chem/phys questions

The data in Table 1 show the effect of increasing pH on the relative rates of reduction of CO2 to CO compared with the reduction of CO to CH4. The researchers would be most likely to conclude that: A. at a low pH, the reduction of H2O outpaces the reduction of CO2. B. the production of H2 gas shows first-order kinetics. C. proton concentration has no effect on the reduction of CO to CH4. D. the rate of reduction of the CO2 radical anion slows at high pH.

***Table 1 shows that at pH = 1, the product ratio of CO to CH4 is small, and the total rate of CO2 reduction activity is low because of a competing hydrogen ion reduction reaction. At higher pH (pH = 3), hydrogen evolution declines, and the rate of CO2 reduction increases, but CO dominates, as the ratio of CO to CH4 produced increases. Since pH is defined as -log[H+], at a pH of 1, the concentration of H+ is 0.1 M (or 1 x 10-1 M), while at a pH of 3, it is 0.001 M (or 1 x 10-3 M). This corresponds to a difference of two orders of magnitude, which is also approximately the difference in the production of H2 gas between the two pH conditions (1.32 mL/min vs. 0.01 mL/min). This corresponds to first-order kinetics, in which the reaction rate is dependent on the concentration of a single reactant (in this case, H+). A. wrong cause not told anything in table about H2O reduction C. wrong cause proton concentration clearly has an effect. D. wrong cause the rate of reduction of CO2 speed up at high pH. B is correct by default cause idk what kinetics can be solved for from this.

which of the following compound-solvent combinations would be expected to show the highest antioxidant activity? A. Compound 1 in benzene B. Compound 2 in water C. Compound 2 in benzene D. Compound 1 in water

All enthalpy changes presented in Table 1 are positive, indicating that all three mechanisms are endothermic for all three compounds in both solvents. In other words, all three mechanisms must overcome an energy barrier to move forward. The lower the barrier, the more thermodynamically favorable the reaction is. Thus, lower H values are associated with higher antioxidant activity. Compound 1 generally shows lower H values than compound 2, and the mechanisms have lower H values in water than in benzene, making choice D correct.

The HAT mechanism is expected to be: A. spontaneous at low temperatures and nonspontaneous at high temperatures. B. nonspontaneous at both low and high temperatures. C. nonspontaneous at low temperatures and spontaneous at high temperatures. D. spontaneous at both low and high temperatures.

C is correct. Spontaneity can be predicted by the equation ΔG = ΔH - TΔS, where ΔG is the Gibbs free energy and ΔS is change in entropy. Negative values of ΔG indicate spontaneity. Table 1 indicates that ΔH is positive for HAT in all circumstances that were studied, and because the HAT mechanism involves splitting a single molecule into a phenoxyl radical and a hydrogen radical, ΔS must also be positive. Therefore, even without knowing the exact numerical value of ΔS, we can recognize that low temperatures will yield a positive ΔG value, meaning that the reaction will be nonspontaneous, while sufficiently high temperatures will cause the (- TΔS) term to "outweigh" the positive ΔH and make the overall reaction spontaneous (as indicated by a negative ΔG).

Which of the following is NOT a functional group present in caffeine? A. Imide B. Amide C. Primary amine D. Tertiary amine

C. Imide, amide, and tertiary amine are all functional groups present in caffeine. All of the functional groups are part of caffeine except for a primary amine, which consists of nitrogen bonded to one R group and two hydrogen atoms. This structure is not part of caffeine.

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? *** appearance changes color but when asking for maintain function, we need an AA switched out for a similar AA. A.Gln62 B.Glu144 C.Ser157 D.His196

C. Serine157 is the only that shows a color change. When a questions mentions results presented in passage go to that figure and find key words

Given the data presented in the passage, what is true regarding the effect of tested target compounds on the cytotoxicity of HCT116 cells? A.It is due to a mechanism of action other than PI3Kα kinase inhibition. Show Explanation B.It is consistent with the effect of treatment on the cytotoxicity of other tested cell lines. Show Explanation C.It cannot be predicted directly from target compound inhibition of PI3Kα kinase in vitro. Show Explanation D.It is negatively correlated with PI3Kα kinase inhibition by the target compound

C. The data shown in Tables 1 and 2 suggest that there is not an obvious correlation between the extent of PI3Kα inhibition by the target compounds and ZSTK474 and the extent of their cytotoxicity against HCT116C. Each Target compound (4b-c and 5a-b) has a greater cytotoxic effect against HCT116 cells than does ZSTK474 (compound 4a), but a smaller inhibitory effect. However, this does not represent, in general, a negative correlation. In Table 2, there is little relationship between the extent of PI3Kα inhibition and cytotoxicity when comparing the measures for any two compounds.

The half-life of 18F is 110 minutes. If 5 grams of FDG remain after 5 hours and 30 minutes, how much energy was emitted from the patient's body in the form of gamma rays from radioactive decay of the FDG? A. 3.57 x 103 keV B. 1.83 x 102 keV C. 1.19 x 1026 keV D. 3.2 x 1026 keV *** somehow you are supposed to know molar mass of FDG is 181 g/mol***

C. This question requires us to determine how many half-lives have occurred after 5.5 hours. 5.5 hr x (60 min/1 hr) = 330 minutes, or 3 half-lives. Working backward from the final amount of 5 g we see that we must have started with 40 g of the original sample: 5 g x 2 (first half-life) x 2 (second half-life) x 2 (third half-life) = 40 g original sample Thus 35 grams of FDG must have undergone positive beta decay. 35 grams x (1 mol/181 grams) = about 1/6 mol of FDG. For each FDG molecule that decayed, 2 gamma rays were emitted. Therefore, 1/3 moles of gamma rays were emitted. (1/3) x (6.022 x 1023) * 511 keV = approximately 1 x 1026 keV.

The reduction of ketone groups, such as those present on CoQ, produces alcohols. In contrast, oxidation of aldehydes produces which functional group? A. None of these; aldehydes cannot be oxidized further. B. Primary alcohols C. Ethers D. Carboxylic acids

D). In organic chemistry, oxidation can be conceptualized as the gain of bonds to oxygen (or the loss of bonds to hydrogen). An aldehyde, which contains a C=O double bond, can gain another bond between carbon and oxygen if it is converted to a carboxylic acid. shout out A*** This option attempts to confuse aldehydes with ketones, which are difficult to oxidize further because their C=O carbon already possesses two carbon-carbon bonds. C-C bonds cannot be broken easily, meaning that ketones cannot easily gain C-O bonds.

If a leukemia patient has a hematocrit of 60%, with all other factors held constant, what will be the effect on the volume flow rate of her blood compared to a patient with normal hematocrit? A. It will increase by a factor of 1.5. B. It will increase by a factor of 2. C. It will decrease by a factor of 1.5. D. It will decrease by a factor of 2.

D. At a hematocrit of 60%, the viscosity (𝞰) of blood is about .008 kg∙cm-1s-1. This is twice as much as normal blood. Therefore, using the following equation: we see that the volume flow rate of her blood will decrease by a factor of 2.

A student uses thin layer chromatography on silica plates to monitor the progress of the reaction below. Does the product have a higher or lower Rf than the starting material?

DLower, because the product is more polar than the starting material The product of the reaction contains an -OHgroup and is therefore more polar than the starting material. Because of the -OH group,the product interacts with silica more strongly than the starting material and thus, has a lower Rf than the starting material.

If 1 mol of a pure triglyceride is hydrolyzed to give 2 mol of RCOOH, 1 mol of R'COOH, and 1 mol of glycerol, which of the following compounds might be the triglyceride?

If the ratio of R to R' in the product mixture is 2:1, then the starting triglyceride must consist of two R-containing esters and one R'-containing ester.

Which of the following characteristic changes in the IR spectrum would indicate the conversion of a fatty acid to an ester? A. Disappearance of a broad peak in the 2500 to 3300 cm-1 region only B. Appearance of a broad peak in the 2500 to 3300 cm-1 region only C. Appearance of a broad peak in the 2500 to 3300 cm-1 region, disappearance of an intense sharp band in the range 1730-1750 cm-1 D. The disappearance of a broad peak in 2500 to 3300 cm-1 region, the disappearance of an intense sharp band in the range 1730-1750 cm-1

In the conversion of a fatty acid to an ester, we lose a hydroxyl group, and neither gain nor lose any other relevant functional groups. Hydroxyl OH bonds in carboxylic acids sound in the range of 2500 to 3300 cm-1, making answer choice (A) the correct answer.

Which of the following statements correctly describes the conversion of pyruvate to lactate, shown below? Pyruvate- + NADH + H+ → NAD+ + Lactate- I. Pyruvate is reduced. II. NADH is the reducing agent. III. Hydrogen ion is a spectator ion. A. I only Show Explanation B. II only Show Explanation C. I and II only Show Explanation D. I, II and III Show Explanation

It should be noted that pyruvate being a reactant and lactate being a product is the same as the reduction half-reaction given in Table 1; therefore, pyruvate is reduced (I). In the given reaction, the NADH is going to NAD+ which is the opposite of the reduction half-reaction given in Table 1. Therefore, NADH is being oxidized, which does make it the reducing agent (II).

which of the following rankings (from most to least favorable) best describes the thermodynamic favorability of the antioxidant mechanisms in benzene? A. SET-PT > SPLET > HAT B. HAT > SPLET = SET-PT C. HAT > SPLET > SET-PT D. SET-PT = SPLET > HAT

Just look at first step of reaction, has to do that first, so only that enthalpy counts and remember more positive, less thermodynamically favored C is correct. The enthalpic changes (ΔH) in Table 1 indicate the enthalpic barrier that must be overcome for a reaction to proceed. As such, mechanisms with a lower ΔH are favored thermodynamically. For all three compounds, HAT shows markedly lower ΔH values in benzene than the other two mechanisms. To distinguish between SPLET and SET-PET, we must observe that although their total ΔH is comparable, SET-PT has a first step with a particularly high ΔH, meaning that it is especially thermodynamically unfavorable. Therefore, the correct ranking is HAT > SPLET > SET-PT.

Which relationship gives the value of R when R3 is adjusted so that the voltmeter reading is zero? A) R=R3 X R1/R2 B) R= R3 + R2-R1 C) R= R3- R2+R1 D) R=R3 X R2/R1

The answer to this question is A, because when the voltmeter reading is zero, the voltage across R is equal to the voltage across R1 and from Ohm's law, IR = I1R1, where I and I1 are the currents through the resistors. Moreover, IR3 = I1R2. By taking the ratio of these two equations, it follows that R/R3 = R1/R2, which is equivalent to R = R3 × R1/R2.

A pair of gamma rays emitted from the same annihilation event collide with sensors, but their collisions occur 0.33 nanoseconds apart. What is the minimum distance the annihilation could have occurred from the center of the machine? A. 5 cm Show Explanation B. 10 cm Show Explanation C. 15 cm Show Explanation D. 20 cm

*** super difficult!! A. Gamma rays travel at the speed of light, so both of the rays involved here must move at 3 x 108 m/s. Since the two rays arrived 0.33 nanoseconds apart, one ray must have traveled 10 cm farther than the other: (0.33 x 10-9 s)(3 x 108 m/s) = 1 x 10-1 m = 10 cm. According to the passage, the apparatus in question is circular, with the sensors positioned along the outer perimeter of the circle. This may be difficult to visualize, so when in doubt, draw it out! Suppose the radius of the machine is 2 m, as shown below. If the annihilation event occurs 5 cm from the center, and if the axis of gamma ray emission goes directly through the center, then one gamma ray will travel 95 cm to reach the sensors on the perimeter of the circle. The other gamma ray, which is propelled in the opposite direction, will have 105 cm to travel to reach the sensors on the other side. This represents a difference of 10 cm, which would explain the 0.33 ns time disparity outlined in the question. It turns out that this is the closest you can get to the center and still have one ray travel 10 cm farther.

