CHEM/Phys

In which organ are sickled red blood cells most likely to be hemolyzed? A. Spleen B. Thymus C. Lymph nodes D. Kidneys

(A) One of the key functions of the spleen is to remove old or deformed red blood cells from circulation. The capillaries of the spleen are much smaller than the diameter of a red blood cell, helping to induce lysis of fragile or old cells. If the spleen is removed, more damaged cells remain in circulation.

ligand

A ligand is a molecule that binds to a central metal atom to form a coordination complex.

Which one of the following processes does NOT take place in the mitochondrion? A. Fatty acid synthesis B. Krebs cycle C. Electron transport D. Oxidation of pyruvic acid

A. Fatty acid synthesis occurs in the cytosol (choice A is correct). The Krebs cycle, electron transport, and the oxidation of pyruvic acid all happen in the mitochondrion (choices B, C, and D are wrong).

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

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

Which of the following is the most likely identity of a gas that effuses four times slower than hydrogen? A. N2 B. O2 C. Cl2 D. Ar

B. Graham's law can be used when the relative rates of effusion of two gases are known to find the molar mass of an unknown gas: The unknown gas has a molar mass of 32 g/mol, which matches the molar mass of O2, making choice B the best answer.

The ΔG° of a reaction increases following an increase in temperature. What impact would this have on the equilibrium constant? A. It increases. B. It decreases. C. It does not change. D. Inadequate information is provided to determine the outcome.

B. If the ΔG° of a reaction increases with increasing temperature, this will cause a decrease in the equilibrium constant (choice B is correct). Qualitatively, as ΔG°increases, the reaction becomes less spontaneous and favors the reactants to a greater degree. With an increase in reactants at equilibrium, the K decreases since K is a ratio of [products]/[reactants]. Quantitatively, ΔG° = -RT ln K, thus as ΔG° increases, K must decrease.

Which of the following substances, when added to a saturated solution containing both aqueous and solid silver chloride, will cause dissolution of the silver chloride solid through complex ion formation with Ag? A. Aqueous silver nitrate B. Aqueous potassium cyanide C. Aqueous sodium chloride D. Aqueous sodium acetate

B. In order to solubilize the solid silver chloride, the aqueous ions must be consumed (Le Châtelier's Principle) causing the equilibrium to shift and dissolve more solid silver chloride. Alternatively, the silver chloride must be converted to a soluble substance, such as a complex ion. Adding neutral salts of silver or chloride is the common ion effect and will actually reduce the solubility of silver chloride, causing more precipitation, not dissolution (eliminate choices A and C). Aqueous potassium cyanide will cause the formation of the complex ion dicyanoargenate, Ag(CN)2- (aq), resulting in the consumption (dissolution) of the solid silver chloride. AgCl(s) + 2 CN-(aq) ? Ag(CN)2-(aq) + Cl-(aq). The addition of sodium acetate will have no effect, since both sodium and acetate salts are known to be completely dissociated in solution (eliminate choice D).

Which of the following changes to an equilibrated system is most likely to change the proportion of reactants and products at equilibrium? A. Increasing activation energy B. Increasing product stability C. Increasing reactant concentration D. Increasing reaction rate

B. Increasing product stability will increase the relative proportion of products at equilibrium (choice B is correct). Changes in reaction rate (and other kinetic changes) are not likely to impact equilibrium (or other thermodynamic properties - choices A and D are wrong). Increasing reactant concentration will temporarily decrease the reaction quotient, Q, but once equilibrium has been re-established, the same proportions of reactants and products will exist (choice C is wrong).

A sample of nitrogen gas is cooled from 80°C to 20°C in a rigid-walled container. What impact would this have on pressure? A. Increase to 1.2 times the original pressure B. Increase by a factor of 4 C. Decrease to 0.8 times the original pressure D. Decrease by a factor of 4

C. Pressure and temperature increase and decrease proportionally, thus a decrease in temperature results in a decrease in pressure (choices A and B are incorrect). As volume, the number of moles of gas, and the ideal gas constant remain unchanged in this process, we can rearrange the ideal gas law to P1/T1 = P2/T2, or P2/P1 = T2/T1. Remembering to convert temperature to Kelvin when using the gas law equations, we can solve for P2/P1 thus: 293 K / 353K ~ 0.8. The pressure therefore also decreases to roughly 0.8 times the original pressure, making choice C correct.

alkaline hydrolysis

In the alkaline hydrolysis of esters and amides the hydroxide ion nucleophile attacks the carbonyl carbon in a nucleophilic acyl substitution reaction.

Refraction Reflection Dispersion Diffraction

It is important to note that when light changes mediums, the frequency of the light does not change. Therefore, all the colors that comprise white light maintain their frequency but have a change in wavelength with a corresponding change in velocity. Since each color has a slightly different wavelength, each color refracts slightly differently. Ultimately, violet light has the largest decrease in wavelen'gth and thus the highest index of refraction. Oppositely, red light has the smallest decrease in wavelength and the lowest index of refraction. The splitting of white light into its components of varying wavelengths is best described by dispersion. Refraction is a component of dispersion but it only explains the way light bends and not the way white light is separated into the full color spectrum. Reflection occurs when light travels from a medium with a higher refractive index to one with a lower refractive index in which none of the light refracts, but instead reflects staying in the same medium as initially. Diffraction is the process of light spreading out after it passes through a narrow opening.

Pauli Principle Aufbau Principle Hund's Principle

Pauli Principle: Describe the carrying capacity of an orbital Aufbau Principle: Describe how electrons are added to or or removed from orbitals of different energy Hund's Principle: Describe how electrons are added to or removed from orbitals of the same energy

Saponification

Saponification is the base-mediated (NOT ACID) hydrolysis of an ester.

The density of a human body can be calculated from its weight in air, Wair, and its weight while submersed in water, Ww. The density of a human body is proportional to: A.Wair/(Wair - Ww). B.(Wair - Ww)/Wair. C.(Wair - Ww)/Ww. D. Ww/(Wair - Ww).

Solution: The correct answer is A. This is a Physics question that falls under content category "Importance of fluids for the circulation of blood, gas movement, and gas exchange." The answer to this question is A because, according to Archimedes' Principle, the ratio of the density of an object to the density of the fluid it is submersed in is equal to the ratio of the weight of the object in air to the difference of submersed weight and weight in air. This question requires Scientific Reasoning and Problem Solving, the correct determination and use of a scientific formula, in order to arrive at the answer.

Why is the velocity of blood flow slower in capillaries than in arteries? A. Capillary walls are more elastic than arterial walls. B. Capillaries have less resistance to blood flow than arteries. C. The total cross-sectional area of capillaries exceeds that of arteries. D. Blood pressure is higher in the capillaries than in arteries.

The answer to this question is C because the high number of capillaries in the body means that the total cross-sectional area of these vessels is larger than any other vessel type in the circulatory system. This causes the velocity of the blood to decrease.

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 = nRT^2 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.

The answer to this question is C since a property of an "ideal" gas is that it is composed of particles that have negligible volume and do not exert intermolecular forces. It is a Knowledge of Scientific Concepts and Principles question since you are asked to recognize a component of the Ideal Gas law.

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

The correct answer is D. This is a Biochemistry question that falls under the content category "Separation and purification methods." The answer to this question is D because 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.

