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MCAT Content / AAMC MCAT Practice Exam 4 Cp Solutions

AAMC FL4 CP [Web]

Exam 4 C/P Solutions: Passage 1

1) First thing we want to do to answer this question is revisit the part of the passage that talks about limestone. The author tells us in Step 1, “The chemist placed a sample of limestone, CaCO3(s), into a sealed chamber and then heated the limestone to 1200 K to generate CO2(g) and CaO(s).” However, there’s a slight change in the question stem. The limestone, CaCO3(s) is only heated to 900 K, but does not decompose and generate CO2(g) and CaO(s). What does that tell us? The reaction is not spontaneous. Let’s jump into some quick background information we’ll use to answer this question:

ΔG = ΔH – TΔS

We don’t get the decomposition the author talks about in Step 1, and for nonspontaneous reactions, ΔG is positive. Additionally, entropy increases if we go from a solid reactant to a gas in our products, which would mean a positive ΔS. Given this information, ΔG can only be positive in our above equation when ΔH is positive and is greater than TΔS (which we know is also a positive number in this circumstance).

  1. positive and less than TΔS. First part of this answer choice is consistent with our breakdown. We mentioned that for nonspontaneous reactions, ΔG is positive. Entropy increases if we go from a solid reactant to a gas in our products, so we have a positive ΔS, and we’re told temperature is 900 K. Given our equation ΔG = ΔH – TΔS, ΔH must be positive and greater than TΔS in order for ΔG to be positive. That means this answer choice is only half right.
  2. positive and greater than TΔS. This answer choice is entirely consistent with our breakdown of the question. We mentioned some key points in the breakdown that we can reiterate here. For nonspontaneous reactions, ΔG is positive. Entropy increases going from a solid reactant to gas in our products, so we have a positive ΔS. We’re told temperature is 900 K. Given our equation ΔG = ΔH – TΔS, ΔH must be greater than TΔS in order for ΔG to be positive. This is going to be our best answer.
  3. negative and less than TΔS. This answer choice is the opposite of what we said in our breakdown. We expect ΔG to be positive and greater than TΔS.
  4. negative and greater than TΔS. This answer choice is only half correct. We expect ΔG to be positive and greater than TΔS. We can stick with answer choice B as our best answer choice.

2) This question has to do with the same part of the passage we focused on for Question 1. We’re asked about limestone being heated during Step 1.

We’re told an equilibrium is established and we can write out the equation here:

CaCO3 (s) ⇌ CO2 (g) + CaO (s)

The equilibrium constant is the ratio of the mathematical product of the concentrations of the products of a reaction to the mathematical product of the concentrations of the reactants of the reaction. Each concentration is raised to the power of its coefficient in the balanced chemical equation. If we have an equation aA + bB ⇌ cC + dD, we can write out the equilibrium constant expression:

However, before we write anything for this specific question, what we need to pay attention to is the state of each reactant and product. Pure solids and liquids are not included when writing out this equilibrium constant. That means we are not including CaCO3 (s) and CaO (s) from our initial equation. That means we are only left with the product CO2 (g). Our equilibrium constant is simply [CO2].

  1. [CaO] We noted in our equation that CaO is a solid. Pure solids and liquids are not included when writing out our equilibrium constant.
  2. [CaCO3] We noted in our equation that limestone (CaCO3) is a solid. Pure solids and liquids are not included when writing out our equilibrium constant.
  3. [CO2] This answer choice matches our breakdown. This is going to be our best answer choice.
  4. [CaO] × [CaCO3] This answer choice incorrectly includes the solid reactant and product instead of excluding these. We can stick with answer choice C as our correct answer.

3) This is another passage-based question in this set, but once again, the test-maker simplifies our lives a bit by telling us where we will find the necessary information to answer our question. We’re focused on Reaction 2:

Specifically, we want to hone in on nitrogen. The oxidation number describes explicitly the degree to which an element can be oxidized (lose electrons) or reduced (gain electrons). The number is the effective charge on an atom in a compound. We can look at nitrogen in our reactants and products.

First our products: The oxidation number of hydrogen is +1 when it is combined with a nonmetal as in CH4, NH3, H2O, and HCl. In uncharged NH3, each hydrogen has an oxidation number of +1 while nitrogen then has an oxidation number of -3.

In our reactants we have protonation of ammonia so we’re looking at NH4+. Once again, the oxidation number of hydrogen is +1 when it is combined with a nonmetal. In NH4+, each hydrogen has an oxidation number of +1 while nitrogen still has an oxidation number of -3. Together that is where we get the +1 overall on NH4+.

  1. Yes; it changed from –3 to –4. This answer choice correctly lists the initial oxidation number as -3, but we reasoned out that the oxidation number does not change after the protonation of ammonia.
  2. Yes; it changed from 0 to +1. This answer choice is incorrectly telling us the charge on the compounds in which nitrogen is found. That’s not what we’re looking for. We’re looking for the oxidation state of N which remains at -3.
  3. No; it remained at –3. This answer choice matches the breakdown of the question exactly. The oxidation state of N does not change; this is going to be our best answer choice.
  4. No; it remained at +1. This answer choice is describing the oxidation number of hydrogen. We’re looking for the oxidation state of N which remained at -3. Answer choice C is our best answer.

4) This is like a pseudo standalone question. Although the test-maker is asking about something that’s tangentially related to the passage, we’re asked about the approximate volume of a gas at STP. The only thing we’ll need to know from the passage is the number of moles of Gas X.

1 mole of any gas at standard temperature and pressure (273 K and 1 atm) occupies a volume of 22.4 L.

  1. 22.4 L. This answer choice matches what we know from our general knowledge. 1 mole of any gas at standard temperature and pressure (273 K and 1 atm) occupies a volume of 22.4 L.
  2. 44.8 L. This answer choice incorrectly lists the volume for two moles of gas at STP. Answer choice A remains superior.
  3. 67.2 L. This answer choice incorrectly lists the volume for three moles of gas at STP. Answer choice A remains superior
  4. 89.6 L. This answer choice incorrectly lists the volume for four moles of gas at STP. We can eliminate answer choices B, C, and D and keep answer choice A as our correct answer.

 

Exam 4 C/P Solutions: Passage 2

5) This question is asking about a specific aspect of the research conducted in the passage.

We’re focused on Paragraph 3 that discusses the cuvettes being identical in terms of optical properties. Why would the researchers make sure the cuvettes are identical in terms of optical properties? It’s simply good research protocol. The spectrophotometer measures the light that passes through the solution, and having mismatched cuvettes can lead to inconsistent results. The researchers have to ensure the cuvettes have identical optical properties so any differences in absorption can properly be attributed to glucose differences, and not because of cuvette differences.

  1. To enable the comparison of the absorption spectra. This is exactly what we’re looking for in an answer. By ensuring the cuvettes were identical in terms of optical properties, the researchers can ensure they are comparing the absorption spectra and not having their results influenced by differences in cuvettes.
  2. To reduce the absorption in the glass walls. This answer choice is out of scope. Having the glass walls be identical does not mean absorption is reduced. Two identical cuvettes might reduce absorption differently than a different set of identical cuvettes. Ensuring the cuvettes were identical in terms of optical properties does not necessarily mean there is reduced absorption.
  3. To decrease the uncertainty in the wavelength. This answer choice is out of scope. Ensuring the cuvettes have identical optical properties does not decrease uncertainty in wavelength, but rather allows the researchers to focus on differences in absorption.
  4. To increase the absorption in the solutions. Reasoning here is going to be similar to answer choice B. Having the glass walls be identical does not mean absorption increases in the solutions. The solution absorption should not be affected by identical optical properties of the cuvettes. Answer choice A remains our best answer choice.

6) Photon energy can be calculated by using the frequency of the light particles emitted from an object and multiplying by Planck’s constant to work out the energy:

E = hf

Where h is Planck constant. This states that the energy of a photon depends directly on its frequency. By substituting wave properties, this equation can also be written as E = hc/λ which helps use utilize the constants we’re given in this particular question. We’re solving for E, we have a value for hc, and we’re told the wavelength at which absorbance occurred in the passage:

hc = 19.8 × 10–26 J•m and λ=625 nm

Use dimensional analysis to convert wavelength to the proper units:

625 nm x (1 m / 109 nm) = 625 x 10-9 m

E = hc/λ = (19.8 × 10–26 J•m) / (625 x 10-9 m) = approximately 3.2 x 10-19 J.

