03/16 Practice Questions

endothermic reaction Dissociation 2x Endothermic 2x

In an ______________________, heat is absorbed by the system from the surroundings as the reaction proceeds. As a result, an endothermic reaction has a positive change in enthalpy ΔH°, which indicates that the products have higher enthalpy than the reactants because ΔH° is evaluated as: ∆H° = H°products - H°reactants However, knowing the relative enthalpy of the reactants and products does not directly give information about the change in entropy ΔS° of the reaction. An endothermic reaction can result in either an increase or a decrease in entropy depending on the change in the extent of disorder within the system. _______________________ reactions are typically endothermic processes because energy must be added to the system (as heat) to dissociate (break) the bonds in the reactants. Furthermore, _____________________________ reactions tend to increase entropy because a larger molecule is broken into two (or more) smaller molecules, which generally increases the disorder of the system. __________________________ reactions have a positive ΔH° and absorb heat from the surroundings as the reaction proceeds because the products have higher energy than the reactants. An ___________________________ reaction can result in either an increase or a decrease in entropy, depending on the change in the extent of disorder within the system.

The genetic code is considered "degenerate" because more than one codon can code for the same amino acid. This occurs because the third position of the mRNA codon and tRNA anticodon can undergo nontraditional base pairing, allowing a single tRNA molecule to bind different codons.

Question 1 The genetic code is said to be degenerate because there are 64 different codons, but translation produces only 20 unique amino acids. The degeneracy of the genetic code is due to which mechanism? A. Exclusion of protein coding regions from mature mRNA B. Errors during tRNA charging C. Ambiguity of the tRNA for the amino acid D. Nontraditional base pairing of the anticodon with the third base of the codon

Distillation is a purification technique that allows for separation of liquid mixtures based on the boiling points of the mixture's components. The three main types of distillation are simple (bp <150°C and >25°C apart), fractional (bp <25°C apart), and vacuum (bp >150°C).

Question 1 A student needs to separate a mixture of chloroform (bp 61°C) and benzene (bp 80°C). What type(s) of distillation would be expected to give the best separation of the two compounds? Simple distillation Fractional distillation Vacuum distillation A. I only B. II only C. I and III only D. II and III only

The compounds presented in this question each consist of a five-carbon chain and differ in the number of polar bonds and nonpolar methyl substituents. Among the four structures, the compound with the greatest number of nonpolar methyl (-CH3) groups and the fewest polar bonds (Choice D) will interact with polar water molecules the least and have the most hydrophobic character. Molecules with many polar bonds that promote dipolar interactions with water are hydrophilic, whereas those with mostly nonpolar bonds that lack attractive dipolar interactions with water are hydrophobic. The overall character of a compound depends on the relative number of each type of bond present.

Question 10 Which of the following molecular structures has the most hydrophobic character?

A van 't Hoff plot trendline in which the y-axis (ln Ka) values decrease as the x-axis (1/T) values increase will have a negative slope. During the dissociation of acetic acid, [H3O+] and [CH3COO−] increase as the temperature increases. As a result, the equilibrium constant also increases. Consequently, the van 't Hoff plot for the dissociation of acetic acid (Figure 1) has a negative slope (−m), which indicates that ΔH° is positive: −m=−ΔH°R mR=+ΔH° Therefore, the dissociation of acetic acid is endothermic.

Question 11 Based on the van 't Hoff plot shown in Figure 1, the dissociation of acetic acid is: A. endothermic, because the y-intercept of the trend line is positive, and therefore ΔS° is positive. B. endothermic, because the slope of the trend line is negative, and therefore ΔH° is positive. C. exothermic, because the slope of the trend line is positive, and therefore ΔH° is negative. D. exothermic, because the y-intercept of the trend line is negative, and therefore ΔS° is negative.

Based on the van 't Hoff plot provided in Figure 1 of the passage, ΔH° and ΔS° are both positive; therefore, the reaction will be nonspontaneous at low temperatures (Number II) but spontaneous at high temperatures (Number III). (Number I) The reaction is not spontaneous at all possible temperatures because it is endothermic at low temperatures. Educational objective:A spontaneous process does not require a continuous input of additional energy. The change in Gibbs free energy (ΔG°) is evaluated as ΔG° = ΔH° − TΔS° and is used to identify whether a process is spontaneous (ΔG° ˂ 0), nonspontaneous (ΔG° > 0), or at equilibrium (ΔG° = 0). Processes in which both changes in entropy ΔS° and enthalpy ΔH° are positive are spontaneous at high temperatures and nonspontaneous at low temperatures.

Question 13 Based on the experimental results in the passage, the dissociation of acetic acid at temperatures outside of the 281 K to 287 K range will be: spontaneous at all possible temperatures. nonspontaneous at only low temperatures. spontaneous at only high temperatures. A. I only B. II only C. III only D. II and III only

Therefore, ΔG° decreases as the pH of the solution decreases with increasing temperature. (Choice B) In Table 1, the pH decreases and Ka increases as the temperature increases. Because ΔG° = −RTlnKa, ΔG° decreases with decreasing pH and increasing temperature. (Choice C) Given that ΔG° = −RTlnKa, ΔG° cannot stay the same because both the temperature and the equilibrium constant are changing in this experiment. (Choice D) Based on the data provided in Table 1, ΔG° cannot equal zero over the experimental temperature range. Educational objective:The change in Gibbs free energy ΔG° for a reaction at a given temperature is related to the equilibrium constant Keq by the relationship ΔG° = −RTlnKeq.

Question 14 Based on the data presented in Table 1, as the pH of the acetic acid solution decreases when going from a temperature of 281 K to 287 K, ΔG° for the dissociation of acetic acid: A. decreases. B. increases. C. is constant. D. is zero.

