamma rays is greater. C. The velocity of the gamma rays is faster. D. The energy of the gamma rays is greater. The intensity I of electromagnetic radiation, such as light, is defined as the amount of power P delivered per unit area A: I=P/A Since P equals the ratio of energy E and time t, I can also be expressed as: I=(E/t)/A Therefore, the intensity of electromagnetic radiation is proportional to the individual energy of each emitted photon and the number of photons n emitted per unit time: I∝n In this question, the drug emits gamma ray photons with a fixed frequency, velocity, and energy. Increasing the dose increases the number of photons emitted, but the frequency, velocity, and energy of each photon remain constant.

(Choice A) 0.1 m results from assuming that the focal length equals the distance to the image. (Choice B) 0.08 m is obtained by calculating the inverse of the sum of the object distance and image distance without converting the value to meters. (Choice C) 0.04 m is calculated from the ratio of the image distance to the object distance. Educational objective:The focal length of a thin lens is calculated from the object distance and image distance using the thin lens equation: 1/f = 1/o + 1/i.

Question 27 What is the focal length of the lens shown in Figure 1 of the passage? A. 0.1 m B. 0.08 m C. 0.04 m D. 0.02 m In a convex lens, also called a converging lens, both surfaces curve outward. The focal length f of a converging lens is the distance from the lens to the point where parallel incident light rays converge. The thin lens equation states that f is related to the object distance o and image distance i by: 1/f=1/o+1/i In this question, o equals 2.5 cm and i equals 10 cm. Substituting these values into the equation above yield: 1f = 1/2.5 cm+1/10 cm = 4/10 cm+1/10 cm = 5/10 cm Hence, the value of f is given by: f = 10 cm/ 5 = 2 cm Therefore, f equals 2 cm or 0.02 m.

(Choices A and B) Both x-rays and gamma rays are electromagnetic radiation composed of photons. (Choice D) The wavelength of gamma rays is shorter than the wavelength of x-rays, not longer. Educational objective:The electromagnetic spectrum consists of different types of radiation ordered by wavelength, frequency, or energy. Electromagnetic radiation with a shorter wavelength has higher frequency and energy.

Question 28 The advantage of using gamma rays for tumor therapy over x-rays is that gamma rays: A. are not electromagnetic radiation. B. are photons. C. have higher energy. D. have a longer wavelength. Electromagnetic radiation includes different types of wave phenomena, such as radio waves, visible light, x-rays, and gamma rays. These different categories of waves are often organized into the electromagnetic spectrum, ordered by decreasing wavelength λ and increasing frequency f. The energy E of electromagnetic radiation equals the product of Planck's constant h and f: E=hf Hence, E is directly proportional to f. In this question, f of the gamma rays described in the passage is 50 × 1018 Hz, and f of the x-rays is 100 × 1015 Hz. Because f is higher for gamma rays than x-rays, E will also be higher for gamma rays than for x-rays. Therefore, the gamma rays used for tumor therapy have a higher E than the x-rays used for imaging.

Therefore, α for the tungsten target is 5 × 10−6 1/°C. (Choice A) 2 × 10−6 1/°C is obtained by using 0.2 mm for the change in length and 0.1 m for the length. (Choice C) 20 × 10−6 1/°C results from calculating the ratio of the length and the change in temperature. (Choice D) 50 × 10−6 1/°C is calculated by using 1 mm for the change in length instead of 0.1 mm. Educational objective:The linear relationship between an object's length and temperature depends on the coefficient of expansion. The coefficient of expansion can be calculated from the lengths of an object at two different temperatures.

Question 29 What is the coefficient of expansion for the tungsten target described in the passage? "To avoid overheating, tube current I is applied only to collect one x-ray image, followed by a cooling period (Figure 2). The tungsten target is 1000 °C and 2 cm long after cooling, but it heats to 2000 °C and increases in length by 0.1 mm when generating x-rays." A. 2×10−6 1℃ B. 5×10−6 1℃ C. 20×10−6 1℃ D. 50×10−6 1℃ Metal objects expand when they are heated. The change in length ΔL of the object equals the product of the coefficient of expansion α for the material, the original length L of the object, and the change in temperature ΔT: ∆L=αL∆T Hence, ΔL is directly proportional to ΔT with a proportionality constant equal to α. Solving this equation for α yields: α=∆L/L∆T In this question, the tungsten target L is 2 cm and its temperature is 1000 °C during the cooling period. When generating x-rays, the target temperature increases to 2000 °C and ΔL is 0.1 mm. The difference between the temperatures while generating x-rays and during the cooling period equals ΔT: ∆T=2000 ℃−1000 ℃=1000 ℃ Substituting these values into the equation for α gives:

Therefore, 2 mC flows into the x-ray tube for each x-ray image. (Choice A) 100 mC is obtained by assuming that the charge equals the current. (Choice B) 20 mC is calculated by using 0.2 s for the time interval. (Choice C) 10 mC results from using 0.1 s for the time interval. Educational objective:The electric current equals the rate of charge flowing through an electric circuit. Electric charge is calculated from the ratio of current and the interval of time.

