TPR 3
At STP, a certain liquid will spontaneously vaporize, even though the reaction is endothermic. The most likely explanation for this is that:
this reaction results in an increase in entropy. A. For a reaction to be spontaneous, the Gibbs free energy (ΔG = ΔH - TΔS) must be negative. If the spontaneous reaction is endothermic (i.e, if ΔH is positive), then ΔS must be positive.
The low chemical reactivity of nitrogen gas is most likely caused by the:
triple bond holding each nitrogen atom in the molecule. Since the two nitrogen atoms participate in a strong triple bond, a great deal of energy is required to dissociate them, resulting in their low reactivity.
The blood vessel that carries the most highly oxygenated blood in the fetus is the:
umbilical vein. The most highly oxygenated blood would be the blood returning from the point of oxygenation to the systemic circulation. In the adult this would be the pulmonary vein, but the point of oxygenation in the fetus is the placenta, not the lungs (D is correct and A is wrong). The ductus arteriosus does carry oxygenated blood but is found after the umbilical vein, so it does not carry the most highly oxygenated blood (B is wrong). The umbilical artery carries deoxygenated blood from the fetus to the placenta (C is wrong).
Which of the following processes do NOT occur at the placenta? 1- Diffusion of amino acids from fetal blood to maternal blood 2- Mixing of maternal and fetal blood and gas exchange 3- Exchange of CO2 and O2 between fetal and maternal blood
1&2 Fetal and maternal blood do not mix (item II does not occur), so the only way substances can cross the placental barrier is if they are lipid soluble or if they are actively transported. Amino acids are charged and cannot cross the placental and fetal capillary membranes (item I does not occur), but CO2 and O2 are lipid soluble and easily cross the barrier (item III does occur).
Which of the following systems would result in the highest total positive pressure against the piston? A. A = 0.5 moles NaCl, B = nothing B. A = 0.6 moles C6H12O6, B = nothing C. A = nothing, B = 0.5 moles NaCl D. A = nothing, B = 0.6 moles C6H12O6
A = nothing, B = 0.5 moles NaCl To produce osmotic pressure against the piston, the solute concentration on the right side of the osmometer must be higher than that on the left side, eliminating choices A and B. The addition of 0.5 mol of NaCl is the equivalent of 1.0 mol of solute (since NaCl is composed of two ions) and will yield a more concentrated solution than the addition of 0.6 mol of C6H12O6, so choice C results in greater osmotic pressure.
In Graves disease, thyroid-stimulating immunoglobulins (TSI) contain an active site that can mimic the active site of TSH to stimulate thyroid hormone production. One would expect TSI to cause: A. an increase in thyroid gland size because of the trophic effect of TSH. B. a decrease in thyroid gland size because of competitive inhibition with TSH. C. no change in thyroid gland size because antibodies only recognize foreign particles. D. an immune response against TSH molecules.
A. an increase in thyroid gland size because of the trophic effect of TSH. A. As described in the passage, goiter is caused by overstimulation of the thyroid gland by TSH. If TSI mimics TSH, then it would produce the same response: enlargement of the thyroid gland (A is correct). It is true that TSI would compete with TSH with binding, but TSI also stimulates thyroid production, indicating that it would have the same effect on thyroid size and TSH and increase, not decrease, the size of the thyroid gland (B is wrong). Antibodies can have virtually any antigen-recognition properties. It is true that B cells do not normally produce self-reactive antibodies, since these are eliminated, but they can be produced in a condition such as described (C is wrong). The antibodies do not recognize TSH, but the TSH binding site, so there is no reason to believe that they will induce an immune response against TSH (D is wrong).
Which of the following best represents O2 saturation vs. O2 concentration for fetal hemoglobin (HbF) and maternal (or adult) hemoglobin (HbA)?
A. Hemoglobin binds oxygen cooperatively, and so has a sigmoidal binding curve (C and D are wrong). Fetal hemoglobin must have a higher oxygen affinity than maternal hemoglobin to be able to bind oxygen at lower partial pressures than maternal hemoglobin (A is correct, and B is wrong).
Which of the following pairs of compounds would be considered tautomers?
A. Only Pair I represents tautomers (two molecules in structural equilibrium). Pair II represents two resonance structures of the same molecule, and Pair III represents two molecules that are identical.
Which one of the following graphs is characteristic of the activity of a radioisotope?
A. Since the rate of decay is proportional to amount of radioisotope present, the activity A decreases exponentially. (The equation is A = A0e-kt, where k is a constant that is inversely proportional to the half-life.) Graph A best illustrates the shape of an exponential decay curve.
What is the effect of adding a small amount of a strong acid to the equilibrium shown in Equation 1? CH3COOH(aq) <-> CH3COO-(aq) + H+(aq) A. The equilibrium shifts to the left. B. The pH shifts to 7. C. The pH decreases dramatically. D. The concentration of CH3COO- increases.
According to Le Châtelier's principle, adding a small amount of acid (H+) to this solution will shift the equilibrium to the left (choice A). This will result in a decrease in the concentration of acetate ion, making choice D incorrect. In addition, since the solution is buffered, it will resist changes in pH, making choices B and C incorrect.
Decreased secretion of aldosterone has all of the following effects EXCEPT: A. increased loss of Na+ in urine. B. decrease in extracellular fluid volume. C. increase in arterial pressure. D. increase in urine volume.
Aldosterone causes increased sodium reabsorption, and because of the rise in systemic Na+, there will be decreased water loss, increased extracellular fluid volume and increased blood pressure. Decreased secretion of aldosterone would cause the opposite: increased loss of sodium (A is true and eliminated), a decrease in the extracellular fluid volume (B is true and eliminated), and an increase in urine volume (D is true and eliminated). It will not cause an increase in blood pressure, however, since more urine is formed and extracellular fluid volume lost (C is false and thus the correct answer choice).