The van 't Hoff plot for the experiment described in the passage is used to find the thermodynamic values for the dissociation of acetic acid. The results are based on the final state of the reactants and products at a given temperature, not the reaction mechanism pathway. Increasing the rate of the reaction by adding a catalyst will not affect the amount of product (H+) that is produced. Therefore, a catalyst will not change the pH of the solution or the thermodynamic parameters measured in the experiment. Thermodynamic values indicate if a reaction is spontaneous or not, but they do not indicate the rate of a reaction because thermodynamic values are state functions that are independent of the reaction pathway.

Question 15 Would changes in the van 't Hoff plot be observed if the reaction rate were increased by adding a catalyst during the experiment? A. No, because the initial concentration of acetic acid does not change. B. No, because ΔG°, ΔH°, and ΔS° are independent of the reaction rate. C. Yes, because the reaction rate and ΔG° depend on temperature. D. Yes, because spontaneity depends on the reaction rate.

Based on the results from the trendline in Figure 1, the dissociation of acetic acid yields a positive ΔH° and ΔS°. A positive ΔH° indicates that H°products is greater than H°reactants, and a positive ΔS° indicates that the molecular disorder of the system increases. (Choice A) Although entropy does increase, the enthalpy of the products is not lower than the enthalpy of the reactants. This would result in a negative ΔH°, which would occur only for an exothermic reaction. (Choices C and D) Entropy increases in this reaction because the dissociation reaction breaks up a larger molecule into two smaller, charged molecules. The increasing entropy is verified graphically on the van 't Hoff plot as a positive y-intercept.

Question 16 During the dissociation of acetic acid at the temperatures tested in Table 1, the molecular disorder within the reaction vessel: A. increases, and the enthalpy of the reactants is greater than the enthalpy of the products. B. increases, and the enthalpy of the reactants is less than the enthalpy of the products. C. decreases, and the enthalpy of the reactants is greater than the enthalpy of the products. D. decreases, and the enthalpy of the reactants is less than the enthalpy of the products.

The question states that an aqueous solution containing a mixture of HNO2 and NO2− (ie, a buffer) has a pH of 6.2. Therefore, the concentration of HNO2 must be less than NO2− (ie, [HNO2] < [NO2−]) because the pH of the solution is greater than the pKa (ie, 6.2 > 4.4). (Choices A, C, and D) The pH of the solution is greater than the pKa of HNO2; consequently, less HNO2 is present in the solution compared to NO2−. Educational objective: The pKa is a measure of the acidity of a particular proton in a molecule. In a solution containing a weak acid (HA) and its conjugate base (A−), when the pH is greater than the pKa, the amount HA is less than the amount of A− and vice versa.

Question 17 Assume HNO2 dissociates in water with a Ka of 4 × 10−5 at 298 K. HNO2(aq) ⇄ NO2−(aq) + H3O+(aq) An aqueous solution containing a mixture of HNO2 and NO2− has a pH of 6.2 at 298 K. Compared to the concentration of NO2−, the concentration of HNO2 must be: A. greater, because the pH of the solution is greater than its pKa. B. less, because the pH of the solution is greater than its pKa. C. greater, because the pH of the solution is less than its pKa. D. less, because the pH of the solution is less than its pKa.

The temperature dependence of Keq depends on whether a reaction is exothermic (ΔH° < 0) or endothermic (ΔH° > 0). An exothermic reaction releases heat (ie, heat is a product), and increasing the temperature (ie, adding heat) causes the reaction to shift toward reactants (ie, Kc decreases) to compensate for the heat gain, in accordance with Le Châtelier's principle. Alternatively, an endothermic reaction absorbs heat (ie, heat is a reactant), and the reaction will shift toward products (ie, Kc increases) with increasing temperature. Therefore, a graph of Keq versus temperature will show Keq decreasing with increasing temperature for an exothermic reaction and Keq increasing with increasing temperature for an endothermic reaction, as depicted in Choice B.

Question 18 Which graph best represents the relationship between temperature and the equilibrium constant Keq of an exothermic and endothermic reaction?

The volumetric flow rate Q of blood flowing through a vessel equals the product of the cross-sectional area A of the blood vessel and the velocity v of the blood: Q=Av Furthermore, A is equal to the product of π and the square of the vessel's radius r. Substituting this relationship into the equation above yields: Q=πr^2v In this question, Q is constant over all the vascular regions in the kidney vasculature model. The equation above can be rearranged to solve for v: v=Qπr^2 Hence, v is inversely proportional to r^2. The value of v for the original vessel r is denoted v0 and equals: v0=Qπr^2 When r increases by 100% (ie, is doubled), the new value of v, denoted v1, equals: v1 = Q/π2r^2 = Q/4πr^2 v1 = 0.25Q/ πr^2 = 0.25 v0 Therefore, increasing r by 100% causes v to decrease by 75% (ie, decrease to one fourth its original value), in agreement with the continuity equation.

Question 19 Researchers determined that the blood velocity in the afferent arterioles of their model was four times larger than physiological measurements. Which of the following changes to their model could correct this discrepancy? (Note: Assume ideal fluid flow.) A.Increase the radius of the afferent arterioles by 100% B.Increase the radius of the afferent arterioles by 75% C.Decrease the radius of the afferent arterioles by 75% D.Decrease the radius of the afferent arterioles by 100% The volumetric flow rate of blood is constant in all the vascular regions of the kidney. The blood velocity equals the ratio of the volumetric flow rate and the blood vessel cross-sectional area.

