NS FL 3

A quick guide to identifying the orbital hybridization of a molecule is to determine the number of regions of electron density around the atom. A region of electron density is defined as either a bond (single, double, or triple) or a lone pair of electrons. Two regions yield a hybridization of sp (common in cases of triple bonds or central atoms with two double bonds). Having three regions of electron density is associated with sp2 hybridization, and having four regions yields sp3 hybridization.

...

Question 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 Show Explanation C. II and III only Show Explanation D. I, II, and III Show Explanation

B...Got it right but confused... deltaS does NOT have to be larger than deltaH only positive. read the question more carefully

Based on passage information, which of the following are post-transcriptional modifications likely involved in the IAV life cycle? I. Splicing II. Polyadenylation III. Glycosylation A. I only B. I and II only Show Explanation C. I and III only Show Explanation D. I, II, and III Show Explanation

B...Got right but guessed Glycosylation is a post-TRANSLATIONAL modification The passage states that splicing is part of the IAV life cycle, and also states that IAV uses host machinery for other post-transcriptional modifications (I). Since IAV genomic material is synthesized in the nucleus, this most likely includes polyadenylation, which is needed to export mRNA from the nucleus to the cytoplasm (II).

How acidic or basic a solution is can be expressed in terms of pH or pOH, which are defined as follows: pH = −log [H+] and pOH = −log [OH−]. For example, a solution with an H+ concentration of 10−4 M will have a pH of 4, and a solution with an OH− concentration of 10−9 M will have a pOH of 9. pH and pOH values can be estimated given a certain concentration using the following shortcut: p(N × 10−M) = (M−1).(10−N), such that a solution with an H+ concentration of 4 × 10−8 will have a pH = (8−1).(10−4) = 7.6.

...

Question 42 If a modern portable defibrillator uses as 12 V battery and a 20 μF capacitor, what is the total charge stored on the plates of the capacitor? A. 0.24 mC Show Explanation B. 24 mC Show Explanation C. 24 C Show Explanation D. 60 C Show Explanation

A...Could not remember equation UNITS BRO UNITS.... Q=VC Capacitance (C, measured in farads) is the amount of charge stored per volt, expressed with the equation C = q/V. This means that charge (q) = VC. (Do you remember the home shopping network, QVC?) The capacitance is given as 20 x 10-6 F. Substituting the voltage (12 V) and capacitance into the equation gives us the charge (in coulombs). (20 x 10-6 F)(1.2 x 101 V) = 24 x 10-5 C = 2.4 x 10-4 C = 0.24 x 10-3 C = 0.24 mC Remember when making the exponent larger, we must make the coefficient smaller by the same factor.

Question 4 If the Gibbs free energy of the liquid-crystalline phase for an aqueous dispersion of DPCC and raffinose is 17.9 kcal/mol, which of the following values is nearest the Gibbs free energy of the associated gel phase? A. 3.5 kcal/mol Show Explanation B. 32.3 kcal/mol Show Explanation C. 566.1 kcal/mol Show Explanation D. 601.9 kcal/mol Show Explanation

A...Got right but guessed deltaG=deltaH-TdeltaS By rearranging Equation 1, ∆G = Gl - Gg, an expression for the Gibbs free energy of the gel phase may be found: Gg = Gl - ∆G. The Gibbs free energy change for the phase transition of raffinose is given by ∆G = ∆H - T∆S. Substituting thermodynamic values from Table 1: ∆G = (16.0kcal/mol) - [(3.15 x 102 K)(5.05 x 10-3 kcal/K•mol)] ∆G = 16.0 kcal/mol - 1.60 kcal/mol = 14.4 kcal/mol Substituting into the rearranged form of Equation 1: Gg = Gl -∆G = 17.9 kcal/mol - 14.4 kcal/mol = 3.5 kcal/mol.