Question 30 According to Figure 2 in the passage, how much charge flows into the x-ray tube during a single image? A. 100 mC B. 20 mC C. 10 mC D. 2 mC Electric current I equals the electric charge Q flowing through an electric circuit over a specific time duration Δt. According to circuit sign conventions, I is positive in the direction of the flow of positive Q and equals the ratio of Qand Δt: I=Q/Δt The unit for I is the ampere (A), which equals a Q of 1 Coulomb (C) flowing through the circuit in 1 s. Furthermore, this equation can be rearranged to solve for Q: Q=I⋅Δt In this question, I flowing into the x-ray tube equals 100 mA, which is 0.1 A, and Δt for a single x-ray image is 20 ms, which is 0.02 s. Substituting these values into the above equation yields: Q=(0.1 A)(0.02 s)=(0.1 C/s)(0.02 s) Q=0.002 C=2 mC

(Choice A) Although oxidation occurs at the anode, NAD+ is the product of NADH oxidation and does not itself become oxidized. (Choice B) An electric potential already exists because NADH spontaneously loses two electrons to form NAD+; no potential needs to be applied. Furthermore, the loss of electrons occurs at the anode rather than the cathode. (Choice C) Galvanic cells are composed of two half-cells, with one half containing the molecule that becomes oxidized at the anode (in this case, NADH) and the other half containing the molecule that becomes reduced at the cathode (in this case, O2). NADH and NAD+ must be in the same half-cell because one is made from the other. Similarly, O2 and H2O2must be together in the other half-cell.

Question 6 Researchers want to measure the reduction potential when NADH is oxidized by oxygen to compare the experimental reduction potential with the calculated reduction potential in the mitochondria. A galvanic cell was constructed as shown in the following diagram. Which of the following statements is true about NAD+/NADH in this experiment? A. NAD+ becomes oxidized at the anode. B. An applied potential causes NADH to lose electrons at the cathode. C. NADH is oxidized at the anode, and NAD+ is reduced at the cathode. D. An electric potential exists because NADH loses electrons at the anode. In a galvanic cell, a spontaneous (ΔG°′ < 0) reduction-oxidation (redox) reaction occurs, producing a flow of electrons between two electrodes connected by a wire. Electrons are generated at the electrode called the anode (ie, where oxidation occurs) and flow through the wire toward the electrode called the cathode (ie, where reduction occurs). According to the question, NADH is oxidized by O2; therefore, NADH is oxidized to NAD+ at the anode. This process produces an electric potential as NADH spontaneously generates electrons that flow from the anode to the cathode. Educational objective:Galvanic cells are electrochemical cells that produce an electric potential as a result of the spontaneous transfer of electrons from the anode to the cathode. In any electrochemical cell, oxidation occurs at the anode and reduction occurs at the cathode.

(Choice A) An increase in the pmf would require increased proton pumping. Shutting down complex I reduces proton pumping. (Choices C and D) Because complex I is involved in proton pumping, the pmf must change when the complex is blocked. Because fewer protons are pumped when complex I is blocked, the pmf must decrease. Therefore, the question and passage give sufficient information to determine the effect of barbiturates on the pmf. Educational objective:Inhibition of any complex in the ETC results in fewer protons being pumped into the intermembrane space of the mitochondria, resulting in a decreased proton motive force and decreased ATP synthesis.

Question 7 Barbiturates are known to inhibit complex I but do not affect complexes II-IV. Based on this fact, how will barbiturates affect the proton motive force? A. The proton motive force increases. B. The proton motive force decreases. C. The proton motive force is unchanged. D. The effect cannot be determined without more information. As stated in the passage, the proton motive force (pmf) arises as hydrogen ions (protons) are pumped from the mitochondrial matrix into the intermembrane space by complexes I, III, and IV. It is the driving force behind ATP synthesis in the mitochondria. When complex I is inhibited, the ETC can still operate from complex II onward as long as FADH2 can continue to provide complex II with electrons. Because complex I normally participates in proton pumping, its inhibition will result in fewer protons being pumped across the mitochondrial membrane. Therefore, the H+ concentration gradient is reduced and the pmf decreases, resulting in decreased ATP synthesis.