Both isoforms shown in the table have pI values below physiological pH (7.4), so at this pH, both are negatively chargedand will bind to an anion-exchange column. When NaCl is added, the chloride anions compete with the negatively charged proteins for binding. Because Isoform 1 has a lower pI than Isoform 2, it is more negatively charged and binds the column more tightly. Therefore, Isoform 2 will elute at a lower salt concentration than Isoform 1. Gradually increasing the salt concentration in the buffer allows each isoform to be collected separately.

Question 2 The physical properties of two isoforms of a protein, both of which bind the same ligand, are shown in the table: Which protocol would yield the best separation of the two isoforms? A. Run the mixture through a cation-exchange column starting at pH 10, then gradually increase the pH of the buffer. B. Pass the mixture through a size-exclusion column with a pore size cutoff of 40 kDa. C. Load the mixture onto an anion-exchange column at physiological pH, then gradually increase the NaCl concentration of the buffer. D. Load the mixture onto an affinity column with the ligand covalently bound to the beads, followed by cleaving the covalent bond to elute. Cation-exchange columns have negatively charged beads that bind positively charged molecules and repel negatively charged ones. At pH 10, both isoforms are negatively charged and would be repelled by the beads, exiting the column together. Both isoforms are approximately the same size (~68 kDa), so they will elute from the size-exclusion column together. Both isoforms bind the same ligand with nearly identical affinity (~3 × 10−6 M), so both would bind to the affinity column. This technique would separate the two isoforms from other proteins but not from each other.

(Choice C) Gases assume the size and shape of their container and therefore have the same volume as the container. This is different from size of the molecules themselves, which is considered to be negligible. (Choice D) The volume occupied by two gases is equal if they share the same pressure, temperature, and number of moles. However, there are more moles of N2 than moles of O2 in a 1.5-g sample. Educational objective:Avogadro's law states that the volume occupied by a gas is directly proportional to its number of moles if the temperature and pressure are constant. When two gases at the same temperature and pressure are compared, the gas with more moles occupies the larger volume.

Question 2 Which gas would occupy more volume at a constant temperature and pressure, 1.5 g of N2 gas or 1.5 g of O2 gas? A. N2 because there are more moles of N2. B. O2 because its molecules are larger. C. Neither gas occupies any volume. D. Both gases occupy equal volumes. Because the masses of both N2 and O2 are equal (1.5 g each), the gas with the lower molar mass has more molecules. Using the periodic table, the molar mass of N2 is calculated to be 28 g/mol, and the molar mass of O2 is 32 g/mol. Therefore, the N2 gas has more moles and occupies a larger volume. (Choice B) The size of a gas' molecules is negligible compared to the volume the gas occupies. In addition, oxygen atoms have a smaller atomic radius than nitrogen atoms, which gives O2 molecules a smaller size than N2 molecules. However, O2 molecules do have more mass in the nuclei than N2 molecules, which decreases the number of moles of O2 in a 1.5-g sample.

Therefore, the renal artery ablation system raises the tissue temperature by 10 °C.

Question 20 What is the temperature change of the renal artery from the RF ablation system described in the passage? (Note: The specific heat of renal artery tissue is 3.3 J/(g∙°C).) "Renal artery ablation destroys the malfunctioning nerves and normalizes blood pressure. Ablation transmits energy, such as radio frequency (RF) or laser energy, to heat the tissue. Researchers tested an RF ablation device and determined it delivered 100 J of thermal energy into 3 g of tissue." A. 1 °C B. 3 °C C. 10 °C D. 30 °C

For each heartbeat, the pulse wave travels through the vasculature, reaching different arteries at different times depending on their distance from the heart. The frequency f of the pulse wave is equal to the heart rate. The wave velocity vdepends on vascular parameters, such as the vessel stiffness. The wavelength λ is the distance between the peaks of the pulse wave. The value of v equals the product of f and λ: v=fλ In this question, ablation of the renal arteries causes an increase in the travel time for the pulse wave to reach the kidney. Therefore, the pulse wave v decreases after renal artery ablation. Because f depends only on the heart rate, it does not change before and after ablation. Furthermore, according to the equation above, λ is directly proportional to v. Therefore, λ must decrease when v is decreased.

Question 21 A clinical study reports that the time required for the pulse wave to arrive at the kidney increases after renal artery ablation. A possible explanation for this could be that: "Renal artery ablation destroys the malfunctioning nerves and normalizes blood pressure. Ablation transmits energy, such as radio frequency (RF) or laser energy, to heat the tissue." A. decreasing blood pressure causes the wavelength of the pulse wave to decrease. B. decreasing blood pressure causes the wavelength of the pulse wave to increase. C. decreasing blood pressure causes the frequency of the pulse wave to decrease. D. decreasing blood pressure causes the frequency of the pulse wave to increase. The pulse wave velocity is equal to the product of the frequency and the wavelength. When the frequency is constant, changes in the pulse wave velocity cause a proportional change in the wavelength.

Pressure refers to a force exerted over an area, such as the atmospheric pressure Patm caused by Earth's atmosphere pushing in on any object. Patm can be measured from the height of a column of mercury in a barometer. The Patm of the air pushing downward equals the hydrostatic pressure Ph of the mercury column, which equals the product of the fluid density ρf, gravitational acceleration g, and the fluid height h: Ph=ρfgh= Patm The equation above can be solved for h, yielding: h=Ph/ρfg In this question, Ph in the mercury barometer equals the maximum blood vessel pressure PB, which is given as 7,000 Pa. From the equation above, h in meters of mercury is given by: h= 7,000 Nm^2 / (14,000kgm^3 x 10ms^2) h= (7 / 140) x (kg·m) / (s^2 ·m^2) x (m^3 kg) x (s^2/ m ) = 0.05m of Hg Converting this value to mmHg yields: h=0.05m of Hg1,000mmHg1m of Hg=50mmHg

