MCAT NS C/P Review

The piezoelectric effect, when generating ultrasound waves, involves the conversion of: passage info: To produce an U/S image, an alternating current is applied across a piezoelectric crystal. The piezoelectric crystal grows and shrinks depending on the voltage applied to it. The an alternating current causes the crystal to vibrate at a high speed and to produce the U/S signal. This is known as the piezoelectric effect. A. chemical energy to mechanical energy. B. electrical energy to potential energy. C. kinetic energy to electrical energy. D. electrical energy to mechanical energy.

** I saw vibration and directly thought its relation to kinetic energy. did not consider initially require current voltage generate current -> the source of energy -> electrical energy => eliminate A and C crystal vibration occur after current hit the crystal -> result energy -> vibration is mechanical energy => eliminate B choice C is wrong - if this answer choice reverse, it will be the correct answer correct answer is D

All of the following phenomena serve to attenuate the ultrasound signal as it passes through the body EXCEPT: A. absorption. B. refraction. C. scattering. D. amplification.

** does not know what attenuation mean attenuation mean reduction of signal. sound signal attenuate d/t reflection, refraction/diffraction, scattering, absorption. amplification is unlikely b/c signal enhancement will happen choice C is wrong -Scattering is when the waves are scattered in all directions in a non-uniform manner. This is especially true for very small objects or rough surfaces and results in less signal being able to propagate. correct answer is D

If an IR absorption spectrum were obtained for the dimer in Figure 1, a strong signal at 1750 cm-1 would most likely indicate the presence of: A. an O-H bond. B. a C=O bond. C. a C≡C bond. D. a C-O bond.

** forget IR frequency C=O bond stretching typically exhibits a frequency range of 1650-1750 cm-1. choice A is wrong - O-H bond stretching in a carboxylic acid: 2500-3300 choice C is wrong - C≡C bond: 2100-2260 choice D is wrong - C-O bond: 1000-1320 cm-1 choice B is correct

Antioxidant activity is an important function of which of the following vitamins? A. Vitamin B1 B. Vitamin E C. Vitamin D D. Vitamin K

** forget about the concept, and thought vitamin A, D, E K are fat soluble so can be eliminated Vitamin E refers to a set of closely-related, lipid-soluble compounds that function as antioxidants choice A is wrong -Vitamin B1, or thiamine, is a coenzyme in metabolic processes involving amino acids and carbohydrates. choice C is wrong - Vitamin D is a lipid-soluble molecule primarily involved in calcium metabolism choice D is wrong - Vitamin K refers to a group of closely-related molecules that contribute to blood coagulation and clotting. correct answer is B

Which of the following compound-solvent combinations would be expected to show the highest antioxidant activity? ** enthalpy values on table 1 on #18 on word doc A. Compound 1 in benzene B. Compound 2 in water C. Compound 2 in benzene D. Compound 1 in water

** forget less enthalpy means more thermodynamically favorable lower the enthalpy, more active of the reaction as more thermodynamically favorable the reaction is - think about the energy barrier needed to overcome to complete the rxn, lower the barrier, more likely the reaction will complete lower ΔH values are associated with higher antioxidant activity. - compound 1 in water has the lowest enthalpy correct answer is D

Which of the following rankings (from most to least favored) best describes the thermodynamic favorability of the antioxidant mechanisms in benzene? ** enthalpy values on table 1 on #18 on word doc A. SET-PT > SPLET > HAT B. HAT > SPLET = SET-PT C. HAT > SPLET > SET-PT D. SET-PT = SPLET > HAT

** forget less enthalpy means more thermodynamically favorable lower the enthalpy, more thermodynamic favorable of the reaction HAT has the lowest enthalpy looking at enthalpy of SPLEt and SET-PT, SET-PT step 1 enthalpy is higher than SPLET step 1 enthalpy, and this is very unfavorable. also, combination of SET-PT is also close but still higher correct answer is C

Cardiac glycosides like ouabain have long been used in the treatment of heart failure.The above reaction of ouabain would most likely result in: ** reaction is shown in #20 on word doc The above reaction of ouabain would most likely result in: A. the formation of 2 ketones. B. the formation of 1 aldehyde and 1 ketone. C. the formation of 1 carboxylic acid and 1 ketone. D. no reaction.

** forget ozonlysis production this reaction is ozonlysis - the reactant is alkene - the product are ketone and carboxylic acid, or ketone and ketone, or ketone and aldehyde - just think about breaking double bond and adding oxygen to the 2 products choice C is wrong - no OH appear on the structure, so no carboxylic formation correct answer is B

L1 and L2 have the structures shown below. What kind of isomers are L1 and L2? ** structure on #19 on word doc A. Enantiomers B. Diastereomers C. Geometric isomers D. Constitutional isomers

** forget the isomer types Constitutional isomers have the same molecular formula but differ in their connectivity choice A is wrong - no mirror image choice B is wrong - no opposite configuration choice C is wrong - geometric isomer is cis and trans around double bond correct answer is D

If researchers failed to take into account the effect of air resistance on the pitch, how would it impact their measurements of the efficiency of energy transfer from the arm to the baseball? A. It would be lower than the actual efficiency, as there was a higher initial velocity at release. B. It would be higher than the actual efficiency, as there was a higher initial velocity at release. C. It would be lower than the actual efficiency, as there was a lower initial velocity at release. D. It would be higher than the actual efficiency, as there was a lower initial velocity at release.

** guess as running out of time Air resistance would decrease the velocity of the ball as it travels from the mound to home plate, where the velocity was recorded. Thus, the measured velocity should be lower than the velocity at release. The decreased velocity will result in a decreased calculated energy for the baseball, leading to a decreased calculation of efficiency for the energy transfer from the arm to the ball. energy efficiency = energy output/energy input and energy is depend on velocity correct answer is A

When Bs1A is assembled as an octamer, what is most likely to be true regarding L76, L77, and L79? passage info: This region, known as the hydrophobic cap, displays numerous solvent-exposed hydrophobic amino acids; however, when Bs1A monomers assemble as an octamer, these amino acids are shielded from polar solvent molecules by solvent-exposed β-sheet faces of other Bs1A molecules. A. They are oriented toward the solvent-exposed exterior of the protein assembly. B. The assembly would incur an entropic penalty if they occupied a solvent-exposed site. C. Unlike in a monomer, they are not situated within the hydrophobic cap. D. Their physiochemical properites are not substantially dependent on their hydrophobicity.

** have no idea for the question Entropic penalty refers to the thermodynamically disfavored requirement of forming a cage of polar solvent molecules around surface-exposed hydrophobic portions of a molecule. - occur when a small molecule/ligand bind to protein to restrict protein activity and its movement => think about cage an entropic penalty would be incurred if L76, L77, and L79, which in the octamer are sheltered from solvent molecules choice A is wrong - according to passage info, these amino acids are away from the solvent choice C is wrong - passage never states this. it states there is a β-sheet sheilding octamer. thus, residues remain in CAP-1 which stay in hydrophobic cap choice D is wrong - The term "physiochemical property" refers to the physical and chemical properties of the leucine residues, both of which are strongly influenced by the hydrophobic side chains. correct answer is B

It is found that in the absence of molecular oxygen, the resulting imidazolinone-containing reactant is not fluorescent. According to Figure 1, what best explains this inability to fluoresce? ** figure 1 on #24 on word doc passage info: The imidazolinone structure, together with the Tyr group, form a conjugated system of π electrons that cause excitation and emission in CP and GFP-like proteins A. The 5-membered ring is not conjugated with the aromatic phenol ring of tyrosine. B. The tyrosine side chain in the final chromophore remains deprotonated. C. The reactant lacks conjugation among any double bonds. D. The more thermodynamically favorable serine keto tautomer is formed.