(Choice A) ATP production requires energy input, which is measured by ΔG°′. The number of ATP molecules that can be produced depends on the energy available. (Choice B) ΔG°′ (Gibbs free energy) depends on E°′ (reduction potential) and is related by the relationship: ΔG°′ = −nFE°′. (Choice D) NADH and FADH2, which are produced by catabolic oxidation of lipids, proteins, and carbohydrates, provide energy for ATP production by passing high-energy electrons through the ETC. This process is also known as electron transduction. Educational objective:Catabolism is an oxidative process that releases energy. During catabolic oxidation of fuel molecules, oxidizing agents such as NAD+ are reduced. Electron transduction from these reduced cofactors provides the energy for ATP synthesis.

Question 8 Which of the following is NOT true concerning catabolism and the function of the ETC? A. The number of ATP molecules produced is dependent on the ΔG°′ of the ETC. B. ΔG°′ of the ETC depends on the overall E°′ of the ETC. C. Catabolism requires NADH to serve as an oxidizing agent. D. Catabolism produces energy in the form of electron transduction. Catabolism breaks down biological fuel molecules such as carbohydrates, lipids, and proteins, releasing energy as they are oxidized. Oxidation cannot occur without reduction; consequently, a different molecule must be reduced in the process. NADH can be oxidized but not reduced, and therefore it serves as a reducing agent, not an oxidizing agent. Instead, NAD+ serves as an oxidizing agent for catabolic reactions and is reduced to NADH in the process. NADH then enters the ETC as a reducing agent after oxidation of fuel molecules, and thus provides the energy necessary for ATP production. The question asks which of the given statements is not true concerning catabolism and the function of the ETC. Choice C provides the correct answer because NADH does not serve as an oxidizing agent in catabolism.

Lipids can be classified based on whether they contain any hydrolyzable linkages (ie, an ester or amide). Lipids containing an ester or amide (eg, fatty acids, phospholipids) are classified as hydrolyzable lipids whereas lipids that do not have an ester or amide (eg, steroids) are known as nonhydrolyzable lipids. Steroids do not contain any hydrolyzable linkages, and as such they are nonhydrolyzable lipids (Number III). (Number I) Steroids contain four rather than only three fused hydrocarbon rings. (Number II) Steroids are made from triterpenes, which in total contain six (not two) isoprene units. Educational objective:Steroids are lipids made from triterpenes, which contain six isoprene units that cyclize to form four fused hydrocarbon rings. Steroids are classified as nonhydrolyzable lipids because they do not contain any hydrolyzable ester or amide linkages.

Question 9 Which of the following is true about the structure and reactivity of steroids? Steroids contain only three fused hydrocarbon rings Steroids are made up of two isoprene units Steroids are nonhydrolyzable lipids A. III only B. I and III only C. II and III only D. I, II, and III Terpenes are precursors to steroids, which are lipids involved in signaling pathways. Individual terpenes are made up of two or more 5-carbon groups known as isoprenes and are classified based on the number of isoprene units present. Terpenes can be joined together in one of three ways. Those containing two isoprenes are known as monoterpenes, and molecules made up of two monoterpenes (four isoprenes) are called diterpenes. Steroids are made up of triterpenes(three monoterpenes, or six isoprene units), which are synthesized by the cyclization of the triterpene squalene, forming cholesterol. Cholesterol itself contains four fused hydrocarbon rings (three 6-membered and one 5-membered rings).

Enantiomers

are nonsuperimposable mirror images in which all stereocenters are inverted.

Diastereomers

are stereoisomers in which at least one, but not all, stereocenters are inverted.

where η is the dynamic viscosity of blood, ρ is the density of blood, and d is the diameter of the vessel through which blood flows. In this question, the diameter of the conduit (a blood vessel) is 2 cm (0.02 m), and the passage states that the density and viscosity of blood are 1,060 kg/m3 and 0.003 kg/m∙s, respectively. Substituting these values into the equation above yields the value of vcritical: Therefore, in this case the blood flow will transition from a laminar to a turbulent flow at a velocity of 85.0 cm/s (Choice B). Educational objective:Reynolds number indicates the transition from laminar to turbulent flows. The critical velocity at which the onset of turbulence occurs can be calculated using the Reynolds number, the conduit diameter, and the density, as well as the dynamic viscosity, of the fluid.