Question 22 If the maximum pressure in a blood vessel is 7,000 Pa, what is the pressure in mmHg? (Note: The density of mercury is 14,000 kg/m3, and gravitational acceleration is 10 m/s2.) A. 0.05 mmHg B. 2 mmHg C. 20 mmHg D. 50 mmHg

Blood circulates through different vascular regions as it perfuses the bodily organs. Blood is pumped out of the heart and typically traverses the following vascular regions: arteries, arterioles, capillaries, and finally veins. Each vascular region has a characteristic vascular resistance that contributes to the overall drop in blood pressure across the circulatory system. In this question, the mean blood pressure in different vascular regions of the kidney is provided in Figure 1. However, the kidney has an unusual vascular architecture with two capillary networks. In an organ with only one capillary network, the arterioles must reduce the blood pressure at the end of the arteries, approximately 90 mmHg, to the value at the beginning of the second capillary network, approximately 20 mmHg. Therefore, the pressure drop across the arterioles is 70 mmHg.

Question 23 Which of the following is closest to the expected drop in pressure in the arterioles for an organ with only one capillary network? A. 100 mmHg B. 70 mmHg C. 40 mmHg D. 20 mmHg

Therefore, n1 equals the product of n2 and a value greater than 1, forcing n1 to be greater than n2: n1>n2 Similarly, n1 equals the product of n3 and a value less than 1, requiring n1 to be less than n3: n1<n3 (Choice A) If n1 were less than n2, θ2 would be less than θ1, but θ2 is 55° and θ1 is 45°. (Choice B) If n1 were greater than n3, θ3 would be greater than θ1, but θ3 is 35° and θ1 is 45°. (Choice C) The ratio of the refractive indices n1n2 is related to the ratio of the angles, θ2θ1. Therefore, n1 is greater than n2and n1 is less than n3, because θ1 is less than θ2 and θ1 is greater than θ3. Educational objective: Snell's law governs the refraction of light when crossing a boundary of two media with different refractive indices. Light traveling from a medium with a higher refractive index into one with a lower refractive index will have a refracted angle that is greater than the incident

Question 24 The refractive indices of the glass optical fiber, polymer lens, and liquid crystal lens described in the passage are denoted as n1, n2, and n3, respectively. The refractive indices of the glass optical fiber and the two focusing lenses are related by which of the following? "The incident angle of the laser in the glass fiber was held constant at 45°. The refracted angles for a polymer lens and a liquid crystal lens were 55° and 35°, respectively." A. n1 < n2 and n1 < n3 B. n1 > n2 and n1 > n3 C. n1 < n2 and n1 > n3 D. n1 > n2 and n1 < n3

The molar mass (the amount of mass in 1 mole of a compound) permits conversions between a given amount of mass and the number of moles that it contains. Stoichiometric mole ratios from a balanced reaction equation relate the number of moles of a chemical species to the moles of another chemical species in a reaction.

Question 25 Consider the reaction between cerium metal and hot water shown below. 2 Ce(s) + 6 H2O(l) → 2 Ce(OH)3(s) + 3 H2(g) How many moles of water are consumed in the reaction if 0.420 g of cerium fully reacts? A. 1.00 × 10−3 mol B. 3.00 × 10−3 mol C. 4.50 × 10−3 mol D. 9.00 × 10−3 mol

Two sequential dilutions were done to make a 100. mL aqueous HCl solution (Solution 3). A sample of concentrated HCl (Solution 1) was diluted by a factor of 20 to make a 50.0 mL stock solution. The volume of the sample of Solution 1 can be calculated by plugging in the DF and the VF of the solution from the first dilution into the DF equation: 20=50.0 mL / VT1 VT1=50.0 mL / 20 = 2.50 mL Therefore, 2.50 mL of Solution 1 was diluted to 50.0 mL by a 1:20 dilution. (Choice A) If VF1 were 5 mL rather than 50.0 mL, a VT1 of 0.250 mL would be obtained. (Choice C) The volume 25.0 mL is the amount of the stock solution (Solution 2) diluted to make Solution 3. (Choice D) A volume of 250. mL would be obtained if 50.0 mL were divided by a DF of 0.2 rather than 20. This volume is not plausible for VT1 because the VF1 is 50.0 mL and the volume transferred cannot be larger than the final volume of the solution.

Question 26 A sample of concentrated HCl (Solution 1) was diluted 1:20 to make a 50.0 mL stock solution (Solution 2). A portion of Solution 2 was then diluted 1:4 to make a 100. mL aqueous HCl solution (Solution 3). What was the volume of the sample of Solution 1 diluted to make Solution 2? A. 0.250 mL B. 2.50 mL C. 25.0 mL D. 250. mL

(Choice A) Introns are found in eukaryotic DNA, not prokaryotic DNA. Introns are DNA regions that do not code for proteins but are removed by the spliceosome during formation of a mature messenger RNA molecule. (Choice C) Although prokaryotes have only a single DNA chromosome, the DNA of both prokaryotic and eukaryotic genomes is still double-stranded, not single-stranded. (Choice D) Histone proteins are restricted to archaeal cells and the nuclei of eukaryotic cells but are not found in bacteria. Histones associate with DNA to facilitate its organization and packaging into chromosomes. Bacterial genes would not have been associated with histones prior to gene transfer to eukaryotes. Archaeal genes do not differ from eukaryotic genes on the basis of histone association.

Question 3 Based on the passage, which of the following best explains how mitochondrial genes differed from eukaryotic genes prior to gene transfer? Mitochondrial genes were: "This endosymbiotic theory of eukaryotic evolution also postulates that endosymbiosis resulted in larger eukaryotic genomes, which originated from the partial transfer of mitochondrial genes to the nuclear genome. On integration into the host genome, mitochondria-derived genes became indistinguishable from the original nuclear genes." A. interspersed with noncoding sequences. B. located on a chromosome without telomeres. C. present on a single-stranded DNA genome. D. associated with histones.