Question 35 Salicylic acid is a benzoic acid derivative which can most directly be obtained though the Kolbe-Schmitt carboxylation (shown below). This is most likely to work on: A. a phenol. B. a phenyl aldehyde. Show Explanation C. a phenyl ether. Show Explanation D. a phenyl ester. Show Explanation

A...Guessed did not know CARBOXYLATION = adding a -COO group... Do not panic if you have not heard of this mechanism; the goal of the MCAT is not to test random mechanisms! We are told that the reaction in question is a carboxylation of the original reactant,indicating that we are adding a -COO group to form salicylic acid. We can see from Reaction 1 that salicylic acid not only has a carboxylic acid functional group,but also a hydroxyl group attached at what is likely the ortho position. Phenol is a benzene ring with a hydroxyl group attached.

Question 2 The instance of nondisjunction for the X chromosome in females over the age of 30 is about one out of every 130 live births. If a woman over 30 gives birth to a viable baby, assuming the risk of nondisjunction from the father is negligible, what is the likelihood that it will have a normal phenotype? A. 99.2% Show Explanation B. 99.6% Show Explanation C. 0.6% Show Explanation D. 0.4% Show Explanation

A...Guessed got right 1/130 use math shortcuts then subtract from 1 This question asks us to find the probability of a mother giving birth to have a phenotypically normal child. The chances of nondisjunction are 1/130 (0.008), or about 0.8%. If nondisjunction occurs, there is a 100% chance that she will pass on either no copies of her X chromosome and have a child with Turner syndrome (XO) or pass on two copies of the X chromosomes and have a baby with Trisomy X (XXX) or Klinefelter syndrome (see the Punnett square below). The only viable monosomy is the XO genotype. A YO genotype is not viable, meaning we cannot include it in our calculations, since the question asks us to assume a viable birth. Therefore, we simply take 1 - (probability of nondisjunction) = 1 - 0.008 = 0.992 = 99.2%. Note that nondisjunction refers to the failure of chromosomes or chromatids to properly separate during anaphase. Nondisjunction can occur during meiosis I or meiosis II. In the diagram below, the left image shows nondisjunction during meiosis II, while the right-hand image shows the same event occurring during meiosis I.

Question 41 The electrical resistance of dry skin is 100 kΩ, but can be lowered to 20 Ω if electrode contact area is large and conducting gel is used on the skin. If Vmax of the defibrillatror 500 V and lasts 0.01 s, what is the maximum possible current delivered to the heart during defibrillation? A. 2.5 × 101 A Show Explanation B. 2.5 × 10-2 A Show Explanation C. 5.0 × 10-2 A Show Explanation D. 5 A Show Explanation

A...Wrong, Could not remember formula LOOK AT WHAT THE EQUATION GIVES YOU AND LOOK AT THE UNITS IN THE ANSWER CHOICES Solving for I given V and R EASY V=IR in amps We are given R, V, and time, and the question asked us to determine current (charge/time, or C/s). We do not need to use the time since we should know Ohm's law, V = IR. If I = V/R and the question asks for the maximum current, we should use the smallest available resistance. I =V/R = (500 V) / (20 Ω) = (50) / (2) = 25 amps = 2.5 × 101 A

Question 48 During strenuous exercise, lactic acid buildup in cells causes the creation of a hydronium complex known as the Eigen cation (H9O4+). If water molecules then experience hydrogen bond attractions to the Eigen cation, this attractive force: A. results in a semi-stable shell of water molecules around the hydronium. Show Explanation B. results in an inability of hydronium to be neutralized by bases the way normal H+ ions would be. Show Explanation C. results in the ability of muscle cells to reverse both the hydronium creation process and the lactic acid creation process once sufficient oxygen is once again made available. Show Explanation D. results in mobility of hydronium within the environment surpassing the mobility of regular water molecules. Show Explanation

A...Wrong, concept wrong A process known as "hydration" or "solvation" occurs when the attractive force of an ion molecule causes a thin shell of water molecules to surround it. In the case of hydronium (H3O+), each of the H atoms attracts the O atom in an H2O molecule due to hydrogen bonding. These H2O molecules cause a "shell" of water molecules to surround the hydronium.