** just thought if no double bond, there would be no fluorescent, but did not thought the system is more specific that it form double bond with tyrosine phenol ring the Cα-Cβ bond of tyrosine is oxidized to a double bond in a reaction which consumes molecular oxygen. This double bond places the 5-membered ring into an aromatic system in conjugation with the aromatic phenol ring of the tyrosine side chain. if O2 is not presence, conjugation would not form choice C is wrong - conjugation still appear in the molecule, ex double bond is the tyrosine ring. the double bond btw tyrosine ring and imidazolinone is more specific correct answer is A

Based on results presented in the passage, researchers hoping to alter the appearance of sgBP while maintaining its function as a CP providing a colored appearance would most logically choose to mutate which sgBP residue? ** table 1 is #23 on word doc A. Gln62 B. Glu144 C. Ser157 D. His196

** no understand what question is asking question is asking modify which residue will still result in colored appearance Table 1 shows that only modification of Ser157 resulted in mutant sgBP chromophores that express a different color. choice A is wrong - a single mutant of Gln62 does not result in color change choice B and D are wrong - both mutant are colorless correct answer is C

The light blue appearance of S157T is most directly attributable to what other phenotypic change versus the wild type? ** Table 1 on #23 on word doc A. Increased λ max B. Decreased λmax C. Decreased ɛ D. Increased ɛ

** not paying attention to the trend table 1 shows negative association between wavelength and color: shorter the wavelength, darker the color - ex: S157C (604 nm; dark blue) -> Q62M (608 nm; blue) -> S157T (611 nm; light blue) But we did not observe such trend in ɛ - ex: Q62M (ɛ=122923; blue); S157T (ɛ=85654; light blue); S157C (ɛ=58029; dark blue) correct answer is A

When situated in the substrate affinity pocket of PI3Kα, the morpholine group oxygen of ZSTK474 is most likely to interact with the side chain of what amino acid residue? passage info: Researchers studying the inhibitor noted that the crystal structure of ZSTK474 in complex with PI3Kα shows one of the two morpholine groups of ZSTK474 extending into the substrate affinity pocket of the enzyme, where its oxygen acts as a hydrogen bond acceptor for a primary amine of the enzyme. A. Aspartate B. Cysteine C. Lysine D. Tyrosine

** not realize passage info provide answer hint we are looking for amino acid with primary amine function group -> lysine correct answer is C

If the majority of the baseball's kinetic energy comes from power generation in the legs and hips, approximately how much energy do the lower extremities produce in the pitch? passage info: average velocity of the ball is 30 m/s and its mass is 150 g, which is equivalent to 0.15 kg. ** efficiency is on #17 on word doc A. 65 J B. 70 J C. 140 J D. 810 J

** running out of time first, calculate KE of the ball: KE = 1/2*m*v^2 = 1/2*0.15*900 = 67.5 J this is the final KE of the ball, but the question is asking for kE come from lower extremities, so we need to use efficiency provided by table 1 to calculate KE from lower extremities, aka initial KE total efficiency = (0.9)(0.89)(0.71)(0.82) = 0.5 so, total energy generated at lower extremities: 67.5 J/ 0.5 = 135J correct answer is C

While evaluating the timing of pitches, the researchers noticed a couple of high-performing athletes who were able to generate higher velocities by increasing the length of their acceleration phase. How did the power production for these athletes compare to those originally mentioned? A. These high-performing athletes had an increased power output due to the higher velocities. B. These high-performing athletes had a decreased power output due to the increased time spent in the acceleration phase. C. These high-performing athletes had an increased power output due to the increased time spent in the acceleration phase .D. There may have been no change in power output between the two groups.

** running out of time, and only think increasing time will decrease power power = work/time work = dKE = 1/2mVf^2 -1/2mVi^2 as work is depend on velocity, increasing velocity will increase work, but since the time increase as well, power will remain constant correct answer is D

In comparison to the cohesive forces between water molecules of the protein solution droplet, how can the strength of interaction between water and oil molecules at an oil-water interface be characterized? A. Weaker, because of the larger surface area of the hydrophobic cap B. Weaker, because they are forces created by induced polarity in nonpolar molecules C. Stronger, because of the predominance of Van der Waals forces D. Stronger, because of sequestration of hydrophobic cap residues from the solvent

** thinking opposite The Van der Waals forces that exist between molecules of water and of oil are predominately of the induced dipole-dipole type. Here, a small, temporary dipole is induced in molecules of oil by the permanent dipole of water, resulting in a weak attraction between the molecules. choice A is wrong - increasing surface area will increase the strength of intermolecular force choice C and D are wrong - bonding between water and oil is weaker than bonding between water and hydrophilic portion correct answer is B

A certain type of tissue is sensitive to radiation with the damage the tissue receives being directly proportional to the charge on the irradiating particle. Which of the following radiation types will cause the least damage? A. Gamma B. Positron C. Beta D. Alpha

** thought gamma is super strong so it must contain charges gamma ray does not have charge; they are high-energy photon choice B and C are wrong - both positron and beta have +1 and -1 charge choice D is wrong - alpha particle is +2 charge correct answer is A

A certain metabolic process in the liver produces NADH as a part of the process. If this process is up-regulated, which of the following effects associated with gluconeogenesis is most likely to follow? passage info: the conversion of lactate to pyruvate is coupled with the reduction of NAD+ to NADH. Pyruvate enters gluconeogenesis after being converted to oxaloacetate by pyruvate carboxylase. A. Intracellular levels of oxaloacetate will increase. B. Plasma glucose concentrations will increase significantly. C. The rate of gluconeogenesis in the liver will decrease. D. Plasma glucose concentrations will decrease significantly.

** thought more NADH will lead to increase rate of gluconeogenesis and glucose production while NADH conc increase, less pyruvate will be converted from lactate as NAD+ reduction will less likely to happen, and gluconeogenesis will slow down choice A is wrong - oxaloacetate is product of gluconeogenesis, and the presence of large amt of NADH will cause less oxaloacetate produced choice B and D is wrong - thought reduce rate of gluconeogenesis cause less glucose to be produced, but plasma glucose require knowledge of dietary glucose and glycogen storage that are not provided in the passage

The frequency used in U/S imaging must be greater than which minimum threshold? A. 1 kHz B. 10 kHz. C. 20 kHz. D. 40 MHz.

** very hard question; no idea and thought need wave velocity equation to calculate frequency ultrasound has frequency above our hearing frequency range, so to answer this question requires the knowledge of hearing range, which is 20 - 20000Hz correct answer is C

According to the information presented in the passage, how would the absorbance at 570 nm change over time for enzyme A if a linear starch was used? A. It would remain relatively constant. B. It would decrease. C. It would increase. D. It would initially increase and then decrease as the starch is consumed.

**running out of time** Based on Figure 3, enzyme A increased absorbance with no change in current. Thus, enzyme A likely breaks down 1,6 linkages, producing more linear polysaccharides, thereby increasing absorbance. However, if the initial starch is an entirely linear polysaccharide, then a debranching enzyme like enzyme A would have no effect on absorbance, and it would remain relatively constant over time. - absorbance related to turn branching polysaccharides to linear polysaccharides correct answer is A

Which pair of enzymes, if used simultaneously, will produce the greatest amount of glucose if the experiment is repeated? ** #12 on word doc passage info: Batteries that run on starch must break down the linear 1,4-D-glucose linkages and the 1,6-D-glucose linkages found at branch points (Figure 2), which is a complex, multi-step process. A A. A and B B. A and C C. B and C D. C and D

**running out of time** 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. greater the absorbance mean more branch plysaccharides is broken down to linear form - enzyme D has the greatest absorbance choice C is wrong - should look at absorbance at 2 mins, not initial absorbance correct answer is D

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? ** #12 on word doc passage info: Iodine selectively binds linear-chain polysaccharides, and when bound, it changes the color of the solution from brown to purple with a characteristic peak absorption wavelength of 570 nm corresponding to the purple color. 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

**running out of time** according to passage info, iodine will bind to linear chain polysaccharides and change color (brown to purple). however, glucose is monosaccharides, so iodine color will remain purple - eliminate A and B 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. - 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. - eliminate D correct answer is C

In 250 mL of the MH solution with the most favorable solubility profile, how many moles of nicotinamide (MW = 122 g/mol) are present? passage info: Researchers investigated a novel method called mixed hydrotropy (MH) to increase the aqueous solubility of Aceclofenac (A), a non-steroidal anti-inflammatory drug. Hydrotropy refers to the addition of a second solute (the hydrotrope) to a solution to increase the solubility of a first solute. researchers investigated the optimal ratio of individual hydrotropes by measuring the solubility of A in several MH combinations using nicotinamide (N), sodium benzoate (B) and trisodium citrate (C). ** table on #22 on word doc A. 3.1 x 10-2 B. 3.1 x 10-1 C. 1.6 x 10-2 D. 1.6 x 10-1

**use solubility of substance A to calculate, should use hydrotrope concentration to calculate as nicotinamide is hydrotrope 40 mg/ml * 15/40 *250 mL = 4000 mg nicotimide 4 g * mol/122g = 0.033 mol correct answer is A

What is the approximate partial pressure of oxygen at 1500 m? at 1500m: P = 641 mmHg = 85 kPa A. 0.09 atm B. 0.11 atm C. 0.15 atm D. 0.18 atm

641 mmHg * 1 atm/ 760 mmHg = 0.84 atm oxygen is 21% in the air 0.84 atm *0.21 = 0.18 atm correct answer is D

An additional experiment on the bovine extracts using a known ADAMTS inhibitor would serve as: A. a positive control. B. a negative control. C. a randomized control. D. a false negative.