We can convert Joules to eV using the constant we’re given in the question stem: 1 eV = 1.6 × 10–19 J

Use dimensional analysis again: (3.2 x 10-19 J) x (1 ev / 1.6 × 10–19 J) = 2eV

We did a small bit of approximating, however, a quick glance at our answer choices shows our calculation corresponds to only to answer choice B. Answer choices A, C, and D all list incorrect values.

7) This is another passage-based question in this set that relies on doing some math, the test-maker simplifies our lives by telling us where we will find the necessary information to answer our question. We’re focused on Table 1:

Table 1 Absorbance of Standard Glucose Solutions and of a Diluted Blood Sample

Absorbance Glucose concentration (mg/dL)
0.00 0
0.06 1.5
0.12 3.0
0.18 4.5
0.20 diluted blood sample
0.24 6.0

The passage also tells us there was a 1/30 dilution ratio, so we have to multiply glucose concentration in Table 1 by 30 to find the actual glucose concentration of the glucose in the blood from which the sample was taken. The diluted blood sample is going to be somewhere between 135 mg/dL (corresponding to absorbance of 0.18) and 180 mg/dL (corresponding to absorbance of 0.24). Only answer choice D falls within this range.

Alternatively, we can note the trend in Table 1. Every 0.06 increase in Absorbance means an increase in glucose concentration of 1.5 mg/dL. Therefore, every 0.02 increase in Absorbance should mean an increase in glucose concentration of 0.5 mg/dL. From 0.18 to 0.20 we should see the diluted glucose concentration jump from 4.5 to 5.0 mg/dL. Multiplying 5.0 mg/dL gets us an exact value of 150 mg/dL, or answer choice D.

8) This is a fairly common question type when it comes to experimental passages. We’re asked about modifications to the research we read about in the passage. By doing so, the test-maker ensures that we have grasped the big picture of the experiment and passage. What do I mean by that? If we don’t understand the initial experiment taking place in the passage, it’s very unlikely that we will be able to explain the effects of different modifications.

Both interpolation and extrapolation involve predicting values related to a data set. Interpolation is used to predict values between known data points, while extrapolation is used to predict values outside the range of data. Note the blood sample in the question stem tested above the range of the standards used in the experiment. We want to get that glucose concentration within the range of the standards used in the experiment so the researchers can use data interpolation as opposed to extrapolation.

  1. Increase the enzyme concentration. This is the opposite of what we’re looking for. We can recall from Paragraphs 1 & 2, that by adding more glucose to Reaction 1, we know we have additional chromogen in solution and optical absorbance is affected by this chromogen. This exacerbates our problem in this situation. We already have a blood sample that tested above the range of the standards used in the experiment, and answer choice A would increase this value.
  2. Increase the oxygen pressure. This answer choice is going to be similar to answer choice A. We can look at Reaction 1 and consider the effects of increased oxygen pressure. The reaction will shift to the right so more reactants are produced. Once again, we know we have additional chromogen in solution and optical absorbance is affected by this chromogen. This once again exacerbates our problem in this situation.
  3. Decrease the content of oxygen acceptor. This answer choice is unreasonable. Removing glucose, the oxygen acceptor, would make it so we don’t get any valuable information from the research. Removing glucose makes quantifying the amount of glucose present essentially pointless.
  4. Dilute the sample with additional solvent. This answer choice is consistent with our breakdown of the question. We said the blood sample in the question stem tested above the range of the standards used in the experiment. We want to get that glucose concentration within the range of the standards used in the experiment so the researchers can use data interpolation as opposed to extrapolation. How can we get that glucose concentration within that range? By diluting the sample with additional solvent and decreasing the concentration of glucose. The absorbance will then fall within the range of the standards and we can use data interpolation. Answer choice D is our best answer choice.

 

Exam 4 C/P Solutions: Questions 9-12

9) This is a standalone question that relies on our general knowledge. We can start by going over distillation before jumping into the specific components of the mixture described. Distillation is a purification method where the components of a liquid mixture are vaporized and then condensed and isolated at different boiling points. We’re using fractional distillation here specifically (as opposed to simple distillation). Fractional distillation is used when there is a more complex mixture with not just fixed boiling points, but ranges of temperatures where they boil. The equipment is more complex as well: a fractioning column filled with beads is used to condense different compounds at different temperatures more efficiently. This allows for obtaining greater purity in the end.

1-chlorobutane and 1-butanol are structurally similar, with 1-butanol having a hydroxyl group where 1-chlorobutane has a chloride. Distillation, like we mentioned, takes advantage of different boiling points which would arise from the hydrogen bonding in 1-butanol. For the MCAT, when we see FON (fluorine, oxygen, nitrogen) atoms, we associate with hydrogen bonding which increases boiling point.

  1. Both 1-chlorobutane and 1-butanol are polar. We have to be careful here because while this is a true statement, does it answer the specific question being asked? Separating the molecules in our question stem via fractional distillation relies on boiling point differences and not whether the molecules are polar or nonpolar. Both molecules being polar is actually a shared property of the two molecules and does not explain the boiling point difference.
  2. Both 1-chlorobutane and 1-butanol are nonpolar. Reasoning here is going to be similar to that in answer choice A. This is actually a factually incorrect statement. Both molecules in the question stem are polar. Separating the molecules in our question stem via fractional distillation relies on boiling point differences and not whether the molecules are polar or nonpolar.
  3. The boiling point of 1-chlorobutane is substantially higher than that of 1-butanol. This is another factually incorrect statement. We mentioned in the breakdown of the question, distillation takes advantage of different boiling points that would arise from the hydrogen bonding in 1-butanol. We expect 1-butanol has a higher boiling point.
  4. The boiling point of 1-chlorobutane is substantially lower than that of 1-butanol. This answer choice matches what we said in the breakdown of the question. Distillation takes advantage of different boiling points that would arise from the hydrogen bonding in 1-butanol. We expect 1-butanol has a higher boiling point. Note one of the keys to answering this question was simply knowing background information about distillation. We said distillation is a purification method where the components of a mixture are vaporized, condensed, isolated at different boiling points. By knowing we’re separating based on boiling points we could have narrowed our answer choices down to C and D. We can stick with answer choice D as our correct answer.

10) This is another standalone question in this set that is going to be answered using general knowledge. A quick glance at our answer choices shows we’ll have to know the properties of different functional groups and which are readily oxidized.

A. A primary alcohol to an aldehyde. Primary and secondary alcohols are readily oxidized. We can show the oxidation of ethanol (a primary alcohol) to form acetaldehyde (an aldehyde) below:

B. A tertiary alcohol to a ketone. Tertiary alcohols (R3COH) are resistant to oxidation because the carbon atom that carries the OH group does not have a hydrogen atom attached but is instead bonded to other carbon atoms. The carbon atom bearing the OH group must be able to release one of its attached atoms to form the double bond. The carbon-to-hydrogen bonding is easily broken under oxidative conditions, but carbon-to-carbon bonds are not. Therefore, tertiary alcohols are not easily oxidized.

C. An aldehyde to a carboxylic acid. Due to the free hydrogen seen in aldehydes, aldehydes can be oxidized to transform the carbonyl group (C=O) to a carboxylic acid group (-COOH). Answer choice B remains the best option.

D. A secondary alcohol to a ketone. Primary and secondary alcohols are readily oxidized. We can show the oxidation of isopropyl alcohol to give acetone (the simplest ketone):

Answer choice B remains our best answer choice.

11)

This is a standalone question that relies on our knowledge of naming hydrocarbons according to the IUPAC. First thing we’ll note is the number of carbons in the parent chain. From left to bottom right in the hydrocarbon, we count 5 carbons. Next, we notice a double bond between the 2nd carbon from the left to the 3rd carbon. We want that double bond to have the lowest possible numbers before numbering the methyl group near the top right:

Going by our numbering system above and given the length of the parent chain (5 carbons) we’re looking at 3-methylpent-2-ene. We have to also consider whether the highest priority groups are on the same side or opposite sides of our double bond between carbon 2 and carbon 3. We note they are on the same side, which means we have Z-3-methylpent-2-ene which corresponds to answer choice A. The incorrect options either use the E-designation which would incorrectly imply the highest priority groups are on the opposite side of our double bond between carbon 2 and carbon 3, or they incorrectly count the length of the parent chain as 4 carbons (ethyl). Only answer choice A is the correct systematic name for the hydrocarbon.