The substrate tert-butyl chloride is a tertiary alkyl halide, meaning the chlorine leaving group is bonded to a carbon with three alkyl substituents (a tertiary carbon) (Number III). Therefore, tert-butyl chloride is sterically hindered and cannot undergo an SN2 reaction with NaI, as described for Experiment 1. Alternatively, breaking the bond between the chlorine and the tertiary carbon (the first step of an SN1 reaction) causes a stable, tertiary carbocation to form (Number II), permitting tert-butyl chloride to go through an SN1 reaction with AgNO3 in Experiment 2. SN1 reactions occur in two steps: carbocation formation and nucleophilic addition. These reactions readily occur with tertiary alkyl halides because they are sterically hindered and form stable carbocations.

Question 3 The substrate tert-butylchloride reacts in Experiment 2 but not in Experiment 1 because: the leaving group of tert-butylchloride is bonded to a primary carbon. a stable carbocation forms in Experiment 2. the leaving group of tert-butylchloride is bonded to a tertiary carbon. an unstable carbocation forms in Experiment 2. A. I and II only B. I and IV only C. II and III only D. III and IV only

To compare the relative nucleophilicity of given anion, the electronegativity of the negatively charged atoms must be considered. The charged carbon atom of CH3(CH2)2CH2− is less electronegative than the nitrogen atom of (CH3)2N−, the oxygen atom of CH3CH2O−, and the fluorine atom of F−. Therefore, electrons on a negatively charged carbon atom are less stabilized and more easily shared with an electrophile, making CH3(CH2)2CH2− the best nucleophile. When comparing the nucleophilicity of atoms of equal negative charge, nucleophilicity tends to increase from right to left across a row of the periodic table as electronegativity decreases. Atoms of higher electronegativity more effectively stabilize negative charge and less readily donate electron density to electrophiles whereas those of lower electronegativity stabilize a negative charge less effectively and more readily donate electrons to electrophiles.

Question 4 Suppose that NaI was replaced with another nucleophile in Experiment 1. Which of the following anions would be the most nucleophilic? A. CH3(CH2)2CH2− B. (CH3)2N− C. CH3CH2O− D. F−

According to the passage, scientists proposed that a prolonged symbiosis period could induce gene transfer from eukaryotes to prokaryotes. In the given example, the symbiont (ie, P. leiognathi) was believed to have acquired the gene coding for Cu/Zn SOD from its eukaryotic host (ie, ponyfish) after an extended period of symbiosis. The passage states that Cu/Zn SOD is typically found in the cytosol of eukaryotes; therefore, scientists concluded that SODs found in P. leiognathi were most likely acquired from ponyfish. However, the discovery of Cu/Zn SODs in other free-living bacterial species with no known eukaryotic symbiotic hosts would argue against the hypothesis of eukaryotic to prokaryotic gene transfer due to symbiosis.

Question 4 Which observation would NOT support the hypothesis of gene transfer from eukaryotes to prokaryotes described in the passage? "Researchers have alternatively proposed that after a prolonged period of symbiosis, the possibility exists of gene transfer from eukaryotes to prokaryotes. This hypothesis was initially supported when copper/zinc (Cu/Zn) superoxide dismutase (SOD), a metalloprotein confined to the cytosol of eukaryotic cells, was found in Photobacterium leiognathi. The free-living bacterium P. leiognathi is also a known symbiont of ponyfish, a small fish species native to the Indian and Pacific Oceans." A. Discovering no difference between the gene sequences of the P. leiognathi and ponyfish SODs B. Discovering P. leiognathi and ponyfish SODs have a similar amino acid sequence in their noncatalytic domains C. Discovering Cu/Zn SODs in other free-living bacterial species with no known eukaryotic symbiotic hosts D. Discovering Cu/Zn SODs in other free-living bacterial species with known eukaryotic symbiotic hosts The endosymbiotic theory explains how primitive eukaryotic anaerobes engulfed ancient aerobic prokaryotes and consequently acquired the ability to produce energy through oxidative phosphorylation.

To determine which of the given alkoxide nucleophiles is the least sterically hindered, the substitution of the carbon bonded to the negatively charged oxygen atom must be considered. More alkyl groups bonded to this carbon atom make it more substituted and, as a result, the nucleophilic site (the oxygen atom) becomes more sterically hindered (crowded). Methoxide has only three hydrogen atoms and no alkyl groups bonded to the carbon adjacent to the negatively charged oxygen. Because this carbon atom does not have any alkyl groups, it is the least substituted and therefore the least sterically hindered nucleophile. Steric hindrance is a characteristic bulkiness in compound structures that either blocks a portion of a molecule so that it cannot react or keeps a reaction from occurring altogether. Decreasing the substitution on (or near) an atom decreases the steric hindrance of the atom.

Question 5 An alkoxide nucleophile is used instead of iodide in Experiment 1. Which of the alkoxides is the LEAST sterically hindered?

P. leiognathi is a unicellular prokaryotic organism, and therefore lacks membrane-bound organelles such as the Golgi body and the smooth endoplasmic reticulum (SER) present in eukaryotes. As a result, in P. leiognathi this signal sequence guides the protein destined for secretion (ie, SOD) directly to protein channels found on the plasma membrane. The channel contains a recognition sequence that interacts with the signal sequence on the polypeptide to induce translocation of the peptide across the plasma membrane and into the extracellular environment. Prokaryotic cells lack membrane-bound organelles and use specialized channels in the plasma membrane to secrete proteins. In contrast, eukaryotic cells contain membrane-bound organelles (eg, nucleus, Golgi body, mitochondria); the rough endoplasmic reticulum and Golgi body are involved in eukaryotic protein secretion.

