Next Step FL3 C/P

Ammonia

NH3

Ammonium

NH4+

azimuthal

angular momentum

quantum numbers

describe its position, the shape of its orbit, its orientation, and its angular momentum.

Rydberg Equation

used to calculate the wavelengths of all the spectral lines of hydrogen E = R (1/n1^2 - 1/n2^2)

common ion effect

-the lowering of the solubility of an ionic compound as a result of the addition of a common ion -occurs when a solution contains a salt that partially dissolves into ions, and an extra quantity of one of these ions is added. This results in the combination of this ion with existing dissolved ions, causing an additional quantity of the salt to precipitate out of solution. In this case, the partially-dissolved salt is Ag3PO4. Adding AgNO3 causes dissolution of extra Ag+ions, which then combine with already-present PO43- ions and precipitate out of solution into the yellow Ag3PO4 precipitate

sp2 hybridization

1. Trigonal planar structure 2. sp2 hybridization creates 3 identical orbitals of intermediate energy and length and leaves one unhybridized p orbital 3. 3 effective pairs of electrons surround the carbon (double bond treated as one effective pair). so it can be bonded to two other atoms and have a lone pair.

ln to log

2.3 × log10(N) = ln(N).

19. A cell's ATP-to-ADP ratio is most nearly equal to the ratio of its fluorescence emission intensities when Y32 is excited at: A. 500 nm versus when it is excited at 420 nm. B. 420 nm versus when it is excited at 500 nm. C. 500 nm versus when it is in its ground state at 500 nm. D. 420 nm versus when it is in its ground state at 420 nm

A is correct. According to the passage, the spectral ratio of the peaks of the excitation spectrum of a Y32-expressing cell directly reflects its cellular ATP-to-ADP ratio. The peak absorbance of the B-state, which is favored in the Mg2+-ATP-bound conformation, is near 500 nm, while the peak absorbance of the A-state, which favors the ADP-bound conformation, is near 420 nm. Thus, the ratio of the fluorescence emission intensities when Y32 is excited at 500 nm versus when it is excited at 420 nm serves as a direct indicator of the cellular ATP-to-ADP ratio.

11. Which of the following best explains the tumor reduction capacities of the isotopes observed in the study? A. GKS-X emits more radiation per unit time, causing more cell death. B. GKS-X emits narrower radiation streams that target the tumor. C. GKS-Co anti-tumor emissions decompose much more slowly than GKS-X emissions. D. GKS-Co does not emit sufficient radiation to kill tumor cells.

A is correct. Table 1 shows that GKS-X is more effective at reducing tumor size (as shown by its lower RTV values over time). We need to choose a statement that can explain this improved ability of GKS-X to kill cancer cells. We must think about the differences between X and 60Co, chief among them the fivefold difference in half-life. Because X has a shorter half-life, it is logical that it would release more radiation in the same period of time because it undergoes decay more quickly. This higher dose of radiation would kill more cells, cancerous and non-cancerous, leading to a larger reduction in tumor size and more undesirable side effects. This corresponds to answer choice A. C: If GKS-Co involved radioactive material that decayed much more slowly than that in GKS-X, we would expect the GKS-Co tumor-reducing effects to catch up to or surpass GKS-X as time passed. Instead, we can see in Table 1 that the disparity gets larger as time post-surgery increases.

51. If sodium sulfate was added to the mixture containing silver ions and the yellow precipitate, what might be observed after a significant amount of time elapsed? A. Little to no silver sulfate formation, because the Ksp of silver sulfate is very large compared to the Ksp of the yellow precipitate. B. Little to no silver sulfate formation, because sodium sulfate cannot be dissolved at all in water. C. Significant silver sulfate formation, because the solubility of sodium sulfate releases sodium ions which catalyze the necessary reactions. D. Significant silver sulfate formation, because the solubility of sodium sulfate releases sulfate ions which catalyze the necessary reactions.

A is correct. This question is asking us to find a possible outcome and explanation for the outcome when adding sodium sulfate to the silver/yellow precipitate mix. This means we need to choose the answer that has a valid outcome and a proper explanation for that outcome. The Ksp, or solubility product, of a substance is defined as the product of each of the substance's dissolved ion concentration raised to the power of its stoichiometric coefficient. For silver sulfate, the dissolution reaction is: Ag2SO4 (s) → 2 Ag+ (aq) + SO42- (aq) which means that the Ksp equation is: Ksp = [Ag+]2[SO42-] A large Ksp suggests that a substance is more soluble than a substance with a lower Ksp. It is not certain, given that the coefficients might be different and might affect the power to which the ion concentrations are raised. However, it provides a better possible explanation than the other answer choice options (and the phrasing of the question prompt, using the words "might explain," notes that there is some uncertainty involved).