One method of isolating polypeptides and proteins from aqueous extracts is freeze drying. The aqueous solution of the polypeptide or protein is frozen. What procedure can be used to remove the water from the frozen sample? A. Sublimation under reduced pressure B. Distillation using steam C. Extraction with organic solvent D. Addition of magnesium sulfate

The goal of most polypeptide and protein extraction methods is to keep the protein from denaturing. Sublimation, the phase change of water from solid to gaseous state, eliminates the water from the sample without raising the temperature and possibly denaturing the protein (choice A is correct). Steam distillation would increase the temp and damage the protein (choice B is wrong), and addition of organic solvents or magnesium sulfate could also denature the protein (choices C and D are wrong).

log(2) = log(3) = log(5) = ln(2) = and ln(x) =

log(2) = 0.3 log(3) = 0.5 log(5) = 0.7 ln(2) = 0.69 and ln(x) = 2.3log(x)

It has been observed that when an alpha particle and a beta particle are given equivalent initial kinetic energies, the beta particle will travel a significantly greater distance in air before losing its energy. Of the following, which is the most likely explanation for this observation? A. The alpha particle's greater size and charge results in a greater number of interactions, which cause it to dissipate its energy faster. B. Beta particles have greater momentum. C. The beta particle's smaller size causes it to have more elastic collisions with air molecules, allowing it to travel farther. D. The greater velocity of the alpha-particle relative to the beta-particle results in an erratic, shorter mean free path for the alpha particle.

A. Choice B is false since the equation relating momentum, p, to kinetic energy, K, is p =. Since the alpha and beta particles have the same K, the one with the smaller mass (the beta particle) has the lower momentum. Choice C is false since a smaller particle would tend to have fewer collisions, not more. Choice D can be eliminated because the greater mass of the alpha particle means it has a lower velocity than a beta particle of equal kinetic energy.The alpha particle would experience a greater electric force and a greater gravitational force than an equally-energetic beta particle, not because of its lower velocity, but because of its greater charge and its greater mass. [In fact, it can be shown the energy lost per unit distance for a particle of charge q and mass m moving through matter is proportional to q2m / K.] Choice A is correct and describes why alpha particles have shorter ranges than beta particles.

A particle with charge q starts from rest at Plate A and accelerates half the distance to Plate B before colliding with a second particle. What is the kinetic energy of the first particle upon collision? (Ignore the effects of gravity.) A. qV / 2 B. qV C. qV / (2d) D. 2qV / d

A. Letting V denote the voltage between the plates, the work done by the electric field on the charged particle as it moves from Plate A to Plate B is qV. However, if the particle moves only half the distance, the work done on the particle will be only half as much: qV / 2. (The work done is indeed proportional to the displacement here because the electric force is constant, due to the uniform electric field between the capacitor plates.) By the work-energy theorem, qV / 2 also gives the particle's kinetic energy at the point in question. Choices C and D can be eliminated because they have the wrong units (units of force rather than of energy).

If an object is traveling in uniform circular motion, and the centripetal force acting on the object is suddenly removed, which of the following best describes the object's subsequent motion? A. The object will continue to travel in a straight line at a constant velocity. B. The object will continue to travel in a straight line, decelerating gradually until it comes to rest. C. The object will continue to travel in uniform circular motion. D. The object will continue to travel in uniform circular motion, decelerating gradually until it comes to rest

A. Newton's First law states that in the absence of external forces, an object will move with constant velocity. Therefore, once the force(s) providing the centripetal force are removed, the object will fly off tangentially from the circular path, traveling in a straight line at constant speed.

A chemist seals a rigid reaction vessel containing a gaseous reaction at equilibrium and begins pumping in helium gas. She fails to see a change in the relative quantity of reactants and products in the system of study. Which of the following best explains these observations? A. Total pressure has increased with no change in the reaction quotient. B. Total pressure has increased with compensatory changes in Q and Keq. C. Total pressure has decreased with no change in the reaction quotient. D. Total pressure has decreased with compensatory changes in Q and Keq.

A. Sealing a rigid container and adding a gas to the vessel will increase total pressure (choices C and D are incorrect). Addition of an inert gas also results in a corresponding decrease in the mole fraction for each gas in the mixture, resulting in no change in partial pressures and therefore no change in Q (choice A is correct and choice B is incorrect). Note that changing reactant or product concentrations cannot change Keq, only a change in temperature changes the value of K.

The following graph of ΔG vs. ln(Q), the natural log of the reaction quotient, may correspond to which of the following reactions paired with its correct corresponding Q? A. ATP hydrolysis; [ADP][Pi]/[ATP]Correct Answer B. NADH synthesis; [NADH]/[NAD+] C. FADH2 oxidation; [FADH2]/[FAD] D. GTP synthesis; [GTP]/[GDP][Pi]

A. This question has two parts. First we must identify which reaction is represented by the graph. Given the relationship: ΔG = ΔG°' + RTlnQ, we expect the y-intercept to correspond to the ΔG°' of the reaction, which we can see corresponds to a large negative number. Thus answer choice A, an exergonic reaction with the appropriate Q, is the best answer. Although answer choice C corresponds to an exergonic process, Q is defined as [products]/[reactants] which for FADH2 oxidation should be [FAD]/[FADH2]. Answer choices B and D correspond to endergonic processes.

2-Fluoro-2-deoxy glucose (shown below as the β-anomer) exists as an equilibrium mixture of α- and β-anomers that are interconverted by mutarotation via the open-chain form. Which of the following statements is true regarding the concentrations of the open-chain forms of glucose and 2-fluoro-2-deoxy glucose in 0.1 M aqueous solutions? A. The concentration of the open-chain form is greater for glucose because the fluorine atom destabilizes the aldehyde. B. The concentration of the open-chain form is lower for glucose because the fluorine atom stabilizes the hemiacetal. C. The concentrations of open-chain forms cannot be determined because they are transition states of mutarotation. D. The concentrations will be the same because both fluorine and oxygen are electronegative atoms.

A. While both fluorine and oxygen have greater electronegativity than carbon, the electronegativity of fluorine is significantly greater than that of oxygen. The chemical properties of a molecule are therefore modified by replacing an oxygen atom with a fluorine atom (eliminate choice D). The open-chain forms of sugars are not transition states, but intermediates in the mutarotation process (eliminate choice C). The carbon atoms of hemiacetals have partial positive charges due to the electron withdrawing effects of two bound oxygen atoms via induction. The greater electronegativity of the fluorine substituent increases the positive charge on the adjacent carbon, which destabilizes the hemiacetal. Choice B can thus be eliminated without evaluating the first part of the statement. Choice A is correct because the carbon atoms of aldehydes have a greater partial positive charge than hemiacetals due to electron withdrawal by the double-bonded oxygen atom via induction and resonance. The increase in partial positive charge on the adjacent carbon due to the presence of the fluorine destabilizes the aldehyde, which reduces the concentration of the open-chain form of 2-fluoro-2-deoxy glucose relative to glucose.

Lipase works to hydrolyze which of the following functional groups? A. Amide B. Ester C. Ether D. Hemiacetal

B. According to the passage, lipase is an enzyme that breaks down fats, or triglycerides, which contain three ester functional groups. Amides are the functional groups present in proteins so eliminate choice A, hemiacetals are present in sugars (eliminate choice D), and ethers are not largely important functional groups in any biologically important molecules (eliminate choice C).