Question 16 Which of the following is least likely to be observed in a patient experiencing hyperventilation? A. Hypoxia Show Explanation B. Net exhalation of CO2 Show Explanation C. Increased blood pH Show Explanation D. Increased hemoglobin O2 affinity Show Explanation

A...got right but wanted to choose D Increased hemoglobin O2 affinity is from loss of CO2 therefore hyperventilation...Hypoxia is oxygen deprivation. Someone hyperventilating would have an increase of O2. THEREFORE someone would NOT be hypoxia This question asks us to determine the effects of hyperventilation. During hyperventilation, there is a loss of CO2 and an increase in O2 in the blood. Hypoxia is another term for oxygen deprivation.

Question 20 Which of the following processes would NOT be a likely target for anti-IAV therapy? A. Proton influx via ion channels in the cell membrane Show Explanation B. Interactions between hemagglutinin and sialic acid-galactose dimers Show Explanation C. Activity of the viral polymerase complex Show Explanation D. Neuraminase activity Show Explanation

A..Got wrong,did not read passage The passage describes proton influx through ion channels in the endosomal membrane, and gives no indication that proton influx via ion channels in the cell membrane is involved in the IAV life cycle.

Question 13 Which of the following types of electromagnetic radiation would have the shortest wavelength? A. Radiation that ejects an electron from an sp orbital Show Explanation B. Radiation that ejects an electron from an sp2 orbital Show Explanation C. Radiation that ejects an electron from an sp3 orbital Show Explanation D. Radiation that excites but does not eject an electron from an sp3 orbital Show Explanation

A..Got wrong...did not know SHORTER WAVELENGTH = HIGHER ENERGY Closer an e- is to the nucleus the harder it is to eject Ejecting an e- from sp3 orbital requires less energy and easier so cannot be C. This question asks us about the wavelength of EMR that ejects an electron from an atom. Shorter-wavelength EMR (such as γ rays)carries much more energy than longer-wavelength EMR (such as radio waves). Therefore, we must look for the answer choice that involves the highest-energy EMR. The closer an electron is to the nucleus, the harder it is to eject. Because sp-hybridized orbitals have the most s character of all of the answer choices, they contain the electrons that are hardest to eject

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

B...Did not know how to do this 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 student assisting with the experiment would observe all of the following about the electron transport chain EXCEPT: A. Electrons are passed from carriers with lower reduction potential to those with higher reduction potential. Show Explanation B. Complex I is also a proton pump. Show Explanation C. All electron carriers are mobile and hydrophobic. Show Explanation D. The electron carriers can transport a maximum of 2 electrons. Show Explanation

B...Got Wrong..Content NADH Dehydrogenase is a proton pump . All electron carriers are mobile and hydrophobic. In the ETC, carriers travel inside the inner mitochondrial membrane, passing electrons from one to another and pumping protons across the inner mitochondrial membrane. Therefore, they are mobile. In order to travel inside the hydrophobic interior of the membrane, we would expect them to all be hydrophobic. However, cytochrome c is a highly water-soluble protein, unlike other cytochromes.

Question 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 Show Explanation B. 37 W Show Explanation C. 46 W Show Explanation D. 331 W Show Explanation

B...Got right but guessed Need to apply all aspects of problem. READ UNITS AND QUESTION 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.

Question 18 Conn's syndrome, also known as primary hyperaldosteronism, is most likely to cause which symptom? A. High renin concentration Show Explanation B. Low blood potassium Show Explanation C. Low blood sodium Hypersecretion of aldosterone will result in high blood sodium. D. Hypotension Show Explanation

B...Got right but was torn between B and C ALDOSTERONE increases BP through Na+ reabsorption from the kidney. K+ secretion therefore increases to balance Na+ reabsoprtion This question is asking you to recall the effects of aldosterone and how it achieves those effects. Aldosterone increases H2O and Na+ reabsorption from the kidney while exchanging K+ ions for Na+ ions. The triggers for and results of aldosterone secretion are shown below.

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

B...Got right guessed Need to determine limiting reagant first using stochiometry. After finding limiting reagent solve for theoretical yield using g/mol of product.. PERCENT YIELD = EXPERIMENTAL/THEORETICAL x 100 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 instep 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%.