6. Which of the following electronic transitions for a hydrogen atom would result in the emission of a photon that would be visible to the human eye? A. n = 1 to n = 3 B. n = 2 to n = 4 C. n = 2 to n = 1 D. n = 4 to n = 2

ASKING ABOUT EMISSION!!!! (i didnt realize and assumed absorption) D is correct. The visible spectrum contains electromagnetic signals with wavelengths ranging from 400 nm to 700 nm. The wavelength of light emitted during a particular electronic transition is determined by the energy difference (ΔE) between the final and initial energy levels. The energy of each level can be determined using Equation 1 and the principal quantum number in question. For light to be emitted at all, an electron must travel from a higher to a lower level, a process that releases energy. This automatically eliminates options A and B; however, we must actually perform calculations to choose between C and D. Begin with choice C, using Equation 1 and approximating the Rydberg constant as 1 ✕ 107 m-1. The energy of the n = 2 level is: E2 = -1 ✕ 107 m-1 / 22 = -0.25 ✕ 107 m-1 Since the energy of the n = 1 level is equivalent to the Rydberg constant, ΔE is: ΔE = E2 - E1 = (-1 ✕ 107 m-1) - (-0.25 ✕ 107 m-1) = -0.75 ✕ 107 m-1 Note that this value is in units of m-1, but due to the passage and the units given in the question stem, we need to convert to nanometers ( 1 nm = 10-9 m). We can ignore the negative sign, since this simply means that energy was released rather than absorbed. 1 / (0.75 ✕ 107 m-1) = 1.33 ✕ 10-7 m = 133 nm This falls in the UV range of the electromagnetic spectrum, below the visible range. This eliminates choice C, and we can choose option D and move on.

amylose and amylopectin

Amylose is a linear polymer of glucose molecules connected by α(1→4) glycosidic bonds; it comprises about 20%-30% of starch. Amylopectin makes up the remaining 70%−80% of starch; it likewise contains glucose molecules connected by α(1→4) glycosidic bonds, but it has branches due to α(1→6) glycosidic bonds every 24−30 units.

32. What is the pH of a 0.10 M aqueous solution of acetylsalicylic acid? A. 1.0 B. 2.3 C. 3.5 D. 4.1

B is correct. Acetylsalicylic acid is a weak acid, with a pKa of 3.5. Therefore, the pH of this solution must be less than the pKa, because the compound is primarily in its acid form and a pH of 3.5 would mean that the concentration of weak acid and conjugate base were equal (a buffer). Choices C and D can be eliminated. A pH of 1.0 for an acid whose concentration is 0.10 M would require complete dissociation (as in a strong acid), eliminating choice A. Alternatively, the pH can be determined from the equilibrium expression: Ka = [H+][A-]/[HA] 10-3.5 = x2/(0.1-x) Since x will be small, we can approximate 0.1-x ~ 0.1, giving: 10-3.5 = x2/(0.1) 10-3.5(0.1) = x 2 10-4.5 = x2 [10-2.25][10-2.25] = x2 10-2.25 = x Since x equals the hydrogen ion concentration, taking the -log(10-2.25) = 2.25. A: This would be correct if acetylsalicyclic acid were a strong acid, but it is weak.

17. How much heat is produced from the complete combustion of 30.0 g of methane, if the enthalpy of reaction is -890 kJ/mol? A. 1.7 x 103 J B. 1.7 x 106 J C. 4.7 x 106 J D. 4.7 x 109 J

B is correct. First, note that methane has a molecular formula of CH4. Thus, its molecular weight is approximately 12 + 4(1) = 16 g/mol. 30 g CH4 x (1 mol/16 g) x (890 kJ/mol) = 1.8 x 103 kJ (if we round 890 to 900 kJ/mol) Note that the answer options have units of joules (J), not kilojoules (kJ), and choice A should be eliminated. Converting the units gives: 1.8 x 103 kJ x (103 J /1 kJ) = 1.8 x 106 J

3. The results of a separate study of the thermodynamic parameters for the interactions of proteins with cyclohexanol and quartenary ammonium salts indicate that the hydrophobic solute-solute interaction is spontaneous, and that ΔH and ΔG have opposite signs. Which of the following must be true for this interaction when temperature is a positive value? I. ΔH > ΔG II. 0 < ΔS III. ΔH < ΔS A. I only B. I and II only C. II and III only D. I, II, and III

B is correct. For the interaction to be spontaneous, the free energy change of the reaction must be negative. The question states that the enthalpy and free energy changes for the interaction have opposite signs. Thus, ΔG is negative and ΔH is positive, making Roman numeral I true. For a reaction to be both endothermic (positive ΔH) and spontaneous (negative ΔG), the reaction must involve an increase in entropy (ΔS is positive). This is according to the equation: ΔG = ΔH - TΔS. Thus, II is also true because ΔS is positive. III: The quantity TΔS, not ΔS, must be greater than ΔH. For this reason, RN III does not have to be true.