Compared to sensitive bacteria, phage-resistant bacteria are most likely to display which of the following characteristics? A. Increased sensitivity to antibiotics that disrupt the cell wall B. The presence of plasmid DNA absent in sensitive cells C. Alteration of the amino acid sequence of a specific protein in resistant cells Correct Answer D. Increased expression of lysosomal enzymes
Alteration of the amino acid sequence of a specific protein in resistant cells C. The results of the experiment are most consistent with the Mutation Hypothesis, since the initial number of phage and bacteria were held constant, but variable numbers of resistant bacteria were observed. If a mutation is responsible, then this will result in a protein with an altered amino acid sequence, or else the mutation would have no effect (C is correct). There is no relationship to antibiotic sensitivity implied (A is wrong). A plasmid is not likely to be responsible since the plasmid would have to spontaneously appear in resistant cells where none was present in sensitive cells (B is wrong). Bacteria do not express lysosomal enzymes since they would lyse themselves (D is wrong).
If samples of equal dimension of the metals in Table 1 were subjected to identical tensile stresses, which metal would undergo the greatest fractional change in length?
Aluminum (relationships in equation)
In oxidative phosphorylation, the production of ATP is linked to electron transport in which of the following ways?
An increase in pH in the mitochondrial matrix drives ATP production. A. Electron transport does not directly drive ATP production. Electron transport pumps protons out of the mitochondria, creating the proton gradient that drives ATP synthesis (A is correct). The pH outside of the inner membrane, including the intermembrane space, is decreased relative to the matrix of mitochondria (the H+ concentration is increased; B is wrong). There is no direct transfer of protons to ATP, only the energy of movement of protons down a gradient to drive ATP synthesis (C and D are wrong).
If the launch speed is kept constant, increasing a projectile's launch angle of elevation from 45° to 60° will have which of the following effects?
An increase in the time in flight, a decrease in the distance traveled, and no change in the final speed of the projectile The total flight time is equal to twice the time required for the projectile to reach its maximum height, which is the point at which its vertical velocity drops to zero. Since vy = v0y - gt, the time required for the projectile to reach its maximum height is given by t = v0y / g, so the total flight time is where θ0 is the launch angle. Now, because sin 60° > sin 45°, the flight time for the projectile launched at θ0 = 60° will be greater than for the one launched at θ0 = 45°. This eliminates choices C and D. To decide between choices A and B, note that choice A includes the statement, "an increase in the...final speed of the projectile." This cannot be true. Assuming no other forces besides gravity, the final speed of the projectile would equal the initial speed (conservation of mechanical energy). And, if dissipative forces were taken into account, then the speed of the projectile launched at θ0 = 60° would experience a greater decrease in its impact speed (compared to the projectile launched at θ0 = 45°) since it is in the air longer. The answer must be B.
Number 20 WTHECK? The reaction will primarily yield: A. 2-chloro-2,3,3-trimethylbutane. B. 1-chloro-2,2,3-trimethylbutane. Correct Answer C. 1,1-dichloro-2,3,3-trimethylbutane. D. 2,3-dicloro-2,3,3- trimethylbutane.
At a quick glance, answer choices C and D can be eliminated since dichlorination will not give the major product. In addition, it is impossible to even form compound D as it requires five bonds to carbon-3. While free radical brominations are selective reactions yielding primarily the tertiary bromides, the major products of chlorinations depend not only on the radical intermediate selectivity, but also on the number of different ways those radicals can be formed. The ratio of primary to secondary to tertiary radical selectivity in chlorination reactions is not great - only 1:4:5. By multiplying this selectivity factor by the number of equivalent hydrogens that will give rise to the radical intermediate, the product distribution can be predicted. In this case, removing any of the nine hydrogens on the tert-butyl group will yield the same primary radical (1 x 9 Hs = 9). Chlorinating that radical will give answer choice B. Substituting a chlorine for the tertiary hydrogen of the molecule would yield answer choice A. While the tertiary radical is roughly five times more likely to form than the primary radical mentioned above, there is only one way to form that radical (5 x 1 H = 5). Therefore the ratio of compounds B:A will be 9:5.
Two substances, I and II, of equal mass are heated at the same rate for the same period of time without undergoing a phase change. If the specific heat of Substance I is twice that of Substance II, then the change in temperature after heating will be: A. twice as large in I as in II. B. twice as large in II as in I. C. four times as large in II as in I. D. the same in I as in II.
B- TWICE AS LARGE IN 2 AS IN 1 B. The specific heat of a substance is the quantity of heat required to raise the temperature of one gram of the substance by one degree. Therefore, if Substance I has a specific heat which is twice that of Substance II, the temperature of Substance I will only be raised by half as much as Substance II after heating for an equivalent time.
Which of the following equilibria is an example of a buffered solution? A. HCl(aq) H+(aq) + Cl-(aq) B. H2CO3(aq) H+(aq) + HCO3-(aq) Correct Answer C. H2O(aq) H+(aq) + OH-(aq) D. CaCl2(aq) Ca2+(aq) + 2 Cl-(aq)
B. H2CO3(aq) H+(aq) + HCO3-(aq) B. The passage states that buffered solutions are commonly made with weak acids and their conjugate bases. The only weak acid given is carbonic acid (choice B). HCl (choice A) is a strong acid, water (choice C) is not acidic, and CaCl2 (choice D) is a salt.