Cortisol is an aromatic hydrocarbon ring-based hormone. Alterations in its structure have produced drugs with greater efficacy and fewer side effects than cortisol itself. This is likely due to: A. increased affinity for all target cell receptors. B. variability amongst target cell receptors responsible for different cortisol activities. C. more accurate dosing than the body can provide. D. increased clearance from plasma by the liver and kidney.

B. Efficacy and side effects are both determined to some extent by receptor binding. To increase efficacy, cortisol must bind with great affinity to the receptors of the cells producing the desired effects; to decrease side effects, it must bind with lesser affinity to the receptors of the cells producing the side effects. Thus, if there was variability among the target cell receptors, then the modified hormones would fit some receptors well and others poorly (choice B is correct). A drug with greater affinity for all receptors may be more efficacious, but would likely not reduce side effects (choice A is wrong). The negative-feedback loops of normal physiology constitute the ideal dosing mechanism, because drug dose is determined directly by drug level. Exogenous dosing could not be more accurate (choice C is wrong). Increased clearance of the drug from the plasma would reduce both efficacy and side effects (choice D is wrong).

Which of the following is NOT true of a magnetic field? A. It can be generated by a moving charge. B. It can accelerate a moving charge. C. It can exert a force on a moving charge. D. It can increase the speed of a moving charge.

B. First, eliminate choices A and D. If the particle is positively charged, it would not be repelled by the negative dee, and if the particle is negatively charged, it would not be attracted to the negative dee. The charge on the particle is determined by the fact that it follows a counterclockwise path in Figure 1. The magnetic force FB provides the centripetal force as follows: Since B points into the plane of the page, q must be positive to cause the magnetic force FB= q(v × B) to point as shown above (this follows from the right-hand rule). Therefore, the answer is B. (Q33 , TPR test 1)

Electronegativity, electron affinity, and ionization energy all increase across a periodic table row and decrease down a periodic table group. Why does acidity not follow this trend? A. Acidity relates to the ability of the nuclear protons to attract hydrogen atoms and not valence electrons. B. Acidity relates to the stability of the conjugate base, and larger atoms form more stable conjugate bases even when they are less electronegative than smaller atoms. C. Acidity is not influenced by nuclear shielding and effective nuclear charge, while the other trends are. D. Acidity increases across a periodic table row like the other trends, but it increases down a group because the electronegativity of the elements approaches that of hydrogen.

B. Given H-X, acidity increases as dissociation into H+ and X-increases. As the electronegativity of X increases across a periodic table row, the dissociation and acidity increase just like the other trends in the question. As the size of X increases, the electrostatic force between H and X decreases. This also makes the dissociation more likely and therefore acidity increases going down the periodic table (i.e., the opposite direction of the other trends listed). Larger atomic anions are more stable conjugate bases because they distribute charge better. The size effect overcomes high values of electronegativity as illustrated by the fact that HI is a strong acid and HF is not. Choice A is eliminated because acidity relates to the ability of an atom to attract valence electrons away from a hydrogen ion, thereby making proton donation (loss of a hydrogen ion) easier. Choice C is eliminated because acidity does increase with increasing effective nuclear charge across the periodic table. Choice D is eliminated because the electronegativity of elements only approaches that of hydrogen when going down the columns on the right side of the periodic table. In addition, this choice does not adequately explain the role of atomic radius in determining acidity.

Two circular pipes both have radii of 3.5 meters. They both steadily eject water horizontally, and the water from each pours onto a measuring grid 5 meters below. The stream from the first pipe lands on the measuring grid 20 cm away from a position directly below the pipe mouth, and the stream from the second pipe lands 5 cm away from a position directly below its pipe mouth. How long would it take the second pipe to pour out as much water as the first pours out in 1 minute? A. 2 minutes B. 4 minutes C. 8 minutes D. 16 minutes

B. In answering this question, it is essential to take advantage of the fact that almost everything is the same between the two flow streams. The given difference is how far they travel horizontally, and the question asks about a difference in time. The task is to determine the relationship between these variables. First, no matter how far the streams fall (5 m or any other distance), they both fall the same distance, so they both take the same amount of time to get to the measuring grid. In the projectile motion equation R = v0xt, if t is constant, then horizontal displacement is proportional to horizontal velocity. The flow speed coming out the second pipe is therefore one-fourth the flow speed coming out the first pipe, since it only travels 5 cm instead of 20 cm. Furthermore, according to f = Av, flow rate is directly proportional to flow speed if area is held constant. Since the two pipes have the same radius, they have the same area. Thus, the flow rate of the fluid coming out of the second pipe is one-fourth what it is in the first pipe. Next, since flow rate is also equal to volume over time, and since they are ultimately going to pour out equal amounts of water (equal volumes), flow rate is inversely proportional to time. The second pipe has one-fourth the flow rate, so it should take four times as much time. Four times as long as 1 minute is 4 minutes.

During Mark's diving lesson his instructor told him to exhale as he rose to the surface to prevent lung collapse. Lung collapse during ascension is typically due to a rupture of the lung wall that permits air to enter the pleural cavity. The rupture of the lung wall is most likely due to: A. decreased water pressure leading to an increase in intrapleural pressure. B. expansion of the air in the lungs according to Boyle's law. C. decreased water pressure leading to a rapid outward expansion of the rib cage. D. increased lung air pressure according to Boyle's law.

B. Just as air volume decreases when water pressure increases on the way down in a dive, air volume will increase as water pressure decreases on the way up in a dive. Thus if a breath of air is taken prior to ascension, and that breath is not exhaled on the way up, that volume of air will rapidly expand to the point where the lung wall can be damaged. There is no reason to assume that decreased water pressure would increase intrapleural pressure; if anything, absolute intrapleural pressure would decrease as well (relative intrapleural pressure would remain the same; choice A is wrong). Decreasing water pressure may allow the rib cage to expand outward, but it is unlikely that this expansion would be so dramatic and so fast as to damage the lungs (choice C is wrong). Again, water pressure decreases during ascension, which means that lung air pressure decreases as well (this is what allows the volume of air in the lungs to expand; choice D is wrong).

If an organic nerve toxin (acting at the neuromuscular junction) renders the ion channel of postsynaptic acetylcholine receptors permanently open, the toxin would be most likely to: A. cause prolonged hyperpolarization of the postsynaptic membrane. B. cause prolonged depolarization of the postsynaptic membrane. C. block the release of vesicles containing acetylcholine. D. prevent muscle contraction.

B. The acetylcholine receptor in the neuromuscular junction is a ligand-gated sodium channel. When it binds acetylcholine, it opens to allow sodium to flow down a gradient into the cell, depolarizing the muscle cell. The more that the channel is open, the greater the depolarization (B is correct). Hyperpolarization would require positive charges to leave the cell, not enter (A is wrong). The toxin acts on the acetylcholine receptor, so there is no reason to believe that it will block acetylcholine release (C is wrong). The result of prolonged depolarization will be excessive excitation of skeletal muscle, not a block on contraction (D is wrong).

Which of the following best describes how the brain determines the location of a sound? A. The angle of the sound wave, with respect to the eardrum, allows the brain to determine the location of the sound. B. Sound waves reach the ear closest to the source of the sound first, and this difference in timing allows the brain to determine the location of the sound. C. The angle of the sound wave, with respect to the auditory hair cells, allows the brain to determine the location of the sound. D. Sound waves closer to the source have a higher frequency, and this difference in frequency allows the brain to determine the location of the sound.