49. Efforts to treat lactic acid buildup in muscles were attempted using dissected muscle specimens in the laboratory. One of these experiments involved ammonium formation from dissolved ammonia. Under conditions of excessive lactic acid: A. the final concentration of ammonium will be higher than otherwise due to higher pH. B. the final concentration of ammonium will be higher than otherwise due to lower pH. C. the final concentration of ammonium will be lower than otherwise due to higher pH. D. the final concentration of ammonium will be lower than otherwise due to lower pH.

B is correct. In an acidic environment, a base such as ammonia (NH3) will dissolve into its conjugate acid, ammonium (NH4+), to a greater extent than would have been the case in a neutral or a basic environment. An environment with lower pH is an environment that is more acidic.

54. A 60-kg runner raises his center of mass approximately 0.5 m with each step. Although his leg muscles act as a spring, recapturing the energy each time his feet touch down, there's an average 10% loss with each compression. What must the runner's additional power output be to account for just this loss, if he averages 0.8 s per stride? A. 0 W B. 37 W C. 46 W D. 331 W

B is correct. This is a multi-step question, though each step is relatively straightforward. The gravitational potential energy at the runner's height is: PE = (60 kg)(10 m/s2)(0.5 m) = 300 J Most of this energy is conserved as the runner hits the ground and his muscles capture the energy as spring potential energy, so the question only asks about the lost energy, amounting to 10%, or 30 J. Don't worry about the mechanics of which foot lands first and how much of the kinetic energy each one absorbs. The question stem doesn't give that kind of information, so there's nothing for it anyway. So, an additional 30 J is needed per stride, and a stride occurs every 0.8 s. Thus: P = (30 J)/(0.8 s) = 40 W This is between answer choices B and C. So is it really 40 W? Well, no, 8 goes into 30 less than 4 times, so the answer should be lower. Also, g is closer to 9.8 than 10, so that also caused the value of energy, used in the power calculation, to be overestimated. 37 W, then, must be the match.

Which of the following best explains why arginine is more basic than lysine? A. The electron-donating groups around the basic nitrogen on arginine make its conjugate base less stable. B. The electron-donating groups around the basic nitrogen on arginine make its conjugate acid more stable. C. The lack of electron-donating groups on lysine make its conjugate acid more stable. D. The lack of electron-withdrawing groups on lysine make its conjugate base more stable.

B is correct. This question is asking us to determine why arginine is more basic than lysine. The reason must be related to how arginine is better able to handle being protonated, as this is the essence of being a base. Since, in its protonated form, arginine has electron-donating groups via resonance with other nitrogens, it is a more stable conjugate acid.

33. A student determined that her yield of aspirin was 3.9 g. What was her percent yield? A. 45% B. 60% C. 78% D. 92%

B is correct. We first need to determine if there is a limiting reagent and then determine the theoretical yield. The moles of acetic anhydride can be determined by using the volume and density information in step 2 of the procedure and the molecular weight given in the passage. 7 mL x 1.08 g/mL x 1 mol/102 g ~ 7 x 10-2 mol We can also calculate the moles of salicylic acid from the number of grams used in step 1 and the molecular weight given in the passage. 5.0 g x 1mol/138 g ~ (5/1.4) x 10-2 ~ 3.5 x 10-2 mol Since the stoichiometry is 1:1 from Reaction 1, the acetic anhydride is in excess and the limiting reagent is the salicylic acid. The stoichiometric ratio between salicylic acid and the acetyl salicylic acid is also 1:1; therefore, the theoretical yield can be calculated using the molecular weight from the passage. 3.5 x 10-2 mol x 180 g/mol ~ 3.5 x 2 x 10-2 x 102 ~ 7 g Since the question indicates that the actual yield was 3.9 g, which is about 4 g, the percent yield is (4/7) x 100 ~ 60%.

ejecting electrons

Because sp-hybridized orbitals have the most s character of all of the answer choices, they contain the electrons that are hardest to eject.