When light moves from air to an unknown medium, its speed decreases to 5/6 of its speed in vacuum. What is the refractive index of this unknown medium?
B. By definition, the index of refraction of a medium is equal to the factor by which the speed of light decreases (from its speed in vacuum) as it passes through the medium. If the speed of light drops to 5/6 of its speed in vacuum, then it has decreased by a factor of 6/5 = 1.2.
Muscle fibers are composed of small contractile units called sarcomeres. During contraction, which of the following occurs within a sarcomere? I. Myosin filaments shorten. 2. Actin filaments shorten. 3. Overlap between actin and myosin filaments increases.
B. During contraction, neither myosin nor actin filaments get shorter. The overlap between them increases to make the sarcomere shorter (I and II are false, and III is true).
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 x 10-5) B. HClO (Ka = 3.5 x 10-8) C. HCN (Ka = 4.9 x 10-10) D. H3CNH3+ (Ka = 2.2 x 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 x 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.
A fiber-optic cable of length 10,000 km is constructed of glass with an index of refraction of 1.5. What is the approximate time a signal will take to pass from one end of the cable to the other? A. 2.2 × 10-2 sec B. 5.0 × 10-2 sec C. 2.2 × 10-1 sec D. 5.0 × 10-1 sec
B. The rate at which light will travel through the glass is v = c / n, where n is the refractive index of the glass. Using the fundamental equation d = vt, we therefore calculate . t=d/v= d/(c/n)= dn/c= 10^4x 10^3(1.5)/3x10^8 m/sec= 5 x10^-2
If фcrit for a glass-air interface is 42°, which of the following best approximates nglass? (Note: sin 42° = 0.67.)
B. Using nb = 1 for the index of refraction of air, Snell's Law gives sin фb = na sin фa. If na sin фa > 1, then there is no фb; that is, the beam experiences no refraction (total internal reflection). The cut-off, then, is determined by na sin фa = 1, that is, by nglasssin фcrit = 1, so nglass= 1/(sin)(angle critical)= 1/sin(42)= 1/0.67= 1/(2/3)= 3/2= 1.5
If the osmotic pressure of the system in Figure 1 is measured to be 80 kPa, approximately how much force must be applied to the piston to precisely counteract the osmotic pressure? (Note: Assume the head of the piston is square and fits exactly in the chamber.)
By definition, pressure is force per unit area: P = F / A; therefore, F = PA. Since the area of the square-head piston is A = (2 m)2 = 4 m2, the required force on the piston must be F = PA = (80 kPa)(4 m2) = 320 kN.
Reduction of acetic acid will initially produce which of the following? A. O2(g) B. CO2(g) C. CH3CHO Correct Answer D. CH3CH2OH
C. Reduction of acetic acid will initially produce the aldehyde given as choice C, where the carbonyl carbon has been reduced from an oxidation state of +3 to +2. Choice D is the result of further reduction to an oxidation state of +1, while choices A and B do not correspond to the reduction of acetic acid.
Which one of the following plots best illustrates the pH of an acidic buffer solution as it is diluted with H 2O?
D. As stated in the passage, the pH of a buffer solution depends on the ratio of the concentrations of acid to base, and not their absolute quantities. Therefore, the dilution of the buffer solution should not affect the pH, as shown in choice D.
A mutation in the thyroid hormone receptor which blocks hormone binding would cause which of the following? A. Atrophy of the thyroid gland B. Elevated secretion of TRH C. Decreased T3 levels in plasma D. Atrophy of the anterior pituitary
Elevated secretion of TRH As shown in Figure 1 and described in the passage, thyroid hormone causes feedback inhibition of TRH and TSH secretion. Thyroid hormone receptor is responsible for hormone signaling, including feedback inhibition by thyroid hormone. If thyroid hormone receptor is defective, this will prevent feedback inhibition, causing TRH and TSH levels to become elevated (B is correct). The absence of feedback inhibition and the high levels of TSH and TRH will cause increased thyroid size and increased thyroid hormone secretion (A and C are wrong). It is unlikely that the pituitary will atrophy, particularly with increased TRH secretion (D is wrong).
All of the following statements regarding enzymes are true EXCEPT that they: A. can be stereospecific in substrate recognition. B. alter reaction equilibria. C. are often regulated by feedback inhibition. D. reduce the activation energy of reactions.
Enzymes lower the activation energy of reactions to speed them up (D is true and eliminated), but they do not alter the potential energy of reactants and products or the reaction equilibrium, only the rate at which equilibrium is reached (B is false and the correct response). Enzymes do have great stereospecificity in the reactions they catalyze (A is true and eliminated) and are frequently regulated by feedback inhibition by products (C is true and eliminated).
Which of the following statements best explains why fiber-optic cables are used instead of copper wiring in real-time data transmission networks?
Fiber-optic cables transmit data at or near the speed of light, whereas copper wiring does not. B. None of choices A, C, or D is mentioned in the passage, but the passage does state that "electromagnetic radiation propagates faster in fiber-optic cables than electrical signals do in copper wiring," so choice B is best.
In a facultative anaerobe, which of the following processes occurs under both aerobic and anaerobic conditions?
Glycolysis Glycolysis occurs under both anaerobic and aerobic conditions A facultative anaerobe can survive through fermentation when oxygen is not available but will use oxidative respiration when oxygen is available. Glycolysis will occur under both aerobic conditions (in which case the pyruvate will go on to enter the Krebs cycle as acetyl-CoA) and under anaerobic conditions (fermentation reduces pyruvate to alcohol or lactate), so C is correct. Fermentation occurs only in anaerobic conditions (A is wrong), while the Krebs cycle and oxidative phosphorylation can occur only in aerobic conditions (B and D are wrong).