12) This is a standalone question that relies on our knowledge of vectors. Vectors are geometric representations of magnitude and direction which are often represented by straight arrows. These arrows start at one point on a coordinate axis and end at a different point. Vectors have a length and direction. Vectors may be added or subtracted graphically by laying them end to end on a set of axes. The graphical method of vector addition is also known as the head-to-tail method. Given the angle in our question stem can vary from 0° to 180°, the vectors can be added head-to-tail to give a resultant vector of magnitudes from 3 units to 13 units. When A and B make an angle of 180° (opposite directions) we have a magnitude of only 3 units. When A and B make an angle of 0° (same direction head-to-tail) we have a magnitude of up to 13 units. The only answer choice that does not fit in this range is answer choice A: 2 units.

 

Exam 4 C/P Solutions: Passage 3

13) This is a passage-related question that involves going back to the passage for key details. After the author explains and shows us Figure 1 with the schematic of the laser device, we get into the function of excimer lasers. We’re told, “Excimer lasers are used in the photorefractive keratectomy (PRK) technique designed to correct vision impairment.” However, we want to make the distinction between the PRK procedure used to correct farsightedness versus the PRK procedure used to correct nearsightedness. The author tells us “To correct nearsightedness, the laser beam is directed onto the central part of the cornea, resulting in a flattening of the cornea (Figure 2B).” Why would this work? Because a patient with nearsightedness has too curved of a cornea and rays from objects are forming images in front of the retina instead of on the retina. A normal eye is not as elongated, and images focus on the retina instead of in front of the retina like we have in nearsightedness. By flattening the cornea, we’re adjusting for this elongation and increasing the radius of curvature so the image properly focuses on the retina instead of in front of the retina.

We want an answer choice consistent with flattening of the cornea.

  1. The density of the cornea is increased. The density of the cornea does not change. Flattening the cornea and removing corneal material will change the mass and volume of the cornea in a relatively equal amount. While mass and volume will change individually, the ratio remains the same. Density stays the same.
  2. The radius of curvature of the cornea is increased. This answer choice is consistent with our breakdown. We have to visualize what happens in nearsightedness. A patient with nearsightedness has too curved of a cornea and rays from objects are forming images in front of the retina instead of on the retina. We want objects to form images on the retina. By flattening the cornea, we’re adjusting for this elongation and increasing the radius of curvature so the image properly focuses on the retina instead of in front of the retina. This is a strong answer choice.
  3. The index of refraction of the cornea is increased. Reasoning here is going to be similar to answer choice A. The index of refraction is not going to change. Answer choice B remains the best option.
  4. The thickness of the cornea at the apex is increased. This is the opposite of what happens in this process. We can reference Figure 2B and what the author tells us in the passage. The laser is removing corneal matter and flattening the cornea. Only answer choice B correctly lists the effect produced by the PRK technique.

 

14) This is a math-centric question that relies on picking out some key numbers and information from the passage. One of the first things we should do in these types of questions is to note the units in our answer choices. Our final magnitude is going to be given in V/m.

The electric field (E) can be worked out by dividing the voltage of the electric field by the distance (d) in meters: E= V/d. This gives us the electric field strength in the necessary V/m units. In Paragraph 1, the author gives us specific information about the electrical discharge produced:

We have our necessary numbers (8.0 kV across a distance of 4.0 mm), but we need to make sure we have the proper units. Let’s use dimensional analysis to get V & m.

(8.0 kV) x (1000 V / 1 kV) = 8000 V

(4.0 mm) x (1 m / 1000 mm) = 0.004 m

E = V/d = (8000 V) / (0.004 m) = 2.0 x 106 V/m

This was a math problem where we did minimal rounding or approximating. We solved for an exact magnitude which corresponds to answer choice A.

15) This is going to be similar to our last question in this question set. It’s a math-centric question that relies on picking out some key numbers and information from the passage. Quick glance at our answer choices shows frequency is given in Hertz, or 1/s. It’s imperative that you know how to convert between units and how different units are related to one another.

The author tells us in Paragraph 2, “Material in excess is removed by vaporization from the surface of the cornea with the use of pulsed laser radiation at time intervals of 250 ms.” The author gives us a duration, or period (T), of 250 ms. Frequency of each pulse is going to be 1/T.

Recall we want our units in Hertz, or 1/s. We can convert our period to seconds using dimensional analysis:

(250 ms) x (1 s / 1000 ms) = 0.25 seconds

f = 1/T = 1/0.25 second = 4 Hz

This was a math problem where we did minimal rounding or approximating. We solved for an exact frequency which corresponds to answer choice B. This passage and question set should really be hammering home the importance of knowing how to manipulate numbers and units on test day!

16) This is a passage-related question that ties into what we touched on in our previous question in the set. The author tells us in Paragraph 2, “Material in excess is removed by vaporization from the surface of the cornea with the use of pulsed laser radiation at time intervals of 250 ms.” We previously solved for the frequency of the pulses that deliver radiation to the cornea, but now we’re actually getting into reasoning. Why was pulsed laser radiation used instead of continuous laser radiation? The author does not provide any more obvious clues that we know we’ll use to answer this question. We can go through each answer choice and decide which is the best, most likely answer.

  1. absorb more radiation. This is the opposite of what we would expect from the use of pulsed laser radiation. Continuous laser radiation, like the name suggests, would provide continuous radiation and more time to absorb radiation. Pulsed laser radiation involves less time interacting between the laser and cornea.
  2. change its index of refraction. This is something we also addressed in a question in this question set. Index of refraction depends on the cornea and is not dependent on the type of laser being used. The index of refraction isn’t going to vary with the use of either laser type.
  3. increase the area exposed to radiation. The area exposed to radiation isn’t going to change. Regardless of the use of pulsed laser radiation or continuous laser radiation, the area exposed to radiation is dependent on the opaque disk described in Paragraph 3, not on the type of laser used.
  4. maintain a lower average temperature. This answer choice sounds the most reasonable. A continuous laser is going to constantly be emitting radiation, while the use of pulsed laser radiation allows for a cool-down period as the laser is not continuously hitting the eye. This cool-down period allows for the eye to maintain a lower average temperature. Answer choice D is going to be our best answer.

 

Exam 4 C/P Solutions: Passage 4

17) This is a pseudo-standalone question that relies on our knowledge of intermolecular forces. Quick glance at our answer choices shows we want an answer that’s related to dipole-dipole forces and London dispersion forces.

Dipole-Dipole interactions are electrostatic interactions that form between opposite partial charges of molecules with permanent dipoles (polar molecules). NH3 is a polar molecule because of electronegativity differences between nitrogen and hydrogen, but also its structure. We can’t forget that NH3 consists of a central nitrogen atom with five outer electrons and three additional electrons from each hydrogen atom. We have three bond pairs and a lone pair; NH3 is asymmetrical. Dipole-Dipole interactions are stronger than London dispersion forces. London dispersion forces are the weakest intermolecular force and exist in all real molecules. Our predicted answer is both forces are present in the molecules of NH3 gas.

  1. dipole–dipole forces only. Dipole-dipole interactions are present in NH3. We noted the electronegativity difference and asymmetrical molecule. However, this answer choice is incomplete. We also expect London dispersion forces which we said exist in all real molecules.
  2. London dispersion forces only. Reasoning here is going to be similar to answer choice A. London dispersion forces do exist among the molecules of NH3 gas, however, this answer choice is incomplete. We want an answer choice that lists both dipole-dipole and London dispersion forces.
  3. both dipole–dipole and London dispersion forces. This answer choice is consistent with our breakdown. Dipole-dipole interactions are present in NH3. We noted the electronegativity difference and asymmetrical molecule. London dispersion forces are the weakest intermolecular force and exist in all real molecules. This is going to be our best answer choice.
  4. neither dipole–dipole nor London dispersion forces. This answer choice is the opposite of our predicted answer. Both forces are present in the molecules of NH3 gas. We can stick with answer choice C as our best answer.

18) First thing we’ll do is find the equilibrium constant which we will set equal to the given Kb value. The equilibrium constant is the ratio of the mathematical product of the concentrations of the products of a reaction to the mathematical product of the concentrations of the reactants of the reaction. Each concentration is raised to the power of its coefficient in the balanced chemical equation. If we have an equation aA + bB ⇌ cC + dD, we can write out the equilibrium constant expression:

However, before we write anything for this specific question, what we need to pay attention to is the state of each reactant and product. Pure solids and liquids are not included when writing out this equilibrium constant. That means we are not including H2O(l) in our expression.