53. When does a runner output the most additional energy to keep the ground reaction forces most nearly vertical and through her body's center of mass? A. When she takes high, bouncing strides and keeps her spine fairly vertical B. When she takes long, low strides and keeps her spine fairly vertical C. When she takes high, bouncing strides and leans her upper half into her run D. When she takes long, low strides and leans her upper half into her run

C is correct. Both the vertical displacement of the runner's steps and the angle of her body from the vertical increase the energy required to realign the ground reaction forces. Recall that tanθ = sinθ/cosθ, so when θ is close to zero, so is tanθ. When θ is close to 90°, tanθ becomes arbitrarily large. The more the runner leans into the run, the greater (or closer to 90°) tanθ is, and the greater the work expended by the runner must be, according to the work equation given in the passage (Wx = Fz tan(θ) Δz). Also, in this equation, the passage states that Δz represents the vertical displacement of a single step. Thus, the higher the vertical displacement ("high, bouncing strides..."), the greater the energy expenditure. A, B: When the runner keeps her spine fairly vertical, θ is close to 0°, meaning that tanθ is close to zero as well. According to the work equation from the passage, the smaller tanθ is, the less work is expended. This question is asking when the runner expends the most energy (in other words, when she must do the most work), rather than the opposite. D: According to the equation Wx = Fz tan(θ) Δz, a large Δz corresponds to a large Wx (work) value. This answer choice describes the opposite, as "long, low strides" indicates a small Δz. A small Δz corresponds to a small work value, indicating that the runner is expending less energy when taking these low strides. This question asks for the opposite — when the runner expends the most energy.

2. Underproduction of pulmonary surfactant in IRDS leads to decreased compliance of alveolar tissue. Based upon this information, which of the following must be true regarding pulmonary surfactant? A. Its adsorption to the water-alveolar interface increases surface tension, preventing alveolar collapse due to intra-thoracic pressure. B. Its absence decreases the minimum radial size of alveoli able to avoid collapse at a given pressure of inspired air. C. Its adsorption to the water-alveolar interface decreases surface tension, decreasing the pressure difference required to inflate the airway. D. Its presence increases the efficiency of gas exchange across the alveolar membrane by decreasing the surface area of the alveolus at a given pressure of inspired air.

C is correct. During inspiration, contraction of the diaphragm and external intercostal muscles leads to expansion of the thoracic cavity and a decrease in intrapleural pressure. This negative pressure, relative to atmospheric pressure at the entry of the upper airway, generates airflow through the respiratory tree and to its terminal extension—the alveoli. The elastic recoil force of the airway and the surface tension of the water lining the airway oppose expansion of the alveoli due to the influx of atmospheric pressure. Pulmonary surfactant adsorbs to the air-water-alveoli interface, reducing surface tension and the total force resisting expansion. This increases pulmonary compliance—a measure of lung volume change at a given pressure of inspired air—and decreases the work required to expand the lungs at a given atmospheric pressure. This is consistent with choice C. B: The absence of pulmonary surfactant increases surface tension, increasing the minimum radial size of alveoli required to overcome the collapsing force of surfactant at a given pressure of inspired air. When surface tension exceeds the average alveolar radii, collapse of the small airways occurs, eliminating choice B.

23. Y32 is pH-sensitive. This can be a problem when employing it to monitor changes in the energy balance of a cell because: A. pH can differ between cells. B. the ATP-to-ADP ratio of a cell depends on pH. C. pH may change over time in the same cell. D. decreased pH favors the CPV B-state.

C is correct. If the biosensor is pH-sensitive, then changes in cellular pH over the course of a measurement could confuse changes in the energy balance of the cell, as reflected by the cell's ATP-to-ADP ratio. A: While this is true, the question specifically asks about how the fact that the biosensor is pH-sensitive could affect measurements made within a single cell over time. B: While the ATP-to-ADP ratio in most cells does to some extent depend on pH, this does not address the question of how pH changes influence the utility of the biosensor in measuring a cell's ATP-to-ADP ratio. D: Decreasing pH would favor the protonated CPV A-state.

47. Phosphorous acid, a common ingredient used for potable water treatment, has a molecular formula of: A. H3PO5 B. H3PO4 C. H3PO3 D. H3PO2

C is correct. The anion in phosphorous acid is phosphite, PO33-. When comparing the -ous acids and -ic acids, the -ous acids will have 1 fewer oxygen atoms than their -ic counterparts. A: This is perphosphoric acid. B: This is phosphoric acid. D: This is hypophosphorous acid.