As electrons are passed from one protein complex to another, the electrons that ultimately reach O2 produce which of the following?
H2O C. The final electron acceptor is oxygen, which is reduced to form water (C is correct).
Which of the following represents the correct order of nucleophilicity of the halides in a polar protic solvent? F-, I-, Br-, Cl-
I- > Br- > Cl- > F- Going down a column in the periodic table, nucleophilicity increases because larger ions are less tightly solvated.
In the absence of oxygen, the oxidation of glucose produces which of the following final products? CO2 NADH ATP
II and III only- NADH and ATP C. Item I is false: CO2 is produced only by the complete oxidation of glucose through PDC and the Krebs cycle (choices A and D are wrong). Note that both of the remaining answer choices include Item III, thus Item III must be true and we can focus only on Item II. Item II is true: in the absence of oxygen, glycolysis can proceed anaerobically (fermentation) and NADH is produced during glycolysis (choice C is correct and choice B is wrong). Item III is in fact true: 2 net ATP are made during glycolysis.
Dinitrophenol allows protons to freely cross the inner mitochondrial membrane. Cells treated with dinitrophenol will display which of the following activities? Krebs cycle Electron transport ATP synthesis
Krebs, ETC, ATP Synthesis (all three) D. If protons freely flow across the inner membrane, the proton gradient will be destroyed. If protons flow across the membrane without passing down a gradient through the ATP synthase, then ATP production will halt. However, ATP production during glycolysis will be unhindered, so statement III is true. The Krebs cycle and electron transport will actually accelerate to attempt to rebuild the proton gradient (I and II are true).
A sample of barium sulfate (BaSO4) incompletely dissolves in water. Which of the following correctly expresses the solubility constant for this reaction? A. Ksp = [Ba2+][SO42-] B. Ksp = [Ba2+]2[SO42-]2 C. Ksp = [Ba2+][SO42-]/[BaSO4] D. Ksp = [Ba2+][SO42-]/[BaSO4][H2O]
Ksp = [Ba2+][SO42-] A. The solubility constant is the product of the concentrations of ions in solution, given by choice A. Choice B is incorrect since there is only one sulfate ion for each barium ion, and choices C and D are incorrect because the undissolved barium sulfate should not be included in the equation.
Does FADH2 result in more or in less ATP production when compared to NADH?
Less, because electrons from FADH2 drive the creation of a smaller proton gradient than the electrons from NADH. ATP production depends only on the number of protons that cross through the ATP synthase to re-enter the matrix. This in turn depends on the size of the gradient (i.e., the number of protons pumped across the inner membrane); a larger proton gradient will result in more ATP produced. Since the electrons from FADH2 enter the chain after the first complex (NADH-Q reductase), FADH2 is ultimately responsible for pumping fewer protons across the inner membrane than NADH. Less total protons pumped means less total ATP made (D is correct). The rate at which the electrons move down the chain towards oxygen does not affect the number of protons pumped; this is determined solely by the number of enzyme complexes (proton pumps) the electrons pass through (A is wrong). The number of protons available to combine with oxygen does not affect the proton gradient across the inner membrane. If anything, this would tend to decrease the gradient, which would reduce ATP production (B is wrong). Since protons are being pumped out of the matrix, the matrix pH is increasing, not decreasing (C is wrong).
Ballistic pendulum: Which of the following is (are) conserved in the collision between the bullet and the block?
Momentum When the bullet hits the block, its kinetic energy is converted to heat as it burrows into the wood, so neither velocity nor kinetic energy is conserved. Momentum, however, is conserved.
Which of the following is the most likely sequence of events in the infection of E. coli by phage T1? A. Replication of viral genome, production of viral DNA polymerase, translation of viral lysozyme, assembly of infectious virus B. Production of viral DNA polymerase, replication of viral genome, translation of viral lysozyme, assembly of infectious virus C. Translation of viral lysozyme, production of viral DNA polymerase, replication of viral genome, assembly of infectious virus D. Production of viral DNA polymerase, replication of viral genome, assembly of infectious virus, translation of viral lysozyme
Production of viral DNA polymerase, replication of viral genome, assembly of infectious virus, translation of viral lysozyme The production of viral DNA polymerase must occur prior to replication of the viral genome since the polymerase is needed to do this (A is wrong). The last step must be production of lysozyme since this will lyse the cell, bursting it open and stopping any further activities required to produce infectious virus (A, B, and C are wrong). The correct events are in D: First express the viral DNA polymerase, which can then replicate the genome. Infectious virus can then be assembled, and released by cell lysis induced by viral lysozyme.
In humans, the fusion of the plasma membranes of a sperm and an ovum is followed first by which of the following?
Release of second polar body from fertilized ovum A. Human ova do not complete the second meiotic cell division, including the formation of the second polar body, until after fertilization (A is correct). This process occurs immediately after fertilization and is a prerequisite for the first cell division of the zygote (C is wrong). Implantation happens a few days later, and gastrulation occurs at a stage many cell divisions after the first cell division (B and D are wrong).
The continual removal of the NH3 from the reaction chamber maximizes the reaction yield because:
Removing a product as it forms creates a stress on the reactants side of the equilibrium, causing the reaction to produce more ammonia. A reaction can be driven forward by the addition of starting materials or the removal of products to perturb the equilibrium. This is an example of Le Châtelier's principle.