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? 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.

A is correct because in the third step of the reaction show O2 is needed which means oxidation occurs somewhere. Looking for the differences between reactants and products shows that there is a double bond made next to the 5 membered ring. This causes another conjugated system to form which is said to "excitation and emission in ... GFP-like proteins." Without the molecular oxygen needed for this oxidation, such conjugation will not arise. Its absence could explain the failure of the molecule to fluoresce.

At what angle relative to the velocity of a red blood cell should the transducer be held during Doppler ultrasound to observe the largest Doppler shift? A. 0ºB. 30ºC. 45ºD. 90°

A is correct. Cosine θ is applied as a correction for the angle between the ultrasound beam and the direction of blood flow. Cos θ = 1 if the beam is parallel to the direction of blood flow, which allows for maximum velocity to be measured. Cos θ = 0 if the beam is perpendicular to the direction of blood flow and zero velocity is measured.

Which specific class of enzymes is primarily responsible for the release of free glycerol from stored triglycerides?A. Lipases B. Carboxylases C. Phosphorylases D. Kinases

A is correct. Lipases are the enzymes that digest lipids (fats). Most dietary fats originally exist in the form of triglycerides.

In 250 mL of the MH solution with the most favorable solubility profile, how many moles of nicotinamide (MW = 122 g/mol) are present? A.3.1 x 10-2 B. A.3.1 x 10-1 C. 1.6 x 10-2 D 1.6 x 10

A is correct. The solution with the most favorable solubility is that with an N:B:C ratio of 15:20:5. In this solution, the concentration of nicotinamide (N) is 15 mg/ml. Since the question states there are 250 mL of solution, the amount of nicotinamide must be 250 x 15 mg, or 3750 mg. We can estimate this value as 4000 mg, which is equal to 4 g.4 g nicotinamide x (1 mol nicotinamide / 122 g) = approximately 3.33 x 10-2

In one trial of this experiment, significant impurities were detected in the extracted caffeine. Based on the results of the experiment, which of the following would be the expected melting point range for this extracted sample? Pure caffeine has mp of 235. A) 195-220 B)245-267 C)233-236 D)190-195 (all in celsius)

A) 195-220 because it is lower than pure melting point and more broad. A melting point range includes the temperature when the first crystal of a compound starts to melt and the temperature when the compound is entirely melted. For a pure compound, the melting point range is narrow. Therefore, melting point determination is sometimes used to identify an unknown pure compound. However, the presence of impurities in a compound lowers and broadens the melting point range. According to the passage, the melting point of pure caffeine is 235°C, making choice A the best answer.

According to the results of the experiments, what was the risk of concussion to Participant 3 when riding Coaster 2? Participant 3= 750N and Coaster 2= HIC of 5.9 A)0% B)10% C)20% D)30%

A) According to the passage, the risk of concussion can be predicted for a given HIC value. According to Table 1, participants averaged a HIC value of 5.9 while riding Coaster 2. Examining Figure 2, we can see that a HIC score of 5.9 corresponds to a concussion probability (risk) of 0%. It is only around HIC values of 8-9 does the risk become non-zero.

In the reaction which forms the metal-bound carboxy-hydroxyl intermediate, the metal-bound CO2• radical anion acts as: A. an Arrhenius base. B. a Bronsted-Lowry base. C. a Bronsted-Lowry acid. D. a Lewis acid.

A) Arrhenius base donates OH so no B) BL base accepts H+ so yes C) BL acid donates H+ so no, CO2 radical can not D) Lewis acid accepts lone pair of electrons so no The bronsted-Lowry base is a proton acceptor. In reaction 2, the CO2• radical anion as a Bronsted-Lowry base when it abstracts a proton from water to generate the metal-bound carboxy-hydroxyl intermediate.

In their published results, the researchers reported that RT112 cells were injected subcutaneously into the right flanks of nude mice on day 1, yet Figure 3 shows the IA-3 was not administered until day 4. Which of the following was the reason for this delay? A. The animals were randomized into two cohorts of four animals each. B. One group received IA-3, while the other group received luciferase vehicle only as control. C. The researchers wished to evaluate the anti-tumor depolymerization effect of IA-3. D. The IA-3 needed time to build up to therapeutic levels in the animals.

A) no. Even if true, this does not explain the need to delay the treatment with IA-3. B)no. Though true, this does not explain the delay. Paragraph 3 outlines how the researchers needed to ensure that luciferase had no effect on tumor growth prior to running the experiment. This means that the luciferase injection acted as the control group in the experiment. C) This is the correct option. The investigators were looking to test if IA-3 had any anti-tumor effects. Thus, the researchers needed to allow detectable tumor to grow before they could test if IA-3 was able to kill the tumors as described in paragraph 2. D)no. This would explain the need for multiple doses administered between days 4 and 10, and Figure 1 does show that the significant anti-tumor effects did not take place until after day 10. However, this does NOT explain the delay of four days prior to injection.

The results shown in Table 1 most strongly support which conclusion? A. Photodynamic therapy is ineffective at killing cancer cells. B. A reduction in mitochondrial membrane potential causes a decrease in proteasome activity. C. Porfirmer is toxic to cancer cells. D. Cellular effects of photodynamic therapy are partially reversible.

A)The last 3 columns of the table show that photodynamic therapy kills up to 50% of cancer cells by 24 hours after treatment. B)Although the mitochondrial membrane potential decreases prior to the decrease in proteasome activity, we can't conclude from this correlation that there is a causal relationship between the two. In this study, both mitochondrial memory potential and proteasome activity are dependent variables. C) According to the third row of Table 1, porfirmer alone does not kill cancer cells. Light activation of porfirmer is required for cytotoxicity. D)The last row of Table 1 shows that mitochondrial membrane potential and proteasome activity recover from an initial decline at 6 and 24 hours, respectively. Therefore, the effect of the photodynamic therapy must be partially reversible in the surviving cells.

If negative charge repulsion in C4S units contributes to the elastic modulus of articular cartilage, given passage information, which of the following is likely observed as a result of OA in articular cartilage? A. Decreased charge repulsion and decreased compressive strength B. Increased charge repulsion and increased compressive strength C. Decreased charge repulsion and increased compressive strength D. Increased charge repulsion and decreased compressive

A. According to paragraph 2, "GAGs ... interact non-covalently with one another to exert outward pressure within the cartilage" and "Loss of C4S is believed to be responsible for changes in the biomechanical properties of cartilage seen in osteoarthritis (OA)." If negative charge repulsion in C4S units contributes to the elastic modulus of the cartilage, decreased C4S content due to OA would then decrease negative charge repulsion and change a mechanical property of the cartilage—namely, its elastic modulus, which is defined in the passage as being a "a measure of compressive strength." This is very nearly the compressive strength mentioned in the question stem. Overall, this is consistent with the positive correlation between elastic modulus and GAG content shown in Figure 2.

According to the results of the experiment, the minimum risk for TBI during the pillow fight occurred during which time interval? A.5.90 sec to 5.91 sec B.5.91 sec to 5.92 sec C.5.92 sec to 5.93 sec D.5.95 sec to 5.96 sec

A. According to the passage and Figure 2, higher TBI risk is associated with larger HIC. HIC is proportional to the average linear force, a product of mass and acceleration. Thus, the smaller the magnitude of acceleration, the lower the risk conferred to the individual. The rate of change of acceleration for any given time interval is the magnitude of the slope of the graph of acceleration vs. time for that interval. Using Figure 1, the rate of change in the interval between 5.90 sec and 5.91 sec is approximately zero, making it the lowest risk period for TBI.

The stress and strain values for four elastic solid materials were used to obtain the following graphs. Which material would have the greatest resistance to linear deformation (strain)? A. 1 Show Explanation B. 2 Show Explanation C. 3 Show Explanation D. 4

A. In solid-state physics, Young's modulus is a measure of the stiffness of an elastic solid material. The slope of the graph (stress/strain) is equal to the Young's modulus. The higher the Young's modulus, the stiffer the solid material is. Graph '1' has the highest slope and therefore is the stiffest. D is wrong because This material has a smaller slope, so it is less stiff than material 1. The order of materials according to their stiffness from the stiffest to the least stiff is as follows: 1 > 2 > 3 > 4.

Which of the following surfactant characteristics would NOT maximize the rate of a nucleophilic reaction with a neutral substrate? A. A relatively high CMC B. A relatively low CMC C. A long hydrophobic tail in a cationic surfactant Show Explanation D. A short hydrophobic tail in an anionic surfactant "A micelle is an aggregate of surfactant molecules dispersed in a liquid. Micelles form only when the concentration of surfactant is greater than the critical micelle concentration (CMC). In a polar solvent, a normal-phase micelle (oil-in-water micelle) forms. However, in a nonpolar solvent, a reverse micelle (water-in-oil micelle) forms, as shown in Figure 2. Self-organized assemblies such as micelles can change the rates of numerous types of reactions, due to their electrostatic and hydrophobic interactions with reactants."

A. Paragraph 4 indicates that organic cosolvents decrease the catalytic effects of a surfactant by increasing its CMC, implying that high CMCs impede catalytic effectiveness. Therefore, a surfactant with a relatively high CMC would be expected to be unfavorable as a catalyst, making it the correct answer to this "NOT" question. B. no Since a high CMC impedes catalytic effectiveness, a low CMC would promote it.

What is the pH of a 0.010 M perchloric acid solution? A 2 B 4 C 7 D 12

A. Perchloric acid is a strong acid that completely dissociates in aqueous solution, so the hydrogen ion concentration is 1.0 x 10-2 M. The pH = -log[H+] = -log[10-2] = 2.

Which of the following compounds will have the highest retention time when run in a gas-liquid chromatography chamber? A. Methyl cyclohexane B. Hexane Show Explanation C. Methylene chloride Show Explanation D. Heptane Show Explanation

A. The compounds listed above are nonpolar, so they will separate mainly based on their boiling points. In gas-liquid chromatography, the sample is volatilized in a hot chamber. The higher the boiling point temperature, the more time the compound will spend at the beginning of the chamber waiting to get volatilized. Retention time is the time it takes for the sample to reach the detector. Therefore, the compound with the highest boiling point, in this case, methyl cyclohexane, will have the highest retention time.

All of the following can be concluded about the heat engines tested EXCEPT: A. if the efficiency of engine 1 increases, the interior area of its heat cycle in Figure 2 will decrease. B. without a change in temperature, no net work could be extracted from the heat cycle. C. isothermal expansion is followed by an adiabatic expansion in Figure 2. D. isothermal compression is followed by an adiabatic compression in Figure 2.

A. The formula for the efficiency of any system is output work/input energy. Therefore, the efficiency of engine 1 is W/QH. Efficiency is increased by increasing W. Since work in Figure 2 is the interior area of the cycle, if W increases, the interior area should increase as well. This makes choice A, which states the opposite, correct for this EXCEPT question.

Administration of a muscle relaxant causes a vessel to increase the radius by 10%. If all else is held constant, the volume flow rate will change to approximately what percent of the original rate? A. Increase to 150% of the original B. Remain at 100% of the original C. Decrease to 65% of the original D. Cannot be determined

A. Volume flow rate is proportional to the radius raised to the fourth power. An increase of 10% means the rate of flow will increase to 1.14 of the original rate. We need not do any calculations here since only choice A says that the flow rate will increase.