A positive control is a control group that is not exposed to the experimental treatment but that is exposed to some other treatment that is known to produce the expected effect. -> known to have a certain effect and can therefore be used to assess whether the experimental methodology was sound The scientists wished to test the effect of the analogs on enzyme activity. Using a known inhibitor is an example of comparing the expected effects of this inhibitor to those of the analog treatment. Thus, it is a positive control. choice B is wrong -> Negative controls are treatments that are known to have no effect correct answer is A

Cis-trans isomerism

A special type of E/Z isomerism in which there is a non-hydrogen group and a hydrogen atom on each C of a C=C double bond: the cis isomer (Z isomer) has the H atoms on each carbon on the same side; the trans isomer (E isomer) has the H atoms on each carbon on different sides of the bond.

closed system

A system in which no matter/mass is allowed to enter or leave, but energy can

isolated system

A system that can exchange neither energy nor matter with its surroundings.

Atherosclerosis, a particular type of arteriosclerosis, usually involves either complete blockage or severe restriction of blood flow due to high cholesterol and buildup of plaque in the arteries. In extreme cases, constricted arteries may collapse under pressure from outside the arteries. This can best be explained by: A. an increase in blood viscosity resulting in an extreme increase in blood flow. B. the necessary increase in fluid velocity through the constricted arterial sections causing a corresponding drop in fluid pressure. C. the necessary decrease in fluid velocity through the constricted arterial sections causing a corresponding increase in fluid pressure. D. the increased work required to push through increasing the potential energy density in constricted sections, causing a corresponding pressure drop.

According to the continuity equation (A1V1 = A2V2, and Q is constant), velocity increases as cross-sectional area decreases. Increasing velocity decreases pressure. If the pressure inside the artery drops far enough, it may fall below the pressure outside the artery, causing it to collapse from external pressure. choice C is wrong -> against continuity equation correct answer is B

When the researchers connected the solution-filled glass plates of the flow chamber to the AC generator, the ITO-coated plates mostly likely functioned as: passage info: Unilamellar vesicles were prepared by electroformation. 1,2-dioleoyl-sn-glycero-3-phosphocholine was dissolved in a chloroform/methanol solution. The solution was then spread on the conductive faces of two glass plates coated with transparent indium tin oxide (ITO). The plates were connected to a 10 Hz alternating current (AC), and the resulting 1.5 V potential difference was maintained between the plates for 3 hours. Concentrations of solution were altered to cause density variations. Finally, vesicles were diluted in a glucose solution of higher osmolality, but lower buoyancy. Osmosis produced partially deflated flaccid deformable objects. Vesicles settled on the substrate, and their buoyancy was controlled and varied. Figure 1 shows partial results and data for selected vesicles. A. a resistor. B. a capacitor. C. a galvanic cell .D. an electrolytic cell.

According to the experimental procedure of paragraph 2, "The plates were connected to an AC generator (10 Hz), and a 1.5 V potential difference between the plates was sustained for 3 h." The paragraph also indicates that ITO is a transparent conducting material. When a voltage is applied, charges will accumulate on the plates, exactly as what occurs on a parallel-plate capacitor. correct answer is A

Which of the following correctly relates the maximum extraction capacity of lauric acid in the given solvents? ** #7 on word doc A. Toluene > cyclohexane > 1-octanol B. Chloroform > dichloromethane > cyclohexane C. Cyclohexane > 1-octanol > toluene D. MIBK > cyclohexane > chloroform

According to the passage, "extraction capacity of HL [is] given by the percentage extraction (%E)." %E should be proportional to log D and to D. Therefore, the maximum extraction capacity of HL in each solvent should be related to the largest log D value reported in Figure 2 for the solvent. Only choice B correctly gives the relative magnitudes of those maximum extraction capacities. correct answer is B

A student theorizes that the differences in ATP-to-ADP ratios between cells detected in an 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.

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. correct answer is D

A cell's ATP-to-ADP ratio is most nearly equal to the ratio of its fluorescence emission intensities when Y32 is excited at: passage info: The protonated chromophore, called the "A-state," is favored in the ADP-bound conformation, and the deprotonated chromophore, which is called the "B-state," is favored in the Mg2+-ATP-bound conformation. ** #10 on word doc 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.

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. choice B is wrong - mistake thinking A state is ATP and B state is ADP correct answer is A

What is the pH of a 0.10 M aqueous solution of acetylsalicylic acid? passage info: Aspirin (180 g/mol; pKa = 3.5) is easily prepared by reacting salicylic acid (138 g/mol; pKa = 2.97) A. 1.0 B. 2.3 C. 3.5 D. 4.1

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). choice A is wrong - pH=1 is strong acid choice C is wrong - pH of 3.5 would mean that the concentration of weak acid and conjugate base were equal (a buffer) alternatively can use calculation to find pH for weak acid: 10^-3.5 = x^2/(0.1) correct answer is C

Doppler effect

An observed change in the frequency of a wave when the source or observer is moving

The equivalence point of the titration of tolbutamide with NaOH was reached by adding 50 mL of NaOH. Which of the following correctly describes the solution during this process? passage info: tolbutamide (C12H18N2O3S; pKa = 5.3) A. The solution had a pH less than 7.1 at the equivalence point. B. After addition of 25 mL of NaOH, the pH of the solution was greater than 5. C. The concentration of the charged form of tolbutamide was greater than neutral tolbutamide during the titration. D. The concentration of the charged form of tolbutamide was less than neutral tolbutamide during the titration.

At half equivalent point, when 25 ml of NaOH is added,, one half of the original tolbutamide present will have been converted to its conjugate base, and their concentrations will be equal. At the half-equivalence point, pH of solution equals the pKa of the analyte as HA= A-. passage states tolbutamide pKa is 5.3 that is greater than 5 choice A is wrong - pH at the equivalent pt will be greater than 7 choice C and D are wrong - The relative concentrations of the charged versus neutral forms of tolbutamide changed throughout the titration. choice B is correct

Which of the following best explains the tumor reduction capacities of the isotopes observed in the study? ** table is #9 on word doc passage info: 60Co, which undergoes β-minus decay and has a half-life of 5 years, X undergoes β-plus decay and has a half-life of 12 months. 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.

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. choice C is wrong - 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. correct answer is A

suicide inhibitors

Binds to the active site, enzyme proceeds with reaction until it tries to form a transition state - at which point the substrate forms a COVALENT instead of temporary bond...

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? Wx = Fz tan(θ) Δz passage info: where F x is the horizontal force required, Fz is the ground reaction force, Δz is the vertical displacement of a single step, and φ is the angle of ground reaction forces with the vertical, related to the angle at which the body is leaning forward (i.e., the tendency to pitch is greater at higher angles, requiring higher counter-acting forces). **#15 on word doc 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

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. choice A and B is wrong -When the runner keeps her spine fairly vertical, θ is close to 0°, meaning that tanθ is close to zero, and work will be 0 choice D is wrong - Δz is small, then work will be small correct answer is C

In a follow-up experiment, two identical gurneys are placed side-by-side on a ramp with their wheels locked to eliminate spinning. Gurney 1 has a dummy placed on it to give it a total mass of 200 kg, while Gurney 2 is loaded with a dummy that makes it only 50 kg overall. If the ramp has a coefficient of friction of μs, which gurney is more likely to slide down the ramp? A. Gurney 1, due to the increased force of gravity B. Gurney 2, due to the reduced force of static friction C. Gurney 1 and Gurney 2 are equally likely to slide. D. Neither gurney can slide unless the wheels are unlocked.