1-Methylcyclohexanol reacts with HBr to form 1-bromo-1-methylcyclohexane. The mechanism for this reaction is likely to be an:
SN1 Reaction A. Since no double bonds are formed or broken, this must be a substitution reaction; this eliminates choices C and D. Since the hydroxyl in the starting material is on a tertiary carbon atom, the mechanism cannot occur by a bimolecular pathway, eliminating choice B. The answer must be choice A.
When a material that obeys Hooke's law is stretched, its change in length, x, is directly proportional to the strength of the force of elongation, F, according to the equation F = kx. Which of the following expressions is equal to the Hooke's law constant, k?
Solving Equation 1 for F yields F = Δl . Comparing this to F = kx, we see that if Δl corresponds to x (the "change in length"), then YA / l0 corresponds to k.
All of the following contribute to speciation EXCEPT: A. geographic isolation of populations. B. genetic diversity. C. natural selection. D. maintenance of Hardy-Weinberg equilibrium.
Speciation requires two populations to become reproductively isolated so that they can no longer interbreed. Geographic isolation is the easiest way to allow two populations to diverge and become reproductively isolated (A is true and eliminated). To become reproductively isolated, the populations must diverge genetically, evolving differently, and this would require genetic diversity in the population to begin (B is true and eliminated). Natural selection could drive the changes in isolated populations that cause them to evolve into two species (C is true and eliminated). If Hardy-Weinberg is maintained, then there can be no changes in the allele frequencies in the gene pool of a population and no evolution, which would not allow speciation (D is false and the correct answer choice here).
Ballistic pendulum experiment: Which of the following assumptions is made in this experiment? A. No energy is consumed by the firing mechanism of the gun. B. The bullet's height does not change while traveling from the gun to the block. C. The mass of the block is greater than the mass of the bullet. D. Some of the bullet's kinetic energy is lost prior to the collision with the block.
The bullet's height does not change while traveling from the gun to the block. Since the quantity being measured, the muzzle velocity v, is set after the gun has fired, the energy consumed by the firing mechanism (if any) is irrelevant. This eliminates choice A. And although C is probably true, the equations given in the passage do not rely on M being greater than m. As for choice D, this is not an assumption on which the procedure is based; it's a source of error that is "ignored by the experimentalists." (Note: If the word "Some" in choice D were replaced by "None," then D would be true.) The answer is B. If the bullet's height changed while traveling from the gun to the block, then the change in gravitational potential energy would alter the bullet's total energy when it struck the block, causing h to change, and, consequently, the calculated value for v. Nowhere in the equations given in the passage is there an allowance for gravitational potential energy.
If the spring in the gun is compressed a distance x from rest before the gun is fired, which of the following expressions will give the value of the gun's spring constant in terms of the measured and calculated values of the experiment?
The elastic potential energy of the compressed spring, (1/2)kx2, is converted to the kinetic energy, (1/2)mv2, of the bullet. Setting (1/2)kx2 equal to (1/2)mv2, we find that k = mv2 / x2.
Measurements of which of the following would be sufficient to calculate the usable work performed by a mechanical engine? A. The rate at which energy is consumed by the engine and the amount of time the engine was running B. The heat produced by the engine and the amount of time the engine was running C. The power output of the engine and the amount of time the engine was running Correct Answer D. The energy consumed by the engine and the amount of time the engine was running
The power output of the engine and the amount of time the engine was running C. Power is defined as work divided by time: P = W / t. Therefore, if the power output P and the time t are known, the work performed can be calculated by the equation W = Pt.
E. coli is a generally harmless bacterium, often used for research purposes. Some strains of E. coli live mutualistically inside human digestive tracts. Sometimes these bacteria can cause infection outside the intestinal tract, most notably in the urinary tract, the biliary tract, and the nervous system. In which of the following structures is E. coli most likely to cause infection? A. Trachea B. Bladder C. Small intestine D. Large intestine
The question states that E. coli can cause disease outside of the intestinal tract (choices C and D are wrong), in the urinary tract, the biliary tract, and the nervous system. The bladder (choice B) is the best answer since this is part of the urinary tract, mentioned as a potential site of infection.
Which of the following expressions gives the value of the muzzle velocity v in terms of the experimentally measured values? 1/2 V2 = gh mv = (m + M)
The second equation given in the passage implies that V2 = gh V = (2gh)^1/2 Substituting this into the first equation given in the passage, we find mv = (m + M) v = (ugh)^1/2 v = [(m+M)/m)] (2gh)^1/2
DNA from resistant bacteria in Dish 6 is extracted and placed on agar with phage-sensitive E. coli. After incubation it is determined that these E. coli are now insensitive to phage T1 infection. The most likely mechanism for their acquisition of resistance is:
Transformation B. The definition of transformation is the process in which naked DNA, not a virus, is taken into a cell and changes the genetic characteristics of that cell. That is the case in this experiment, with extracted DNA making cells resistant to the virus (B is correct). Transduction is mediated by a virus, conjugation involves direct transfer of DNA between bacteria, and sexual reproduction does not apply to bacteria (A, C, and D are wrong).
The vitelline layer in sea urchins is analogous to which of the following layers in the human ovum?
Zona Pellucida- both located just outside the plasma membrane B. The vitelline layer is analogous to the zona pellucida in human ova, since both are protective acellular layers located just outside the plasma membrane (B is correct). The corona radiata is a region of surrounding supportive cells (A is wrong), and the vitelline layer is not a membrane (C is wrong). The ovum, like all human cells, does not have a cell wall (D is wrong).
An ectopic TSH-secreting tumor is suspected in a hyperthyroidic (high thyroid-hormone concentration) patient with goiter. Administration of a high dose of thyroid hormone confirms this, due to the fact that: A. [TSH] in the plasma decreases dramatically. B. [TRH] in the pituitary circulation increases. C. [TRH] in the pituitary circulation remains constant. D. [TSH] in the systemic circulation remains high.