B. The angle of the sound wave with respect to the ear is not pertinent to determining the location of the sound (choices A and C can be eliminated). Sound waves reach the ear closest to the source of the sound first, and the further ear second. This time gap allows the brain to determine the location of the sound (choice B is correct). Choice D is incorrect; the frequency of the sound is not relevant to determining the location, nor is there a difference from one ear to the next.

Two resistors R1 and R2 are connected in series to a battery with no internal resistance. The current through each is measured. A third resistor, R3, is then connected in parallel to R2. How does this affect the currents throughR1 and R2? A. I1 and I2 both increase. B. I1 increases and I2 decreases. C. I1 decreases and I2 increases. D. I1 and I2 both decrease.

B. The combination of R2 and R3 has less resistance than R2 alone. This is because, in parallel, 1 / Rparallel = 1 / R2 + 1 / R3 > 1 / R2 and therefore, Rparallel < R2. The equivalent resistance of the entire circuit is R1 + Rparallel, which has now decreased. Since the current supplied by the battery is V / Req, where V is the battery's voltage, decreasing Req increases the total current. The current flows through R1; therefore, I1 must also increase. This eliminates choices C and D. Increasing the current through R1 would also increase the voltage across it. This means that there would be less voltage available to the parallel portion of the circuit. Less voltage across R2 means less current through it.

A scientist is experimenting with enzymes found in the human body. Of the following, which would make the most appropriate buffer for such work? A. C6H5COOH (Ka = 6.5 × 10-5) B. HClO (Ka = 3.5 × 10-8) C. HCN (Ka = 4.9 × 10-10) D. H3CNH3+ (Ka = 2.2 ×10-11)

B. The human body has a pH of 7.4. It is best to choose a buffer whose pKa value is close to the pH that one is trying to maintain. Since HClO has a Ka value of 3.5 × 10-8, we know that its pKa is a little less than 8 (in fact, it's about 7.5). Therefore, of the choices listed, HClO (choice B) would be the best here.

An unknown compound has the chemical formula C3H8O2. All of the following functional groups could be present EXCEPT? A. hemiacetal B. ester C. diol D. acetal

B. The index of unsaturation of C3H8O2 is 0. Esters have a π bond, which has a degree of unsaturation of 1, therefore choice B could not be a functional group in the unknown compound.

Which of the following represents a deviation from the assumptions of the kinetic molecular theory of gases when observed in real gases? A. Kinetic energy is conserved upon collisions. B. Particles are instantaneously polarized. C. Molecules remain in constant random motion. D. The temperature of the gas depends on its average kinetic energy.

B. The kinetic molecular theory of gases describes the behavior of gases but relies on many assumptions, including: the total volume of gas molecules is negligible, all collisions are elastic (choice A is incorrect), particles are in constant random motion (choice C is incorrect), particles span a distribution of speeds, and gas particles do not interact appreciably (no IMFs). Instantaneous polarization of molecules is characteristic of London dispersion forces and would contribute to deviations from predicted behavior (choice B is correct). Temperature is a function of average kinetic energy of particles and is true of any sample (choice D is incorrect).

The 10 kg dancer leaps into the air with an initial velocity of 5 m/s at angle of 45° from the floor. How far will she travel in the air horizontally before she lands on the ground again? A. 1.25 √2 m B. 2.5 m C. 2.5 √2 m D. 5 m

B. The question is asking for the range R of the projectile (in this case, a dancer). Use v0x as the initial velocity in the x-direction and v0y as the initial velocity in the y-direction. So range = (v0x)(time in the air), or in this case, R = (5 m/s)(cos 45°)(total time). To find the time in the air, consider the time it takes to get to the top of the projectile parabola. The time to the top of the parabola can be calculated since the velocity in the y-direction at the top of the parabola is zero. So 0 = v0y - g(time to top) and time to the top = v0y / g = (5 m/s)(sin 45°)/(10 m/s2) = √2/4 s. Since the time in the air is twice the time to the top, the time in the air = 2(√2/4 s) = √2/2 s. Plug this into the original equation for range, so range = (5 m/s)(cos 45°)(√2/2 s) = 5(√2/2)(√2/2) = 5(2/4) = 2.5 m. The correct answer is choice B.

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

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

When aqueous solutions of AgNO3 and Ba(OH)2 are combined, one should expect to observe: I. no visible change since all ion combinations create soluble compounds. II. the formation of a precipitate since AgOH is insoluble. III. the formation of a precipitate since Ba(NO3)2 is insoluble. A. I only B. II only C. III only D. II and III only

B. The reaction suggested is the double displacement: AgNO3(aq) + Ba(OH)2(aq) → Ba(NO3)2(aq) + AgOH(s). Two common ionic solid solubility rules are that nitrate compounds are soluble, while most Pb, Hg, and Ag solids are insoluble. Because the AgOH forms a precipitate, Roman numeral I is false while Roman numeral II is true. Since all nitrates are soluble, Roman numeral III is also false, making B the best answer.

Creatine phosphate is a high-energy buffer that helps maintain the level of available high-energy phosphates such as ATP. During intense muscular exertion, creatine phosphate replenishes the muscle's ATP by releasing a relatively larger amount of energy in transferring its phosphate group to ADP. ATP is able to accept this phosphate group from creatinine phosphate and serve as a convenient carrier of energy because: A. it releases more energy upon hydrolysis than any other molecule in the body. B. it has phosphate groups with an intermediate transfer potential. C. it has such a specific and limited role. D. it is present in selected cells.

B. What makes ATP a particularly good energy carrier is that it has an intermediate transfer potential. This allows higher energy phosphate carriers (like creatine phosphate) to drive the synthesis of ATP by transferring their phosphate groups to ADP (choice B is correct). ATP's hydrolysis does not release the largest amount of energy of any molecule in the body, since creatine phosphate can drive ATP synthesis (choice A is wrong). ATP has a varied role in the body, as an energy carrier and also as a nucleotide for RNA and DNA (choice C is wrong), and ATP is present in every cell in the body (choice D is wrong)

Which of the following statements explains why the boiling point of HF is abnormally high when compared to the boiling points of other Group 7A hydrides? A. The H-F bond is much less polar than the bonds between H and the other halogens. B. HF has the lowest molecular mass of the Group 7A hydrides. C. HF is affected by hydrogen-bonding interactions to a much greater degree than the other Group 7A hydrides. D. HF has the highest vapor pressure among the Group 7A hydrides.

C As in any boiling point trend, the reason that HF has an anomalously high boiling point is a function of intermolecular forces. One commonly employed definition of hydrogen bonding is interactions between an H bound to an F, O or N, with a lone pair of electrons on another F, O or N. This is a way of denoting a very strong dipole interaction. Since HF can participate in intermolecular hydrogen bonding, while HCl, HBr, and HI cannot, it experiences greater intermolecular forces, requiring more energy to escape from a condensed phase. Choice A is an incorrect statement, as the polarity of HF is greater than in the other hydrogen halides. Choice B is a correct statement, but does not explain the phenomenon, as normally compounds with lower molecular weights have lower boiling points than comparatively heavy compounds. Choice D is an incorrect statement, as HF has a lower vapor pressure (is less volatile, has a higher boiling point) than the other hydrogen halides.