The first step of the formation of the imidazolinone ring of sgBP is most likely accomplished by the: A.attack of Gly64 amide nitrogen on the electrophilic Gln62 carbonyl carbon. B.attack of Gln62 amide nitrogen on the electrophilic Gly64 carbonyl carbon. C.attack of Tyr63 phenolic hydroxyl group on the electrophilic Gln62 carbonyl carbon. D.attack of Tyr63 phenolic hydroxyl group on the electrophilic Gly64 carbonyl carbon.

A. the attack of Gly64 amide nitrogen on the electrophilic Gln62 carbonyl carbon. PAY ATTENTION TO GLY VS GLN. the heterocycle in the Aequorea fluorophore is likely formed in step 1 by the attack of the nucleophilic amide nitrogen of glycine on the electrophilic carbonyl carbon of serine. In the QYG chromophore of sgBP, serine is replaced by glutamine. According to the first paragraph of the passage, "CP chromophores form an imidazolinone structure through autocyclization and dehydration reactions, shown in steps 1 and 2 of Figure 1." Given this, choice A, the attack of the nucleophilic amide nitrogen of glycine 64 on the electrophilic carbonyl carbon of glutamine 62, is most likely to occur during the first step of the formation of the imidazolinone ring of sgBP.

Which of the following situations would present Participant 3 with the greatest risk for TBI? A. A linear acceleration of 10 m/s2 for 2 ms in a car crash B. A single punch to the head causing acceleration of 100 m/s2 for 3 ms C. Riding a roller coaster of 9 m/s2 sustained for 40 s D. Total body acceleration of 15 m/s2 for 1 ms

According to the second paragraph of the passage, G-forces alone do not predict TBI. Besides the force, the time duration over which the force is applied is important in determining the severity of a brain injury. The second paragraph also states that the body can endure a large G-force if it occurs over a very short time period. Using Equation 1 from the passage, we can calculate the value of HIC for each of these activities. The acceleration values are given for all the choices. To incorporate acceleration, we rewrite the equation into HIC = m*a*t, where "m" represents the mass of the brain (which, since the subject is the same, can be assumed to be equal for all cases). Therefore, the impulse will be: A: HIC = m (10 m/s2) (0.002 s) = (0.02 m) Ns B: HIC = m (100 m/s2) (0.003 s) = (0.3 m) Ns C: HIC = m (9 m/s2) (40 s) = (360 m) Ns D: HIC = m (15 m/s2) (0.001 s) = (0.015 m) Ns According to Figure 2, the risk of TBI increases with HIC, making the conditions in choice C have the highest probability of brain injury.

Surfactant-substrate interactions can involve which of the following intermolecular forces? I. London dispersion forces II. Hydrogen bonding III. Ion-dipole interactions A. I only Show Explanation B. I and II only Show Explanation C. II and III only Show Explanation D. I, II, and III

All three of these intermolecular forces can be involved in surfactant-substrate interactions, depending on the surfactant and the substrate.

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? A.Compared to hydrogen bonding at pH 4.5, the ion-dipole interaction at pH 10 destabilizes the sgBP chromoprotein barrel conformation. B.Compared to hydrogen bonding at pH 4.5, the ion-dipole interaction at pH 10 decreases the maximum normalized absorbance. C.Compared to hydrogen bonding at pH 4.5, the ion-dipole interaction at pH 10 promotes peak absorbance of higher energy radiation. D.Compared to hydrogen bonding at pH 4.5, the ion-dipole interaction at pH 10 increases the wavelength of maximum absorption.

Answer is C. When pH is 10 the cysteine and serine are in deprotonated form so there is more ion-dipole interactions that causes a "protonation state-dependent ion-dipole" interaction that "accounts for greater strength of ion-dipole" bonds relative to hydrogen bonds." The pH of 10 promotes the maximum absorption of shorter-wavelength, higher energy radiation.

Aspartame (L-α-aspartyl-L-phenylalanine 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. hexose. B. phenol. C. dipeptide. D. amino acid.

As mentioned in the name there are two amino acids so this is a dipeptide!! slow down and read all options

What is the voltage generated by an electrochemical cell at equilibrium? A. The voltage must be positive. B. The voltage must be zero. C. The voltage must be negative. D. All REDOX reactions are at equilibrium, irrespective of their voltage.

At equilibrium, the free energy change is zero. Therefore, the REDOX potential is zero as well, from ΔG = -nFE. When considering Equation 2, if Q = Keq, then the -0.06/n log K term cancels the E°. Choice B is the best answer. C.

Participant 1 rides a fourth roller coaster as shown below. What is the minimum ramp height H if the ride at the top of the loop maintains the minimum speed needed to stay on the track throughout the loop? (Note: Neglect friction.)

At the top of the loop, the gravitational and the normal forces (if any) point downward toward the center of the loop. Therefore, the net force causing centripetal acceleration is the sum of the gravitational force and the normal force. When the centripetal force is the minimum amount needed for the ride to stay on track, the normal force is zero. Note: At the top, the normal force can be zero but the gravitational force cannot. Let's designate the initial height where the ride starts as H. At the very beginning, the energy of the ride is the gravitational potential energy. Therefore, Ei = mgH. At the top of the loop, the ride includes both gravitational and potential energy. Therefore, Ef = mgh + ½ mv2 = mg (2r) + ½ mv2. The net acceleration at top (centripetal acceleration) = v2/r. Since we concluded that the normal force at the top is zero, the net acceleration at the top is the gravitational acceleration, g. Therefore, v2 = gr. Substituting this equation into the 'Ef' equation, we obtain the following: Ef = 2mgr + ½ mgr = (5/2) mgr To obey conservation of energy, Ei = Ef. Therefore: mgH = (5/2) mgr. We conclude that H is 5r/2. Since r is 8 m, H is 5(8 m)/2 = 20 m.

The conversion between glucose and pyruvate strongly favors the formation of pyruvate, and yet the gluconeogenic pathway is able to utilize several shared enzymes with glycolysis to create glucose from pyruvate. It is able to do this primarily because: A) the formation of glucose from the gluconeogenic precursors is strongly favored by the increasing entropy of the reaction B) the formation of glucose, fructos-6-phosphate, and PEP through gluconeogenesis-specific enzymes push the equilibrium of reaction catalyzed using shared enzymes to favor gluconeogenesis C) the activity of the enzymes with two or more functions is reversed when those enzymes are altered by kinases. D)enzymes in gluconeogenesis bypass reversible, equilibrium-state steps common to both pathways.

B is correct. Production of PEP, glucose, and fructose 6-phosphate by gluconeogenesis-specific enzymes that bypass irreversible steps of glycolysis push the equilibrium of reversible reactions in the direction of glucose production. A- glycolysis and gluconeogenesis are reliant on changes in enthalpy through breaking and forming high-energy bonds, rather than entropic changes. C- enzymes in glycolysis or gluconeogenesis do not depend on the nature of their phosphorylation D- the equilibrium-state steps common to both pathways are irreversible, not reversible

Which of the following boundaries would give a brighter ultrasound image at the boundary but the poorest overall resolution? A. Fat/muscle B. Muscle/bone C. Muscle/blood D. Water/fat

B is correct. The passage states that if the ΔZ is very large, all of the ultrasound will be totally reflected at the boundary. The imaging will be poor because too much sound was reflected back, and there was not enough left to be able to penetrate further and continue imaging. If ΔZ is small, a small amount of sound will be reflected back, which would allow enough sound left to continue through for further imaging. Muscle and bone have the largest difference in impedance (7.8 - 1.7 = 6.1) of all the combinations given.

Three students in physics lab are given four capacitors, each with a capacitance of 4 μF, and are told to construct a circuit with the maximum possible total capacitance. Student 1 attaches all four capacitors in series. Student 2 attaches all four capacitors in parallel. Student 3 only inserts a single capacitor in his circuit. Which of the following is true? A. Student 1 and Student 3 will create circuits that have the same total capacitance. B. Student 2 will create the circuit with the highest capacitance. C. Student 2 and Student 3 will create circuits that have the same total capacitance. D. Student 3 will create the circuit with the highest capacitance.

B is correct. The total capacitance of capacitors in parallel is simply the sum of their individual capacitances, so student 2 would create a circuit with 16 μF and the highest capacitance. Student 3, by only using a single capacitor, would create a circuit with 4 μF capacitance. When capacitors are connected in series, the reciprocal of the total capacitance is found by adding the reciprocals (like resistors in parallel), so student 1 would only create a circuit with 1 μF capacitance. *** capacitors in series add inverse, if in parallel add whole numbers****

Which of the following is most likely the net charge on a valine molecule in the human body? A. -1 B. 0 C. +5 D. +1

B is correct. This question is asking us to remember what factors contribute to the net charge on a valine molecule. Being an amino acid, valine has an acidic carboxy group that will be deprotonated at physiological pH. It also has a basic amine group that will be protonated at physiological pH, as well as a neutral side chain.

The Arrhenius equation, k = Ae-Ea/(RT), expresses the relationship between the rate constant, k, and the temperature of a reaction. According to this equation, which of the following will increase the rate of a reaction? A. Decreasing the pre-exponential factor, A B. Increasing the temperature, T C. Increasing the activation energy, Ea D. Decreasing the temperature, T

B is correct. While this question introduces the Arrhenius equation, you can answer it simply by remembering the principle in general chemistry that increasing temperature increases the rate of reaction.

What will be the difference in hematocrit between a person living at a higher elevation and a person living at a lower elevation? A. The person living at a higher elevation will have a higher hematocrit because of decreased atmospheric nitrogen. B. The person living at a higher elevation will have the higher hematocrit because there is less oxygen in the air. C. The person living at a lower elevation will have the higher hematocrit because of the higher atmospheric pressure. D. The person living at a lower elevation will have a higher hematocrit because there is more oxygen in the air.

B. A person living at a higher elevation will produce more red blood cells because there is less oxygen in the air. The kidney has receptors that read the oxygen levels in the blood and produce erythropoietin in order to produce more red blood cells. This will increase the hematocrit.

Following the procedure of the experiment, how much total caffeine could you expect to extract? To extract caffeine, a No Doz® tablet (10 g each) was crushed and added to 100 mL of distilled water and boiled for ten minutes. density of 1.23 g/cm3 A. 6.50 g B. 9.84 g C. 11.72 g D. 12.97 g

B. Caffeine is largely nonpolar, so that it will be more soluble in dichloromethane than in water. Therefore, it will dissolve more in the dichloromethane layer. Using the partition coefficient equations and the caffeine properties provided by the passage, we must add up the amount of caffeine dissolved in each layer in both the first extraction and the second extraction. First extraction: According to paragraph 1, the solubility of caffeine at standard temperature is 0.14 g/mL in dichloromethane and 0.02 g/mL in distilled water. Multiplying each number by 100 ml of dichloromethane and water used in the extraction results in 14 g / 100 mL and 2 g / 100 mL, respectively. Ψ ≅ solubility of a solute in organic layer / solubility of a solute in aqueous layer: Ψ = (14 g / 100 mL)/ (2 g / 100 mL) = 7 A tablet weighing 10 g was used. If we assume that X grams dissolve in dichloromethane, then (10-X) g dissolve in the aqueous layer. Therefore: Ψ = (X g / 100 mL of CH2Cl2) / ([10 - X] g/ 100 mL of H2O) = 7 X = 70 - 7X 8X = 70 X = 70/8 = 8.75 g dissolved in dichloromethane If 8.75 grams dissolve in dichloromethane, then 10 g - 8.75 g = 1.25 is dissolved in the aqueous layer, which is used in the second extraction. Ψ = (X g / 100 mL of CH2Cl2) / ([1.25-X] g / 100 mL of H2O) = 7 X = 8.75 - 7X X ≅ 1 g Total caffeine extracted ≅ 8.75 g + 1 g = 9.75 g Choice B is the closest answer. Notice that calculations result in weight slightly less than that of a tablet. Tablets use binders, so the actual amount of caffeine is slightly lower.