[TSH] in the systemic circulation remains high. In a normal individual, high levels of thyroid hormone would suppress the release of TRH by the hypothalamus and TSH by the anterior pituitary. Injection of even more thyroid hormone should suppress the release of TRH and TSH even further. If this does not occur, and TRH and/or TSH levels remain high, it is likely that something outside the normal axis is involved. The passage states that ectopic tumors do not respond to normal feedback inhibition pathways; so, if an ectopic TSH-secreting tumor were present, the injection of thyroid hormone would not reduce the level of TSH in the system (D is correct and A is wrong). Since the tumor only secretes TSH, TRH levels should respond normally to the additional thyroid hormone—that is, they should go down due to negative feedback (B and C are wrong).
Fetal circulation differs from adult circulation in many ways. One of the main differences is that in the former:
a pathway exists for blood to circumvent the lungs. The passage states that fetal hemoglobin has a higher affinity for oxygen (A is wrong) and that there is some blood flow through the pulmonary circuit (C is wrong). The ductus arteriosus does carry oxygenated blood in the fetus but closes shortly after birth so that this vessel is not present in adults (D is wrong). The purpose of the ductus arteriosus is to help shunt blood past the inactive fetal lungs (B is correct).
When carbon dioxide in a closed container is subjected to external pressures less than 650 atm, the deviation from ideality is primarily due to the fact that:
actual gas pressure is less than calculated gas pressure. At external pressures less than 650 atm, Figure 1 shows that the PV / (RT) ratio for CO2 is less than 1. Therefore, the actual gas pressure is less than the calculated gas pressure (which corresponds to a ratio of 1).
A decrease in the temperature of a gas sample in a closed container will always result in an increase in the:
attraction between gas molecules. Since the temperature of a gas is proportional to the average kinetic energy of its molecules, decreasing the temperature will result in a decrease in kinetic energy. This decrease in kinetic energy will result in an increased potential energy, or an increased attraction between the individual molecules.
During the exponential phase of bacterial growth, bacteria are reproducing by:
binary fission. During the exponential phase of bacterial growth, the population is increasing in size. The only choice which allows an increase in population size is D, binary fission. Sexual reproduction in bacteria is called conjugation and serves only to increase genetic diversity; it does not increase the size of the population (eliminating A and C). Transduction is also a means to create genetic diversity. It occurs when a lysogenic virus excises from the genome and takes a portion of the genome with it. Any new hosts infected by that virus will be given the new genes. This does not increase the size of the population (eliminating B).
For SN1 reactions, the best leaving groups are those that:
can stabilize the negative charge of the leaving group. Leaving groups must be stable as anions
In transposition of the great arteries, a congenital birth defect, the aorta of the newborn is connected to the right ventricle and the pulmonary artery is attached to the left ventricle. This would result after birth in circulation of: A. oxygenated blood through the systemic vasculature and deoxygenated blood through the pulmonary vasculature. B. deoxygenated blood through the systemic vasculature and oxygenated blood through the pulmonary vasculature. Correct Answer C. oxygenated blood through both the systemic and pulmonary vasculature. D. deoxygenated blood through both the systemic and pulmonary vasculature.
deoxygenated blood through the systemic vasculature and oxygenated blood through the pulmonary vasculature. B. In the normal circulation, the right ventricle pumps blood through the pulmonary artery to the lungs. Blood returns from the lungs to the left atrium and is pumped by the left ventricle through the aorta to the rest of the body. If blood from the left ventricle passes to the pulmonary artery instead of the aorta and is then returned to the left atrium, oxygenated blood will loop around through the pulmonary system without passing to the rest of the body. At the same time, the deoxygenated blood from the systemic circulation will pass to the right atrium, then back out to the systemic circulation through the aorta rather than the pulmonary artery, causing deoxygenated blood to loop through this system (B is correct, and A is wrong). C and D are wrong since they have all blood either oxygenated or deoxygenated.
T3 and T4 most likely exert their effect on target cells by: A. binding to a transmembrane protein and activating a second messenger system. B. diffusing across the cell membrane and binding to a nuclear receptor. C. entering the cell via facilitated transport and binding to a nuclear receptor. D. entering the cell via iodine channels and binding to a cytosolic receptor.
diffusing across the cell membrane and binding to a nuclear receptor. The passage states that T3 and T4 are lipid soluble; in fact, although they are peptide hormones, T3 and T4 have the same mechanism of action as the steroid hormone, diffusing through the plasma membrane to bind to an intracellular receptor that goes into the nucleus to regulate transcription (B is correct and A is wrong). They do not require a protein to cross the plasma membrane, so facilitated transport is not involved (C and D are wrong).
Networks built of copper wiring experience signal degradation. This occurs because copper wires:
dissipate some of the electrical energy that flows through them as heat.
The negligible percent yield of ammonia at low pressure and high temperature suggests that the decomposition of ammonia gas is an: N2(g) + 3 H2(g) 2 NH3(g) ΔH°rxn = -91.8 kJ/mol
endothermic and spontaneous reaction. The ΔH of the formation of ammonia is given as -91.8 kJ/mol. Therefore, the decomposition of ammonia must have a ΔH of +91.8 kJ/mol, and must be endothermic. Since this decomposition limits the yield of ammonia, it must be a spontaneous process, competing with the formation of ammonia. Note that the decomposition of ammonia produces 4 molecules (1 nitrogen gas molecule and 3 hydrogen gas molecules) for every two molecules of ammonia, so the change in entropy (ΔS) of the decomposition is positive. Therefore, at high temperatures, the reaction will be entropy driven, and the Gibbs free energy (ΔG = ΔH - TΔS) will be negative due to the TΔS term.