C is correct. In order for the gurney to slide down the ramp, the force pulling it downward (mgsinƟ) must be greater than the static frictional force (μsFN = μsmgcosƟ). The net force on each gurney is thus Fnet = (mgsinƟ) - (μsmgcosƟ). Since net force is equal to the product of mass and acceleration, we can rewrite this equation as ma = (mgsinƟ) - (μsmgcosƟ), where an acceleration greater than 0 means the gurney will slide down the ramp. The mass of the gurney is present in all terms and can be canceled, meaning that it is not a factor in whether the gurney slides. Thus, both gurneys have an equal likelihood of slipping down the ramp, regardless of the fact that they have different total masses. choice D is wrong -> if the force of gravity acting along the ramp is greater than the static frictional force, the gurneys will slide down the ramp, even if their wheels are not capable of spinning. correct answer is C

If a flat Petri dish containing a single layer of cells suspended in viscous culture medium is tapped, some cells will collide into each other. If cell 1 collides into cell 2 on such a plate, which of the following describes what happens to cell 2 after the collision, assuming that it undergoes an elastic collision and experiences drag from the medium? I. Cell 2 continuously accelerates. II. Cell 2 moves with decreasing speed. III. Cell 2 moves with constant speed. IV. Cell 2 decelerates until it reaches a velocity of 0 m/s. A. I only B. II only C. III only D. II and IV only

Cell 2 can only accelerate while it is being pushed/in contact with cell 1, so statement I is false. Since the cells experience drag from the medium, cell 2 will decelerate after the collision, making statement II correct and statement III incorrect. The drag will cause cell 2 to continuously decelerate until it comes to rest. correct answer is D

competitive inhibition

Competitive inhibition is a process by which a chemical substance has a shape that fits the active site of an enzyme and competes with the substrate, effectively inhibiting the enzyme. - increases Km -> it now takes more substrate to ensure half of the active sites are occupied - Vmax does not change

isomer

Compounds with the same formula but different structures.

negative control

Control group where conditions produce a negative outcome. Negative control groups help identify outside influences which may be present that were not accounted for when the procedure was created. - no result is expected - ensure no response to the test - used to identify the influence of external factors on the test

Conversion of pyruvate into glucose requires enzymes present in: A. the interstitial fluid only. B. the mitochondria only. C. the cytosol only. D. both the mitochondria and the cytosol.

Conversion of pyruvate to glucose requires its initial conversion into oxaloacetate, in a reaction catalyzed by pyruvate carboxylase in the mitochondria. Oxaloacetate (OAA) is then decarboxylated and phosphorylated by cytosolic or mitochondrial forms of phosophoenolpyruvate carboxykinase (PEPCK). After transport of either OAA in the form of malate or PEP directly from the mitochondria, the remainder of gluconeogenesis takes place in the cytosol. correct answer is D

A scientist uses an ultrasound device mounted to a vehicle to measure fluid flow underground. The device makes use of the Doppler effect to track fluid movement in the water table. Which of the following scenarios is most likely to produce a readable Doppler shift? I. The fluid is flowing at a velocity twice that of the sound-emitting device, in the same direction as the device is moving. II. The fluid is flowing at the same velocity and in the same direction as the sound-emitting device is moving .III. The fluid is not moving at all. A. I only B. III only C. I and II only D. I, II, and III

Doppler shift will be registered only if the fluid is moving relative to the source of the sound (the device) - some component of the fluid's velocity must exist in the same directional plane as the wave's velocity (otherwise, the device will register the fluid as not moving at all) - this component must be different from the velocity of the sound source. II is wrong - doppler shift will not occur if moving in the same velocity correct answer is A

Assuming a 95% yield for each coupling step, what would be the final yield for synthesizing a 10 amino-acid length peptide? A. 30% B. 60% C. 80% D. 95%

Each time we add an amino acid to the chain, we get a 95% pure yield for that step. Here, imagine that each coupling step has a 95% yield and that we are running the cycle 9 times (since no coupling step is needed for the first of the 10 amino acids), with each cycle being independent from the preceding cycle. Thus, a 10-amino acid peptide would be synthesized in a (0.95)9 = 0.63 ≈ 60% final yield. This can be assured because the passage allows us to assume a 100% yield in each unprotection step. There is no need to calculate it all out. A crude estimate of the yield would be that 9 steps, in each of which 5% is lost (corresponding to a 95% yield), would result in 45% being lost. However, the actual yield will be somewhat higher, because each step does not involve exactly a 5% reduction from the starting percentage. 60% is the best approximation. simple way: (0.9)^9 = 0.36 so (0.95)^9 must be greater than 0.36, so the only answer is B correct answer is B

What is the exclusion limit of the SEC column used to create the calibration curve shown in Figure 2? ** #11 on graph A. 10 kDa B. 100 kDa C. 600 kDa D. 1000 kDa

Figure2 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. correct answer is D

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

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). III is not correct - ΔS could be smaller than ΔH with high temperature to cause dG <0 correct answer is B

positive control

Group expected to have a positive result, allowing the researcher to show that the experimental set up was capable of producing results. - give predictive result - ensure success and validity of the test

When 2 moles of hydrofluoric acid are added to 100 mL of water, the resulting solution has a pH equal to 4. What is the percent dissociation of HF? A. 0.0005% B. 0.05% C. 0.5% D. 5%

HF is a weak acid and it does not fully dissociate in water. percent dissociation of HF is simply the percent of the original acid concentration that has dissociated into H+ and F- ions. This value is equal to [H+]/[HF] x 100%. pH=4 -> [H+] = 10^-4 M [HF] = 2mole/0.1L = 20M percent dissociation = [H+]/[HF] = (10^-4)/20 = 0.5 x 10^-6 = 0.0005% correct answer is A

What is the approximate pH of a saturated aqueous solution of hydrochloric acid whose molarity is 10.6 M? A. -1 B. 1 C. 7 D. 13

Hydrochloric acid is a strong acid and completely dissociates in aqueous solution. In this solution, the hydronium ion concentration is 10.6 M, which can be approximated as 10 M to make the math easier. The pH is the -log of the hydronium ion concentration: -log[10] = -log[101] = -1. While the typical pH range is normally thought of as ranging from 0 to 14, if the concentration of hydronium ion is greater than 1 M, negative pH values are possible. It is also possible to have pH values greater than 14, i.e. if the hydroxide concentration is greater than 1 M. choice B is wrong -> This pH would result from a saturated HCl solution with a concentration of 0.1 M, not 10.6 M. correct answer is A

Which regions of Figure 1 are most likely to be explained by the movement of sodium ions into a cardiac neuron? I. P II. Q III. R IV. T passage info: Electrocardiograms (ECGs) can detect the electrical activity of the heart. Figure 1 shows a hypothetical ECG cycle. The peak-to-peak voltage changes average 1-2 mV. The P wave is associated with atrial depolarization. After a short interval, the Q and R portions are observed, which indicate ventricular contraction. The final portion, the T wave, is associated with repolarization. In some individuals, an additional repolarizing U-signal is detected after the T wave **#16 on word doc A. I only B. II and III only C. I, II, and III only D. I, II, III, and IV

I is correct - the P wave is associated with the depolarization and contraction of the atria. A rest potential is created when sodium ions are pumped to the outside surface of the membrane of a cell. - When ion channels open, the sodium ions spontaneously flow back into the cell, resulting in depolarization and an action potential. II and III are correct - the passage states that the Q and R waves are associated with the contraction of the ventricles. Like atrial contraction, ventricular contractions result from depolarization, which would involve the inward movement of cations like sodium ions. IV is wrong -The passage states that the T and U waves occur due to repolarization, which takes place when positive ions are pumped back out of the cell to return to rest potential prior to the next heartbeat. correct answer is C

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.