37. A second student repeated the experiment using glucose and the equivalent enzymes of glycolysis instead of starches. How would his results compare to those shown in Figure 3? A. Purplish color with higher initial current B. Purplish color but no current because of incomplete glycolysis C. Brownish color with higher initial current D. Brownish color with lower initial current

C is correct. The first difference between the first student and the second is the use of glucose, a monosaccharide (shown below), as the fuel. Paragraph 2 states that iodine binds selectively to linear-chain polysaccharides. As a result, the iodine will NOT bind the glucose, causing the solution to remain brown. We can eliminate choices A and B. Next, we must determine the effect of using glucose and the glycolytic enzymes on the current generated by the battery. From Figure 1, we can infer that the amount of current generated should be directly related to the production of NADH. Unlike in the biostarch battery, glycolysis does not require the breakdown of starch. Thus, NADH is generated directly and more quickly via the conversion of glyceraldehyde 3-phosphate (GAP) to 1,3-bisphosphoglycerate (1,3-BPG). The enzymes should produce 1 NADH per GAP molecule, or 2 NADH per glucose molecule (since each unit of glucose breaks down into two GAP molecules). The original experiment only produced 1 NADH per glucose conversion to phosphogluconate, so we can conclude that the second student should observe a higher initial current than the first. D: We would expect the initial current to be higher, not lower, when using glucose and glycolytic enzymes. As described above, glycolysis produces more NADH per glucose molecule than the starch → glucose 6-phosphate → 6-phosphogluconate conversion. More production of NADH means that more electrons will be passed along to AQDS and the electrodes of the apparatus, and current will be greater.

If a 3-kg rabbit's leg muscles act as imperfectly elastic springs, how much energy will they hold if the rabbit lands from a height of 0.5 m and its legs are compressed by 0.2 m? A. -0.6 J B. 0 J C. 10 J D. 14.7 J

C is correct. This is a tricky question, because the passage states that nature has no perfectly elastic springs, which means energy cannot be completely conserved. This does not, however, mean that all the energy is lost, nor does it make sense for the potential energy stored in a spring to be negative (ruling out answer choices A and B). Then, to distinguish between choices C and D, we need to obtain an estimate of what the value would be if energy were completely conserved, since that could serve as a baseline. The potential energy of the rabbit at the peak of its height can be estimated as PE = (3)(10)(0.5) = 15 J. Since that value was calculated using the estimate of 10 m/s2 for g, instead of 9.81 m/s2, we should recognize that it is an overestimate. That is, the actual value that would be obtained if no energy was lost would be slightly less than 15 — therefore, answer choice D, 14.7 J, seems about right for how much we need to correct for our estimation of g. That is, choice D corresponds to the value that we would calculate if no energy was lost. Finally, we are left with choice C, 10 J, which corresponds to a non-trivial loss of energy due to inefficiency, which is what the question stem suggests that we should expect (because of the wording "imperfect spring"). A: The spring wouldn't store negative energy. B: This would imply that all of the energy was lost and that no energy was stored in the spring (muscle) after landing. The passage tells us that the springs are imperfect, but not that they would dissipate all energy D: This value corresponds to what would be expected if this were a perfect spring (i.e. if all energy was conserved).

10. Question 10 A hospital purchases brand-new GKS-Co and GKS-X machines. Five years after installation, what is the expected ratio of the total atomic mass of material in the Co machine to that in the X machine, assuming both machines start with the same mass of radioactive material? A. 1:16 B. 1:5 C. 1:1 D. 5:1

C is correct. When reading questions, be careful not to read too quickly. In this case, fast but inefficient reading will lead us to assume that it is asking about the percentage of a certain isotope that is left after radioactive decay. However, the question is asking about atomic mass. While β-decay does cause a nuclear transmutation of protons to neutrons (β-plus) or neutrons to protons (β-minus), the atomic mass lost in these processes is negligible. This means that whether after one (Co) or five (X) half-lives, the atomic mass will be the same in both samples. A: This is the ratio of undecayed nuclei between the two samples. The GKS-Co machine has 1/2 of its nuclei undecayed and the GKS-X machine has 1/32 of its nuclei undecayed. The question, however, asks about all of the nuclei, both decayed and undecayed. B, D: These choices both involve the differences in half-live between 60Co (5 years) and X (1 year). However, since the question is asking about undecayed and decayed material, this distinction does not matter.

absolute configuration

describes the exact spatial arrangement of groups of atoms independent of other molecules. R or S

1. In the presence of a carbohydrate, which of the following statements regarding the enthalpy and entropy of the gel-to-liquid-crystal transition in DPPC is true? A. The entropy of transition decreases, while the enthalpy of transition increases. B. The entropy of transition increases, while the enthalpy of transition decreases. C. Both the entropy and the enthalpy of transition increase. D. Both the entropy and the enthalpy of transition decrease.