Question 58 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. Show Explanation B. The electron-donating groups around the basic nitrogen on arginine make its conjugate acid more stable. Show Explanation C. The lack of electron-donating groups on lysine make its conjugate acid more stable. Show Explanation D. The lack of electron-withdrawing groups on lysine make its conjugate base more stable. Show Explanation

B...Got right...Guessed MORE STABLE CONJUGATE ACID = STRONGER BASE 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. The resonance structures of arginine at this position are shown below. Note that the backbone amino and carboxylic acid groups are deprotonated, meaning that this is the structure of arginine at relatively high pH (albeit not high enough to deprotonate the side chain).

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 Show Explanation B. 1.7 x 106 J Show Explanation C. 4.7 x 106 J Show Explanation D. 4.7 x 109 J Show Explanation

B...Got wrong..did not know how to use formula...LOOK AT UNITS kJ/mol to solve for J 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

Question 6 A defect in the absorption of certain nutrients in the small intestine has been linked to an abnormally-delayed initiation of differentiation. If the levels of HER2, PTEN, and Akt protein can be estimated accurately, what is the earliest point in development at which the defect could be evaluated? A. 14 weeks Show Explanation B. 16 weeks Show Explanation C. 20 weeks Show Explanation D. 24 weeks Show Explanation

B...wrong, did not read the question correctly. ASKING for differentiation...PTEN refers to apoptosis and cell death From Figure 1, it is clear that the initiation in differentiation results from the downstream signaling from Akt, and, thus, differentiation would not occur until the levels of Akt expression increase. From Figure 2, it should be noted that Akt levels are first noticed in the small intestine during week 16; thus, you would not expect differentiation to occur before week 16.

Question 17 Which of the following is NOT considered an organic acid? A. Folic acid Show Explanation B. Carbonic acid Show Explanation C. Ascorbic acid Show Explanation D. Citric acid Show Explanation

B..Got right but confused An organic compound must contain carbon and hydrogen in its formula. Furthermore, there must be a covalent bond between a carbon and hydrogen atom in the molecular structure. Organic acids are weak acids, generally having formulas of R-CO2H, with the acidic hydrogen bonded to an oxygen atom. Even knowing this information, you may be intimidated, as you may not be familiar with the structures of all of these choices. However, you can begin with the most familiar compound, carbonic acid. The Lewis structure of this compound is shown below. From this structure alone, we can see that neither hydrogen atom is directly bound to carbon, so carbonic acid is not an organic acid and must be the correct answer here.

Question 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. Show Explanation B. Its absence decreases the minimum radial size of alveoli able to avoid collapse at a given pressure of inspired air. Show Explanation C. Its adsorption to the water-alveolar interface decreases surface tension, decreasing the pressure difference required to inflate the airway. Show Explanation 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. Show Explanation

C...Got it right but guessed Inspiration-->diapgragm and muscles contract increasing volume of lung cavity...decreasing pressure AIR FLOWS INTO LUNGS BECAUSE OF NEGATIVE PRESSURE Expiration-->diaphram relaxes lung volume decreases...pressure increase PUSHES AIR OUT OF THE LUNGS Asthma increases resistance making it harder to breath Parasympathetic Stimulation and Histamine narrow bronchioles, increasing resistance and decreasing air flow Epinephrine dilates bronchioles and increases air flow Pulmonary surfactant adsorbed 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. In general, surfactant molecules are amphipathic, meaning that they contain both hydrophobic and hydrophilic regions. The diagram below shows surfactant molecules surrounding a micelle of oil. The hydrophobic tails of the surfactant molecules mix well with the hydrophobic oil, while the hydrophilic heads point away from the oil droplet.

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 Show Explanation B. 1:5 Show Explanation C. 1:1 Show Explanation D. 5:1 Show Explanation

C...Got right but almost got wrong. QUESTION IS ASKING THE EXPECTED RATIO OF TOTAL ATOMIC MASS AFTER 5 YEARS DOESN"T SAY ANYTHING ABOUT DECAY SO DON'T ANSWER DECAY 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.