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. choice A is wrong - the question specifically asks about how the fact that the biosensor is pH-sensitive could affect measurements made within a single cell over time. correct answer is C

Noncompetitive inhibition

In noncompetitive inhibition, the inhibitor binds with the enzyme at a site other than the active site (the allosteric site) and inactivates the enzyme by altering its shape. - combine with enzyme or enzyme-substrate complex - Vmax decrease - Km is the same -> Since these inhibitors do not compete with the substrate, their activity is unaffected by substrate concentration

What is the mechanical advantage of using the ramp, as opposed to lifting the gurney straight up? (Ignore any effects of friction.) passage info: In a second experiment, the gurney brakes are unlocked and the gurney is pushed from rest to the top of a ramp, then brought to a stop. The ramp makes an angle of 30° with the horizontal, and the top of the ramp is 5 meters above ground level. Once at the top of the ramp, the gurney is released and allowed to roll downward. A. 2 B. 1.5 C. 1 D. 0.5

Inclined planes provide a mechanical advantage of greater than 1 MA = hypotenus/height = 10/5 = 2 correct answer is A

Which of the following most closely approximates the pKa of phenolphthalein? passage info: 1 L samples from each broth were then inoculated with A. aceti grown in 5%, 10%, or 15% bacterial concentration growth media. The acidity of each broth sample was evaluated by acid-base titration with a standardized solution of 0.1 N sodium hydroxide, using phenolphthalein as an indicator. The acetic acid content of the fermentation products were measured 24, 48, and 72 hours after inoculation and reported as titratable acidity (%TA) in Table 2. A. 2.4 B. 4.8 C. 6 D. 9.3

Indicators function in acid-base titrations to identify, via color change or a similar mechanism, that an expected pH, and thus the titration endpoint, has been reached. In order to function in this way, an indicator must undergo a color change near the desired pH. This typically occurs because of a reversible change in the protonation state of the indicator. It is desirable then that the pKa of a chosen indicator be within ±1 unit of the target pH. For the titrations performed in the study of acetic acid, a weak acid, and sodium hydroxide, a strong base, the endpoint of the titration will occur at a pH greater than 7 correct answer is D

Vinblastine is a microtubule-disrupting drug that inhibits tubulin polymerization. Which of the following processes would be directly inhibited upon vinblastine treatment? I. Phagosome transport to the lysosome II. Mitosis III. Meiosis IV. Electron transport A. II only B. I and IV only C. II and III only D. I, II, and III only

Microtubules are used in the transport of vesicles and the positioning of organelles within the cell. As part of this role, they form structures that assist in the transport of phagosomes (vesicles that contain particles that have been engulfed via phagocytosis) to the lysosomes of the cell, to which the phagosomes fuse. -> I. is correct correct answer is D

The flow rate of stomach content emptying is 100 cm3/sec. Patients who undergo gastric bypass surgery will increase this rate to almost 1600 cm3/sec. Assuming the flow of stomach contents approximates Poiseuille's Law, what change to their gastrointestinal connection would explain this, provided no other changes occur in the conditions of stomach content flow? A. The new connection is 2 times longer. B. The new connection is 4 times longer C. The new connection radius is 2 times larger. D. The new connection radius is 4 times larger.

Poiseuille's law quantify the flow: Flow = ΔP*π*r^4 / 8*L*η the flow increase in the 16-fold change, and only doubling the radius will reach the new flow rate -> 2^4 = 16 Mistake: don't remember Poiseuille's law correct answer is C

Pentane is a straight-chain hydrocarbon with the molecular formula C5H12. How many additional structural isomers can be constructed using this molecular formula? A. 0 B. 1 C. 2 D. 3

Since the molecular formula follows the CnH2n+2 formula, we can only build alkanes. This makes the job of finding isomers much easier. We can make a tertiary carbon, 2-methylbutane. We can make a quaternary carbon, 2,2-dimethylpropane. And that's it. correct answer is C

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.

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). choice A is wrong -The stability of arginine's conjugate base is not in question. correct answer is B

optical isomers (enantiomers)

Stereoisomers that are non-superimposable mirror images of each other - chiral center have opposite configuration

Cefixime displays what inhibitive behavior on PBP? passage info: Cephalosporins function by inhibiting the bacterial enzyme PBP. The structural similarity between the core structure and D-alanyl-D-alanine, the terminal amino acid residue of the PBP substrates N-acetyl-glucosamine and N-acetyl-muramic acid, facilitates the irreversible binding of cephalosporin to the active site of PBP in a manner that is not competitive inhibition. A. Competitive inhibition B. Non-competitive inhibition C. Uncompetitive inhibition D. Suicide inhibition

Suicide inhibition occurs when an enzyme binds the inhibitor (structurally a substrate analogue) and forms an irreversible complex with it, usually through a covalent bond. This can involve the inhibitor being chemically modified by the enzyme during the normal course of catalysis to produce a reactive group that is specifically responsible for the formation of the irreversible inhibitor-enzyme complex. correct answer is D

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

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. choice A is wrong - H3PO5 is perphosphoric acid. choice B is wrong - H3PO4 is phosphoric acid choice D is wrong - H3PO2 is hypophosphorous acid correct answer is C

A child is sliding a toy block (with mass = m) down a ramp. The coefficient of static friction between the block and the ramp is 0.25. When the block is halfway down the ramp, the child pushes down on the block perpendicular to the plane, halting it. What is the minimum force the child must apply to keep the block from starting to slide down the ramp? **#1 on word doc A. mg sin θ B. 0.25 mg cos θ + mg sin θ C. mg sin θ - 0.25 mg cos θ D. [(mg sin θ) / 0.25] - mg cos θ

The gravitational force pulling the block down the ramp is mg sin θ. (This is the typical value for the gravitational force that acts on an object to drag it down an incline.) To stop the block from sliding down the ramp, we must have an equal and opposite frictional force. Since these forces are equal and opposite, we can set them equal to each other as follows: Ff = mg sin θ Now, remember that frictional force is equal to the product of the appropriate coefficient of friction and the normal force: Ff = μFN Ff = 0.25 x FN = 0.25 mg cos θ The block itself has a mass m and thus generates a normal force of FN = mg cos θ. Again, this is the standard value for the normal force on an object positioned on an inclined plane. From this information alone, we may be tempted to pick C, which is the difference between the gravitational force and the frictional force. If the child were pushing the block upwards along the plane in a parallel fashion, this choice would be correct. However, the child is actually pushing down on the car, perpendicular to the plane. Thus, the force exerted by the child (Fa) will add to the force created by the mass of the car itself and alter the value for the normal force. Our total FN = mg cos θ + Fa. Substituting, we get: Ff = 0.25 x (mg cos θ + Fa) = mg sin θ mg cos θ + Fa = (mg sin θ) / 0.25 Thus, the force with which the child must push down on the car is Fa = [(mg sin θ) / 0.25] - mg cos θ. correct answer is D

Why did the researchers choose to study pediatric, rather than adult, thyroid cancer cases? A. Thyroid cancer is not otherwise present in children. B. Thyroid cancer is not otherwise present in adults. C. Children receive a higher relative dose of I-131 at the same contamination level. D. Children receive a lower relative dose of I-131 at the same contamination level.

The most likely reason for the use of pediatric cancer cases is that these cases are more likely to be caused by I-131 than those in adults. Because children weigh less than adults, the same quantity of external I-131 would result in a higher concentration of the isotope in the body, making DNA damage and cancer more likely. choice A is wrong -> this choice is too extreme and passage never discuss ; thyroid cancer is the third most common cancer in kids correct answer is C

In order to find the concentration of metal ions needed to determine the distribution coefficient of copper (II), what calculation could the researchers have performed in addition to the procedure described in the passage? A. Average the concentrations of metal ions in the aqueous and organic phases after extraction. B. Subtract the concentration of metal ions in the aqueous phase after extraction from the concentration before extraction. C. Subtract the concentration of metal ions in the organic phase after extraction from the concentration before extraction. D. Average the concentrations of metal ions in the organic phase before and after extraction.