Which of the following best describes the velocity profile of laminar flow from left to right in a stationary tube?

B. Laminar flow is due to shear forces (friction) between the fluid and the solid surface of the tube. This results in layers having a gradient of velocities, in which the flow is the fastest in the middle of the tube (where friction is low) and slowest near the surface (where friction is high).

Thalassemias are blood disorders in which the synthesis of either α-globin-like or β-globin-like chains are defective. Thalassemia involving non-expression of which protein chain is least likely to produce physical manifestations prior to birth? A. α globin B. β globin C. γ globin D. ζ globin

B. Of the four proteins listed, Figure 1 shows that at all points prior to birth, at least two other globin chains are present at greater levels than β globin. Given this, β globin will be present in the smallest concentration, if at all, relative to the content of Hb proteins synthesized.

Approximately half of hemoglobin A's oxygen binding sites are occupied at what partial pressure of O2? A. 20 mmHg B. 30 mmHg C. 50 mmHg D. 80 mmHg

B. One-half of Hb's oxygen binding sites are filled at its P50 value. In Figure 2, Y for Hb A is near 50% at a pO2 of approximately 30 mmH

The Stobbe condensation mechanism shares a similar reaction scheme with the Claisen condensation. Which of the following correctly identifies a step in the Claisen condensation? A. Alkoxy group elimination results in generation of an ester. Show Explanation B. The enolate anion attacks the carbonyl through a nucleophilic acyl addition mechanism. C. The overall reaction is exergonic. Show Explanation D. Dehydration takes place via an E2 mechanism.

B. Step 1: An α-proton is removed by a strong base, resulting in the formation of an enolate anion, which is made relatively stable by the delocalization of electrons. Step 2: The carbonyl carbon of the (other) ester is attacked by the enolate anion/nucleophile.Step 3: The alkoxy group is then eliminated, and the alkoxide removes the newly formed doubly α-proton to form a new, highly resonance-stabilized enolate anion. Aqueous acid is added in the final step to neutralize the enolate and any base still present. The newly formed β-keto ester or β-di-ketone is then isolated. D wrong cause : Claisen condensation, much like aldol condensation, occurs via a modified mechanism different from E1 and E2.

In order to measure the ∆G, researchers needed to denature the proteins. Which of the following steps would be LEAST suitable for this procedure? A. Increasing temperature Show Explanation B. Irradiating the protein with non-ionizing radiation Show Explanation C. Adding a concentrated chaotropic agent Show Explanation D. Lowering the pH

B. This answer choice specifically says that the protein will be irradiated with nonionizing radiation, which does not break bonds. Therefore, this method would not be able to denature the proteins. Irradiating the proteins with X-rays or gamma rays certainly can denature them, but these are forms of ionizing radiation. not C because A chaotropic agent is any molecule in an aqueous solution that can disrupt the hydrogen bonding network including the bonds between water molecules. As such, these agents (urea is one example) are capable of denaturing proteins by disrupting the hydrogen bonds that stabilize their configurations.

Which of the following is a plausible scenario for the work of CoQ in the electron transport chain? A. CoQ10 is reduced by Complex I and later oxidizes Complex III. B. CoQ10 oxidizes Complex I and then is later oxidized by Complex III. C. CoQ10 reduces Complex I and later reduces Complex III. D. CoQ10 is oxidized by Complex I and is later oxidized by Complex III

B. This question is asking us to identify the function of CoQ. From the passage we learn that CoQ takes electrons from Complex I and II and passes them to Complex III. Since it takes electrons from Complex I, it is reduced and it oxidizes Complex I. Later, when it passes electrons to Complex III, CoQ is oxidized and reduces Complex III. Co Q takes electrons from complex I = coq oxidizes complex I; coq is reduced (GER). complex III takes electrons from Coq = complex III oxidizes CoQ; coq is oxidized (LEO) *** shout out A is confusing but coq does not oxidize complex III, complex III oxidizes CoQ.

The UV absorption of A in MH solutions and in aqueous solutions both peak at 314 nm, suggesting... A) A contains C=O double bond B) no significant structural changes in A occur during MH solution preparation C) A in solution would be red D) significant

B. UV radiation is higher-energy, higher-frequency EMR than visible light or IR radiation. The passage states that UV absorbance spectroscopy was performed in order to assess whether the MH preparation process had changed the structure of A. The fact that A's peak UV absorbance remains the same in aqueous and MH solutions suggest that the structure is identical.

The area enclosed by the four steps shown in Figure 2 represents: A.the work done on the engine during one complete cycle. B.the work done by the engine during one complete cycle. C.the thermal energy change of the surroundings during one complete cycle. D.the heat transferred out of the engine during one complete cycle.

B. W=PVPressure-volume diagrams are a visualization tool for the study of heat engines. Since work is done only when the volume of the gas changes, the diagram gives a visual interpretation of the work done. Since the internal energy of an ideal gas depends upon its temperature, the PV diagram along with the temperatures calculated from the ideal gas law determine the changes in the internal energy of the gas so that the amount of heat added can be evaluated from the first law of thermodynamics. For a cyclic heat engine process, the PV diagram will be a closed loop. The area inside the loop is a representation of the amount of work done by the engine during a cycle.

Towards the end of a game, a team's pitcher begins to lose his focus and shortens his stride, uncoupling the kinetic chain at his hips. How will this impact his pitch? A. It will make his shoulder more injury-prone as a result of the increased energy being dissipated during the follow-through from the inefficient kinetic chain. B. It will decrease the velocity of the pitch due to a decrease in energy transfer through the torso. C. It will decrease the velocity of the pitch as a result of the increased energy being dissipated during the follow-through from the inefficient kinetic chain. D. It will decrease the velocity of the pitch due to a decrease in the ability to generate force in the legs.

B. uncoupling the kinetic chain at the hips will decrease the efficiency of the energy transfer from the legs and hips to the lower torso and result in overall less energy transfer through the upper body to the upper extremities and ball. This will decrease the velocity of the pitch.

Under anaerobic conditions, bacteria can sometimes derive energy from the oxidation of sulfur-containing species. Sulfur in which of the following compounds would be LEAST likely to be oxidized by an anaerobic bacteria? A. H2S (aq) Show Explanation B. SO42- (aq) C. S2O32- (aq) Show Explanation D. S8 (s)

B.. check oxidative states of each sulfur. in B it is most positive so it has no electrons to give. Of the answer choices, the oxidation number for sulfur would need to be in a reduced form. The highest oxidation state for sulfur is +6, corresponding to the loss of all of its valence electrons. The oxidation number for sulfur (x) in sulfate, SO42-, can be determined by assuming that the oxygen atoms are oxides having a -2 charge and that the sum of the oxidation numbers must equal the overall charge of the ion, giving -2 = x + 4(-2) -2 = x - 8 +6 = x The oxidation number of sulfur in sulfate is +6, and choice B is the best answer. The oxidation number of sulfur in hydrogen sulfide (H2S), thiosulfate (S2O32-) and yellow elemental sulfur (S8) are -2, +2 and 0, respectively, all of which could be oxidized and be a source of energy for anaerobic bacteria.

Given two segments of a blood vessel with the following information: If the volume flow rate through segment 1 is 3 times that of segment 2, with all else being held constant, what is the value of P3? A. 40 B. 70 C. 80 D. 90

Because we are given the information on volume flow rate in the question stem, our first step is to use Equation 1: (p1-p2)/R= (P2-P3)/R The information given to us in the question stem indicates that none of the parameters affecting R (viscosity, length, or radius) will change, so we can cancel out R and solve as follows: Plugging in the values for P1 and P2, we get P3 = 80. Thus, choice C is the correct answer.

A certain metabolic process in the liver produces NADH as a part of the process. If this process is upregulated, which of the following effects associated with gluconeogenesis is most likely to follow? 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.

C is correct. The key to this question is that we do not know which process the question stem refers to; we only know that it produces NADH. We are given no reason to assume that this process is the same as that described in the question stem, in which "the conversion of lactate to pyruvate is coupled with the reduction of NAD+ to NADH." Since we cannot assume that this is the process being discussed, the only information available fo

Which of the following compounds is amphiprotic? A. Acetic acid, HC2H3O2 B. Sodium acetate, NaC2H3O2 C. Sodium bicarbonate, NaHCO3 D. Sodium carbonate, Na2CO3

C is correct. The prefix "amphi-" means "both." Therefore, an amphiprotic species is one that can act as both an acid or a base. Sodium bicarbonate (choice C) dissolves in aqueous solution to produce sodium ions and bicarbonate ions. The former ion is neither acidic nor basic, but the bicarbonate ion, HCO3-, can act as a Bronsted-Lowry acid by loss of a hydrogen ion and can act as a B-L base by accepting a hydrogen ion to form carbonic acid, H2CO3.

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

C is correct. Ultrasound is defined as sound with a frequency above the human range of hearing ("ultra" = "above"). To answer this question, then, we need to know what the human range of hearing is. This range is 20 Hz to 20 kHz, so anything greater than 20 kHz qualifies as ultrasound.

Which of the following correctly represent the units of the Boltzmann constant? A) kg/s B)N/K C)J/K D)K/J*s

C) J/K because we know the Boltzmann constant has units of (m2•kg)/(s2•K). A joule has units of kg•m2/s2. Dividing by K gives us units for the Boltzmann constant.

How is Solubility Enhancement Ratio (SER) in Table 1 calculated? A) by dividing A's hydrophilic solubility by its hydrophobic solubility B) by multiplying A hydrophilic solubility by its hydrophobic C) by dividing the solubility of A in MH by its solubility in water D) by multiplying the solubility of A in MH by its solubility in water

C) Table 1 shows that the higher the solubility of A in a given solution is, the higher the SER. In fact, the SER is a direct reflection of A's solubility in a given solution versus that of water. Using the solubility of A in water from the passage, we can see that SER is calculated by dividing the solubility of A in the mixed hydrotrope by the solubility of A in water.

What is the most likely purpose of the long tail on Coenzyme Q10? A. To allow CoQ10 to be soluble in the mitochondrial matrix Show Explanation B. To act as a flagellum that is used for propulsion Show Explanation C. To allow CoQ10 to stay in and move in the hydrophobic portion of the inner mitochondrial membrane Show Explanation D. To allow CoQ10 to interact with hydrophobic regions on Complexes I and II

C) This question is asking us to identify the purpose of the hydrophobic region of CoQ10. Based on the information in paragraph 2, we can infer that CoQ acts as an electron transporter between two complexes embedded in the inner mitochondrial membrane and that it resides in the membrane itself. CoQ can only do this if it has a large hydrophobic region.

Industrially-synthesized trans fats were developed for use as cheap substitutes for animal saturated fats in food products, such as baked goods, that are generally stored at room temperature. Which of the following statements provides a reasonable hypothesis as to why they might be suitable for this purpose? A. The hydrocarbon tails of trans fats stack more densely than those of comparable cis fats, making them more likely to be liquids at room temperature. Show Explanation B. The hydrocarbon tails of trans fats stack less densely than those of comparable cis fats, making them more likely to be liquids at room temperature. Show Explanation C. The hydrocarbon tails of trans fats stack more densely than those of comparable cis fats, making them less likely to be liquids at room temperature. Show Explanation D. The hydrocarbon tails of trans fats stack less densely than those of comparable cis fats, making them less likely to be liquids at room temperature.