D is correct. According to the experimental data summarized in Table 1, addition of any carbohydrate leads to an entropy change, ΔS, for the phase transition of an aqueous dispersion of DPPC that is smaller than the entropy change in the absence of a carbohydrate (7.22 cal/K•mol). A, B, C: The above info allows us to eliminate choices B and C. Also, according to Table 1, addition of any carbohydrate leads to an enthalpy change, ΔH, for the phase transition of an aqueous dispersion of DPPC that is smaller than the enthalpy change in the absence of a carbohydrate (22.9 kcal/mol), eliminating choice A and making choice D the correct answer.

20. A student theorizes that the differences in ATP-to-ADP ratios between cells detected in the experiment could be due to variations in Y32 expression rather than differences in the cells' metabolic conditions. Does information presented in the passage support this possibility? A. Yes, Y32 fluorescence intensity is proportional to the number of sensor molecules excited. B. Yes, Y32 expression levels can influence the cellular ATP-to-ADP ratio. C. No, the shape of the fluorescence excitation spectrum of Y32 is influenced by the bound state. D. No, the spectral ratio intrinsically normalizes for the amount of biosensor.

D is correct. According to the passage, the spectral ratio of the peaks of the excitation spectrum of a Y32-expressing cell directly reflects its cellular ATP-to-ADP ratio. The fact that a ratio is used is the key to the idea of intrinsic normalization. The peak absorbance of the B-state, which is favored in the Mg2+-ATP bound conformation, is near 500 nm, while the peak absorbance of the A-state, which favors the ADP-bound conformation, is near 420 nm. Thus, the ratio of the fluorescence emission intensities when Y32 is excited at 500 nm versus when it is excited at 420 nm serves as a direct indicator of the cellular ATP-to-ADP ratio. If, as stated in the passage, the fluorescence intensity is proportional to the number of sensor molecules excited, then excitation at both 500 nm and 420 nm would change proportionally with changes in biosensor expression, and would have no impact on determining the ratio in which ATP and ADP are present in the cell. A: While this is true, it does not take into account the proportional increased fluorescence corresponding to both the ATP- and ADP-sensitive peaks. The ratio of ATP to ADP in the cell will not be affected. B: There is no indication in the passage that Y32 expression levels can influence the cellular ATP-to-ADP ratio. C: While this is true, it does not provide a way to distinguish between the two possibilities described in the question stem (i.e. variations across cells, rather than variations within a cell's metabolic activity).

More complete fractionation of proteins using an SEC column could be achieved by using a: A. nonpolar solvent. B. vacuum across the column. C. larger protein sample concentration. D. longer column.

D is correct. As with other forms of chromatography, increasing the column length will enhance the resolution of the column, leading to more completion fraction by SEC. This is because the material of the matrix provides the physical means of separating the proteins. If the proteins come in contact with a longer length of matrix, the differences in retarding forces experienced by the proteins will have a greater cumulative influence on the migration of the proteins, lengthening the differences in their retention times. A: There is no passage information to suggest a relationship between solvent (buffer) polarity and the function of the SEC column. B: Applying a vacuum will increase the flow rate of buffer through the column. Increasing the flow rate will only increase the rate at which eluent is collected; it will not improve separation. C: Increasing the concentration of protein in the sample will likely decrease the resolution of the column, by increasing the chance of overlapping within the fractions of collected eluent.

25.What is the exclusion limit of the SEC column used to create the calibration curve shown in Figure 2? A. 10 kDa B. 100 kDa C. 600 kDa D. 1000 kDa

D is correct. Figure 2 indicates that the exclusion limit of the SEC column is at log (MW) = 6. If true, the molecular weight can be found by taking the inverse log of 6, 106 Da = 1000 kDa. C: If log (MW) = 6, as Figure 2 indicates, then MW = 106 Da = 103 kDa = 1000 kDa, not 600 kDa.