The passage defines the distribution coefficient as "the ratio between the concentration of metal in organic and aqueous phase following extraction." In order to determine D, the researchers must know the concentrations of metal ions in the organic phase and of the metal ions remaining in the aqueous phase following the extraction. According to the description provided by the passage of the researchers' procedure "after the two phases were separated completely, concentrations of the metal remaining in the aqueous phase were determined spectrophotometrically at 820 nm." This measurement provides the concentration of metal ions remaining in the aqueous phase directly. The concentration of metal in the organic phase after the extraction is not directly known. However, the passage also indicates that the concentration of metal contained in the aqueous solution prior to extraction is known—"researchers performed a series of extractions of a known concentration of copper (II) nitrate in aqueous solution." In order to determine the concentration of metal ions in organic solvent after extraction, the researchers could use the known concentrations of metal in aqueous solution before and after extraction—the difference, found by subtracting the value, should equal the concentration of metal ions in the organic phase after extraction. correct answer is B

Following extraction into chloroform, the spectrophotometric absorbance at 820 nm of copper in aqueous solution will be greatest at which pH? ** #7 on word doc passage info: After the two phases were separated completely, concentrations of the metal remaining in the aqueous phase were determined spectrophotometrically at 820 nm. A. 5.45 B. 5.35 C. 5.25 D. 5.15

The passage states that spectrophotometric absorbance is used to measure the concentration of metal ion remaining in aqueous solution following extraction. This value should be maximized when the extent of the extraction is minimized. In Figure 2, this should correspond to the aqueous solution pH in which log D is smallest for the extraction in chloroform. Of those pH values given as answer choices, log D is smallest at pH 5.15. correct answer is D

energy efficiency

The ratio of the amount of work done to the total amount of energy introduced to the system energy efficiency = energy output/energy input

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. passage info: The radiation is strong enough to cause molecular electronic transitions by exciting electrons to higher energy levels in molecular orbitals. In some molecules the beam passes through, the radiation is sufficiently strong to produce molecules with unpaired valence electrons by way of electron ejection. A. I only B. I and II only C. II and III only D. I, II, and III

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). correct answer is D

The hemoglobin (Hb) dissociation curve at high altitude has a distinct shape from that at atmospheric pressure. Which of the following best explains this shape? A. Dissolved oxygen in the blood serves to increase pH. B. The primary structure of Hb is changed by oxygen binding C. Dissolved oxygen forms a chain of atoms that is easier for Hb to bind. D. Homotropic regulation by oxygen occurs.

The sigmoidal shape of the curve implies that as each oxygen molecule binds to Hb, the affinity of Hb for oxygen goes up. Homotropic regulation is when a molecule serves as a substrate for its target enzyme, as well as a regulatory molecule of the enzyme's activity. O2 is a homotropic allosteric modulator of hemoglobin. The four subunits of hemoglobin actually bind to oxygen cooperatively, meaning the binding of oxygen to one of the four subunits will increase the likelihood that the remaining sites will bind with oxygen as well. This is the cause of the sigmoidal curve shown in the figure. Cooperative binding in hemoglobin is also depicted below. choice A is wrong -> changing pH has nothing to do with Hb structure choice B is wrong -> primary structure of Hb is determined by amino acid sequence, not oxygen concentration choice C is wrong -> oxygen does not form chain correct answer is D

Based upon the researchers' analysis of the extraction equilibria, the organic phase extractant forms what complex or complexes in low-polarity organic solvents and high-polarity organic solvents, respectively? ** structure and chem equation is #8 on word doc A. Complex A; complex A B. Complex A; complex B C. Complex B; complex A D. Complex B; complex B

The two chemical equations shown in the passage indicate the structure of the organic phase extractant complex formed in both low-polarity and high-polarity organic solvents. In low-polarity organic solvents, copper (II) is coordinated with the organic phase extractant as part of a complex consisting of two dimers. Both dimers contain one molecule of lauric acid (HL) and another of its conjugate base, laurate (L). This is consistent with the structure of complex A shown in Figure 1. In high-polarity organic solvents, copper (II) is coordinated with the organic phase extractant as part of a complex consisting of two monomers of laurate. This is consistent with the structure of complex B shown in Figure 1.

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

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. correct answer is A

If a 65-kg man undergoes a turning acceleration of 5 m/s2 during a running turn, what is the magnitude of force experienced by the foot due to the ground? A. 325 N B. 650 N C. 750 N D. 1075 N

To turn while running, the foot must push off the ground, applying a shearing force while simultaneously supporting the weight of the body. When faced with problems like this, always consider the forces involved. Here, we must account for the normal force exerted by the ground on the foot; this is a vertical force which occurs as a result of the runner's weight. We also must consider the acceleration force, which (since the person is turning) is horizontal. These two force vectors are perpendicular and will form a right triangle. F_normal = mg = 650N F_turning = 65 * 5 = 325 N because F_n and F_t form right triangle: V_ground = (650^2 + 325^2)^1/2 = (58)^1/2 * 10^2 = 750N correct answer is C

continuity equation

V1A1 = V2A2 within a closed system, the flow rate of a liquid is constant, which indicates that the velocity of the fluid (v) is inversely proportional to the cross-sectional area that it is flowing through.

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

When the leg "absorbs" kinetic energy, it is converted to elastic potential energy. PE (U) = (1/2) k*x^2 k = 2U/ x^2 = 2*125 / (0.1*0.1) = 25000 N/m correct answer is D

Positional isomers

a given functional group in different locations (e.g., 1-pentanol vs. 2-pentanol).

open system

a system in which exchanges of matter or energy occur across system boundaries

The findings of the experiment support which of the following conclusions? ** figure on #3 on word doc A. ADAMTS-5 is mainly responsible for the aggrecan degradation during human OA. B. ADAMTS-5 has a higher specificity constant than ADAMTS-4 C. The ADAMTS-5 analog acts as an agonist D. Both analogs act as ADAMTS inhibitors.

according to study result, ADAMTS-5 inhibited cartilage degradation as cartilage mixture is higher than no ADAMTS analog, and combination of ADAMTS-5 and ADAMTS-4 increase the protective effect. Thus, it can be concluded that both analog inhibit degradation choice A is wrong -> ADAMTS-5 stop degradation, and we can;;t make this conclusion b/c we did not compare the effect of ADAMTS-5 and ADAMTS-4 separately choice B is wrong -> we don't know choice C is wrong -> ADAMTS-5is antagonist correct answer is D

homotropic effects

allosteric interactions that occur when several identical molecules are bound to the protein; - e.g., the binding of aspartate to ATCase - ex: oxygen bind to hemoglobin and change its affinity

If a hydrogen NMR spectrum were obtained for a double-stranded amidine dimer, would the amide N-H proton resonances experience any shielding effects? A. Yes, they would exhibit upfield shifts due to shielding effects. B. Yes, they would exhibit downfield shifts due to deshielding effects. C. No, they would not experience any shielding effects. D. Yes, they would exhibit downfield shifts due to shielding effects.

amidine dimer, which contain amide, should be held together by hydrogen bond, just like carboxylic acid. when hydrogen bond is formed, electron density around hydrogen will decrease as electron density around nitrogen will increase - thus the proton is deshield (focus on H only) - in opposite, addition of electron density will create upshield effect the deshielding effect will create chemical shift downfield shift for N-H proton due to reduce electron density correct answer is B

vitamin E

antioxidant. Essential for protection of cell structure, especially of red blood cells

Atrial fibrillation is the most common heart arrhythmia, causing palpitations, fainting and chest pain. According to the information in the passage, what is most likely to be observed in the ECG during atrial fibrillation? passage info: Electrocardiograms (ECGs) can detect the electrical activity of the heart. Figure 1 shows a hypothetical ECG cycle. The peak-to-peak voltage changes average 1-2 mV. The P wave is associated with atrial depolarization. After a short interval, the Q and R portions are observed, which indicate ventricular contraction. The final portion, the T wave, is associated with repolarization. In some individuals, an additional repolarizing U-signal is detected after the T wave A. Missing P waves and an irregular rate B. Complete absence of P waves and a regular rate C. Complete absence of R waves and an irregular rate D. Missing R waves and a regular rate

arrhythmia is an irregular time period between heartbeats, not necessarily a rate that is too slow or too fast. Since the question asks about atrial fibrillation, and the P wave is attributed to contraction of the atria, the correct answer should show the P wave being affected. This does not mean that the atria are not contracting at all (eliminate choice B), but it indicates that blood is not efficiently being pumped from the atria to the ventricles. choice B and D are wrong - arrhythmia si irregular HB choice C is wrong - Absence of an R wave would be indicative of ventricular problems, and is more dangerous problem than AFib. The R wave is a strong, sharp electrical signal coming from the ventricles. Complete absence of an R wave suggests ventricular contraction is likely not occurring or fibrillating at best correct answer is A