We can write out the expression and set equal to Kb:

Kb = 1.76 × 10–5 = [NH4+][OH] / [NH3]

NH3 acts as a weak base, so we can use [NH3] = 10 M which we get from our question stem. Our reactant (weak base) does not dissociate as much, the concentration of our products can be considered equal: [NH4] = [OH]. We can plug in these values:

Kb = 1.76 × 10–5 = [NH4+]2 / 10 M

Multiply both sides by 10:

1.76 x 10-4 = [NH4+]2

Take the square root of both sides:

[NH4+] = 1.33 x 10-2 M

We’re asked which of the answer choices listed is closest to our calculated answer. Answer choice B is closest to 1.33 x 10-2 M.

19) This is a passage-based question that relies on understanding the demonstration presented in the passage. The author gives us a great visual in the passage where we’re told NH3 gas fills an inverted 12-oz aluminum can. The can is sealed and placed sealed-side down in a tray containing water. Once the seal is removed, NH3 rapidly dissolved in the water, and the atmospheric pressure crushed the can. This question is testing whether we understand the reasoning behind the crushing of the can and if substituting CH4 for NH3 would have the same effect. Our answer choices involve describing CH4 as polar or nonpolar; this question is going to tie into something we talked about earlier in the question set as well where we discussed the intermolecular forces in NH3.

NH3 is a polar molecule because of electronegativity differences between nitrogen and hydrogen, but also its structure. We can’t forget that NH3 consists of a central nitrogen atom with five outer electrons and three additional electrons from each hydrogen atom. We have three bond pairs and a lone pair; NH3 is asymmetrical. On the other hand, CH4 is a symmetrical molecule. Why does that make a difference? Take the molecule tetrachloromethane, for example.

The chlorine atoms are more electronegative than the carbon atom, and the electrons are drawn toward the chlorine atoms, creating dipoles. However, these carbon-chlorine dipoles cancel each other out because the molecular is symmetrical, and CCl4 has no overall dipole. While ammonia gas is polar and dissolves in water, methane is nonpolar and would not dissolve in the water in the tray.

  1. occur because CH4, being polar, would dissolve in the water in the tray. This is the opposite of what we said in our breakdown of the question. CH4 is nonpolar and would not dissolve in the water in the tray.
  2. occur because CH4, being nonpolar, would dissolve in the water in the tray. First part of the answer choice is true, but CH4 is nonpolar and would not dissolve in the water in the tray like we saw with ammonia.
  3. not occur because CH4, being polar, would not dissolve in the water in the tray. The reasoning here would be sound if CH4 were actually a polar molecule. This is a factually incorrect statement.
  4. not occur because CH4, being nonpolar, would not dissolve in the water in the tray. This answer choice matches our breakdown of the question exactly. While ammonia gas is polar and dissolves in water, methane is nonpolar and would not dissolve in the water in the tray.

20) This is a passage-based question and we can pull up Equation 1 here to start:

This is an open-ended question, so quick glance at our answer choices show we’re looking for a description in terms of periodic table classification. Calcium is present in both of the compounds in the question stem. Calcium is an alkaline earth metal. Alkaline earth metals are elements found in Group 2:

  1. an alkali metal. Alkali metals are found in Group 1 of the periodic table. Calcium is found in Group 2.
  2. an alkaline earth metal. This answer choice correctly describes Group 2 of the periodic table where we find calcium. This is going to be our best answer choice.
  3. a metalloid. Metalloids separate the metals and nonmetals and share characteristics of metals and nonmetals. Answer choice B remains the best option.
  4. a transition metal. Transition metals are metallic elements in groups 3-12 characterized by metallic bonding, colored compounds, varying oxidation states, and catalytic ability. Calcium is not a transition metal. Answer choice B is our best answer choice.

 

Exam 4 C/P Solutions: Passage 5

21) This is a passage-based question and one that I always recommend you should be looking to answer or at least consider during your readthrough of any experimental passage. The conclusion is one of the most important aspects of any experiment and being able to identify the conclusion and its significance helps show you understand the experiment itself. Before we jump into the conclusion, we should identify the purpose of the experiment. Why was this research being conducted, and why did these scientists take the time to do this research? We’re told in Paragraph 3, “researchers were interested in determining if a member of this family of compounds, Compound 1 (Figure 1), could selectively sensitize cancer cells to the potent chemotherapeutic agent camptothecin (Compound 2, Figure 1).” That’s our purpose, so our conclusion should be related to this purpose. We can dive into Figure 2 and see if we can make any conclusions or general observations:

If we have no drug, we have very low levels of apoptotic cells which is expected. However, we see that adding Compound 1 alone also means no apoptosis. As we get into Compound 2, we see Compound 2 induced apoptosis as time goes on. The biggest induction of apoptosis happens when we have a combination of Compound 1 and Compound 2. While Compound 1 alone did not induce apoptosis, when Compound 1 is combined with Compound 2, we see exaggerated effects and higher levels of apoptosis. This conclusion is strengthened by Table 1:

Cell line CC95 CC95 Combination
Acute leukemia 44 nM 0.6 nM
Fibrosarcoma 86 μM 6.7 μM
Cervical carcinoma 53 μM 5.2 μM
Noncancerous fibroblast 15 μM 16 μM

We see a decreased concentration of compounds needed to induce apoptosis in cancer cells when we have a combination of Compound 1 and Compound 2 versus when we have Compound 2 alone. Once again, we can say when Compound 1 is combined with Compound 2, we see exaggerated effects and higher levels of apoptosis in cancer cells specifically. The noncancerous fibroblast cells did not see an increase in apoptosis. In fact, the noncancerous cell line saw an increase in CC95 in the combination group.

  1. Compound 1 selectively sensitizes cancer cells to chemotherapeutic action by Compound 2. This answer choice is consistent with what we see in our passage. While Compound 1 alone did not induce apoptosis, when Compound 1 is combined with Compound 2, we see exaggerated effects compared to Compound 2 alone, and higher levels of apoptosis. Compound 1 is what sensitizes cancer cells to chemotherapeutic action by Compound 2 and we see that through increased apoptosis.
  2. Compound 1 sensitizes all cells to chemotherapeutic action. We saw in Table 1 that the noncancerous cell line saw an increase in CC95 in the combination group. This tells us the statement in answer choice B is false.
  3. Compound 2 induces apoptosis in cancer cells only in conjunction with treatment with Compound 1. We saw in Figure 1 that Compound 2 induced apoptosis over time. While treatment with Compound 1 led to increased apoptosis in cancer cells, Compound 2 still induced apoptosis when Compound 1 was not present.
  4. Compound 2 inhibits the NF-κB signaling pathway, which leads to apoptosis. The passage tells us Compound 1 is a member of the imidazoline family of compounds that inhibits the NF-κB signaling pathway. This answer is factually incorrect. We can stick with answer choice A as our best answer choice.

22) This is essentially a pseudo-standalone question. The only information we need from the passage is the concentration of the acute leukemia CC99 Combination value.

Cell line CC95 CC95 Combination
Acute leukemia 44 nM 0.6 nM
Fibrosarcoma 86 μM 6.7 μM
Cervical carcinoma 53 μM 5.2 μM
Noncancerous fibroblast 15 μM 16 μM

We need to convert 0.6 nM to μM which simply involves dimensional analysis. It’s imperative that you know how to convert between units and how different units and prefixes are related to one another.

1 M = 109 nM = 106 μM so we can multiply units to get to our preferred μM units.

(0.6 nM) x (1 M / 109 nM) x (106 μM / 1 M) = 0.6 x 10-3 or 6.0 x 10-4.

This was a math problem where we did minimal rounding or approximating. We solved for an exact concentration which corresponds to answer choice C.

23) This question is actually similar to Question 21 in the question set because we’re asked to analyze the data from the passage. We can look at Table 1 which showed the results of the CC95 Studies on the four cell lines.