C. As shown in the image below, the hydrocarbon tails of trans fats are very similar to those of saturated fatty acids, meaning that they stack together more densely than the corresponding cis fats. This allows choices B and D to be eliminated. The denser, more stable stacking of trans fats means that they are more likely to be solids—and therefore less likely to be liquids—at room temperature. Thus, A is incorrect and C is correct. This property of trans fats makes them useful for inclusion in commercially produced baked goods because the stable solid state of trans fats contributes to longer shelf life. Unfortunately, trans fats have also been closely associated with an elevated risk of cardiovascular disease.

In the passage, the researchers attempted to model chemical bonds and intermolecular forces as springs. Suppose the chemical potential energy in a single disulfide bond was modeled as elastic potential energy stored in a stretched spring, with a spring constant of 2 N/m. Before stretching, the spring length was 10 Å. What would the length of the stretched spring be? (A disulfide bond has a bond dissociation enthalpy of 54 kJ/mol.) A. 3 x 10-13 m Show Explanation B. 3 x 10-10 m Show Explanation C. 1.3 x 10-9 m Show Explanation D. 2.3 x 102 m

C. C is the only possible answer as A and B are smaller than the original spring length and D is much too large for a bond. To tackle this question, we first need to figure out how much chemical potential energy is contained in a single disulfide bond. To do this, we have to divide the value of 54 kJ/mol by Avogadro's number. We will round Avogadro's number to 6 × 1023 to make the math a little easier: 54 kJ/mol / 6 × 1023 bonds/mol = 9 × 10-23 kJ/bond = 9 × 10-20 J/bond Recall that elastic potential energy has the following equation: PE (spring) = ½ kx2 where k is the spring constant and x is the displacement in the spring. We can rearrange this equation to isolate the displacement: Now we'll substitute our values into this equation: But don't forget, we have to find the total stretched length of the spring. The passage tells us that 1 Å = 10-10 m, so our spring has a length of 10-9 m. Summing that value and the displacement of the spring gives us a grand total of 1.3 x 10-9 m for the length of the stretched spring.

Before coordination to protoporphyrin, what is the ground state electron configuration of Co2+? A. [Ar]4s23d5 B. [Ar]4s23d7 C. [Ar]3d7 D. [Ar]4s24d5

C. During the ionization of transition metals, electrons from 4s subshell orbitals are generally removed before those from 3d subshell orbitals. This is because it is generally true that, when occupied by electrons, electrons of 4ssubshell orbitals are higher energy than those in 3d subshell orbitals.

Hb Gower 1 is the most common hemoglobin produced during the first month following conception. Based on passage information, the most likely subunit composition of the Hb Gower 1 protein is: A.α2γ2. B.α2ε2. C.ζ2ε2. D.α2β2.

C. Figure 1 indicates that only ζ and ε are present in meaningful quantities for the first month following conception. Consequently, Hb Gower 1 is most likely to be composed of ζ and ε chains.

A horizontal force of 100 N is applied to a 50-kg box that is accelerating at 1 m/s2 on a rough, horizontal surface. What is the work done by kinetic friction if the box is moved 4 m? A. 0 J Show Explanation B. -150 J Show Explanation C. -200 J Show Explanation D. -400 J

C. Fnet = ma = Fapplied - Fk, where Fk is the force of kinetic friction. Using this formula: (50 kg)(1 m/s2) = 100 N - Fk 50 N = 100 N - Fk Fk = 50 N To calculate work, we use W = Fd cos θ. W = (50 N)(4 m)(cos 180º) = -200 Nm = -200 J.

The intravenous solutions used to correct fluid imbalances in trauma patients are aqueous solutions of glucose. Approximately how much glucose is needed to make 1.0 L of a 100 mM solution? A. 18 mg Show Explanation B. 34 mg Show Explanation C. 18 g Show Explanation D. 34 g

C. From the molarity and the volume of the solution, we can calculate the number of moles and then the mass of glucose (C6H12O6, MW = 180 g/mol) needed to prepare the solution. 1.0 L x 0.10 mol/L x 180 g/mol = 18 g glucose. D.

Surfactants exert catalytic effects through which of the following mechanisms? A. Making the catalyzed reaction more energetically favorable B. Changing the equilibrium constant of the catalyzed reaction to favor the products C. Reducing the activation energy of the reaction D. Covalently transferring a reactive functional group to a reactant

C. In general, catalysts speed up reactions by reducing the activation energy, which applies to both the forward and backward reactions. Furthermore, catalysts are neither produced nor consumed in a reaction.

What is the normality of a 0.015 M solution of phosphoric acid assuming complete dissociation? A. 0.015 N B. 0.030 N C. 0.045 N D. 0.060 N

C. The chemical formula of phosphoric acid is H3PO4. Normality, in the context of acids, refers to the number of moles of protons per liter of solution (in other words, to the "molarity of protons"). Normality can be calculated by multiplying the molarity of the solution by the number of protons per molecule of acid (here, 3). (0.015 M solution) x (3 protons per molecule) = 0.045 N D. 0.060 N

The molecule below is: I. aromatic II. non-aromatic III. a heterocycle A) I only B) II only C) I and III only D) II and III only

C. The molecule has 6 pi electrons, so according to Huckel's rule it is aromatic. It is also a heterocycle, a ring of atoms of more than one kind, so both I and III are correct.

An object made of silicon (specific heat = 710 J/kg°C) absorbs 3500 J of heat while increasing its temperature from 43°C to 53°C. What is the approximate mass of the object? A) 350 g B) 400 g C) 500 g D) 2050 g

C. The most efficient way to find this answer is by using the equation q (the heat in or out of a system) = mcΔT. Because this is a change in temperature, either centigrade or Kelvin temperatures can be used. Substituting in numbers, we get: 3500 J = m x 710 J/kg°C x (53°C - 43°C) Approximating values and rearranging gives: m = 3500 J / [700 J / kg°C] *10°C ≈ 3500/7000 kg = 0.5 kg = 500 g

What does a negative ∆∆G imply about a mutation's effect on protein structure? A. The mutation makes the protein more likely to unfold, because the ∆G of unfolding is increased. A decrease in the ∆∆G of unfolding would make the protein more likely to unfold. B. The mutation makes the protein less likely to unfold, because the ∆∆G of unfolding is increased Show Explanation C. The mutation makes the protein more likely to unfold, because the ∆∆G of unfolding is decreased. Show Explanation D. The mutation makes the protein less likely to unfold, because the ∆∆G of unfolding is decreased

C. Unfolding a protein requires energy, so if we require less energy to unfold a mutated protein than a wild-type protein (i.e. a negative change in ∆G), then the mutated protein is less stable, or easier to unfold, than the wild-type protein.

In addition to its X-ray-attenuating properties, researchers specifically chose CA4b from a host of other potential contrast agents. What is the most likely reason for its selection? A. Its irreversible binding of chondroitin sulfate. B. It mimics the properties of natural agents. C. Its charge character at neutral pH. D. It significantly increases the thickness of cartilage samples.

CA4b is a cationic contrast-enhancing agent that, in the CECT technique, can be used to assess the GAG content of cartilage. Additionally, it is written in the passage that "GAG content is positively correlated with contrast in CT imaging of cartilage samples treated with ... CA4b." This correlation is likely because the extent of binding by cationic (positively charged) CA4b, and, thus, the extent to which contrast enhancement occurs, depends upon the GAG content to which CA4b binds. For this reason, the positively-charged CA4b is central to its function as a contrast-enhancing agent in the CECT technique.

A scientist wished to prepare a buffer for an experiment to be conducted at pH 9.7. Which of the following organic acids would be the best choice for this experiment? A.acetic acid (pKa = 4.76) B. Carbonic acid (pKa = 6.35) C. Tricine (pKa = 8.05) D. Taurine (pKa = 9.06)

D To construct the best possible buffer, we should choose the organic acid with the pKa closest to the pH at which the experiment will take place (9.7). This gives us answer choice D. Note that an ideal buffer should have a pKa within 1 pH unit of the expected experimental conditions.

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.

D is correct. 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.

The piezoelectric effect, when generating ultrasound waves, involves the conversion of: A. chemical energy to mechanical energy. B. electrical energy to potential energy C. kinetic energy to electrical energy D. electrical energy to mechanical energy.

D is correct. The piezoelectric effect begins with voltage generating a current through the crystal (electrical energy) and culminates in the crystal vibrating (mechanical energy). Mechanical energy is made up of kinetic and potential energy, as it is associated with the motion and position of an object.

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. *** attenuation is weakening of U/S signal.***

D is correct. The question stem asks about attenuation, which is a weakening of the U/S signal. Sound energy is attenuated as it passes through the body because parts of the signal are reflected, scattered, absorbed, refracted or diffracted. Amplification does the opposite of attenuation; it makes the signal stronger. Since this is an "EXCEPT" question, amplification is our answer.

What is the pH of a 0.010 M sodium hydroxide solution at 25°C? A. 1 B. 2 C. 7 D. 12

D is correct. You should immediately notice that choice D is the only option that is basic. To answer this question using math, however, first note that sodium hydroxide is a strong base that completely dissociates in aqueous solution. Therefore, the hydroxide ion concentration of this solution is also 0.010 M, or (in scientific notation) 1 x 10-2 M. Taking the negative logarithm of the hydroxide concentration gives a pOH of 2. Since pH + pOH = 14 (at 25ºC), we can calculate the pH by subtracting the pOH from 14, which yields pH = 14 - 2 = 12.

Which of the following best explains what would result if CoQ accepted only one electron instead of two? A) There would be no change. B) There would be a negative charge on the molecule. C) It would be ubiquinol D) There would be a free radical

D) There would be a free radical Normally, when two electrons are added, the original ketone oxygen will have a negative charge as well and grab a H+. However, if only one electron is added, there will be 5 valence electrons on that oxygen and it will be a free radical.

Addition of which of the following amino acids would MOST effectively interfere with sodium hydroxide's ability to induce an in vitro saponification reaction (performed under approximate biological conditions) with triacylglycerols? A. Leucine B. Valine C. Proline D. Arginine

D. The hydroxide anion operates as both a base and a nucleophile in saponification; we can therefore interfere with its action by providing an alternate stronger nucleophile, or by neutralizing it to prevent it from acting as a base. The question specifies that the reaction is happening under conditions that approximate biological activity, so we should assume that arginine is present in its physiologic (positive) form. Hydroxide is a very strong base, so it would be able to deprotonate the positive arginine side chain. This would neutralize at least some of the hydroxide ion in solution, thereby interfering with its activity

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 throwing the ball only in the horizontal direction.) ***reaching its maximum velocity (an average of 30 m/s) that is perpendicular to the arm and shoulder.... home plate 18 meters away.*** A. 2.0 m above the ground Show Explanation B .1.8 m above the ground Show Explanation C .0.5 m above the ground. D. 0.2 m above the ground

D. This means the ball's flight time is: (18 m) / (30 m/s) = (18/30) s = 3/5 s = 0.6 s The ball is released from a position 2 m off the ground (0.2 m from the pitcher's mound and 1.8 m from the pitcher). To calculate the distance the ball falls during 0.6 s, we can use the equation d = v0t + 1/2at2: d = (0 m/s)(0.6 s) + 1/2(10 m/s2)(0.6 s)2 d = 1/2(10)(0.36) = 1/2(3.6) = 1.8 The ball has fallen 1.8 m from an initial height of 2.0 m. Thus, the catcher must hold the glove 0.2 m above the ground to catch the pitch.