14. A person pushes horizontally on a 50-kg crate, causing it to accelerate from rest and slide across the surface. If the push causes the crate to accelerate at 2.0 m/s2, what is the velocity of the crate after the person has pushed the crate a distance of 6 meters? A. 1 m/s B. 2 m/s C. 3 m/s D. 5 m/s

D is correct. Since the crate is initially at rest, the Vi = 0 m/s, the acceleration a = 2.0 m/s2, and the displacement = 6 m. The appropriate kinematic equation is vf2 = vi2 + 2ad. Substituting the variables gives: vf2 = 02 + 2(2)(6) vf2 = 24 vf = ~5 m/s

38. Which pair of enzymes, if used simultaneously, will produce the greatest amount of glucose if the experiment is repeated? A. A and B B. A and C C. B and C D. C and D

D is correct. The greatest glucose production per mass of starch will occur when there is the greatest rate of starch breakdown. For complete breakdown, this requires both enzymes that break the 1,4 linkages in linear sequences and those that break the 1,6 linkages at branch points. The enzymes that are most active at the 1,4 sites will cause the largest amount of current, as they will produce the most glucose 6-phosphate from the available linear portions of the starch. The enzymes most active at the 1,6 linkages will result in the largest absorbance of light at the peak wavelength (570 nm). Based on paragraph 3, iodine binds linear polysaccharides, so as the 1,6 linkages are broken, more linear polysaccharides are formed; this will lead to more iodine-starch interactions and an increased absorbance. Thus, the combination that maximizes total breakdown should have the maximum current and maximum absorbance at 570 nm, which matches to enzymes C and D, respectively. A, B: Enzyme A produces no current at all (0.0 mA) and does not have the highest absorbance at 570 nm of the enzymes listed. As such, we should not include it in the pair of enzymes we choose. C: This answer is tempting, as enzymes B and C are the two enzymes that produce the largest amount of current! However, current is only one aspect of this question, which specifically asks about glucose production, not current production. Paragraph 2 mentions that starch-powered batteries must break down both "the linear 1,4-D-glucose linkages and the 1,6-D-glucose linkages found at branch points." We thus need to include an enzyme that will break 1,6 linkages; otherwise, starch will not be fully broken down and less glucose will be produced. The passage tells us that iodine selectively binds linear (unbranched) polysaccharides, so the greater the absorbance of 570-nm light, the "more purple" the solution and the more linear the starch molecules must be. Thus, the enzyme that produces the highest absorbance must be the enzyme which breaks the 1,6 linkages at branch points. Initial absorbance values are very similar across the four solutions, but absorbance at 2 minutes is clearly highest in the solution that contains enzyme D. Thus, enzyme D should be included in our answer.

9. The high-energy radiation produced by the γ rays has sufficient energy to: I. generate free radicals. II. excite electrons to higher energy levels. III. eject electrons from molecular orbitals. A. I only B. I and II only C. II and III only D. I, II, and III

D is correct. The second paragraph tells us that the radiation is strong enough to cause molecular electronic transitions by exciting electrons to higher energy levels in molecular orbitals. This indicates that that the radiation can either excite or eject electrons (depending on its energy) and can create free radicals (atoms with unpaired valence electrons). These properties correspond to Roman numerals I, II and III.

55. An artificial leg designed for use by runners is spring-based, to mimic the compression required of a muscle during hard running. For safety reasons, it was determined that the leg should be able to absorb as much as 125 J of kinetic energy without compressing more than 10 cm, or the runner would be likely to stumble. What should the spring constant be? A. 250 N/m B. 2,500 N/m C. 12,500 N/m D. 25,000 N/m

D is correct. When the leg "absorbs" kinetic energy, it is converted to elastic potential energy. While this process involves the loss of some energy as heat, we can assume that it is a perfectly elastic process here for the sake of simplicity. Thus, the leg needs to hold up to 125 J of elastic PE. The formula for potential energy contained in a spring is PE = (1/2) kx2. Since PE and KE are interconverted as the individual runs, we can write this formula as KE = (1/2) kx2, which can be rearranged to yield k = 2 KE/x2. k = (2)(125 J) / (0.10 m)2 k = (250 J) / (.01 m2) k = 25,000 N/m

cellulose

In contrast, a polysaccharide known as cellulose plays a major structural role in the cell walls of plants. Like starch and glycogen, it is a polymer of glucose, but unlike them, it incorporates the β-anomer of glucose, with glucose subunits connected by β(1→4) glycosidic bonds. This seemingly small structural distinction makes a world of practical difference, as humans lack the necessary enzymes to digest cellulose.

Ka (Enzyme Kinetics)

Km is not the only way to measure enzyme affinity. The association constant (Ka) can also be defined, using the mathematical formalism of equilibrium constants, as [ES]/[E][S], where [ES] is the concentration of the enzyme-substrate complex, [E] is the concentration of the enzyme, and [S] is the concentration of the substrate. The dissociation constant (Kd) is then the inverse of Ka, and can be defined as [E][S]/[ES].