Question 5 SOD mRNA in P. leiognathi encodes a signal sequence that directs the transport of the SOD protein for secretion. This signal sequence will direct SOD proteins to which structure in P. leiognathi? "SODs are antioxidant enzymes that serve as the cell's first line of defense against reactive oxygen species (ROS)." A. Smooth endoplasmic reticulum B. Plasma membrane C. Mitochondrial outer membrane D. Golgi body Although eukaryotic protein synthesis of secreted proteins (eg, in ponyfish) may direct proteins from free ribosomes in the cytoplasm to the rough ER (RER) and then to the Golgi body prior to secretion, the question asks about SOD secretion in P. leiognathi. P. leiognathi is a prokaryote that does not have membrane-bound organelles like the RER and Golgi. According to the question, SOD mRNA in P. leiognathi encodes a signal sequence that directs the transport of the SOD protein to the extracellular environment. Signal sequences are short amino acid sequences typically located at the N-terminus of a polypeptide, and are used by both eukaryotic and prokaryotic cells to direct proteins to precise destinations within the cell.

The inductive effect occurs when electron density is donated through sigma bonds. Carbocations are stabilized by electron donating groups because they donate electrons to the positively charged carbon; they are destabilized by electron withdrawing groups because they pull electrons away from the carbocation, creating two adjacent positive charges.

Question 6 Compounds 1 and 2 are both intermediates that could form by a reaction with AgNO3 in ethanol, but Compound 1 is less stable than Compound 2 because: A. Compound 1 contains an electron donating group that destabilizes the carbocation. B. there is an inductive effect from the fluorine atoms that destabilizes the carbocation. C. the methyl groups on the structure withdraw electrons from the positively charged carbon. D. the electronegative fluorine atoms cause the carbocation to rearrange into a secondary carbocation.

Exposing eukaryotic ponyfish cells to a spindle fiber toxin interferes with the formation of the spindle fibers, and therefore impairs the separation of sister chromatids. Because the sister chromatids cannot separate effectively, nondisjunction during mitosis is expected. As a consequence, affected ponyfish cells would not show an equal distribution of genetic material (genes) because nondisjunction can result in one daughter cell having a different number of chromosomes than the other. The spindle fiber toxin would not influence the copy number of Cu/SOD genes or delay the separation of daughter cells in P. leiognathi (ie, a prokaryote) because prokaryotes divide by binary fission, a simple method of asexual reproduction that results in two identical daughter cells and does not involve spindle fibers. Therefore, a toxin that targets eukaryotic microtubules is unlikely to adversely affect prokaryotic cell division.

Question 6 Ponyfish cells containing P. leiognathi symbionts were exposed to a spindle fiber toxin that specifically inhibits microtubule polymerization. Given this information, which of the following would most likely result as a consequence of toxin exposure? A. P. leiognathi daughter cells with multiple copies of the Cu/Zn SOD gene B. Ponyfish daughter cells containing the same copy number of the Cu/Zn SOD gene C. Delayed separation of P. leiognathi cells during binary fission D. Nondisjunction in somatic ponyfish cells undergoing nuclear division Most eukaryotic cells (except germ cells) undergo cell division via mitosis. In contrast, prokaryotic cells duplicate via binary fission, a simple form of reproduction that does not involve the separation of chromosomes by spindle fibers. Nondisjunction in mitosis or meiosis can occur when sister chromatids fail to separate properly during anaphase.

(Choice A) Both eukaryotic and prokaryotic mRNAs can be bound by multiple ribosomes in the cytoplasm. (Choice B) Ponyfish mRNA is translated by the 80S ribosome whereas P. leiognathi mRNA is translated by the 70S ribosome. In this scenario, no information is given to suggest that ribosome affinity is a factor for SOD synthesis. (Choice C) In eukaryotic (eg, ponyfish) cells, transcription of mRNA and its modification (eg, addition of the 5′ cap, 3′ poly-A tail, splicing) typically occur in the nucleus. The modifications, which result in a mature mRNA molecule, increase the stability of the transcript and prevent its degradation in the cytoplasm. Because the nucleus is separated from the cytoplasm by a nuclear envelope, mRNA must first be transported through nuclear pores to the cytoplasm, where it is translated by ribosomes.

Question 7 Investigators used a ribosome profiling technique to measure the level of SOD protein synthesis in P. leiognathi and ponyfish cells during ROS accumulation. After introduction of ROS, investigators found that P. leiognathi cells accumulated ribosome/SOD mRNA complexes at a faster rate than did ponyfish cells. What is the best explanation for this result? A. P. leiognathi SOD mRNA is capable of being bound by more than one ribosome. B. P. leiognathi SOD mRNA has a greater affinity for ribosomes than ponyfish mRNA. C. mRNA transcribed from the ponyfish SOD gene undergoes immediate degradation. D. mRNA transcribed from the bacterial SOD gene is being translated simultaneously. Prokaryotic cells have no nucleus; therefore, transcription and translation occur simultaneously in the cytoplasm (ie, translation begins before the mRNA is fully transcribed). In contrast, in eukaryotic cells transcription and post-transcriptional modifications occur in the nucleus but translation occurs in the cytoplasm. Prokaryotic cells have no nucleus or nuclear envelope, and due to the lack of this physical barrier between DNA and the ribosomal machinery, transcription and translation occur simultaneously in the cytoplasm. Ribosome profiling was used to measure the level of SOD translation by detecting the number of SOD mRNA/ribosome complexes found within ponyfish (a eukaryote) and P. leiognathi (a prokaryote). The question states that P. leiognathi cells have a greater number of ribosome/SOD mRNA complexes compared to ponyfish cells. This rapid protein synthesis is likely due to the fact that all SOD mRNA transcripts in P. leiognathi can quickly interact with ribosomes even before being fully transcribed (ie, transcription and translation are coupled).