Cell line CC95 CC95 Combination
Acute leukemia 44 nM 0.6 nM
Fibrosarcoma 86 μM 6.7 μM
Cervical carcinoma 53 μM 5.2 μM
Noncancerous fibroblast 15 μM 16 μM

The key to answering this question is noting the relative concentrations shown in the table. What I mean by that is the “greatest enhancement” is going to be measured in percent decrease in concentration needed to induce apoptosis in the combination group. We can actually add an additional column to the table above so our calculations are easier to see. We can simply divide the CC95 column concentration by the CC95 Combination column concentration to see how much of an enhancement we get as a result of sensitization by Compound 1:

Cell line CC95 CC95 Combination x Increase
Acute leukemia 44 nM 0.6 nM 73.3
Fibrosarcoma 86 μM 6.7 μM 12.8
Cervical carcinoma 53 μM 5.2 μM 10.2
Noncancerous fibroblast 15 μM 16 μM 0.9

The acute leukemia cell like shows the greatest enhancement of chemotherapeutic activity as a result of sensitization by Compound 1. Answer choice A is our best answer choice.

24) This answer choice is once again asking us to interpret the data from the passage. Figure 2 showed us apoptic leukemia cells as a function of time under four different treatments, so we can rely on the information in the Figure to answer this question. Note we’re looking into apoptosis 20 hours after treatment.

We’ve got the chemotherapeutic agent camptothecin (Compound 2) and the sensitizing agent (Compound 1). We know that when Compound 1 is combined with Compound 2, we see exaggerated effects and higher levels of apoptosis in cancer cells specifically. Why is that? Because Compound 1 functions to inhibit the NF-κB signaling pathway which normally down-regulates apoptosis. If Compound 1 works its magic, Compound 2 can come in and effectively increase apoptosis much more effectively without the NF-κB signaling pathway preventing apoptosis in some cells. Ideally, we have some combination of the two compounds, with Compound 1 being administered prior to Compound 2.

  1. Administration of both Compound 1 and Compound 2 simultaneously. While this would lead to some level of apoptosis, we would want Compound 1 to be administered first. Compound 1 functions to inhibit the NF-κB signaling pathway which normally down-regulates apoptosis. This is a decent answer choice to start.
  2. Administration of Compound 1 followed by Compound 2 after 0.5 h. This sounds exactly like what we’re looking for. Recall in our breakdown we said if Compound 1 works its magic, Compound 2 can come in and effectively increase apoptosis much more effectively without the NF-κB signaling pathway preventing apoptosis in some cells. This is going to be our best answer choice so far.
  3. Administration of Compound 2 followed by Compound 1 after 0.5 h. Compound 1 functions to inhibit the NF-κB signaling pathway which normally down-regulates apoptosis. We would want Compound 1 to be administered first.
  4. Administration of Compound 2 followed by Compound 1 after 1 d. A few things I don’t like about this answer at first glance. Compound 1 functions to inhibit the NF-κB signaling pathway which normally down-regulates apoptosis. We would want Compound 1 to be administered first. Adding Compound 1 after an entire day would likely mean a higher number of cells will be exposed to Compound 2 prior to being exposed to Compound 1. Answer choice B remains our best answer choice.

 

Exam 4 C/P Solutions: Questions 25-28

25) This is a standalone question that relies on our general knowledge. We’ll identify the bond between the C and O atoms in a carbonyl group, then explain the type of bonding interaction. Quick glance at our answer choices shows we’re looking for an answer in terms of σ and/or π bonds. Sigma (σ) and Pi (π) bonds form in covalent substances when atomic orbitals overlap. A sigma (σ) bond is a covalent bond whose electron density is concentrated in the region directly between the nuclei. A pi (π) bond is a bond formed by the overlap of orbitals in a side-by-side fashion with the electron density concentrated above and below the plane of the nuclei of the bonding atoms. Double and triple bonds can be explained by orbital hybridization. In general, single bonds between atoms are sigma bonds. Double bonds are comprised of one sigma and one pi bond. Triple bonds are comprised of one sigma bond and two pi bonds. Carbonyl groups contain a C=O double bond; double bonds are comprised of one sigma and one pi bond. Our best answer choice is going to be answer choice B: One σ and one π only.

26) This is a standalone question that relies on us knowing periodic trends and how willing different atoms are to gain one electron. Generally, nonmetals have a more positive electron affinity than metals. Atoms, such as Group 7 elements, whose anions are more stable than neutral atoms have a higher electron affinity.

Keep in mind, the electron affinities of the noble gases have not been conclusively measured, so they may or may not have slightly negative values. Noble gases have a full valence shell, so they’re generally unwilling to gain or lose electrons. This is why Group 7 elements have higher electron affinities.

  1. Ar. Argon is a noble gas which we mentioned means it’s not likely to gain an electron readily.
  2. Br. Bromine is a group 7 element which typically means the highest electron affinity. This is going to be our best option for the time being.
  3. Co. Cobalt is a transition metal and would be more likely to gain an electron. While this is our second-best answer choice so far, answer choice B remains superior.
  4. Na. Sodium is a group 1 metal that would much more likely lose a single electron than gain an electron. Answer choice B is going to be our best answer choice.

27) To answer this question, we have to be able to read a phase diagram. A phase diagram is just a visual representation of phases and their most stable ranges of pressures and temperatures. Phase diagrams are substance-specific so this looks different than the phase diagram of water which we’re used to studying. This diagram will help point out a few key points:

This question is asking about the point that is separating CO2 gas and CO2 solid. We want an answer choice that best describes Point A.

  1. Boiling point. Boiling point is between points B and C where we transition from liquid to gaseous CO2.
  2. Critical point. The critical point is actually identified in the water phase diagram and it corresponds to Point C in our question stem.
  3. Melting point. The melting point is found between CO2 solid and CO2 liquid.
  4. Sublimation point. Sublimation is the transition directly from a solid to a gas without passing through the CO2 liquid state. That happens at Point A in our phase diagram. Answer choice D is our best answer.

28) An atom can be at its lowest energy state, ground state, or gain energy to move to excited states, higher energy levels, in discrete steps. The electrons in an atom tend to be arranged in such a way that the energy of the atom is as low as possible, also known as the ground state. When those atoms are given energy, the electrons absorb the energy and move to a higher energy level. A photon is a source of energy that electrons can absorb to make a quantum jump to a higher energy level. We can write out the following that corresponds to the change in energy.

Note what I’ve outlined above. Change in energy (transition between energy levels) is inversely related to wavelength. That means the shortest transition in the figure will correspond to our correct answer (longest wavelength). Transition 2 is a transition between the closest possibly energy levels, meaning answer choice B is our correct answer choice.

 

Exam 4 C/P Solutions: Passage 6

29) This is a passage-based question, but it is almost like a pseudo-standalone question where the only thing we’ll need from the passage is Figure 1.

We have a nitrogen atom that is bonded to three carbon atoms. Best way to go about these types of questions is to find key identifiers that will allow us to easily eliminate incorrect answer choices. Looking at pyrrolizidine, there are no double or triple bonds, and no other atoms (single nitrogen, no oxygen, and the rest are carbons and hydrogens).

  1. An amide. Amide groups have the general formula R1(C=O)NR2R3 which right away eliminate this answer choice. We have no double bonds and no oxygen.
  2. An amine. Amine groups have the general formula R3N which is exactly what we have in Figure 1. This is going to be our best answer choice for now.
  3. An imine. Imine groups have the formula R2C=NR. Once again, we have no double bonds in Figure 1, so we can eliminate this answer choice.
  4. A carbamate. Carbamate has the formula R1O(C=O)NR2R3. Right away we can eliminate this answer choice. Reasoning is going to be the same as we used for answer choice A. We have no double bonds and no oxygen. Only answer choice B correctly lists a functional group present in pyrrolizidine.

30) This question is similar to Question 29 from this same question set. We can attack this question the same way where we pull up the necessary part of Figure 2, then we’ll use our general knowledge to actually answer the question.

Hydride reagents such as LiAl4 and NaBH4 participate in nucleophilic addition with aldehydes/ketones, reducing the oxygen in the C=O bond to a hydroxyl (OH) and attaching a proton to the functional group. That’s exactly what is taking place in this reaction.

Make sure to orient yourself correctly as we go through Step 3. In Compound 2 we have our ketone, we reduce the oxygen in the C=O bond to a secondary alcohol, and the hydroxyl attacks the carbonyl in the ethyl ester, forming a lactone and displacing C2H5OH. The only answer choice consistent with this mechanism is answer choice A. Note that answer choices C and D both imply reduction of an ester. Recall hydride reagents such as NaBH4 participate in nucleophilic addition with aldehydes and ketones, but not esters. Answer choice A is our best answer choice.