Which of the following correctly lists the floating lag times for the three trials in increasing order? (note: assume that the mixing of the drug and polymer does not change the density of either component)] Passage: The time taken for the tablet to rise to the surface of the dissolution media (floating lag time) and total duration that the tablet remained on the surface (total floating time) were recorded. * Table 1 on #1 on NS c/p doc A. Trial 1 < Trial 2 < Trial 3 B. Trial 2 < Trial 1 < Trial 3 C. Trial 3 < Trial 2 < Trial 1 D. Trial 2 < Trial 3 < Trial 1

based on the passage, floating time is the amt of time tablet rise to the surface of the media, and is related to the density of the object -> less dense object will rise to the surface more quickly Looking at Table 1, pills in trial 3 has the smallest density while trial 1 has the largest density -> floating time: trial 3 < trial 2 < trial 1 or calculate the density in each trial (though take more time) Trial 1: 1/2 * 0.2 + 1/2 * 0.6 = 0.4 Trial 2: 1/5 * 0.2 + 4/5 * 0.3 = 0.28 Trial 3: 1/9 * 0.2 + 8/9 * 0.1 = 0.11 correct answer is C

vitamin K

blood clotting and coagulation

When in vivo studies were performed on the three drug-polymer combinations, it was found patients in the polaxemer-188 group experienced the most stomach pains after administration. Given the results of the study, this is most likely due to: ** figure 1 in #2 in NS word doc A. elevated drug concentration causing inflammation of the gastric mucosa. B. tablet fragments causing inflammation of the gastric mucosa. C. glipizide/polaxemer-188 matrices being the smallest of the 3 tested. D. increased pH of the stomach contents.

choice D is wrong -> passage never discuss pH of each drug, so we don't know any of drug content will alter pH of stomach choice A is wrong -> based on figure 1 , drug 3 has lowest percent of drug release choice C is wrong -> passage states that reduce polymer size (reduce drug size) leads to increase solubility of the drug (increase drug release) -> drug 3 should have the largest size b/c it has the smallest drug release b/c drug 3 has slowest solubility, the pill must remain partially undissolved for a longer time than the other medications, and the resulting solids could cause the most inflammation. correct answer is B

What is the relationship of these molecules to each other? ** #13 on word doc A. Meso compounds B. Enantiomers C. Diastereomers D. Same compound

compound on left has S, S configuration while compound on right has R, R configuration answer is so obvious....don't overthnik choice A is wrong - Meso compounds have an internal plane of symmetry, choice C is wrong - both configurations are reversed choice D is wrong - configuration are not the same correct answer is B

Which of the following is a list of the functional groups shown on the protected amine? ** #6 on word doc A. Amide, imide, ester B. Ester, imine, carboxylic acid C. Ether, ester, amide D. Carboxylic acid, ether, imide

compound show (right to left) 1. carboxylic acid 2. ether 3. amide 4. amine 5. ester correct answer is C

heating solution will increase or decrease gas solubility

decrease gas solubility

structural isomers

differ in the covalent arrangements of their atoms

Aspartame (N-L-alpha-aspartyl-L-phenylalanine 1-methyl ester) is a very well-known artificial sweetener found in the large majority of non-sugar containing food products. This compound is classified as a(n): A. ketone. B. phenol. C. dipeptide. D. amino acid.

dipeptide = aspartic acid and phenylalanine choice D is wrong - if it's amino acid, we will only see either amino acid correct answer is C

The acceleration of the gurney during the initial 6 seconds of Experiment 1 is: passage info: A medical student is asked to push the gurney horizontally. During the experiment, the force exerted on the gurney is increased until the tires begin to slide along the floor without rotating. Once the gurney begins to slide, the pushing force is decreased to the lowest force needed to maintain a fixed speed. Figure 1 shows data from a force sensor attached to the gurney, which records the student's applied force over time. (The floor has a coefficient of static friction, µ.) ** #5 on word doc A. zero. B. nonzero and in the direction of Fapplied. C. constant and greater than 0. D. rising as Fapplied increases.

during the initial 8s gurney did not move, so there is no acceleration choice A is correct

How many moles of captopril were present in the original analyte solution tested? A. 7.5 × 10-5 moles B. 1.5 × 10-4 moles C. 7.5 × 10-3 moles D. 1.5 × 10-2 moles

equivalent point occur when approximately 7.5 mL of 2 x 10-2 M NaOH

Based on the passage results, how would the mass and stiffness be expected to vary along the length of the membrane? passage info: The researchers hypothesized that the membrane is a resonant structure, with each location along its axis acting like an individual spring with unique characteristics, giving each location a specific resonant frequency to which it is most sensitive. ** data on #21 A. They would both be at a maximum value at the base and decrease along the membrane to a minimum at the apex. B. They would both be at a minimum value at the base and increase along the membrane to a maximum value at the apex. C. The stiffness would be highest at the base, while mass would be highest at the apex. D. The stiffness would be lowest at the base, while mass would be lowest at the apex.

figure 2 shows frequency is highest at base and lowest at apex, and as passage state membrane is like individual spring, so we can adopt frequency equation, f = (1 / 2 π)(√k / m) using the equation, we can conclude that base has smaller mass and greater stiffness (aka k, spring constant/stiffness) compared to apex as its frequency is higher and more vibration correct answer is C

The λmax of S157C at pH 4.5 is greater than at pH 10. How does the proposed bonding of deprotonated C157 with Y63 at pH 10 account for this observation? passage info: a mechanism (Figure 3) was proposed for the protonation state-dependent ionic interaction of Y63 of wild-type sgBP and S157C chromophores at pH 11.6 and pH 10, respectively. This mechanism accounts for the greater strength of ionic bonds relative to hydrogen bonds. ** figure 3 is #25 on word doc A. Compared to hydrogen bonding at pH 4.5, ionic bonding at pH 10 destabilizes the sgBP chromoprotein barrel conformation. B. Compared to hydrogen bonding at pH 4.5, ionic bonding at pH 10 decreases the maximum normalized absorbance. C. Compared to hydrogen bonding at pH 4.5, ionic bonding at pH 10 promotes peak absorbance of higher energy radiation. D. Compared to hydrogen bonding at pH 4.5, ionic bonding at pH 10 increases the wavelength of maximum absorption.

figure 3 shows pKa of cys side chain is 8.37, and at pH=4.5 cys will remain protonated, so the maximum absorbance wavelength at pH 4.5 is longer because the Y63-C157 interaction predominately consists of hydrogen bonding that the ion-dipole bonding which predominates at pH 10 promotes the maximum absorption of shorter-wavelength, higher-energy radiation. choice A is wrong - no evidence choice B is wrong - figure 2 show the opposite choice D is wrong - wavelength at PH=10 decreases based on figure 2 correct answer is C

Chain isomers

have different arrangements of the carbon 'skeleton. - subtype of structural isomer

uncompetitive inhibition

inhibitor binds only to enzyme-substrate complex locks substrate in enzyme preventing its release (increasing affinity b/w enzyme and substrate so it lowers Km) Lower Km and vmax

when temperature decrease, air is able to hold more or less water

less water in the air in temp decrease, because solubility decrease

structural isomers (constitutional isomers)

molecules that have the same molecular formula but differ in the way the atoms are connected

is 1,4-dichlorobenzene polar or nonpolar

non polar because the plane of symmetry that the two chlorine substituents are positioned opposite each other, causing their dipoles to cancel out overall.

which substance has higher Rf value with thin-layer chromatography using a silica plate

nonpolar substance because it move faster on silica plate Silica is a highly polar molecule; thus, when chromatography is done on a silica plate, a polar substance will move slower on the plate due to polar-polar interactions between the substance and the plate. This means that the Rf value of a polar substance will be smaller than that of a nonpolar substance since the substance will move less on the plate compared to the solvent front than a nonpolar substance would. Remember, Rf is equal to the distance the compound of interest traveled along the plate divided by the distance traveled by the solvent front. A sample TLC plate, with those distances labeled, is shown below. Note that A, B, and C represent arbitrary compounds in this image.