The surface temperature of the sun (5770 K) is much greater than that of the earth (298 K). As a result, solar photons have a much higher energy than Earth photons. Given that information, how does the frequency and velocity of an Earth photon differ from that of a solar photon? A. The frequencies of the two photons would be identical; the speed of the solar photon would be greater. Show Explanation B. The frequency of the solar photon would be lower; the speed of the photons would be identical. Show Explanation C. The frequency of the Earth photon would be lower; the speed of the solar photon would be greater. Show Explanation D. The frequency of the Earth photon would be lower; the speed of the photons would be identical.

D. In a question like this, where the stem is in fact two questions for the price of one, it is easiest to tackle the question separately. Recall Planck's law: E = hf. Since Planck's constant is a constant (which you do not need to memorize), what this equation really says is that the greater the energy of a photon, the higher its frequency. Therefore, Earth photons will have a lower frequency than solar photons. The speed of photons is the speed of light, and it does not depend on the energy of the photons. Therefore, the speed of the photons would be identical.

In addition to a drug's solubility in hydrophilic media, which of the following is likely to be an important factor in determining its oral bioavailability? A)PKa B)PKB C) solubility enchancement ratio D) membrane permeability

D. Orally-administered drug must be absorbed across the membranes of the gums, stomach, or small intestine to be effective. Therefore, membrane permeability must be important because if it is low, the drug will not cross these membranes into the bloodstream.

According to the experimental results, the elastic modulus of a sample of articular cartilage with mean CECT attenuation of 1500 HU is nearest: A) 0.1Mpa B) 0.25Mpa C)0.3 Mpa D) 0.5 Mpa

D. Pay close attention to the caption of Figure 2! CECT attenuation is represented by the gray line. The point on that line that corresponds to 1500 Hounsfield units corresponds to a GAG content of about 6% (as shown below). Next, we need to remember what the question is asking for: elastic modulus in units of MPa. If we look at the black line, which is a plot of E (in MPa) versus GAG content, we can see that a 6% GAG content corresponds to an E value of 0.5 MPa.

Which of the following is most likely to undergo positive beta decay? A. 14C Show Explanation B. 13C Show Explanation C. 17O Show Explanation D. 22Na

D. Positive beta decay, also known as positron emission, occurs when the proton-to-neutron ratio is too high. Thus, we are looking for the answer choice with the highest proton-to-neutron ratio. Of the options given, 22Na has the highest ratio, as it has 11 protons and 11 neutrons (1:1). Because choice D has the highest proton-to-neutron ratio among the answer choices, it is the most likely to undergo positive beta decay.

Measurements are made in two arteries with the same diameter and the same flow rate. One artery is in the head and the other is in the leg. In a person standing upright, which of the following statements is true? A. The speed of the blood in the head is greater than the speed of the blood in the leg. B. The speed of the blood in the leg is greater than the speed of the blood in the head. C. The pressure of the blood in the head is greater than the pressure of the blood in the leg. D. The pressure of the blood in the leg is greater than the pressure of the blood in the head.

D. The gravitational potential energy of the blood in the head is greater than the gravitational potential energy of the blood in the leg. As a result, the blood in the leg will have a greater pressure due to the conversion of the gravitational potential energy into the kinetic energy of molecular motion. The increased pressure is the result of the increased number of collisions between the molecules in the blood of the leg and the surface of the blood vessel. ***Poiseuille's law is used to describe laminar flow of incompressible fluids through a long cylindrical tube. Poiseuille's law contains five variables: the flow rate (Q), the pressure drop between both ends of the tube (ΔP), the radius of the tube (r), the length of the tube (L), and the viscosity (η). It can be written in two equivalent forms, both of which are given below: ***Bernoulli's equation is essentially conservation of energy for fluids, and is given below: P1 + ½ ρv12 + ρgh1 = P2 + ½ ρv22 + ρgh2 Here, the term that includes velocity (v) is analogous to the kinetic energy of the fluid, while the term that includes height (h) is analogous to the fluid's potential energy. The remaining term is pressure (P). Importantly, this equation indicates that when height is constant (as in a horizontal pipe system), an increase in velocity corresponds to a decrease in pressure, and vice versa. A final important relationship is given by the continuity equation (v1A1 = v2A2). This equation states that 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.

Which of the following statements correctly describe the methods used in the experiment? I. The retention factor in a TLC procedure depends on the solvent system, temperature, and the adsorbent. II. A polar compound will exhibit a smaller retention factor on a normal phase TLC plate than a less polar compound. III. Anhydrous methanol has greater eluting strength than pentane when used as solvents in a TLC procedure. A) I only B) I and II only C) II and III only D) I, II, and III

D. The retention factor is the distance the compound migrated divided by the distance of the solvent front. TLC depends on the differential affinity of a compound for the stationary vs. the mobile phase. Therefore, depending on how polar the solvent and the adsorbent are, a compound will move on a TLC plate at a certain rate (I). Temperature also affects the rate of movement. In normal phase TLC, a polar compound will be attracted to the adsorbent through electrostatic interactions and, therefore, will not move as far on a TLC plate, whereas a nonpolar compound will have more affinity for the mobile phase and will move further (II). Eluting strength is a less familiar concept but depends on how strongly a compound adsorbs onto the adsorbent. Since typical adsorbents in normal phase TLC are highly polar, eluting strength increases with increasing solvent polarity. Methanol is more polar than pentane and therefore has a greater eluting strength (III). A schematic of a TLC plate is shown below.

Which of the following gases would decrease the pH when dissolved in an aqueous solution? A. Ammonia Show Explanation B. Oxygen Show Explanation C. Nitrogen Show Explanation D. Carbon dioxide

D. When carbon dioxide dissolves in water, it slowly establishes equilibrium between the dissolved carbon dioxide and carbonic acid (H2CO3), which is a weak acid that would lower the pH. Carbonic acid anhydrase speeds up this reaction in biological systems. Acid-base buffers confer resistance to a change in the pH of a solution when hydrogen ions or hydroxide ions are added or removed to solution. An acid-base buffer typically consists of a weak acid and its conjugate base. The most important buffer to know for the MCAT is the bicarbonate buffer system, which is shown below. H2O (aq) + CO2 (g) → H2CO3 (aq) → H+ (aq) + HCO3- (aq) Carbonic acid (H2CO3) has the conjugate base of HCO3-. Buffers work because the concentrations of the weak acid and its salt are large compared to the number of protons or hydroxide ions added or removed. When protons are added to the solution from an external source, some of the bicarbonate in the buffer is converted to carbonic acid, using up the protons added; when hydroxide ions are added to the solution, protons are dissociated from some of the carbonic acid in the buffer, converting it to bicarbonate and replacing the protons lost.

Benedict's test typically involves exposure of a carbohydrate to a solution of CuO in ammonia (NH3). Glucose yields a positive Benedict's test, but sucrose does not. Which of the following best explains this fact? A. Sucrose contains a hemiacetal group, while glucose does not; this classifies sucrose as a reducing sugar. Show Explanation B. Sucrose contains a hemiacetal group, while glucose does not; this classifies glucose as a reducing sugar. Show Explanation C. Glucose contains a hemiacetal group, while sucrose does not; this classifies sucrose as a reducing sugar. Show Explanation D. Glucose contains a hemiacetal group, while sucrose does not; this classifies glucose as a reducing sugar.

D.Benedict's test is intended to identify "reducing sugars," or sugars with the capacity to serve as reducing agents. Specifically, sugars with hemiacetal groups can undergo mutarotation, allowing them to be oxidized by CuO. The process of mutarotation requires ring opening, which occurs at a hemiacetal group. Thus, all monosaccharides are reducing sugars because they undergo mutarotation. Polysaccharides have a reducing end, but the glycosidic linkage between multiple sugar units prevents isomerization to the aldehyde from occurring. This means that sucrose gives a negative test. The hemiacetal group on glucose, shown below, marks it as a classic reducing sugar. Remember, a hemiacetal consists of a carbon atom directly attached to one -OR and one -OH group. The same carbon atom is also attached to a hydrogen atom and an R group.

Which of the following strategies would be suitable for converting TX-100 into an anionic surfactant at physiological pH? A. Reacting TX-100 with a weak base Show Explanation B. Adding PCC to TX-100 in a non-aqueous solution Show Explanation C. Treatment with KMnO4 Show Explanation D. Titrating TX-100 with concentrated sulfuric acid

Figure 1 shows that the hydrophilic head of TX-100 is an alcohol (-OH). KMnO4 is a reagent that is commonly used to oxidize alcohols to carboxylic acids. Since carboxylic acids are generally sufficiently acidic to dissociate at physiological pH (i.e. they have a pKa < 7), this reaction would convert TX-100 into an anionic surfactant at pH levels in the physiological range.

Use of CA4b is likely to increase the contrast obtained in CECT imaging because: A. cationic agents bind poorly to chondroiton sulfate in articular cartilage. B. the elements in CA4b have larger atomic numbers than the elements in cartilage. C. collagen and other components of articular cartilage are opaque to X-rays. D. articular cartilage is too thick to be imaged by traditional modalities.

From the passage: "X-ray attenuation, which is responsible for contrast in CECT imaging and is enhanced by the use of contrast agents, generally increases with the atomic number of atoms composing". Given this, it's reasonable to conclude that the contrast agent used in CECT (CA4b) may increase contrast in cartilage imaging because it is composed of elements with much greater atomic numbers than those found in cartilage.

Which of the following reagents is/are likely to be used to form the Britton-Robinson buffer solution(ph=3.8) used in the experiment? I. a polyphonic weak acid II. Na2HPO4 III. NH3

I and II only. Buffer needs to be weak acid or acidic salt made in reaction. Buffer solutions, intended to resist changes in pH, consist of a weak acid or weak base and its corresponding salt. The passage states that the buffer solution is acidic (pH = 3.8), so we can assume that a base is not present in the solution.

Based on its structure, naturally occurring hematoporphyrin most likely forms a complex with which of the following? A. Cl- B. O2- C. O2 D. Fe2+

If you have had a few biochemistry courses or are deep into your MCAT content, you might recognize that hematoporphyrin is a close cousin of heme - the iron-binding cofactor that allows hemoglobin to carry oxygen. If not, how might you figure that out? Look carefully at the pyrroles (the 5 membered nitrogen-containing aromatic rings). Those pyrroles can be deprotonated at physiological pH, resulting in a negative formal charge that will de-localize across the entire large ring due to its conjugated structure. Since the ring has a negative charge, we are looking for the central ion or molecule to carry a postive charge. The only option in this list is Fe2+.

The enzyme glutathione reductase catalyzes the conversion of the molecule GSSG to GSH, which is essential in order for cells to counteract the effects of reactive oxygen species in the cytoplasm. Glutathione reductase cleaves a single disulfide bond within the GSSG molecule, resulting in two GSH molecules. Solely using this given information, how does the oxidation number change from a single GSSG molecule to a single GSH molecule? A. +2 to 0 Show Explanation B. +1 to 0 Show Explanation C. 0 to -1 Show Explanation D. 0 to -2

In a disulfide S-S bond, the two sulfur atoms share their covalent electrons equally and therefore have an oxidation number of zero. When the disulfide bond gets reduced to two -SH sulfhydryl groups, S-H covalent electrons spend their time more around the sulfur in the GSH molecule, as opposed to the hydrogen. This is because sulfur is more electronegative than hydrogen. As a result, the sulfur in GSH has an oxidation state of -1. All else being equal, this suggests that a GSSG molecule has an oxidation state of 0, while a GSH molecule has an oxidation state of -1.

In the condensation mechanism shown, what is the most likely role of the diethyl succinate in the reaction? A. Weak base B. Electrophile source C. Strong base D. Enolate source

In the Stobbe condensation example, diethyl succinate is condensed with benzophenone using tert-butoxide (a strong base) followed by aqueous acid. The diethyl succinate (the molecule on the right in step 1) is the enolate source. This enolate will act as a nucleophile to attack the electrophilic benzophenone.