31) Once again, we have a similar question setup in this question-set that relies on the Figures from the passage. We can pull up the relevant part of Figure 2 here.

This ties into our previous question that discussed the preparation of Compound 3 from Compound 2. Now we’re going to identify the intermediates formed in Step 4, which involves going from Compound 3 to Compound 4, but with four equivalents of base (OH). We know one equivalent of base gets us to Compound 4, but the key difference here is the four equivalents of OH. OH can be used to hydrolyze esters, meaning one equivalent of base is going to attack the carbonyl carbon in Compound 3. This gets us to answer choice C. However, we have to recall that we have four equivalents of base, not just one. Our second equivalent functions to hydrolyze carbamate like we saw in Step 4 in the passage when there was no excess base. The two additional equivalents of base are excess. Looking at our answer choices, only answer choice D matches our breakdown. Note, answer choices A and C are possible if we had limited base (one equivalent), but the additional equivalents get us to answer choice D.

32) Once again we’re focused on a specific step from our passage. We can pull up Step 5 from Figure 3 here.

First clue for what happens in Step 5 is going to be the positive charge on the nitrogen in Compound 4. Na2CO3 is going to deprotonate the positively charged amine which, now that it has a lone pain, can act as a nucleophile to displace bromine from BrCH2CO2C2H5. An SN2 reaction takes place and we get the formation of compound 5.

33) This is actually something we touched on in a previous question in this set. Question 31 discussed Step 4 in the passage which involved carbamate in Compound 3 being cleaved by a hydroxide to a secondary amine.

We can go through each of the steps in Figures 2 and 3. Carbamate is R2NCO2C2H5 and is present until Step 4 in the passage where it is cleaved.

  1. Step 2. Carbamate is R2NCO2C2H5 and is present until Step 4 in the passage.
  2. Step 3. Reasoning here is the same as answer choice A. Note the orientation of the compound is a bit different than previously. Carbamate is R2NCO2C2H5 and is present until Step 4 in the passage.
  3. Step 4. This answer choice matches our breakdown. Carbamate is R2NCO2C2H5 and is present until Step 4 in the passage where it is cleaved.
  4. Step 6. While structurally similar to carbamate (R2NCO2C2H5), there is no carbamate in Step 6. We can stick with answer choice C as our best option.

 

Exam 4 C/P Solutions: Passage 7

34) This is a passage-based question where we will go back to the passage for some key information about compounds 1-3. We’re told, “Compounds 1–3 bind to a site on RT that is not the active site, and the binding can occur during any stage of substrate binding.” This should sound familiar because binding can occur during any stage of substrate binding. This should sound like mixed inhibition.

Mixed inhibitors bind at a site on the enzyme other than the active site, so they do not prevent the substrate from binding. Their name is inspired by the fact that they can bind to either the enzyme alone or the enzyme-substrate complex.

  1. Competitive. In competitive inhibition, an inhibitor molecule is similar enough to a substrate that it can bind to the enzyme’s active site to stop it from binding to the substrate. It “competes” with the substrate to bind to the enzyme.
  2. Mixed. Mixed inhibition is a type of enzyme inhibition in which the inhibitor may bind to the enzyme whether or not the enzyme has already bound the substrate. In other words, binding can occur during any stage of substrate binding. This is going to be our best answer choice.
  3. Product. Products are not a type of inhibitor, but are rather the substances produced in a reaction.
  4. Irreversible. Irreversible inhibitors bind to a target enzyme in a chemical reaction that is not reversible. This is not what the author describes in this passage. We can stick with answer choice B as our best answer choice.

35) This is a passage-based question and the test-maker tells us exactly where we’re focusing our efforts. We can pull up the reaction scheme in Figure 1. The author tells us, “Figure 1 is a reaction scheme that shows inhibitor binding to RT, RT with bound RNA template (TP), and the RT/TP/dNTP complex along with the associated Michaelis–Menten parameters.”

Figure 1 Reaction scheme of RT and inhibitor binding model (Note: I represents any of the three inhibitors, and RT/P represents the enzyme bound to the product of the reaction.)

We want to know the mechanism of substrate binding to RT. Enzymes can bind to substrates in 3 ways: ordered, random, or ping-pong (double-displacement) mechanisms.

  • Ping-pong mechanism (double-displacement reaction), is characterized by the change of the enzyme into an intermediate form when the first substrate to product reaction occurs.
  • In an ordered mechanism, one particular substrate has to first bind to the enzyme, followed by the other substrate.
  • In random sequential reactions, the substrates and products are bound and then released in no preferred order, or “random” order.
  1. Random order. In random sequential reactions, the substrates and products are bound and then released in no preferred order, or “random” order. We can see from Figure 1 that TP binds before dNTP in the reaction scheme. That implies there is a preferred order. We don’t have any RT/dNTP complexes prior to the RT/TP complexes.
  2. Ordered. In an ordered mechanism, one particular substrate has to first bind to the enzyme, followed by the other substrate. Reasoning for this answer choice is going to be along the same lines as answer choice A. We can see from Figure 1 that TP binds before dNTP in the reaction scheme. That implies there is a preferred order. We don’t have any RT/dNTP complexes prior to the RT/TP complexes meaning there’s an order in which one substrate binds. This is going to be a better answer than answer choice A.
  3. Ping-pong. Ping pong mechanism is characterized by the change of the enzyme into an intermediate form when the first substrate to product reaction occurs. First substrate transforms into product and transfers a group to the enzyme. The first product is released and a second substrate will enter and receive the group that was transferred to the enzyme from the first substrate. Ternary structures are not formed. In the reaction scheme in the passage, only one product is formed. We can stick with answer choice B as our best answer.
  4. Double-displacement. This is the same answer as answer choice C. There can only be one correct answer to every MCAT question, so this gives us confidence that both C and D are incorrect as they are essentially the same answer. Answer choice B is our correct answer.

36) This is a passage-based question that relies on revisiting the passage for key information. I don’t always advocate going back to the passage, but for specific details like we need in this question, it’s better to flip back for a few seconds than try and memorize every detail in your first readthrough.

We’re told the binding pocket contains side chains of valine and phenylalanine that make the strongest interactions with compounds 1-3.

Valine and phenylalanine have nonpolar side chains, with phenylalanine having an aromatic side chain. These are hydrophobic amino acids. We can go through the 4 different types of interactions listed and see which one is the primary type.

  1. Covalent. This would only be a good answer choice if we were dealing with cysteine. In this case these nonpolar amino acids cannot form covalent bonds.
  2. Hydrogen bonds. Amino acid side chains with hydrogen donor and/or acceptor atoms are polar. Valine and phenylalanine side chains do not form hydrogen bonds.
  3. Ionic. Charged amino acids can form ionic bonds. Valine and phenylalanine side chains do not form ionic bonds.
  4. Hydrophobic. This is going to be our best answer choice. We can reference our amino acid chart above (which you should have committed to memory)! Valine and phenylalanine have nonpolar side chains, with phenylalanine having an aromatic side chain. These are hydrophobic amino acids. We can stick with answer choice D as our best answer.

37) This is another passage-based question. First thing we’re going to do here is pull up the compounds we’re asked about.

Next, we can consider what we know about π-bonds. A pi (π) bond is a bond formed by the overlap of orbitals in a side-by-side fashion with the electron density concentrated above and below the plane of the nuclei of the bonding atoms. Double and triple bonds can be explained by orbital hybridization. In general, single bonds between atoms are sigma bonds. Double bonds are comprised of one sigma and one pi bond. Triple bonds are comprised of one sigma bond and two pi bonds. We can simply count the number of double bonds (one π-bond) and triple bonds (two π-bonds) and note how many pi bonds each compound has.

Compound 1 has a total of 6 π-bonds, while Compound 3 has a total of 7 π-bonds. The only answer choice consistent with our breakdown is answer choice B.

 

Exam 4 C/P Solutions: Passage 8

38) This is a passage-related question that relies on us using math. We’re looking to find an approximate concentration. One of the first things we should do in these types of questions is to note the units in our answer choices. Our answer will be in either μM or mM. It’s imperative that you know how to convert between units and how different units are related to one another. In this specific question, we can go back to the passage to pick out the key information we’ll need to find our approximate concentration.