how many proton and electron are transferred in FAD -> FADH

one electron and one proton - reduction is gain of 1 electron

A patient presents in the emergency department having ingested a large quantity of tolbutamide. Intravenous administration of which of the following compounds is most likely to increase the rate of urinary excretion of the drug? passage info: The uncharged state of a weak acid, such as tolbutamide (C12H18N2O3S; pKa = 5.3), is capable of diffusion across mucosal membranes, such as those found lining the stomach and ureters, while the charged form cannot diffuse. A. KCl B. NaHCO3 C. NH4ClO4 D. NaCl

passage states uncharged states will be absorbed, so we need to keep tolbutamide in charged state in order to remove it from pt's body. to do so we need to keep tolbutamid dissociate continuously by reducing H+ concentration or add base to solution NaHCO3 is the salt of the conjugate base of carbonic acid, so it is base choice A and D are wrong - they are neutral compound, and Cl- will only act as as a base in extremely low pH as it is conjugate base of HCl, a strong acid choice C is wrong - ClO4- is also the conjugate base of a strong acid (HClO4), so it will also fail to act as a base under biological conditions. Additionally, NH4+ is acidic. Because of this, NH4ClO4 is acidic overall, not basic. correct answer is B

extraction

separate two or more compounds in solution by manipulating their solubility properties using acid-base chemistry an extraction setup will have a layer of a less-dense organic (i.e. nonpolar) solvent on top of a layer of water, which is a highly polar compound. Nonpolar compounds in solution will move to the organic layer, while polar (or charged) compounds will be in the aqueous layer. For example, imagine we had a flask containing valeric acid (CH3(CH2)3COOH) and aniline (C6H5NH2) suspended in a layer of diethyl ether floating above a layer of water, and wanted to separate out the valeric acid. We could add a strong base (for example, NaOH), which would deprotonate the valeric acid to form CH3(CH2)3COO−. This charged molecule would be much more likely to move to the water layer, which can then be extracted. This process may have to be repeated to ensure that the products are pure.

single/double/triple bond formula

single: CnH2n+2 double: CnH2n triple: CnH2n-2

Which of the following properties explains why sound wave velocity increases as it travels from air to a liquid medium at the same temperature? A. Kinetic energy of the medium B. Volume of the medium C. Density of the medium D. Bulk modulus of the medium

sound velocity = √(Κ/ρ) K: bulk modulus speed of sound increases with the stiffness (the resistance of an elastic body to deformation by an applied force) of the material; since solids are considerably more stiff than liquids and especially gases, sound travels fastest through them. Since density is present in the denominator of the equation, sound waves moving through two equally-stiff solids will travel faster in the less dense material. But when comparing solids to gases, density has a smaller effect than stiffness, because the extent to which a typical solid is more stiff than a gas far exceeds the ratio of their densities. correct answer is D

diastereomers

stereoisomers that are not enantiomers with only one opposite chiral center when multiple chiral center are present

Functional isomers

structural isomers that have different functional groups - ex: aldehyde and alcohol group but have the same formula

which inhibitor is irreversible

suicide inhibitor competitive, uncompetitive, and noncompetitive inhibitor are reversible

vitamin D

sunshine vitamin; calcium and phosphate absorption from the gastrointestinal tract

If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher than the rest of the baseball field, at what height would the catcher need to hold his glove to catch the pitched ball? (Note: neglect air resistance, estimate the acceleration due to gravity as 10 m/s2, and assume the pitcher is only throwing the ball horizontally.) according to passage x=18m and horizontal velocity is 30 m/s A. 2.0 m above the ground B. 1.8 m above the ground C. 0.5 m above the ground D. 0.2 m above the ground

t = X/V_ox = 18/30 = 0.6s d=Vo*t + 1/2*a*t^2 = 1.8 m total height is 2.0 m, so 2-1.8 = 0.2 m correct answer is D

what can affect equilibrium constant, like Ksp

temperature, and changing concentration will not affect Ksp

Compound 2083 is a potent ADAMTS-5 receptor antagonist.Which of the following contains only functional groups that are found on compound 2083? * compound structure is #4 on word doc A. Amide, imine, thioether B. Amine, amide, ester C. Amine, disulfide, imide D. Phenyl chloride, sulfoxide, thiol

the compound contains: amide: R(C=O)NR amine: R-NR2 imine: R=NR' thioether: R-S-R' phenyl chloride thioketone: R(C=S)R' choice D is wrong - This molecule contains no thiol (R-SH) or sulfoxide [R-S(=O)-R'] group. correct answer is A

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

the question is asking for power output for the loss energy potential energy = 0.5*10*60 = 300J 10 % loss energy = 30J so need additional 30J per stride power = 30J/0.8s = 37W correct answer is B

Early research into the mechanism of antibiotics theorized that it was implausible that the amide in penicillin would convert to an ester. Which of the following explains this doubt? A. Amides possess better leaving groups than esters. B. Esters possess poorer leaving groups than amides. C. Amides are less reactive than esters. D. Serine contains no good nucleophiles.

the relative reactivities of these carboxylic acid derivatives, starting with the most reactive: acyl halides > anhydrides > esters and acids > amides. It is not thermodynamically favorable for a molecule to transform from a less reactive (more stable) form, like an amide, to a more reactive (less stable) form, like an ester. - to convert from less reactive to more reactive molecule, requires converting a carboxylic acid derivative back to a carboxylic acid and then generating a new carboxylic acid derivative. correct answer is C

When the outer hair cells vibrate, they generate a standing wave in the closed organ of Corti, which amplifies the frequency of interest at particular locations along the basilar membrane. How will the length of the enclosed space within the organ of Corti vary along the length of the basilar membrane to explain this? A. The length of the enclosed space in the organ of Corti will increase linearly from the base to the apex of the cochlea. B. The length of the enclosed space in the organ of Corti will decrease linearly from the base to the apex of the cochlea. C. The length of the enclosed space in the organ of Corti will decrease exponentially from the base to the apex of the cochlea. D. There should be no change in the length of the enclosed space of the organ of Corti, as length has no impact on the frequency of standing waves.

the resonant frequency decreases from the base to apex along the basilar membrane with a linear relationship, so, in order for the standing wave to produce constructive interference, its frequency will also have to decrease linearly along the membrane from base to apex. If the frequency must decrease, then the wavelength has to increase. Thus, in order to increase the wavelength, λ, it would be necessary to increase the length, L, of the enclosed space within the organ of Corti. choice B and C are wrong - they are opposite as the length should increase choice D is wrong - length and frequency will change correct answer is A

The distance between the ears for an average adult male is 20 cm. If a wave with a frequency of 17,000 Hz hits both the right and left ear and has a phase difference of 5π radians between the two, what is the difference in distance between the origin of the sound and each ear (velocitysound in air = 343 m/s)? A. 2 cm B. 5 cm C. 8 cm D. 20 cm

then the difference in distance from the origin of the wave to the right and the left ear: 5π rad * λ/2π rad = 2.5λ to find wavelength: v = f * λ λ = v/f = 340 / 17000 = 2cm difference in distance between the origin of the sound and each ear: 2.5λ = 2.5*2 = 5cm correct answer is B

According to the description in the passage, Earth is: passage info: Earth has had negligible mass exchange with the rest of the universe. In an attempt to understand the driving forces behind the biogeochemical cycles, researchers created a model of the earth-atmosphere-ocean system and calculated the energy and entropy flux between system components on the basis of a steady-state assumption. A. a reducing system. B. an isolated system. C. an open system. D. a closed system.

there is no mass exchange but energy exchange in Earth, so this is closed sys correct answer is D

During the anterior drawer test, the joint translation occurs in 0.2 s. What is the average velocity of the talus during translation? passage info: The standard talar test has a degree of subjectivity and is not as quantifiable as a new approach that uses mechanical isolation of the joint. By applying the force via a mechanical stirrup and combining the talar tilt test with the anterior drawer, consistent values have been measured. For an anterior force of 98 N and a torque of 13 Nm, the mean talar-tilt translation was 48°, and the mean anterior-drawer translation was 5.7 mm. A. 2.85 x 10-3 m/s B. 2.85 x 10-2 m/s C. 3.0 x 10-2 m/s D. 3.0 m/s

v = d/t = 5.7 mm / 0.2s = 2.85 * 10^-2 m/s correct answer is B

vitamin A

vision

On the penicillin heteroatom ring, what is the absolute configuration of the carbon labeled 1? ** #3 on word doc A. + B. - C. S D. R

we assign priority based on atomic number, and sulfur's atomic number > oxygen atomic number 1: N 2: C-S 3: COOH 4: H clockwise- R counterclockwise-S correct answer is C

Assuming a mass of 0.04 g, what is the power exerted by the electrical forces on a fragment of fiber if the fragment has a velocity of 200 mm/s, 10 ms after ejection? A. 5 x 10-10 W B. 1 x 10-8 W C. 2 x 10-5 W D. 8 x 10-5 W

work = dKE = 1/2*mv_f^2 - 1/2*mv_i^2 power = work/time = 1/2*0.04*(200*10^-3)^2 / 10*10^-3 = 8*10^-5 correct answer is D