Which of the following will contribute to the level of FDG signal seen during a PET scan of the colon? I. Blood flow to the tissue II. Availability of oxygen to the tissue III. GLUT4 transport A. I only Show Explanation B. II and III only C. I and II only This question requires us to think about whether the three listed factors would affect FDG's location in the scanned area. Statement I is logical since if there is less blood flow to an area, less FDG will be present in that tissue. Similarly, when less oxygen is available to a tissue, less oxidative respiration occurs, resulting in a smaller need for FDG. This makes statement II correct, as this tissue would have lower amounts of FDG per unit time and would show less activity. D. I, II, and III

No. GLUT4 is an insulin-regulated glucose transporter found primarily in adipose tissues and striated muscle (skeletal and cardiac). The presence of GLUT4 transporters in tissue would not affect the amount of FDG in a GI tissue such as that in the colon (III). C. yes. This question requires us to think about whether the three listed factors would affect FDG's location in the scanned area. Statement I is logical since if there is less blood flow to an area, less FDG will be present in that tissue. Similarly, when less oxygen is available to a tissue, less oxidative respiration occurs, resulting in a smaller need for FDG. This makes statement II correct, as this tissue would have lower amounts of FDG per unit time and would show less activity.

Which of the following structures correctly represents a phenoxyl radical formed from compound 3?

Passage information indicates that phenoxyl radicals are formed from -OH groups, allowing choices B and C to be eliminated, since they both show a radical formed from a C=O (carbonyl) group. Choice D reflects removal of a single electron from an -OH group, corresponding to the mechanisms shown in the passage.

Which of the following 1H-NMR signals would allow for researchers to differentiate between colchicine and IA? A. Large peaks at 7 ppm Show Explanation B. Several peaks between 0 and 5 ppm Show Explanation C. A doublet at 8.5 ppm Show Explanation D. A signal at 3.4 ppm

Peaks around 5-8.5 ppm are indicative of amide (R(C=O)NR2) hydrogens. Only the colchicine has an amide group. The doublet (signal splitting) is due to the hydrogen on the nearby sp3 carbon in the cycloheptane of colchicine.

If Reaction 1 releases 188.6 J, what is the ∆H of Reaction 2? A. 188.6 J B. 377.2 J C. -188.6 J D. -377.2 J

Reaction 2 is simply Reaction 1 run in reverse with the stoichiometric coefficients doubled. So when we're told that Reaction 1 releases 188.6 J, that means its enthalpy change is -188.6 J. The reverse reaction would have an enthalpy change of +188.6 J, and if you double the stoichiometry, you must double this value to get +377.2 J.

The light blue appearance of S157T is most directly attributable to what other phenotypic change versus the wild type? . A. Increased λmax B. Decreased λmax C. Decreased ɛ D. Increased ɛ

Read the chart to find the molar absorptivity does decrease but lower molar absorptivity leads to dark blue and light blue so can't be the answer. the λmax decreases to give a dark blue and increases to give a light blue. So answer must be A, increased λmax

What is the ratio between the maximum and the minimum sound intensities that produce this particular loudness? A) 10^6 B) 10^5 C) 10^4 D) 10^3

The answer to this question is B because the maximum sound intensity level is 80 dB and the minimum level is 30 dB, which correspond to intensities Imax = I0 × 108 and Imin = I0 × 103, based on the definition of the decibel units. Their ratio is then Imax/Imin = 105.

the time dependence of the potassium current through a cell membrane channel subject to a constant 80-mV depolarization voltage is shown. What is the minimum electrical resistance of the ion channel during the time interval shown?

The answer to this question is C because the electrical resistance R of a conductor through which a current I passes when subject to a voltage V is given by Ohm's law: R = V/I = 80 mV/(400 × 10-12 A) = 200 MΩ.

What is the kinetic energy of a photoelectron produced in the energy meter of the PAC device when the frequency of an incident photon that is NOT absorbed in the solution is f = 5.0 × 1015 Hz? (Note: Use h = 4.1 × 10-15 eV•s.) A) 23.9eV B) 20.5 eV C) 17.1 eV D) 12.3eV

The answer to this question is C because the kinetic energy of a photoelectron is equal to hf - 3.4 eV = 20.5 eV - 3.4 eV = 17.1 eV.

Compound 1 is a stronger acid than Compound 2 because the anion of Compound 1 is better stabilized by: A resonance effect. B dehydration. C an inductive effect. D hydrogen bonding between OH and CO2-.

The carboxylate ion formed from Compound 1can be stabilized by hydrogen bonding with the hydroxy group. The carboxylate ion formed from Compound 2 is not in a suitable orientation to hydrogen-bond with the hydroxy group. Thus, Dis the best answer.

Because pH affects the reaction, a student carefully creates a suitable aqueous solvent for the reaction. After bubbling CO2 through the solution, he checks the pH and is surprised to find that it is not the same as the original value. What is the most likely cause for this? A. The pH will be more alkaline because the pH of a solution will naturally move to neutral over time. B. The pH will be more acidic because of the hydrogen gas that is generated C. The pH will be more alkaline because CO2 will combine with H2O to create bicarbonate. D. The pH will be more acidic because CO2 will combine with H2O to create carbonic acid.

The combination of CO2 and H2O to make carbonic acid (H2CO3) is a reaction that you must know for the MCAT. Carbonic acid and its conjugate base (bicarbonate) play a large role in maintaining the pH of the human body. While this reaction is sped up by carbonic anhydrase, it will still occur even in the absence of this enzyme, albeit at a slower rate.

How many grams of hydrogen gas are required to completely react with 32 g of oxygen gas to form hydrogen peroxide? A 0.5g B 1.0g C 1.5 g D 2.0 g

The formation reaction for hydrogen peroxide is: H2 (g) + O2 (g) → H2O2 (l) Moving on to the stoichiometry calculation: 32 g O2 x (1 mol/32 g) x (1 H2/1 O2) x 2 g/mol = 2 g H2

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? A. 65 J B. 70 J C. 140 J D. 810 J

The kinetic energy of the baseball can be calculated using the equation 1/2 mv2. Paragraph 2 tells us that the average velocity of the ball is 30 m/s and its mass is 150 g, which is equivalent to 0.15 kg. KE = (½)(0.15 kg)(30 m/s)2 KE = (½)(0.15)(900) KE = (0.15)(450) = 67.5 J The ball ends up with 67.5 J of KE, but the question asked for how much energy the lower extremities generated (eliminate choice A). Table 1 shows us the overall efficiency of each energy transfer. Using these, we can calculate how much energy was needed to end up with 67.5 J of KE in the ball. The overall efficiency of the kinetic chain can be calculated by multiplying the efficiencies of each step, and we can round the numbers to make our calculations easier. Efficiency = (0.9)(0.9)(0.7)(0.8) = 9*9*7*8 x 10-4 = (81)(56) x 10-4 ≈ (80)(60) x 10-4 = 4800 x 10-4 = 0.48 ≈ 0.5 Thus, the total energy generated in the lower extremities in order to ensure that 67.5 J makes it to the ball is 67.5 J / 0.5 = 135 J. If you are uncomfortable choosing a number that is not exactly what was calculated, you only need to remember that we estimated by rounding up to 0.5. Thus, our denominator in the final calculation, 67.5/0.5, is a bit larger than it should be, making our final value smaller than it would be had we not rounded or used a calculator (exact value of 67.5/ 0.48 =140 J). Thus, we round up and choose choice C.

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? A) aspartate B) cysteine C) lysine D)tyrosine

The passage says that 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. SOOOO we need an AA in the Enzyme to be able to donate an H from an amine group. C. is the correct answer because Lysine (K) has an amino group that can donate an H to an H bond.

Compared to the cytotoxic activity of ZSTK474 against the cell lines tested, the cytotoxicity of each target compound was: A. enhanced against at least one cell line tested. Show Explanation B. enhanced against all cell lines tested. Show Explanation C. reduced against all cell lines tested. Show Explanation D. reduced against at least one cell line tested.

The passage tells us that ZSTK474 is compound 4a. We should also realize that the lower the amount of the drug it takes to kill 50% of the cells (i.e. smaller IC50 values), the greater the cytotoxic effects against the tested cell lines. Table 1 shows that the IC50 values (i.e., the cytotoxicity) of each target compound were smaller than compound 4a (ZSTK474) in at least one, but not all, tested cell lines.

While evaluating the timing of pitches, the researchers noticed a couple of high-performing athletes were able to generate higher velocities by increasing the duration 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.Comparisons of power output between the original two groups cannot be made without additional information.

The power output is given by the equation P = W / t, where W is the work done in joules (the net-accelerating energy output) and t is the time period over which the work was performed. From the information provided, we only know that v and t increased, but not by how much. If the velocity increased, then the acceleration (and therefore the work and power) may have increased, but we are told that the time spent in the acceleration phase increased as well. Thus, there may be no significant difference in power output between the two groups because these two increases could potentially cancel each other out.

Compound A is shown below. Compound A Which of the following structures has a configuration that is identical to that of Compound A? _ A B C D

The relative positions of the four substituents around the chiral carbon atom are the same in the figure in D as in Compound A. Also, the -OHand -COOH groups extend away from the viewer, and the hydrogen and -CH2OH group extend toward the viewer both in Compound Aand in the figure in D. Thus, D is the best answer.

Which of the following reagents would be involved in the first synthetic step to chemically derive an amide from oleic acid? A. Phosphorous pentachloride B. Aminobutane C. Octadecanol D. 2-aminopentane

This question is a bit tricky. Remembering the hierarchy of relative reactivities of carboxylic acid derivatives is relatively easy, but applying the concepts under the time constraints of the MCAT is more challenging. Amides are the most stable (least reactive) of the carboxylic acid derivatives because the amino substituent is a very poor leaving group. In order to synthesize amides however, one cannot simply use the amino substituent as a nucleophile; as a relatively strong base, it will instead preferentially participate in an acid-base neutralization reaction with the acidic hydrogen of the organic acid. Instead, one must first create an acid chloride (or an acid anhydride, which is not an option in this question) to eliminate the confounding acidic hydrogen. Afterwards, one can proceed with a normal nucleophilic substitution. Aminobutane and 2-aminopentane are both suitable as amino substituents but are therefore unsuitable as a first step. The addition of octadecanol would be a poor choice of a first step; absent rigorous reaction conditions, it's unlikely anything productive would happen at all.

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.

This question requires us to examine the experimental setup and determine what effect air resistance would have on the scientists' calculations. 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.

Which of the following is the structure of ubiquinol? *** passages shows ubiquinone

This question requires us to think about what will happen to ubiquinone when it is reduced with 2 electrons accompanied by H+. Above all else, the molecule will want to maintain a conjugated pi system; in other words, our choice should contain an aromatic ring. When two electrons are added to one of the ketone oxygens, the pi electrons from the ketone bond will be transferred into the ring, pushing the pi electrons from one of the double bonds onto the adjacent single bond, pushing the pi electrons from the other ketone bond to the oxygen and a H+will be grabbed from solution to create an alcohol. The two electrons that were added to the original oxygen will grab a H+ as well and we'll get choice D. Alternatively, the last sentence of the passage tells you that ubiquinol is a diol, which means it must have two alcohol groups. As we know this structure is based on hydroquinone (another diol), it seems likely that they would be in similar positions.

he conversion of cefixime, an antibiotic, to one of its metal complexes is shown below. Which technique is most useful in monitoring the progress of cefixime metal complex preparation? A. Mass spectrometry B. IR spectroscopy C. Gas chromatography D. UV-VIS spectroscopy

This technique detects conjugated pi systems and metal-to-ligand charge transfer species, both of which exist here. The appearance of a peak corresponding to the charge transfer component will appear using UV-VIS spectroscopy. UV spec : in analytical chemistry for the quantitative determination of analytes, such as transition metal ions, highly conjugated organic compounds, and biological macromolecules.