Question 7 The transition from N to O offers an exception to the trend for first ionization energy due to Hund's rules of spin pairing. If this spin pairing was not present, what would be the expected first ionization energy for the p-orbital on an oxygen atom? A. 400 kJ/mol This is the energy difference between the expected and actual IE value for oxygen. B. 1300 kJ/mol Show Explanation C. 1700 kJ/mol Show Explanation D. 2000 kJ/mol Show Explanation

C...Got wrong, READ CAREFULLY. Doesn't ask for actually IE value but if spin pairing was not present. In Figure 1, the trend exception can be seen in the dip between group 5 (N) and group 6 (O). If we examine the trend for the first 6 elements, we can see that ionization energy goes from about 500 (Li) to 1500 (N).This works out to a slope of about 200 kJ/mol per element. If spin pairing did not occur, we would most likely expect this trend to continue, and the IE for oxygen would be about 1500 + 200 = 1700 kJ/mol. The actual value shown is about 1300 kJ/mol, meaning spin pairing allows for an IE difference of about 400 kJ/mol. This energy represents the amount of destabilization that occurs for the p-type orbital containing two electrons of opposite spin.

Question 36 What is the standard cell potential for the starch battery created? A. -1.136 V Show Explanation B. -0.916 V Galvanic (spontaneous) cells do not have negative standard cell potentials, so this choice can be eliminated. We know the starch battery is meant to act like a galvanic cell because it is described as "generat[ing] sufficient power," meaning that it must run spontaneously. C. 0.916 V Show Explanation D. 1.136 V Show Explanation

C...Wrong, did not know that it was a galvanic cell Emf=Ered(cat)-Ered(an) The passage states that the cell is meant to act like a galvanic cell, meaning that it proceeds in a spontaneous fashion. Since galvanic cells always have cell potentials that are greater than 0, we can eliminate any negative options (choices A and B) immediately. To decide between C and D, we need more information, namely the reduction potentials of the species involved. Paragraph 1 gives E°reduction of O2 as 0.816 V and E°reduction of AQDS as -0.10 V.

Question 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 Show Explanation B. When she takes long, low strides and keeps her spine fairly vertical Show Explanation C. When she takes high, bouncing strides and leans her upper half into her run Show Explanation D. When she takes long, low strides and leans her upper half into her run Show Explanation

C...Wrong, hard question did not truly understand passage and what it was asking 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.

Question 52 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 Show Explanation B. 0 J Show Explanation C. 13 J Show Explanation D. 14.7 J Show Explanation

C...Wrong, stupid question Wants us to assume that no elastic springs truly exist so the energy drop is MORE significant than you THOUGHT This is a tricky question, because the passage states that nature, i.e., biology, 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. So, if energy were completely conserved, what would it be? Potential energy of the rabbit at the peak of its height is PE = (3)(10)(0.5) = 15 J, or, if we're more exact and use 9.81 for g, slightly less than that. 14.7 J seems about right. 13 J is clearly a little on the low side, but this is exactly what's needed for this question, since the question stem is dropping a big hint that some energy is lost.

Question 56 A balloon has a volume of 3.0 L at 25°C. What is the approximate volume of the balloon at 50°C? A. 1.5 L Show Explanation B. 2.0 L Show Explanation C. 3.3 L Show Explanation D. 6.0 L Show Explanation

C...Wrong..did not convert to Kelvin K IS THE UNITS FOR THERMO Charles' law states that there is a direct relationship between the volume of an ideal gas and its temperature, when pressure is constant. Note that the temperature must be in Kelvin! We can approximate the initial temperature as 300 K and the final temperature as 320 K. (3.0 L)/(300 K) = V2/(320 K) [(3.0)(320)] / (300) = (3.0)(1.1) = 3.3 L

Question 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 Show Explanation B. Purplish color but no current because of incomplete glycolysis Show Explanation C. Brownish color with higher initial current Show Explanation D. Brownish color with lower initial current Show Explanation

C...wrong, did not know how to solve second part BIOCHEMISTRY KNOWLEGE. Glucose from Glycolysis produces 2NADH while this cell only produces 1 so higher initial current 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.