Third Law of Thermodynamics

No system can reach absolute zero Entropy at absolute zero is zero

oncogenes versus tumor suppressor genes

The basic difference between them is that oncogenes function to promote abnormal growth and proliferation, leading to cancer, while tumor suppressor genes function to prevent tumorigenic properties. Oncogenes can arise from the mutation of other genes, termed proto-oncogenes. If not mutated, proto-oncogenes do not promote cancer, but certain mutations or inappropriately elevated gene expression can effectively turn them into oncogenes.

ways to estimate entropy

Reactions that increase the number of moles of substances in the system (or produce more gas particles) typically increase the entropy of the system. Entropy generally increases when a solid or liquid is dissolved in a solvent. Entropy increases when the solubility of a gas decreases and it escapes from a solvent. Entropy generally increases as molecular complexity increases (KOH vs. Ca(OH)2) due to the increased movement of electrons.

SDS-PAGE

SDS-PAGE is an electrophoretic technique which involves the binding of the anionic detergent SDS to a polypeptide chain. SDS binding denatures and imparts an even distribution of charge per unit mass to the protein, resulting in fractionation by approximate size alone during electrophoresis. when the highly anionic SDS associates with the polypeptide backbone, the intrinsic charge of the polypeptide becomes negligible in comparison to the negative charges due to SDS. Since the protein is now highly negative, it will travel toward the positive end of the gel apparatus. Otherwise, SDS-PAGE functions similarly to standard gel electrophoresis. The larger the protein, the more hindered it is as it moves down the gel, and the shorter the distance it travels toward the positive end. In contrast, smaller proteins can travel through the pores of the gel more easily, so they migrate farther toward the positive pole.

Solvents for SN1 and SN2:

SN1: polar protic SN2: polar aprotic

Log values

Some log values to be familiar with are log(0.01) = -2, log(0.1) = -1, log(1) = 0, log(3) ≈ 0.5, log(10) = 1, and log(100) = 2

Ionization energy

The 2p orbital is slightly higher in energy than the 2s orbital, making it slightly easier to ionize this electron. (beryllium versus boron)

Bohr model

The Bohr model sees electrons as orbiting the nucleus in spherical shells that vary in their distance from the nucleus. Positively-charged protons in the nucleus exert an attractive force on electrons, preventing them from straying, although they remain in constant motion with quantized angular momentum. The electrons closer to the nucleus experience the greatest attractive force, so the closer an electron is to the nucleus, the greater its stability and the lower its energy level. In contrast, electrons located farther from the nucleus will have a higher energy and be less stable.

relative configuration

The Spatial arrangement of groups in a chiral molecule compared to another chiral molecule. D or L

glycosidic bond

bond formed by a dehydration reaction between two monosaccharides -when the anomeric carbon of one sugar reacts with a hydroxyl group in another sugar. -ether is formed Moreover, the formation of a glycosidic bond transforms the hemiacetal or hemiketal found at the anomeric carbon into an acetal or a ketal, because the -OH group that is characteristic of a hemiacetal/hemiketal is transformed into a second -OR group, which defines an acetal/ketal.

springs

he potential energy stored in a spring can be expressed as PEelastic = ½kx2, where k is a spring constant that is specific for each spring and can be thought of an indicator of its stiffness and x is how far the spring has been stretched or compressed. Thus, the more a spring is compressed, the more energy will be stored in it, and energy increases nonlinearly with compression or extension. However, the force needed to compress or stretch a spring by a distance of x is a linear relationship: F = kx, where k is the spring constant. This is known as Hooke's law. Additionally, springs can be used to generate periodic motion. The time T that separates adjacent peaks on a graph of periodic motion is known as the period, and for a mass on a spring undergoing periodic motion, T = 2π√(m/k), where m is the mass and k is the spring constant.

Zeff trend

increases up and to the right

nitrogen and oxygen ionization energies

nitrogen has a higher first IE than oxygen because of spin pairing. the lower oxygen IE represents the amount of destabilization that occurs for the p-type orbital containing two electrons of opposite spin

sigma and pi bonds

sigma bonds have lower energy, but are stronger and more stable. pi bonds have higher energy, but decreased strength and stability. A single bond—consisting of two electrons—between two atoms will form a sigma (σ) bond, which has an overlapping region of electron density. Pi (π) bonds occur between two parallel p orbitals and are weaker than σ bonds. Double bonds consist of one π bond and one σ bond, while triple bonds include two π bonds and one σ bond.

ionic radius

size of an ion. cations smaller, anions bigger

glycogen

structurally, glycogen is similar to amylopectin in that it contains chains of glucose molecules connected by α(1→4) glycosidic bonds, with intervening α(1→6) glycosidic bonds that create branches; the main difference is that glycogen is more heavily branched than amylopectin, with branches occurring every 8−12 units.