An object is floating in a fluid of 1.2 specific gravity. If the volume of the fluid displaced by the floating object is 5 × 10-3 m3, what is the object's mass? A. 2.4 kg B. 4.2 kg C. 6.0 kg D. Cannot be determined from the information given

C. Because the object is floating, the object's weight is balanced by the buoyant force; that is, mg = ρfluidVsubg, or, after canceling the g's, m = ρfluidVsub. With ρfluid = 1.2ρwater = 1200 kg/m3 and Vsub = 5 × 10-3 m3, we find that m = ρfluidVsub = (1200 kg/m3)(5 × 10-3 m3) = 6 kg.

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

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

Which of the following reaction(s) occur(s) in the same cellular location as fatty acid oxidation? Citric acid cycle Glycolysis The decarboxylation of pyruvate to acetyl-CoA A. I only B. III only C. I and III only D. I, II, and III

C. Fatty acid oxidation occurs in the mitochondrial matrix. Thus, Item I is true: the citric acid cycle also occurs in the matrix of the mitochondria (choice B can be eliminated). Item II is false: glycolysis occurs in the cytosol (choice D can be eliminated). Item III is true: the conversion of pyruvate to acetyl-CoA also occurs in the matrix (choice A can be eliminated and choice C is correct).

If the circuit is closed and a neutral molecule with a dipole moment is oriented between the plates so that the dipole is at a 45° angle to the plates, then the molecule will experience: A. no net force and no net torque. B. a net force but no net torque. C. no net force but a net torque. D. both a net force and a net torque.

C. If the molecule is neutral, its overall charge is zero, so the force on the molecule must also be zero (eliminating choices B and D). Now, if we treat the molecule as containing two equal but oppositely-charged ends, +q and -q, then each of these will experience an electric force (in opposite directions), and, as a result, the electric field exerts a torque on the molecule. (See the figure below.)

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

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

Which of the following could affect any of the dissociation constants for hemoglobin? Decreasing the temperature Adding NH3 Increasing the [O2] A. I only B. III only C. I and II only D. I, II, and III

C. K′ is constant for any equilibrium at a certain temperature, and changing the temperature altersK′. Therefore, Item I is correct and choice B can be eliminated. According to the passage, lowering the pH alters the dissociation equilibrium of hemoglobin, preferentially stabilizes its tense form, and lowers its affinity for O2. This indicates that the hemoglobin-oxygen dissociation equilibrium as a whole and its dissociation constant are pH dependent. Ammonia, NH3, is a weak base and will affect the pH of the system and alter K′. Item II is correct and choice A can be eliminated. Adding [O2] will temporarily disturb the system's equilibrium, but the system will return to equilibrium according to Le Châtelier's principle until Q = K′. K′ will remain constant with the addition of [O2] which makes Item III false and choice C the correct answer.

If NADPH were fully hydrolyzed to its component bases, phosphates, and sugars, what type of monosaccharide would result? A. A three-carbon triose B. A hexose C. A pentose D. An α-D-glucose

C. NADPH (nicotinamide adenine dinucleotide phosphate) is a nucleotide and thus contains ribose, which is a pentose (choice C is correct; eliminate choices A, B, and D).

Perceiving the color of a pH indicator requires what type of visual processing? A. Depolarization of cone cells to trigger hyperpolarization of bipolar neurons. B. Depolarization of cone cells to trigger depolarization of bipolar neurons. C. Hyperpolarization of cone cells to trigger depolarization of bipolar neurons. D. Hyperpolarization of cone cells to trigger hyperpolarization of bipolar neurons.

C. The answer choices contain pairs of information so proceed with a 2x2 question elimination strategy. When cone cells are not exposed to light and are thus not processing the visual stimuli of color, they are depolarized. Since the question is asking about perceiving color, the cone cells need to be hyperpolarized (eliminate choices A and B). When cone cells hyperpolarize, the bipolar neurons are no longer being inhibited and can then depolarize in order to transmit signal to the ganglion and eventually the optic nerve (choice C is correct; eliminate choice D).

What are the most stable configurations of 1,2- and 1,3-diisopropylcyclohexane, respectively? A. trans, trans B. cis, cis C. trans, cis D. cis, trans

C. The most stable configurations are those in which both substituents occupy equatorial positions. 1,2-Diisopropylcyclohexane equatorial, equatorial is trans. 1,3-Diisopropylcyclohexane equatorial, equatorial, is cis.

Which of the following salts, when dissolved in water, will decrease the pH of the solution? NH4Cl KHSO4 NaHCO3 A. I only B. II only C. I and II D. I, II, and III

C. The question is asking for the acidic salt(s). The conjugate acid of a weak base will lower the pH, while the conjugate base of a weak acid will raise it. Conjugates of strong acids or bases have no effect on pH. NH4+, the conjugate of the weak base NH3, is acidic. Cl-, the conjugate of the strong acid HCl, will not affect pH. Thus, Item I is an acidic salt (eliminate choice B). K+ and HSO4- are conjugates of a strong base and strong acid, respectively. Generally this would indicate a neutral salt. However, the HSO4- ion is acidic in its own right and will lower the pH, so Item II is also true (eliminate choice A). Na+ is the conjugate of a strong base and HCO3- is the conjugate of a weak acid, so Item III is a basic salt and will raise the pH of the solution (eliminate choice D).

Mark found that during his trip to Vancouver his appetite increased significantly and he ingested a larger quantity of food; however, he did not gain any weight. A quick calculation showed that the increase in appetite could not be accounted for by an increase in work during skiing compared to diving. All of the following could be reasons for Mark's increased appetite EXCEPT: A. decreased Krebs cycle activity due to decreased PO2. B. increased glycolysis due to anaerobic respiration. C. increased beta-oxidation of fatty acids due to increased activity. D. decreased oxidative phosphorylation due to high altitude.

C. The question states that the increase in appetite is not due to increased work; choice C is not true, and therefore the correct answer choice. Mark's increase in appetite is most likely due to the fact that at high elevations there is less oxygen available for aerobic respiration (choices A and D can be reasons for the increased appetite and can be eliminated), thus the body relies more heavily on anaerobic respiration (glycolysis alone) to produce ATP (choice B can be a reason and is thus eliminated). Because anaerobic respiration is inefficient, Mark must eat more food to compensate.

A cook picked up a pan on the stove and, realizing it was too hot, quickly threw it into a nearby pot on the counter filled with water (at 99.9°C). After treating his hand for the burn, he found approximately one liter of water had vanished. Assuming no heat exchange with the surroundings, approximately how much energy did the pan lose in the vaporization of the water? [Note: ΔHvap(water) = 40.6 kJ/mol] A. 40 kJ B. 400 kJ C. 2200 kJ D. 4000 kJ

C. This question involves the calculation of the energy required for the vaporization of water. One liter of water, weighing one kilogram, contains 55.5 moles (1000 g / 18 g/mole = 55.5 moles). Thus here we would expect 40.6 kJ/mol * 55.5 moles to result in approximately 2200 kJ of energy required to vaporize one liter of water (choice C is correct). Note that if the water had been at a temperature below boiling, more energy would have been required to heat the solution before vaporization could have occurred.

Damage and destruction of hepatocytes, secondary to a hepatitis virus infection for example, is known to result in serum transaminase elevation. Which of the following most accurately describes the mechanism for this finding? A. Hepatocyte destruction results in increased ammonia levels due to urea cycle interruption leading to a reactive increase in transaminase enzyme production. B. Regeneration of hepatocytes results in increased production of transaminase. C. The transaminase enzymes within the liver are expelled due to increased cell membrane permeability during cell destruction. D. Transaminase serum elevation is specific for hepatitis virus infection due to viral induction of increased transcription and translation of several enzymes, including viral enzymes.