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

D...Got wrong, didn't read graph carefully. GRAPH STATES log(mW)=kDa SO NEED TO TAKE INVERSE LOG OF 6 10^6= 1000kDa

Question 44 What is the net charge on a phenylalanine molecule at pH 1? A. -1 Show Explanation B. 0 This would be the charge on a phenylalanine molecule at or around physiological pH (7.4). C. +0.5 Show Explanation D. +1 Show Explanation

D...Wrong , guessed pH=1 highly acidic MEANING THINGS WILL BE PRONATED This question is asking us to remember what factors would contribute to the net charge on a phenylalanine molecule. Being an amino acid, phenylalanine has an acidic carboxy group that will be protonated at a pH of 1 (remember, such a pH is highly acidic). Additionally, it has a basic amino group that will also be protonated. Finally, it contains a neutral toluene side chain. In total, the charge will be (0 from the carboxy group) + (+1 from the amino group) + (0 from the side chain) = +1. The image below depicts phenylalanine at this pH.

Question 46 How many liters of carbon dioxide at STP are produced by reacting 100 g of calcium carbonate with an excess of hydrochloric acid? A. 0.0 L Show Explanation B. 11.2 L Show Explanation C. 22.4 L Show Explanation D. 44.8 L Show Explanation

D...Wrong, calculated wrong 22.4L=1molSTP The balanced chemical reaction is: CaCO3 (s) + 2 HCl (aq) → CO2 (g) + H2O (l) + CaCl2 (aq) From the periodic table, the formula weight of calcium carbonate is 40 + 12 + 3(16) = 100 g/mol. 100 g of calcium carbonate therefore represents one mole, and based on the reaction, this will produce one mole of carbon dioxide gas. Remember, one mole of any gas at STP (standard temperature and pressure) has a volume of 22.4 L! Thus, the volume of gas produced will be 22.4 L at STP. 100 g CaCO3 x 1 mol/100 g x 1 CO2/1 CaCO3 x 22.4 L/mol = 22.4 L CO2

Question 15 Which of the following particles is expected to have the LEAST mass? A. An alpha particle Show Explanation B. A beta particle Show Explanation C. A positron Show Explanation D. A gamma particle Show Explanation

D..Got right but confused A GAMMA PARTICLE IS A PHOTON WHICH HAS NO MASS. Alpha Decay: emmits He4,2) lowers atomic mass by -4 atomic number by -2 Beta (-) decay: ejects e-...Increases atomic # Beta (+) decay: ejects positron...decreases atomic # Electron Caption: e- shifts to lower, decreases atomic # Gamma decay: photon is emitted, no change to atomic #

Question 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 Show Explanation B. n = 2 to n = 4 Show Explanation C. n = 2 to n = 1 Show Explanation D. n = 4 to n = 2 Show Explanation

D..Got right but confused VISIBLE SPECTRUM IS 400-700nm Using the formula given in the passage En = -RH/n2 Where RH= 1.097*10^7 m-1 CALCULATE EMISSION OF C AND D...C DOES NOT FALL IN VISIBLE SPECTRUM 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 (∆E) 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. Using equation 1 and approximating the Rydberg constant as 1 x 107 m-1. The energy of the n = 4 level gives: E4 = -1 x 107 m-1 / 42 = -1/16 x 107 m-1 ∆E = E2 - E4 = (-4/16 x 107 m-1) - (-1/16 x 107 m-1) = -3/16 x 107 m-1 To make the fraction easier to work with, round 3/16 to 3/15, or 1/5. Again, we must convert wavenumbers to nanometers: 1 / (1/5 x 107 m-1) = 5 x 10-7 m (5 x 10-7 m)(1 x 109 nm/m) = 500 nm While this is also an approximation, it falls well within the visible spectrum. The actual color of the n = 4 to n = 2 transition of hydrogen is teal (blue-green), with a wavelength of 486 nm.