​​​​​​​Question 12 Blood flow in the aorta becomes turbulent at a Reynolds number of approximately 6,000. If the diameter of the aorta is approximately 2 cm, what is the average velocity of blood at this Reynolds number? A. 8.50 cm/s B. 85.0 cm/s C. 101 cm/s D. 170 cm/s A flowing fluid can be conceptualized as a collection of packets traveling within the same conduit (ie, a pipe or container through which fluid flows). The flow of these packets can be described as either laminar or turbulent. In the case of laminar flow, the packets move together in a layered, uniform manner, and the flow is smooth. In cases of turbulent flow, the motion of the packets becomes highly nonuniform, and the resulting flow is more chaotic. The transition from laminar to turbulent flow in a fluid is indicated by the Reynolds number Re of a flow. Large values of Reynolds number (ie, Re > 2,000) are associated with the onset of turbulence because, like the onset of turbulence, Redepends on both the characteristics of the fluid and the fluid flow. In blood flows, the value of Reynolds number at which the transition between laminar and turbulent flow takes place is measured to be about 6,000. A rearrangement of Equation 2 may be used to express the velocity (vcritical) associated with the transition:

When hydrogen forms a bond to carbon, the more electronegative carbon atom takes the electron from hydrogen and is reduced (Number I). The loss of a bond to hydrogen or formation of a bond to electronegative oxygen results in carbon losing electrons, which is oxidation (Number II). A simple way to determine the oxidation state of an organic molecule is to count the number of C-O and C-H bonds. For example, an aldehyde is in a higher oxidation state than an alcohol because it has two C-O bonds in the form of a double bond compared to the single C-O bond in alcohols. Alkanes are more reduced than alkenes because they have more bonds to hydrogen. For organic molecules, oxidation involves the loss of electrons by forming bonds to oxygen or losing bonds to hydrogen; reduction is the opposite process. The compound that becomes oxidized is the reducing agent, and the compound that becomes reduced is the oxidizing a

​​​​​​​Question 3 For biological processes, which of the following is true of oxidation-reduction reactions? Reduction often involves the formation of bonds to hydrogen. Oxidation often involves the formation of bonds to oxygen. Reducing agents undergo a change to a lower oxidation state. A. I only B. I and II only C. II and III only D. I, II, and III

Educational objective:The overall reduction potential of a redox pair is the difference in standard reduction potentials of the oxidizing and reducing agents. This value indicates how likely a particular compound is to gain or lose electrons. More positive values indicate a tendency to accept electrons, and more negative values indicate a tendency to lose electrons.

​​​​​​​Question 4 What is the net reduction potential for the ETC? A. −1.136 V B. −0.496 V C. 0.496 V D. 1.136 V Redox reactions occur when one or more electrons are transferred from one atom to another. The atom that loseselectrons is oxidized (ie, electron donor) and the atom that gains electrons is reduced (ie, electron acceptor). Redox reactions are composed of a reduction half-reaction and an oxidation half-reaction. The net standard reduction potential(E°′) for a redox reaction is equal to the sum of E°′ for each half-reaction: E°′net reaction = E°′reduction + E°′oxidation Because half-reactions are typically written as reductions, one of the half-reactions must be reversed to make it oxidation, and thus the sign of its E° value is flipped. The passage states that O2 is the final electron acceptor (ie, O2 is reduced). Therefore, the net reduction potential is calculated as the sum of the standard reduction potential of O2 (+0.816 V) and the standard oxidation potential of NAD+(+0.320 V): E°′net reaction = 0.816 V + 0.320 V = 1.136 V (Choice A) This value, which is the correct magnitude but the wrong sign, is the potential for the reverse reaction in which electrons are transferred from H2O to NAD+. (Choices B and C) Omitting the negative sign before −0.320 V results in an answer of 0.496 V. If the sign is omitted and the electron donor and acceptor are reversed, then the answer becomes −0.496 V.

(Choice A) The reduction potentials for each electron carrier must be less than that of oxygen for the process to be spontaneous. (Choice B) For a process to be spontaneous, the Gibbs free energy ΔG°′ must be negative. Gibbs free energy is inversely proportional to standard reduction potential, as described by the equation, ΔG°′ = −nFE°′. (Choice D) The reactions in the ETC move spontaneously from one complex to the next. However, the spontaneity of a reaction does not provide a direct correlation to the rate of a reaction. Educational objective:Standard reduction potentials (E°′) indicate whether a molecule spontaneously gains or loses electrons and are related to the Gibbs free energy (ΔG°′). In a spontaneous process, E°′ is positive, ΔG°′ is negative, and electrons flow from molecules with low reduction potential to molecules with high reduction potential.

​​​​​​​Question 5 If the ETC is a spontaneous process, which of the following MUST be true about the redox reactions within each of the four complexes? A. The electron-carrying molecules in the ETC have E°′ values greater than that of oxygen. B. The overall free energy ΔG°′ of the ETC is positive. C. The E°′ values increase as electrons are passed from one carrier to the next. D. The redox reactions in complex IV are faster than the reactions in complex I. Standard reduction potentials (E°′) are indicative of whether a molecule will spontaneously gain or lose electrons. In a spontaneous process, E°′ is positive. The ETC is a series of spontaneous processes, so all the standard reduction potentials must fall within the range of NAD+ (E°′ = −0.320 V) and O2 (E°′ = 0.816 V), and each step in the chain must have a larger E°′ than the previous step.