The researchers want to use narrow-spectrum LEDs to make their lamp more efficient. Assuming that the energy of a photon absorbed by porfirmer is transferred without loss to oxygen, what wavelength of light should the researchers select? (Note: Planck's constant is 6.626 x 10-34 J∙s) A. 1000 nm B. 1250 nm C. 2500 nm D. 3000 nm

To answer this question, we need to work backwards by calculating the amount of energy required to excite one oxygen atom from the triplet state to the singlet state. That energy is given in the passage as 94.6 kJ/mol. We can divide that number by Avogadro's number to figure out the energy required for 1 atom to transition: (9.46 x 104 J/mol) / (6.02 x 1023 mol-1) = 1.57 x 10-19 J. To make that calculation doable without a calculator, use 9 instead of 9.46 and 6 instead of 6.02. By simplifying the fraction, you get that 9/6 = 3/2 = 1.5. Then you can deal with the exponents (4 - 23 = -19). Since the question says that no energy is lost in the transition from porfirmer to oxygen, we can assume that 1.5 x 10-19 J is the energy of the photon that we need. But how can we get the wavelength of the photon? Recall that E = hf relates the energy of the photon to its frequency, and v = fλ relates frequency, wavelength, and speed. The speed of the photon will be the speed of light (c = 3 x 108 m/s). Rearranging the second equation for frequency gives f = v/λ. Now we can substitute for f in the first equation, giving E = hv/λ. Rearranging this equation for wavelength gives λ = hv/E. λ = (6.6 x 10-34 J?s)(3 x 108m/s)/(1.5 x 10-19 J) λ = 13.2 x 10-7 m λ = 1.32 x 10-6 m λ = 1320 nm We should look for an answer that is slightly lower than that since we estimated the energy of the photon to be lower than it actually is.

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

Vitamin B1, or thiamine, is a coenzyme in metabolic processes involving amino acids and carbohydrates Vitamin E refers to a set of closely-related, lipid-soluble compounds that function as antioxidants. Additional possible functions of vitamin E are under investigation, but for the purposes of the MCAT it is primarily considered an antioxidant. Vitamin D is a lipid-soluble molecule primarily involved in calcium metabolism. Vitamin K refers to a group of closely-related molecules that contribute to blood coagulation and clotting.

Antioxidant activity is an important function of which of the following vitamins?

Vitamin E. Vitamin E refers to a set of closely-related, lipid-soluble compounds that function as antioxidants. Vitamin B1, or thiamine, is a coenzyme in metabolic processes involving amino acids and carbohydrates. Vitamin D is a lipid-soluble molecule primarily involved in calcium metabolism. Vitamin K refers to a group of closely-related molecules that contribute to blood coagulation and clotting.

When NADH is oxidized, what happens to the hybridization of the nitrogen of the nicotinamide ring, and the C4 carbon atom opposite this nitrogen? A. The nitrogen is converted from sp3 to sp2 hybridization, whereas the C4 carbon is converted from sp2 to sp3 hybridization. B. The nitrogen is converted from sp2 to sp3 hybridization, whereas the C4 carbon is converted from sp3 to sp2 hybridization. C. Both the nitrogen and the C4 carbon are converted from sp2 to sp3 hybridization. D. Both the nitrogen and the C4 carbon are converted from sp3 to sp2 hybridization.

We first need to examine the ring nitrogen atom in Figure 1. In the NADH, there are three single bonds shown. You also need to recognize that there is a lone pair of electrons on the nitrogen (not shown) to complete the octet, and therefore there are four electron domains, corresponding to sp3 hybridization.

Which one of the engines in the study has the highest efficiency? A. Engine 1 B. Engine 2 C. Engine 3 D. Engine 4

We need to calculate the efficiency for every engine in order to choose the one with the highest efficiency. We should know that for any engine, efficiency is defined as work output/work input. Therefore, for the engines in the study, the efficiency is calculated using the following formula: η (efficiency) = W/QH Table 1 gives us the values for QH and QC. Therefore, it is easier to rewrite the above equation by taking into account the first law of thermodynamics (conservation of energy), which tells us that QH = QC + W, using Figure 1 of the passage. In the above equation, we replace W with QH - QC: η = (QH - QC) /QH = 1 - (QC/QH) Next, the efficiency of every engine can be calculated: Engine 1: η = 1 - (QC/QH) η = 1 - ((3.81 x 106 J) / (5.68 x 106 J)) This is approximately 1 - 4/6 = 1 - ⅔ = 1 - 0.67 Actual η = 0.33 Engine 2: η = 1 - (QC/QH) η = 1 - [(8.30 x 106 J) / (9.21 x 106 J)] This is approximately 1 - 9/10 = 1 - 0.90 Actual η = 0.10 Engine 3: η = 1 - (QC/QH) η = 1 - [(2.32 x 105 J) / (5.32 x 105 J)] This is approximately 1 - 3/6 = 1 - 0.50 Actual η = 0.56 Engine 4: η = 1 - (QC / QH) η = 1 - [(1.20 x 105 J) / (3.10 x 105 J)] This is approximately 1 - ⅓ = 0.67 Actual η = 0.61 Thus, engine 4 has the highest efficiency (0.61).

What is the value of the Keq for the following reaction under standard biological conditions? Pyruvate- + NADH + H+ → NAD+ + Lactate- A. 1 x 10-7 B. 5 x 10-5 C. 2 x 104 D. 1 x 107

We start by finding Eº for this reaction. The oxidation of NADH generates +0.32 V and the reduction of pyruvate generates -0.19 V. Thus the reaction listed generates +0.13 V. We can use Equation 2 to determine the equilibrium constant, since the E will be zero and Q = Keq. The number of electrons transferred in the overall REDOX reaction is n = 2. E = E° - (0.060/n) log Q 0 = 0.13 - (0.06/2) log K 0.13/0.03 = log K (0.12 + 0.01) /0.03 = log K (4 + ⅓ ) = log K 10(4 +1/3) = K 101/3 x 104 = K 2 x 104 ≈ K The value of the equilibrium constant is a little larger than 2 x 104, so choice C is the best answer.

Consider the chemical reaction NH4+(aq) + NO3-(aq) → N2(g) + 2H2O(l). The rate law for this reaction is: A. rate = k[NH4+][NO3-]. Show Explanation B. rate = k[NH4+]2[NO3-]2. Show Explanation C. rate = k[NH4+]2[NO3-]. Show Explanation D. rate = k[NH4+][NO3-]2.

With rate law problems, we should first identify two trials where only one of the reactants changes concentration. Between trials 1 and 3, we can see that the concentration of nitrate doubles while the rate quadruples. This means that the reaction is second order with respect to nitrate (eliminate choices A and C). Between trials 1 and 2, the nitrate concentration stays the same while the ammonium concentration increases by a factor of 1.5. We can see that this causes the reaction rate to increase by a factor of 1.5, indicating that the reaction is first order with respect to ammonium (eliminate choice B).

Anionic surfactants most likely reduce the rate of nucleophilic reactions of hydroxide ion with neutral compounds by: A. preferentially interacting with the nucleophile, thereby inhibiting it from reacting with the substrate. B. acting as a nucleophile and undergoing undesired cross-reactions with the substrate. C. forming a negatively charged surfactant-substrate complex that repels the nucleophile. D. more readily forming micelles that sequester the substrate from the nucleophile.

YIKES A)An anionic substrate would experience electrostatic repulsion from the hydroxide ion, so it would not preferentially interact with it. B)The degree to which nucleophilic cross-reactions might occur between the surfactant and substrate most likely depends on the specific structure of the surfactant. However, passage information suggests that anionic surfactants are generally much bulkier than the hydroxide ion, meaning that they would be poorer nucleophiles. Furthermore, this possibility is not specifically supported by passage information in the way that choice C is. C)The passage directly tells us that cationic surfactants increase the reaction rate between the negatively charged hydroxide ion and neutral substrate by forming positively charged surfactant-substrate complexes. We can therefore infer that electrostatic attractions between the positively-charged surfactant-substrate complexes and the hydroxide ion promote the reaction. We are also told that anionic surfactants reduce the reaction rate; similar logic suggests that electrostatic repulsion may occur between negatively-charged surfactant-substrate complexes and the negatively-charged hydroxide ion, which would inhibit the reaction. D)Based on passage information, it is logical to hypothesize that sequestration of the substrate by incorporation into micelles might be the reason why surfactants only increase reaction rates up to a certain threshold, but this factor would not be expected to differ between anionic and cationic surfactants.

Which extraction procedure will completely separate an amide from the by-product of the reaction between an amine and excess carboxylic acid anhydride?

add 0.1 M NaOH (aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the ether layer by evaporating the solvent. The answer to this question is A because the by-product of the reaction will be an acidic carboxylic acid and the excess unreacted starting material will also be acidic. Extraction with aqueous base will hydrolyze and extract both of these into the aqueous layer, leaving the neutral amide in the ether layer.

What amino acid in the Aequorea fluorophore would be designated as "X" in the X-Tyr-Gly GFP fluorophore?*** this is name in figure so look at figure for the unknown AA!!! A.Glutamine B.Glycine C.Threonine D.Serine

answer is D, if you look at figure of Aequorea then you see a tyrosine group, a serine group and a glycine. Do NOT confuse example mentioned in passage with question about figure

Given passage information, which of the following compounds is most likely present in greater concentration in the maternal blood relative to the level at which it is found in the blood of a fetus at 20 weeks of gestation? A. 2,3-BPG B. O2 C. αγ dimers D. Hb A2

d. According to the passage, Hb A2 is a minor hemoglobin synthesized only by adults. As a result, it should be found in greater concentration in the maternal blood than in the fetal blood.

Which of the following is a list of straight-chain hydrocarbons, each of which could have one triple bond? A) C4H10; C3H6; C2H2 B.)C5H8; C3H4; C2H4 C.)C5H10; C3H6; C2H4 D)C8H14; C6H10; C2H2

he formula for a straight-chain alkane is CnH2n+2. A double bond will replace two of those hydrogens with an extra C-C bond, giving a formula of CnH2n. A triple bond will replace four of those hydrogens with two extra C-C bonds, giving a formula of CnH2n-2. Only choice D has all molecules that fit the formula CnH2n-2.

In comparison to piperazine, the inductive effect should cause the pKb of methyl-piperazine to be: A.lower, because of electron donation by the methyl substituent. B.lower, because of electron withdrawal by the methyl substituent. C.higher, because of electron donation by the methyl substituent. D.higher, because of electron withdrawal by the methyl substituent.

note*** small pKB means stronger base which means more likely to accept proton. Methyl-piperazine is an electron donating group which makes it a better base (worse acid since it wants to hold onto the H+) and therefore better base means lower pKb. so A is correct. The pKb is a measure of basicity, where a smaller pKb value corresponds to a stronger base. Methyl-piperazine is a stronger base than piperazine because of the inductive effect generated by the electron-donating N-methyl group of methyl-piperazine. This effect increases the availability of electron charge density available to donate, thereby increasing the strength of N-methyl piperazine to act as a Lewis base.

Based on the ray diagram and distances shown in Figure 1, the focal length of the lens is: A) 3 B) 4 C)8 D)16

thin sense equation f = (12 cm × 4 cm)/(12 cm + 4 cm) = 3 cm.