C. Transaminases, as well as any cellular enzymes or other cellular contents, are contained within cells by a cell membrane. Dysfunction or increased permeability of the membrane would result in the leakage of the enzymes into the serum (choice C is correct). Transaminase enzymes are only released into the serum during cell damage; otherwise they are contained within cells, and the passage provides no information to indicate that they might normally be secreted from cells (choices A and B are wrong). While it is true that virus infection can induce transcription and translation of many enzymes, it is unlikely the virus would encode a human transaminase (choice D is wrong).

Which of the following would increase the initial velocity of all enzymatic reactions? A. Increasing temperature to 70°C B. Maintaining pH at 7.0 C. Decreased substrate concentration D. Decreased Michaelis-Menten constant

D. A decreased Km would indicate increased substrate affinity for the associated enzyme and would therefore increase initial reaction velocity (choice D is correct). Although increasing the temperature can increase reaction kinetics, increasing to an extreme temperature can cause the enzyme to denature, and 70ºC is a high temperature (choice A is wrong). It is important to note that every enzyme does not have the same optimal pH at which they perform best. Enzymes of the stomach, for example, can work best at a pH as low as 2.0 (choice B is wrong). Decreasing the amount of substrate would decrease the velocity of enzymatic reactions (choice C is wrong).

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

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

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. 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).

Systolic blood pressure approaches 0 mm Hg when it reaches the: A. capillaries. B. arteries. C. aorta. D. right atrium

D. Blood from the systemic circulation at the end of its circuit through the body enters the heart through the right atrium, and at this point its pressure is near 0 mm Hg (choice D is correct). Although blood pressure is relatively low in the capillaries, it is not 0 mm Hg, or it would never circulate all the way back to the heart (choice A is wrong). Blood leaves the heart through the aorta on its way to the systemic circulation and is at its highest pressure at that point (choice C is wrong); from the aorta the blood travels through many arteries, and the pressure in these vessels is still high (choice B is wrong).

An object is placed in front of a concave mirror between the focal point and the mirror. If the object is then moved closer to the mirror, the image would be: A. larger and farther from the mirror. B. larger and closer to the mirror. C. smaller and farther from the mirror. D. smaller and closer to the mirror.

D. Conceptually, an object placed directly at the focal point would produce no image (i.e., an image infinitely far away). Therefore it makes sense that moving away from the focal point would produce an image closer to the mirror. This would eliminate choices A and B. Also, when an object is placed between the focal point and the mirror, the image is virtual and larger than the object. As the object moves closer to the mirror, the image would get smaller and eventually become the same size as the object when the object is touching the mirror. The answer is therefore choice D. Using equations, recall that 1 / o + 1 / i = 1 / f, where f is positive for a concave mirror and remains constant. If o decreases, the (1 / o) term must increase, which means the (1 / i) term must decrease to compensate. Since i is negative for virtual images, decreasing the (1 / i) term means becoming less negative, which means that the magnitude of i decreases, which corresponds to the image moving closer to the mirror. As far as magnification goes, if we solve the above equation for i, we get i = of / (o - f). Magnification = -i / o = f / (f - o). Since o < f, then decreasing o would increase the denominator (f - o), and thus decrease the magnification. Another option would be to choose two object distances (say o = f / 2 and o = f / 4) and calculate the image distances and magnifications.

Respiratory rate is regulated by many factors. Which of the following conditions is most likely to cause a decrease in breathing rate? A. Metabolic acidosis B. Low O2 concentration in the blood C. High CO2 levels in the blood D. High plasma pH

D. Consider the following equilibria: CO2 + H2O ?H2CO3 ? HCO3-. In the blood, CO2 from the tissues is quickly converted to carbonic acid, which then dissociates into hydrogen ions (lowering the pH) and bicarbonate ions. Excess CO2 thus promotes metabolic acidosis, which is quickly compensated for by increasing the respiratory rate. This helps to eliminate CO2 from the blood and restore pH to normal (choices A and C are wrong). Low O2concentrations in the blood would lead to an increase in respiratory rate (choice B is wrong); furthermore, blood O2 levels affect the respiratory rate only in situations where O2 availability is low, such as at high altitudes (above 7000 feet). Of the choices given, the most likely condition to induce a decrease in respiratory rate is high plasma pH (choice D is correct) since a reduction in respiratory rate would lead to an accumulation of CO2 in the blood, and a corresponding increase in H+.

Which of the following is an intermediate of triacylglycerol degradation into glycerol and free fatty acids? A. Acyl-CoA B. Acetyl-CoA C. Water D. Diacylglyceride

D. Fatty acids are removed from the glycerol backbone one by one; therefore, the triacylglycerol will first become a diacylglyceride and a free fatty acid (choice D is correct). Acyl CoA and acetyl CoA are intermediates and products in later steps of the overall degradation process (choices A and B are false). Water is required for hydrolysis of the ester bond, but it is not considered an intermediate since it is consumed by the reaction.

From one point in space, Point S, to another, Point T, electric potential increases continuously from 100 V to 200 V. Which of the following must be true of the electric field near these points? A. Field lines point away from both S and T. B. Field lines point towards bothS and T. C. Field lines point from S to T. D. Field lines point from T to S

D. First, positive charges undergo a force from higher electric potential to lower electric potential. This happens because all things undergo forces from higher potential energy to lower, and from ΔPE= qΔφ, a positive charge means that q is positive, so ΔPE is negative (the positive charge is going to lower potential energy) when Δφ is negative (the positive charge is going to lower electric potential). Second, electric field lines point in the direction that a positive charge would undergo a force (by definition). Thus, electric field lines would point in the direction of lower electric potential, from 200 V to 100 V (that is, from T to S). Choice C is backwards and can be eliminated. The question stem doesn't describe the rest of the surroundings, so we have no way of knowing whether choices A or B would be mostly true; neither one would be completely true, because at least one field line (coming from T to S) points away from T, eliminating choice B, and at least one field line (coming from T to S) points towards S, eliminating choice A.

What is the most stable conformation of trans-1-ethyl-3-isopropylcyclohexane? A. Both substituents are in the equatorial position. B. Both substituents are in the axial position. C. The isopropyl group is axial, and the ethyl group is equatorial. D. The ethyl group is axial, and the isopropyl group is equatorial.

D. In trans-1,3-disubstituted cyclohexanes, one substituent will always be axial and one equatorial; this eliminates choices A and B. It is more favorable (more stable) to have the larger substituent in the equatorial position. An ethyl group with two carbons is smaller than an isopropyl group with three carbons. The isopropyl group will therefore be equatorial and the ethyl group will be axial, so choice C can be eliminated.

Lipolysis is followed by β-oxidation, in order to ensure the cell can harvest ATP from lipid molecules. Which of the following is true of this process? A. An isomerase and a reductase are required for complete oxidation of a monounsaturated fatty acid. B. Lipolysis of a DAG will generate twice the amount of fatty acids and glycerol, compared to lipolysis of a MAG. C. Unlike cell respiration, fatty acid catabolism starts in the mitochondria and finishes in the cytosol. D. The cell requires more than twice the number of NAD+electron carriers compared to FAD electron carriers, in order to harvest ATP from fatty acids.