Observations for the SN2 reaction conditions (Table 1) show that 1-bromobutane (a primary halide) formed a precipitate whereas tert-butyl chloride (a tertiary halide) did not react, even after heating. Conversely, results for the SN1 reaction conditions (Table 2) record that tert-butyl chloride formed a heavy precipitate, but 1-bromobutane only formed a trace amount of precipitate after heating. This data demonstrates that SN1 and SN2 reactions are dependent on substrate substitution because primary and tertiary alkyl halides react only in certain conditions. In particular, experimental results show that nucleophilic substitution reactions (SN1 and SN2) are dependent on substrate substitution.

Question 7 Which of the following statements is NOT supported by the results in Tables 1 and 2? A. In Experiment 2, crotyl chloride forms a stable carbocation. B. Primary alkyl halides react more readily in SN2 reactions than in SN1 reactions. C. Nucleophilic substitution reactions are not dependent on the substrate (alkyl halide) substitution. D. Bromide is a better leaving group than chloride and enhances reactivity in less favorable reactions.

The passage states the reaction with NaI is an SN2 reaction, indicating a concerted reaction occurs (ie, no carbocation formation or the possibility of a hydride shift). 3-Bromo-2,4-dimethylpentane is a secondary alkyl halide with two isopropyl groups bonded to the electrophile for the nucleophilic substitution reaction. Although secondary alkyl halides can undergo SN2 reactions, the large bulky isopropyl groups will hinder the nucleophile's access to the electrophile. Therefore, 3-bromo-2,4-dimethylpentane will react very slowly with NaI at room temperature, forming only trace precipitate because the electrophile is too sterically hindered.

Question 8 If 3-bromo-2,4-dimethylpentane (shown below) was used as a substrate in the experiments described in the passage, what would the observation be for reaction with NaI at 25°C? Experiment involving Nal; is an Sn2 reaction A. Precipitate formation, because the substrate has a tertiary electrophile B. Precipitate formation, because the electrophilic carbon is not sterically hindered C. Trace precipitate, because the methyl groups hinder the addition of iodide to the electrophile D. No reaction, because SN2 reactions cannot happen at sp3 carbons

The enzyme ATP synthase harnesses the potential energy of the ETC-generated proton gradient to generate ATP from ADP and inorganic phosphate. It is stated in the question that P. leiognathi can utilize the ETC in anaerobic environments because it can use compounds other than O2 as the final electron acceptor. Although ATP synthase does not directly need O2 to generate ATP, the formation of the proton gradient by which it functions is oxygen-dependent in ponyfish cells but not in P. leiognathi. Consequently, under anaerobic conditions, ATP synthase is active in P. leiognathibut not active in ponyfish cells.

Question 8 In anaerobic environments, P. leiognathi can produce energy by using an inorganic final electron acceptor other than oxygen in the electron transport chain. Under anaerobic conditions, which of the following is active in P. leiognathi but is NOT active in ponyfish cells? "The free-living bacterium P. leiognathi is also a known symbiont of ponyfish, a small fish species native to the Indian and Pacific Oceans." A. Glycolysis B. Gluconeogenesis C. Fermentation D. ATP synthase In eukaryotes, cellular respiration continues when pyruvate is imported into the mitochondria and converted into acetyl-CoA, which then enters the Krebs cycle (citric acid cycle) to generate the necessary electron carriers (NADH and FADH2) for the electron transport chain (ETC). ETC protein complexes pump protons across the inner mitochondrial membrane, generating a proton gradient with O2 as the terminal electron acceptor. In comparison, prokaryotes have no mitochondria; instead, the ETC is located on the plasma membrane. Fermentation metabolizes pyruvate under anaerobic conditions to oxidize NADH to NAD+. Fermentation yields lactic acid in the muscles of ponyfish and alcohol in bacteria. Gluconeogenesis is the process by which noncarbohydrate carbon sources are converted into glucose. Because neither process requires O2, fermentation and gluconeogenesis are active in both P. leiognathi and ponyfish cells.

Latent heat is the amount of heat 1 mole of a substance must absorb (or release) to produce a change in phase (eg, solid to liquid, gas to liquid) while maintaining a constant temperature. When a substance in the liquid phase is at its boiling point, the latent heat energy that goes into the system is used to break the attractive intermolecular forces between the liquid molecules rather than causing an increase in temperature. Consequently, the energized liquid molecules can escape the surface of the liquid and enter the gas (ie, vapor) phase. Therefore, the amount of heat absorption needed to convert 1 mole of a substance in the liquid phase to the gas phaseis described by the latent heat of vaporization (ΔHvaporization). The latent heat of vaporization of a substance describes the heat required to convert 1 mole of that substance from the liquid phase to the gas phase.

Question 9 The latent heat of vaporization describes: A. the amount of heat a substance must release to transition from the liquid to the gas phase. B. the amount of heat a substance must absorb to transition from the liquid to the gas phase. C. the amount of heat a substance must release to transition from the solid to the gas phase. D. the amount of heat a substance must absorb to transition from the solid to the gas phase.