The author makes our lives slightly easier by giving us the exact equation we will be using to solve for concentration c:

A = εbc

We have the following:

A = 0.7

At a pH of 6, ε (M–1cm–1) = 1367

Path length b of the cell used for all experiments = 1 cm

0.7 = (1367 M–1cm–1) (1 cm) (c)

Solve for c:

c = 5.12 x 10-4 M

Recall our answer will be in either μM or mM so we can use dimensional analysis.

c = 5.12 x 10-4 M = 512 μM = 0.512 mM. Answer choice B is the approximate concentration that matches our calculated value.

39) This is a fairly common question type when it comes to experimental passages. We learned in Paragraph 2 that, “UV–VIS spectroscopy was used to measure the concentration p-nitrophenol produced from a p-nitrophenol phosphate substrate.” AAMC likes to ask questions about modifications or limitations to the research we read about in the passage. By doing so, the test-maker ensures that we have grasped the big picture of the experiment and passage. What do I mean by that? If we don’t understand the initial experiment taking place in the passage, it’s very unlikely that we will be able to explain the effects of different limitations and modifications. In this specific case, the spectrophotometer contained a detector that had low sensitivity and was unable to measure high absorbance samples. We related absorbance to other variables later in Paragraph 2.

Using Beer’s Law, A = εbc, what conditions do we need to get low absorbance? We need some combination of low molar absorptivity (which is dependent on pH as shown in Table 1), path length should be kept small, and concentration should be as low as possible. Note, the lowest molar absorptivity values happen at the lowest pH values in Table 1. Higher pH values mean higher molar absorptivity values.

  1. The experiments done at high pH should be diluted relative to those at low pH, or the path length of the high pH experiments should be decreased. This is a strong answer choice to start. Diluting the experiments done at high pH will decrease concentration which we said was necessary for experiments done at high pH. Furthermore, decreasing path length will also keep absorbance low.
  2. The experiments done at low pH should be diluted relative to those at high pH, or the path length of the high pH experiments should be decreased. This is the opposite of what we’d need in terms of the different pHs. Experiments done at low pH will already have the lowest molar absorptivity values. We are more concerned with the experiments done at high pH as we want to decrease concentration. This answer choice is half correct, however, as path length should be decreased for high pH experiments. Answer choice A remains superior.
  3. The experiments done at high pH should be diluted relative to those at low pH, or the path length of the low pH experiments should be decreased. This answer choice is only half correct. Diluting the experiments done at high pH will decrease concentration which we said was necessary for experiments done at high pH. Path length should be decreased in the high pH experiments. Answer choice A remains superior.
  4. The experiments done at low pH should be diluted relative to those at high pH, or the path length of the low pH experiments should be decreased. This answer choice is the opposite of what we would need to happen. Diluting the experiments done at high pH will decrease concentration which we said was necessary for experiments done at high pH. Furthermore, decreasing path length will also keep absorbance low. This answer choice incorrectly addresses the experiments done at low pH which will already have the lowest molar absorptivity values. We can stick with answer choice A as our correct answer.

40) This can be treated almost like a pseudo-standalone question that relies on our general knowledge. We’ll have to rely on our knowledge of buffers. Buffers are a weak acid or base used to maintain the pH of a solution near a chosen value and which prevent a rapid change in pH when acids or bases are added to the solution. For any weak base/conjugate acid pair, the buffering range is a pH that is pKa ± 1. That means that given a pKa of 8.07, Tris would have been an acceptable buffer for any experiments at pH between 7.07 and 9.07.

pH ε (M–1cm–1) kcat (min–1) kcat/KM (min–1M–1)
6.0 1367 1.87 1.83
7.5 13435 1.06 0.76
8.5 17736 0.68 0.64

Only two of the experiments from our passage fall within the acceptable range given the pKa for Tris. That means only answer choice C matches our breakdown of the question.

41) First thing we’ll want to do as we go through this question is to pick out the necessary information we need from the passage. We’re given some background information about G3PPs in Paragraph 1. The author says, “The active site of G3PP contains Asp14 and Asp16, which act as a nucleophile and general acid, respectively, during catalysis.” Nucleophiles are typically electron-rich and can donate electrons to form covalent bonds. What this tells us is that the substrate is covalently attached to a nucleophile, or Asp14 in this situation. We know it’s unlikely to be Asp16 which is the general acid Quick glance at our answer choices shows that we were looking to pick between Asp14 and Asp16 (which we’ve done at this point), and also decide between the phosphorus and oxygen atom of the phosphate group. The nucleophile attacks an atom on the substrate-either the isotopically labeled oxygen or phosphorus.

The oxygen that we see here is going to be the one that is doing the attacking, but it’s not going to attack another oxygen. Rather, we expect it will attack a phosphorus.

The substrate is covalently attached to Asp14 through the phosphorus atom of the phosphate group. Answer choice A is our best option.

42) This is another question in this set that we can treat essentially like a pseudo-standalone question. The only information we’ll need from the passage is the lowest pH in the experiments shown in Table 1:

pH ε (M–1cm–1) kcat (min–1) kcat/KM (min–1M–1)
6.0 1367 1.87 1.83
7.5 13435 1.06 0.76
8.5 17736 0.68 0.64

Given a pH of 6.0, we want to calculate the [H3O+] in the solution which we can do using our external knowledge. The pH of a solution is the negative logarithm of the hydrogen-ion concentration:

pH = -log[H3O+] = 6

Solving for [H3O+] we get 1.0 x 10-6 M

Note, our answer choices are given in μM so we can use dimensional analysis. It’s important to review your prefixes and converting units if you are not comfortable doing so.

(1.0 x 10-6 M) x (106 μM / 1 M) = 1.0 μM

This was a math problem where we did minimal rounding or approximating. We solved for an exact concentration which corresponds to answer choice C.

43) This is another passage-based question that relies on us reading Table 1:

pH ε (M–1cm–1) kcat (min–1) kcat/KM (min–1M–1)
6.0 1367 1.87 1.83
7.5 13435 1.06 0.76
8.5 17736 0.68 0.64

We’re asked the effects of increasing pH on catalytic activity and the maximum velocity of the reaction. Let’s start with catalytic efficiency. Catalytic efficiency is given by Kcat/Km and measures how well an enzyme is able to “grab” its substrate and turn it into product. This means you want the greatest number of reactions happening, for the least amount of input of time/substrate. That’s basically what kcat/km says. Right away, we can say that as pH increases, catalytic efficiency decreases.

Kcat, the catalytic constant, measures the number of molecules of substrate turned over into product per unit time (usually seconds). Km is our Michaelis constant. It’s used to determine the affinity between an enzyme and substrate. The higher your kcat is, the more reactions happen per unit time (i.e. Vmax). All of this is taking place when conditions are optimal, and you have enough substrate. So, Kcat is positively proportional to efficiency. We can also see in Column 2 that this is consistent with the data. As pH increases, catalytic efficiency (kcat/km) decreases and maximum velocity of the reaction decreases.

  1. Both the catalytic efficiency and the maximum velocity of the reaction increase. This is the opposite of our breakdown of the question. We said as pH increases, catalytic efficiency (kcat/km) decreases and maximum velocity of the reaction decreases.
  2. The catalytic efficiency increases with pH, but the maximum velocity of the reaction decreases. This answer choice is only partially correct. Maximum velocity of the reaction decreases, and catalytic efficiency also decreases as pH increases.
  3. The catalytic efficiency decreases with pH, but the maximum velocity of the reaction increases. This answer choice is also only partially correct. As pH increases, catalytic efficiency (kcat/km) decreases and maximum velocity of the reaction decreases.
  4. Both the catalytic efficiency and the maximum velocity of the reaction decrease. This answer choice is consistent with our breakdown of the question. As pH increases, catalytic efficiency (kcat/km) decreases and maximum velocity of the reaction decreases. Answer choice D is our best answer.

 

Exam 4 C/P Solutions: Questions 44-47

44) This is a standalone question that asks about a phenomenon that we’ve probably all seen. Small insects can walk on water, but when humans try, or if we see a bigger insect or animal try, it’s not possible. We typically attribute this to the surface tension of water. Surface tension is a contractive tendency of the surface of a liquid that allows it to resist an external force.

The model of a liquid surface acting like a stretched elastic sheet can effectively explain surface tension effects. For example, some insects can walk on water, like we’re discussing in this question, as humans would walk on a trampoline. The insects “dent” the surface as shown in Figure (a).