D. It is true that cells require more than twice the number of NAD+electron carriers compared to FAD electron carriers, in order to harvest ATP from fatty acids (choice D is correct). As an example, catabolism of a ten-carbon saturated fatty acid will require four rounds of β-oxidation. This will generate four FADH2, four NADH, and five acetyl-CoA. The acetyl-CoAs will enter the Krebs cycle and will generate 15 more NADH and 5 more FADH2 (as well as 5 GTP). In total then, 19 NADH are generated (requiring 19 NAD+) and 9 FADH2are generated (requiring 9 FAD). β-oxidation of a monounsaturated fatty acid requires only an isomerase, not a reductase (choice A is wrong). Lipolysis of a DAG will generate two fatty acids and one molecule of glycerol. Lipolysis of a MAG will generate one fatty acid and one molecule of glycerol, so the number of fatty acids generated is doubled, but not the amount of glycerol generated (choice B is wrong). Fatty acid catabolism starts in the cytosol, and is mostly performed in the mitochondrial matrix (choice C is wrong).

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

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

Most biological unsaturated fatty acids are cis and contain non-conjugated double bonds. Because of this, additional steps are required in β-oxidation. These most likely include: A. Changing a single double bond to trans via a reductase enzyme in the mitochondrial matrix. B. Combining two double bonds via a reductase enzyme (which uses NAD+ as a reducing agent), then changing the resultant trans double bond to cis via an isomerase enzyme. C. Moving a single double bond down the fatty acid chain via a translocase enzyme in the mitochondrial matrix. D. Combining two double bonds via a reductase enzyme (which uses NADPH as a reducing agent), then changing the resultant cis double bond to trans via an isomerase enzyme.

D. Monounsaturated fatty acids require an isomerase enzyme to move the double bond during β -oxidation (eliminate choice A). Polyunsaturated fatty acids require both an isomerase and a reductase enzyme to complete β -oxidation; this also requires the reducing agent NADPH (choice D is correct). Note that NAD+ is an oxidizing agent (not a reducing agent) because it oxidizes another molecule, and is thus reduced itself (eliminate choice B). β -oxidation does not require a translocase enzyme (eliminate choice C).

Phalloidin is a poison which affects the depolymerization of actin. A cell treated with phalloidin would likely exhibit difficulties with: A. separation of chromosomes during mitosis. B. cell-cell adhesion. C. flagellum motility. D. motility via pseudopodia.

D. Pseudopodial mobility is coordinated by the polymerization and depolymerization of actin and would thus be affected by phalloidin poisoning (choice D is correct). Separation of chromosomes during mitosis is coordinated by the spindle apparatus, made of microtubules (choice A is wrong). Cell-cell adhesion is mainly coordinated by intermediate filaments and desmosomes (choice B is wrong). Flagella and cilia are microtubule-based (choice C is wrong).

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

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

A pure sample of which of the following substances will exhibit the strongest intermolecular interactions? A. O2 B. NO2 C. CO2 D. H2O

D. Since none of the molecules shown are ionic, they will experience only London dispersion, dipole-dipole, and hydrogen bonding intermolecular interactions. Hydrogen bonding is the strongest interaction of the possiblities. Water is the only molecule listed that can exhibit hydrogen bonding, and so has the strongest intermolecular interactions. Note that this is the reason that, of the molecules shown, water is the only one that exists as a liquid at STP; the other molecules are gases.

Which of the following is most likely to inhibit and activate, respectively, isocitrate dehydrogenase, a key enzyme in the Krebs cycle? A. NADH; ATP B. AMP; NADH C. NAD+; ATP D. ATP; NAD+

D. The Krebs cycle produces 1 FADH2, 3 NADH, and 1 GTP per molecule of acetyl CoA. For regulated enzymes in the Krebs cycle, we expect any product of the cycle to inhibit these enzymes by feedback inhibition. Thus we should expect NADH, ATP (a GTP analog), and FADH2 to inhibit isocitrate dehydrogenase, eliminating answer choices B and C. We also expect the "low energy" versions of these molecules, FAD, NAD+, and ADP/Pi to activate this enzyme, making answer choice D the best option.

If a drug inhibited the activity of the Na+/K+ ATPase in neurons, this would result in: A. increased cellular glucose uptake via secondary active transport. B. increased cellular glucose uptake via primary active transport. C. decreased cellular glucose uptake via primary active transport. D. decreased cellular glucose uptake via secondary active transport.

D. The Na+/K+ ATPase is used to generate a Na+gradient which is then used for glucose uptake by cells via secondary active transport (choices B and C are wrong). Uptake of glucose is mediated by the Na+-glucose cotransporter. Inhibiting the Na+/K+ ATPase would destroy the Na+ gradient and in turn decrease glucose uptake (choice A is wrong and choice D is correct). Remember that secondary active transport does not directly use ATP, but instead relies on the gradient established by the direct use of ATP.

A certain fatty acid chain can be degraded to produce 158 ATP. If these fatty acids are the only ones found in a triacylglycerol, which of the following statements best describes the energy produced from complete catabolism of the triacylglycerol? A. Less than 474 ATP, since ATP is required when attaching CoA to the fatty acid chains B. Less than 474 ATP, since catabolism of these three chains accounts for total energy production C. More than 474 ATP, since fatty acid breakdown yields additional ATP as acyl CoA D. More than 474 ATP, since other products of the breakdown are not accounted for in the calculation

D. The energy yield of 474 ATP (158 x 3) only accounts for the breakdown of three fatty acids, while ignoring breakdown of the glycerol backbone, which yields additional energy in its conversion to G3P and subsequent entry into glycolysis (choice D is correct and choice B is wrong). While the statement in choice A is true, this is already accounted for in the 158 ATP generated from the catabolism of the fatty acid, so it cannot be used to explain deviations from that value in itself. ATP and acyl CoA are two different molecules (choice C is wrong).

Mg^ (2+) + 2e- → Mg E° = -2.37 V Cu^ (2+) + 2e- → Cu E° = 0.34 V Given the reduction potentials above, what is the value of Gibbs free energy at standard state (ΔG°) for the galvanic cell Mg|Mg^(2+)||Cu^(2+)|Cu? (F = 96,500 C/mol e-) A. 391 kJ B. 196 kJ C. -391 kJ D. -523 kJ

D. The standard notation for an electrochemical cell is anode|anode ion||cathode ion|cathode. Magnesium is the anode and is oxidized. Copper is the cathode and is reduced. The total potential of the cell is obtained by adding the voltages of each half reaction: Mg → Mg2+ + 2e- E° = 2.37 V (oxidation) Cu2+ + 2e- → Cu E° = 0.34 V (reduction) E = 2.37 V + 0.34 = 2.71 V or 2.71 J/C At standard state, cell potential (E°) is related to Gibbs free energy by ΔG° = -nFE°, where n is the number of moles of electrons transferred and F is Faraday's constant, given in the question. Galvanic cells are always spontaneous (ΔG < 0), which eliminates choices A and B. Using approximation: ΔG° = -nFE° ΔG° ≈ -(2 mol e-)(105 C/mol e-)(2.5 J/C) ≈ -5 × 105 J ≈ -500 kJ

1H NMR spectroscopy can provide all of the following information EXCEPT: A. the connectivity of atoms in a molecule. B. the number of nonequivalent hydrogens in a molecule. C. the chemical environment of the hydrogen atoms in a molecule. D. the splitting patterns of equivalent hydrogen atoms.