(Choice A) This structure depicts the peptide Ser-Ala with an extra carboxyl group attached to the N-terminus and an extra amino group attached to the C-terminus. (Choice B) This structure depicts the peptide Ser-Ala rather than Ala-Ser. Although the alanine residue is shown on the left side of the molecule, it is part of the C-terminal residue rather than the N-terminal residue. (Choice D) In this structure, the carbonyl and α-carbons have their positions switched relative to where they should be. In each residue, the α-carbon (to which the side chain is linked) should be between the nitrogen and the carbonyl carbon. Educational objective:The backbone of an amino acid consists of an amino group linked to an α-carbon, which is in turn linked to a carboxyl group. The α-carbon is also linked to a side chain that determines the identity of the amino acid.

Question 9 What is the structure of a dipeptide with the sequence Ala-Ser? The backbone of an amino acid consists of an amino group (-NH3+, under physiological conditions) bonded to a carbon (the α-carbon), which is in turn bonded to a carboxyl group (-COO−, under physiological conditions). The α-carbon is also bonded to a hydrogen atom (which is often omitted when drawn as a skeletal structure) and to a side group whose structure determines the identity of the amino acid. The carboxyl group of one amino acid can react with the amino group of another to form a peptide bond that links the two amino acid residues of the dipeptide. The residue that retains a free amino group is called the amino-terminal or N-terminal residue, and the residue that retains a free carboxyl group is called the carboxy-terminal or C-terminal residue. By convention, the sequence of a peptide is always written with the N-terminal residue on the left and the C-terminal residue on the right, regardless of how the structure of the peptide is drawn. The question asks for a peptide with the sequence Ala-Ser. Therefore, the correct structure must have an alanine residue at the N-terminus and a serine residue at the C-terminus. Only the structure shown in Choice C depicts a correct dipeptide backbone structure with an N-terminal alanine and a C-terminal serine.

arterioles arterioles

The ___________________ are responsible for reducing the blood pressure from the major arteries (high blood pressure) to the capillary networks (low blood pressure). As a result, the ______________________ are the primary pressure-reducing vascular region of the circulatory system.

∆G°=∆H°−T∆S° ΔG° = ΔH° − TΔS°, where T is given in Kelvin.

The change in Gibbs free energy ΔG° for a reaction is related to the changes in enthalpy ΔH° and entropy ΔS° at temperature T (measured in Kelvin) according to the formula:______________________________ The Gibbs free energy is a state function that describes the spontaneity of a reaction. It is related to the temperature, enthalpy, and entropy of the reaction via the equation _______________________

less than greater than

The dissociation of a weak acid (HA) in water can be represented by the general equilibrium reaction: HA(aq) +H2Ol⇄KaH3O+aq+A-aq The extent to which an acid dissociates in water is characterized by the acid dissociation constantKa, which is also expressed as the pKa: pKa=-logKa Without a calculator, the negative logarithm can be approximated using the relationship: -log(m×10^-n) ≈ n-0.m where n is a whole number and m is a number between 1 and 10. A solution's acidity is measured by its pH, which depends on the molar concentration of H3O+. The relationship between pKa and pH can be quantitatively described by the Henderson-Hasselbalch equation: pH = pKa + log ([A-]/[HA]) If the pH of a solution is ______________ the pKa (ie, the solution is more acidic), the equilibrium is shifted toward a greater concentration of the protonated acid (ie, [HA] > [A−]). Conversely, if the pH is _____________ the pKa (ie, the solution is more basic), the equilibrium is shifted toward a greater concentration of the deprotonated conjugate base (ie, [HA] < [A−]). The dissociation of HNO2 in water (Ka = 4 × 10−5) proceeds according to the equilibrium reaction: HNO2aq+H2Ol⇄KaH3O+aq+HNO2-aq As such, the pKa is: pKa=-log4×10-5=4.4 Without a calculator, the pKa of HNO2 is approximated as: pKa≈5-0.4=4.6≈4.4

decreases increases (Choice A) This graph suggests Keq is temperature independent, which is not true. (Choices C and D) As temperature increases, Keq will decrease for the exothermic reaction and increase for the endothermic reaction. Educational objective:

The equilibrium constant Keq expresses the relationship between products and reactants at equilibrium. When temperature increases, keq ______________________ for exothermic reactions and ______________ for endothermic reactions.

Methyl (-CH3) groups are EDG because the carbon atom is more electronegative than the adjacent hydrogen atom, creating a partial negative charge on carbon. This carbon can then donate electrons to the carbocation to stabilize it. The trifluoromethyl (-CF3) group is an EWG because the fluorine atoms are more electronegative than the adjacent carbon and draw electrons toward themselves, creating a partial positive charge on the carbon next to the positively charged carbocation. Because Compound 1 has one EWG and Compound 2 does not contain any EWG, Compound 1 is less stable than Compound 2 due to the inductive effect from the fluorine atoms on -CF3.

The inductive effect occurs with the donation of electron density through sigma bonds. Electronegative atoms or electron withdrawing groups (EWG) pull electrons away from an adjacent atom, creating a dipole with a partial negative charge on the electronegative atom and a partial positive charge on the adjacent atom. Carbocations are stabilized by electron donating groups (EDG), such as alkyl groups, because they donate electrons toward the positively charged carbon. EWGhave the opposite effect: They tend to destabilize carbocations because the electronegative atom pulls electrons toward itself, creating a partial positive charge on the adjacent atom. With two adjacent positive charges, the carbocation is not stable.

rate increase the rate does not

Thermodynamic values indicate if a reaction is spontaneous or not, but they do not indicate the ___________ of a reaction. Enthalpy, entropy, and Gibbs free energy are state functions, which mean they describe the energetic differences between the reactants and the products in their current state. These values are independent of the chemical pathway that a reaction takes to get from reactants to products. A catalyst is a chemical species that is added to a reaction to __________________________ of the reaction by stabilizing the transition state and decreasing the activation energy of the reaction. A catalyst ____________________ change the amount of products produced and does not change the relative energies of the reactants and products.