In one case study, the acrosomal process does not extend following fusion of the acrosomal vesicle with the sperm plasma membrane. This could best be explained by the:
failure of intracellular H+ concentration to decrease. B. Following fusion of the acrosomal vesicle, calcium and sodium enter the sperm while potassium and protons exit the sperm. The decrease in H+ concentration allows actin to polymerize and the acrosomal process to extend. Failure of the H+ concentration to decrease would prevent the acrosomal process from extending (B is correct). Potassium leaves the cell during the activation process, rather than being absorbed (A is wrong). Sulfated polysaccharides are always present in the egg jelly and are not formed during the activation process (C is wrong). The influx of calcium is a normal part of the activation process and would stimulate, not block, formation of the acrosomal process (D is wrong).
A semipermeable membrane separates an aqueous solution of 0.003 M NaCl from an aqueous solution of 0.0025 M BaF2. With this apparatus, one would observe that water crosses the membrane:
from the NaCl solution to dilute the BaF2 solution. Osmotic pressure is a colligative property, which means it depends only on the concentration of solute particles, and not their identity. Therefore, a 0.003 M solution of NaCl will have an effective molality of 0.006 M (since NaCl is composed of two ions) and a 0.0025 M solution of BaF2 will have an effective molality of 0.0075 M (since BaF2 is composed of three ions). Osmotic pressure will drive movement of water from the lower concentration solution (NaCl) to the higher concentration solution (BaF2). Note that this movement of water acts to dilute the BaF2 solution.
In order for the mechanism outlined in Figure 1 to produce fresh water, the pressure applied by the piston must be:
greater than the osmotic pressure. Osmotic pressure will promote the movement of water from areas of low solute concentration (fresh water) to high solute concentration (salt water). For the purification system in Figure 1 to produce fresh water, the pressure on the piston must exceed the opposing osmotic pressure.
Goiter development resulting from iodine deficiency is most likely due to: A. accumulation of T3 and T4 in the thyroid gland. B. high levels of TSH. C. the build-up of lipids in thyroid hormone-producing cells. D. low levels of TRH.
high levels of TSH. B. Iodine is required for thyroid hormone production, so in the absence of sufficient iodine, thyroid hormone levels will fall, and feedback inhibition will not occur. TSH and TRH levels will increase, causing goiter (B is correct, and D is wrong). In the absence of iodine, T3 and T4 cannot be made, so they will not accumulate (A is wrong). There is no reason to believe that lipids build up in these cells (C is wrong).
When pathogens invade the tissues, they trigger an inflammatory response which causes blood vessels to dilate. This mechanism process is vital because it: A. increases the flow of immune cells into the tissue. B. causes blood vessels to rupture. C. isolates the affected tissue from the body's defenses. D. allows histamines to be released.
increases the flow of immune cells into the tissue. To improve the immune response and healing, the flow of blood into an infected tissue is increased. Along with the increased blood flow, more white blood cells enter the tissue and can help to remove pathogens (A is correct). Rupture of blood vessels would not improve healing but would make it worse (B is wrong). C is wrong: This would not isolate the tissue, but would increase its circulation. Although histamines do cause vasodilation and increased blood flow, increased blood flow is not required for histamines to be released (D is wrong).
Decreasing the size of a container of a real gas would always result in:
increasing the gas pressure on the container walls. Since the number of moles of the gas, n, is constant, decreasing the volume of the container will always result in an increase in the gas pressure on the container walls.
The current at parallel resistors
is split. .so if total current is 2 amps.. the current through each parallel resistor is 1 amp
Soon after birth, pressure will increase significantly in the:
left ventricle C. At birth, the resistance through the pulmonary circulation decreases dramatically as the lungs inflate, and resistance increases through the systemic circulation as a result of the loss of blood flow through the placenta. The increase in systemic resistance means that the pressure in the left ventricle increases (C is correct). Meanwhile, the pressures in the pulmonary artery, the right atria, and the right ventricle decrease as a result of inflation of the lungs (A, B, and D are wrong).
Increased intermolecular attractions cause the ratio PV / (RT) to decline because individual molecules:
lose kinetic energy to potential energy and strike the side of the container with less force Intermolecular attractions increase the potential energy between molecules, decreasing their kinetic energy. Note that the molecules do not undergo any reactions, eliminating choices A and D. Choice C is incorrect because intermolecular attractions do not result in electron repulsion
A scientist wants to separate a mixture of amino acids in a solution buffered to pH 6.0 by using electrophoresis. The mixture is composed of lysine, glycine, and aspartic acid, which have isoelectric points of 9.7, 6.0, and 3.0, respectively. The expected order of the amino acids I, II, and III is:
lysine, glycine, aspartic acid. A. First, it is clear that glycine will be amino acid II on the electrophoresis gel since its isoelectric point matches the pH of the solution. This eliminates choices B and D. Since opposite charges attract each other, the positively-charged amino acid at pH 6.0 will be amino acid I since it will be attracted to the minus charge. Lysine, with an isolelectric point of 9.7, is the positively-charged amino acid at pH 6.0, so the answer is A.
In order to maintain the maximum yield of ammonia when raising the temperature of the reaction, scientists should:
maintain N2 and H2 at high pressure. In order to maximize the yield of the reaction while increasing temperature, the pressure should be kept as high as possible. Maintaining the compression of the gases (choice A) will maintain a high pressure. However, increasing the volume (choices B or C) would decrease the pressure and lower the overall yield, and lowering the temperature of the reaction (choice D) would also lower the yield.