D. This question is asking for the FALSE statement. The number of signals in a 1H NMR spectrum indicates how many different types of hydrogens a molecule has (eliminate choice B). The splitting pattern of each signal provides insight into the connectivity of the atoms since nonequivalent neighboring hydrogens split the signals of their neighbors (eliminate choice A). The chemical shift of a signal generally indicates what types of groups are near the hydrogens in the molecule (eliminate choice C). While 1H NMR provides information about the splitting patterns of hydrogen atoms in a molecule, splitting effects only apply between nonequivalent hydrogens (not equivalent hydrogens). This answer choice is false, thus choice D is correct.

Some headpieces also employ noise reduction technology to eliminate sounds not coming from the earpieces of the stethoscope. How might such technology work? A. Moving parts in the earpieces dampen noise not coming from the earpieces via the Doppler effect. B. The earpieces that cover the ears are made of a material in which sound travels at a very different speed than it does in air, thereby shifting the frequency out of the range that humans can hear. C. The noise reduction system in the earpieces constructively interferes with the sound from outside the earpieces, making the unwanted sound harder to hear. D. The total energy of the sound waves coming from outside the earpieces is reduced by adding wave energy out-of-phase with the unwanted sound waves.

D. Use Process of Elimination. Eliminate choice A because the Doppler effect shifts frequency, not intensity, so it cannot dampen sounds. Eliminate choice B because one of the fundamental rules of waves is that frequency does not change when the wave travels from one medium to another, so having a different medium between the air and the ear would not change the frequency. Eliminate choice C because constructive interference would increase the intensity of the sound, not decrease it. Choice D is correct because adding wave energy out-of-phase with the incoming sound would cause destructive interference, decreasing the amplitude and therefore the energy (and thus loudness) of the outside sound.

Neutralization reactions are always .........

Exothermic

Compared to the wild-type LipA, what is the change in net charge in variant XI at pH 7? variant XI: R33G, K112D, M134D, Y139C, I157M · A. +4 · B.+3 · C. −3 · D.−4

The answer is D because the amino acid substitutions replace amino acids with charges of +1, +1, 0, 0, and 0 with ones that have charges of 0, −1, −1, 0, and 0 so the net charge goes from +2 to −2.

Which statement correctly describes how enzymes affect chemical reactions? Stabilization of: A. the substrate changes the free energy of the reaction. B. the transition state changes the free energy of the reaction. C. the substrate changes the activation energy of the reaction. D. the transition state changes the activation energy of the reaction.

The answer is D because the stabilization of the transition state, not the substrate, provides binding energy that is used to lower the activation energy.

Which of the following animal pairs best illustrates the outcome of convergent evolution? A. The dolphin and the shark B. The domestic sheep and the mountain goat C. The polar bear and the panda bear D. The light-colored and the dark-colored forms of the peppered moth

The answer to this question is A because convergent evolution is defined as a process whereby distantly related organisms independently evolve similar traits to adapt to similar needs.

Assume that a certain species with sex chromosomes R and S exists such that RR individuals develop as males and RS individuals develop as females. Which of the following mechanisms would most likely compensate for the potential imbalance of sex-chromosome gene products between males and females of this species? A. Inactivation of one R chromosome in males B. Doubling transcription from the S chromosome in females C. Inactivation of the R chromosome in females D. Doubling transcription from the R chromosomes in males

The answer to this question is A because males of the species have two copies of the R chromosome, whereas females have one R chromosome and one S chromosome. Therefore, in order to compensate for a potential imbalance of sex chromosomes between females and males, one R chromosome should be inactivated in males.

An ultrasound examination could show the motion of a fetus. In order to image this motion, the ultrasound examination devices requires what minimal information? A. The speeds of the sound and of the moving object. B. The speed of the sound, and the frequencies of the sound waves emitted and observed. C. The speeds of the sound and of the moving object, and the frequencies of the sound waves emitted and observed. D. The speeds of the sound and of the moving object, and the frequencies and wavelengths of the sound waves emitted and observed.

The answer to this question is B because the Doppler effect is used with ultrasound waves to provide fetal images. The Doppler effect relates the frequency of the ultrasound wave as detected by a moving detector to the frequency of the wave when the source is stationary, the speed of the source, and the speed of the detector. Three of the four quantities involved in the effect are required.

Which of the following will decrease the percentage ionization of 1.0 M acetic acid, CH3CO2H(aq)? A. Chlorinating the CH3 group B. Diluting the solution C. Adding concentrated HCl(aq) D. Adding a drop of basic indicator

The answer to this question is C because HCl is a strong acid that will increase the amount of H+ in solution and thus decrease the percentage of CH3CO2H that ionizes. It is a Scientific Reasoning and Problem Solving question because you are asked to reason using a scientific principle (Le Châtelier's principle) to identify that adding a strong acid to a solution of weak acid will decrease the amount of ionization of the latter.

Which of the following types of orbitals of the central atom are involved in bonding in octahedral compounds? A.sp B.sp3 C.p D.d2 sp3

The answer to this question is D because octahedral compounds have six σ bonds and no lone pairs. According to valence bond theory, the central atom requires the hybridization of six atomic orbitals, d2sp3.

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.

The answer to this question is D because the 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.

Which of the following is LEAST likely to occur in a muscle fiber that is metabolizing anaerobically? A. Fermentation of glucose to lactic acid B. Phosphorylation of ADP within the electron transport chain C. Depletion of stores of glycogen D. Acidification of the cytoplasm

The correct answer is B. This is a Biochemistry question that falls under the content category "Principles of bioenergetics and fuel molecule metabolism." The answer to this question is B because the electron transport chain only provides ATP aerobically. Lack of oxygen effectively shuts down the Krebs cycle and the electron transport chain.

The advantage of the Doppler ultrasound technique over the standard ultrasound technique is that it also allows: A. distinguishing between fluids and tissue. B. measuring the blood flow. C. measuring the tissue density. D. measuring the heart wall thickness.

The correct answer is B.The answer to this question is B because Doppler ultrasound takes advantage of the Doppler Effect in which a sound wave emitted by, or reflected from a moving object, will change its frequency based on the relative speed of the object, in this case, the flow of blood in a vessel.

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.

The correct answer is C. This is a General Chemistry question that falls under the content category "Unique nature of water and its solutions." The answer to this question is C because, 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.

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.

The correct answer is D. This is a Biochemistry question that falls under the content category "Principles of chemical thermodynamics and kinetics." The answer to this question is D because 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 Voversus substrate concentration data to calculate KM and Vmax using Michaelis−Menten enzyme kinetics.

The effects of urea And beta mercaptoethanol on protein

Urea disrupts the hydrophobic effect and hydrogen bonding due to it's carbonyl oxygen and amine group. It would effect secondary and tertiary both, but obviously not effect covalent interactions like disulfide bridges. This is why beta mercaptoethanol and urea are both used in combination to denature proteins. ****unfolding will cause entropy to increase, not decrease, due to the disordering effect of unfolding

If light enters any substance with a higher refractive index (such as from air into glass) it .....(a)........ . The light bends ....(b).... the normal line. If light travels enters into a substance with a lower refractive index (such as from water into air) it ....(c)..... The light bends ......(d).... from the normal line.

a) slows down, (b)towards c)speeds up, (d) away