What is the characteristic valence-shell electron configuration for atoms in the silicon family?
ns2 np2 As indicated on the periodic table, the atoms from Group IVA (column 14) contain four electrons in their valence shell, with two in the s shell and two in the p shell. The valence shell of carbon is 2s22p2, while that of silicon is 3s23p2, both of which are specific examples of a general ns2np2 valence shell.
What is the pH of a solution containing equal concentrations of acetic acid and acetate ion?
pH= pkA= -logKa= -log(1.8 x 10^-5)= 5- log 1.8= 4.7
As electrons flow within the complexes of the electron transport chain, each intermediate carrier molecule is:
reduced by the preceding molecule and oxidized by the following molecule. B. The high-energy electrons flow downhill (energetically) from reduced carriers to more oxidized carriers. In each redox reaction one partner is reduced and another oxidized (C and D are wrong). The intermediate molecules are first reduced by the electrons, but then oxidized back to their original state by the next member of the chain down the line (B is correct, and A is wrong).
The events that transpire during the fusion of the acrosomal vesicle with the sperm plasma membrane are most similar to those that occur in the:
release of neurotransmitters from synaptic terminals. B. The acrosomal vesicle is a membrane-bound vesicle that fuses with another membrane, the plasma membrane, to release its contents. A similar process occurs to release neurotransmitter at synapses, fusing two membranes to release the contents of a vesicle into the exterior of the cell (B is correct). Phagocytosis is an uptake mechanism, not a release mechanism (A is wrong). Absorption of nutrients is mediated by proteins in membranes, not by membrane fusion (C is wrong), and steroids passively diffuse through membranes (D is wrong).
In the absence of any other forces, the semipermeable membrane described above allows:
solvent to move from areas of low solute concentration to areas of high solute concentration.
The product(s) of the bromination reaction is (are):
trans-1,2-dibromocyclohexane. A. As the passage explains, only trans-1,2-dibromocyclohexane will be formed.
With battery voltage and resistance given, how to find current?
I= V/R
In which of the following are the acrosomal enzymes synthesized? (protein synthesis NEVER happens in nucleus)
In which of the following are the acrosomal enzymes synthesized?
Emphysema often results in the destruction of pulmonary capillaries and an increase in the resistance of the pulmonary vascular system. Which of the following would most be likely to accompany this condition?
Increased pressure in the pulmonary artery B. If pulmonary resistance increases, then the pressure in the pulmonary artery, the right ventricle, and the right atrium will increase as a consequence (B is correct). The left atrium receives blood from the pulmonary vein, which will not have increased pressure (A is wrong). The resistance in the pulmonary circulation would reduce, not increase blood flow through the pulmonary vein (C is wrong). The loss of capillaries will reduce the volume of blood in the lungs (D is wrong).
The acidity of hydrogen halides increases as one moves down the periodic table. Which of the following properties of halides is most useful in explaining this trend?
Ionic radius B. Moving down the periodic table, the ionic radii of the atoms increase. As the ionic radius increases, the stability of the corresponding anion also increases (since the negative charge is distributed over a larger volume). Therefore, since larger halides are more stable as anions, larger hydrogen halides are more likely to undergo deprotonation, and are stronger acids.
Acetaldehyde shows a strong absorption band when analyzed by IR spectroscopy because:
the carbonyl group has a dipole moment. C. The strong dipole moment of the C=O double bond is what gives acetaldehyde its strong IR absorbance.
The human ear can comfortably respond to a range of sound intensity levels from 0 dB to 120 dB. If the decibel level, b, is related to the intensity, I, by the formula b = 10 log (I / I0), then the human ear can comfortably respond to an intensity range that varies by a factor of:
1.0 × 10^12 For each increase by 10 in the decibel level, the intensity increases by a factor of 10. The range from 0 to 120 dB corresponds to twelve such increases in the decibel level, and, correspondingly, to twelve factor-10 increases. Thus, the range of intensities varies by a factor of 1012.
Basically when there is no total internal refraction, critical angle is 1, so n glass is
1/(sin)(critical angle)
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?
4.2 kg 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 = 1.2ρH2O = 1200 kg/m3 and Vsub = 5 × 10-3 m3, we find that m = ρfluidVsub = (1200 kg/m3)(5 × 10-3 m3) = 6 kg
How to find resistance of two resistors in parallel
R12= R1R2/(R1+R2)
How to find total resistance of resistors in series
R= R1 + R2
The friction present in the ratchet system will have what effect on the experimental results?
The calculated muzzle velocity will be less than the actual muzzle velocity If the ratchet system were frictionless, the block would be able to swing a little farther upwards, giving a greater value of h, and consequently, a greater value for the muzzle velocity, v. Therefore, because friction is present, h will not be as great as it should be, giving a calculated muzzle velocity v that is less than the actual value.
Which of the following statements is true concerning the intermediate in the reaction between 2-bromo-2-methylpropane and water?
The three remaining bonds attached to the carbon are in a planar arrangement. Since this reaction is SN1, the carbocation intermediate is sp2. Choice B is best.
Total internal reflection occurs when
angle of incidence is greater than critical angle
Based on Figure 2, the critical angle between Media a and b: A. is less than 45°. B. is equal to 45°. C. is greater than 45°. D. cannot be determined.
is less than 45°. Figure 2 shows that if the angle of incidence is 45°, then the beam is totally internally reflected. Since total internal reflection occurs when the angle of incidence is greater than the critical angle, 45° must be greater than the critical angle. (Or, said the other way, the critical angle must be less than 45°.)