  1. Hydrostatic pressure. Hydrostatic pressure refers to the pressure exerted by a fluid (gas or liquid) at any point in space within that fluid, assuming that the fluid is incompressible and at rest. This pressure applies to objects that are submerged, but the insect is not actually immersed when walking on the free surface of water.
  2. Viscosity. Viscosity is a quantity expressing the magnitude of internal friction in a fluid, defined based on laminar flow. That friction prevents relative motion between water and objects and is not how insects walk on the free surface of water.
  3. Buoyancy. Buoyancy force is caused by the pressure exerted by the fluid in which an object is immersed. The insect is not actually immersed when walking on the free surface of water.
  4. Surface tension. We said in our breakdown of the question, surface tension is a contractive tendency of the surface of a liquid that allows it to resist an external force. This is going to be our best answer choice.

45) This is a standalone question that involves working with half-lives. Half-life is the time required for half of the nuclei in a sample of a specific isotope to undergo radioactive decay. In this specific case, half-life is 2.0 hours. If we observe the isotope for 6.0 hours, that means 3 half-lives have passed (2.0 hours/half-life x 3 half-lives). If at the end of 3 half-lives only 24 grams remain, we can solve for the original amount by working backwards and multiplying by 2 three times:

24 grams after 3 half-lives (6 hours)

48 grams after 2 half-lives (4 hours)

96 grams after 1 half-life (2 hours)

192 grams originally

This was a math problem where we solved for an exact value. Quick glance at our answer choices show they are not close to one another. We didn’t do any rounding or approximation, so we can compare all 4 of our answer choices at once. Our calculation matches answer choice D: 192 g.

46) This is a standalone question where the test-maker presents us with a visual in the question stem. When I am reviewing, one of the ways I like to attack these types of questions is to predict what the answer will be, go through the answer choices and define each one, then find the best answer choice. This question sounds like it is describing diffraction at first glance. Diffraction is the collision of light waves through a small slit, just like we have in the question stem. The light is encountering an obstacle and essentially bending around the obstacle. The pattern of dark and bright fringes is a classic example we associate with diffraction.

A. refraction. Refraction is when a light ray changes its direction in a new medium with different properties depending on the incident angle and the changes in speed. This is not what is described in the question stem.

B. polarization. Polarization can occur in two forms. Linear polarization where the electric field of light is confined to a single plane along the direction of propagation. Or circular polarization, where the electric field of the light consists of two linear components that are perpendicular to each other, equal in amplitude but have a phase difference of π/2. The resulting electric field rotates in a circle around the direction of propagation and, depending on the rotation direction, is called left- or right-hand circularly polarized light

C. diffraction. This is consistent with our prediction from our initial breakdown of the question. We said the dark and bright fringes are consistent with what we traditionally see when discussing diffraction and slit experiments. We can stick with answer choice C as our best option for now.

D. reflection. Reflection is the property of a propagated wave being thrown back from a surface. When light is bounced off of a material, such as a mirror, this is called a reflection. This is not what is described in the question stem. We can stick with answer choice C as our best option.

 

47) This is a standalone question the relies on us knowing the properties of fatty acids. The relatively long nonpolar hydrocarbon chains on fatty acids makes them hydrophobic, so fatty acids are generally considered to be insoluble in water. Longer hydrocarbon chains have a lower solubility, while shorter hydrocarbon chains are the most soluble in water. That means the saturated fatty acid that is the most soluble in water from our answer choices is going to be the shortest fatty acid. We can actually compare all of our answer choices at once here because the only difference is the length of the alkyl chain. Answer choice A has the shortest alkyl chain and is the most soluble in water.

 

Exam 4 C/P Solutions: Passage 9

48) This is a passage-based question that focuses on two different nucleotides we saw in the passage. We can go through some quick background information before jumping into key information from the passage.

Adenine and guanine are classified as purines, and have a primary structure consisting of two carbon-nitrogen rings. Cytosine, thymine, and uracil are classified as pyrimidines which have a single carbon-nitrogen ring as their primary structure.

In Figure 2, we have the keto and enol form of guanine (top) and thymine (bottom). Keto–enol tautomerism refers to a chemical equilibrium between a keto form (a ketone or an aldehyde) and an enol (an alcohol) as they are tautomers of each other. The keto form is more stable, it is the predominant form, but we can look at this situation specifically in the context of the passage. A pKa of 9.2 and a pKa or 9.7 are both greater than what we see at physiological conditions. At physiological conditions, or a pH of around 7.4, we expect protonation and the keto form to be preferred. Answer choice A is going to be our correct answer.

Note that AAMC’s explanations may incorrectly say that pKa of 9.2 and 9.7 are less than pH of 7.2. That likely should have been more* than pH of 7.2.

49) This is a passage-related question and we can focus on Paragraph 3 where the author talks about β− decay and phosphorus-32.

The author mentions the reaction product was measured by the detection of β− decay after purification. Beta radiation occurs when a neutron turns into a proton releasing an electron.

To calculate the atomic properties of an atom before and after radioactive decay, two important quantities are used – the atomic mass number A (number of protons + neutrons), and the atomic number Z (number of protons). We can see the equation for Beta- decay above. These numbers must balance before and after decay.

We start with 32P that undergoes β− decay:

3215P → 3216S + e

Note, the atomic mass number stays the same, but the atomic number increases as we have one more proton. The atom that corresponds to an atomic number of 16 is Sulfur.

  1. Si. This answer choice incorrectly lists silicon which has an atomic number of 14, not 16. This answer would be obtained by incorrectly using beta+ decay.
  2. Al. This answer choice could be produced by alpha decay. We can see from our visual that beta- decay involves a neutron turning into a proton, releasing an electron.
  3. S. This answer choice matches our breakdown of the question. 3215P 🡪 3216S + e . This is going to be our best answer choice.
  4. Cl. That answer choice would be the result of successive beta- decays, not the single one described in the passage. We can stick with answer choice C as our correct answer.

50) This is a question that AAMC likes to ask across multiple sections of the exam. If there is one thing you should make sure to review constantly before your exam, it’s amino acids.

We’re going from D (aspartate) to a different amino acid in our DNA polymerase. We’re told we want a variant that will most likely retain catalytic activity. Aspartate has a negatively charged side group, just like E (glutamate). We can actually see it in Figure 1 as well. Theoretically, the highest chance of retaining catalytic activity is by substituting D for E because both have the negative charge.

  1. D429A. Alanine has a neutral side chain. Substituting an amino acid with this neutral side chain for the charged aspartate is not ideal. We would likely see a change in activity.
  2. D429E. This answer choice correctly shows a negatively charged amino acid substituting for a negatively charged amino acid. This would mean the highest chance of retaining catalytic activity. This is going to be our best answer choice.
  3. D429K. Lysine has a positively charged amino acid. Substituting an amino acid with this positively charged side chain for the negatively charged aspartate will likely decrease catalytic activity. Answer choice B remains superior.
  4. D429F. Phenylalanine has a neutral, aromatic side chain. Substituting for the negatively charged aspartate will likely affect catalytic activity. Only answer choice B correctly showed a negatively charged side chain substituting for another negatively charged side chain. We can stick with B as our correct answer.

51) This is a passage-related question and the author asks about a specific set of details from the passage. We’re asked about the two types of G•T mismatches discussed in the passage. The author says, “G•T mismatches occur in two different ways. In the wobble base pair, the G•T mismatch appears as an offset A•T base pair. In addition, the G•T mismatch can mimic a G•C base pair if the enol tautomer of T is present.” We want to know the hydrogen bonds present between an A•T base pair and a G•T base pair that mimics a G•C base pair. The final part of this question relies on our general knowledge.

A can form two hydrogen bonds only with T. G can form three hydrogen bonds only with C. That means we expect 2 hydrogen bonds in the wobble pair, and 3 in the enol T mismatch.

  1. 2 in the wobble pair, 2 in the enol T mismatch. This answer choice correctly lists the number of hydrogen bonds in the wobble pair, but we mentioned the G•T mismatch can mimic a G•C base pair if the enol tautomer of T is present. A G•C base pair involves three hydrogen bonds.
  2. 3 in the wobble pair, 2 in the enol T mismatch. This answer choice switches the number of hydrogen bonds we’re looking for in our correct answer; we expect 2 hydrogen bonds in the wobble pair, and 3 in the enol T mismatch.
  3. 2 in the wobble pair, 3 in the enol T mismatch. This answer choice correctly lists the number of hydrogen bonds in the two types of mismatches. The wobble pair is an A•T base pair (2 hydrogen bonds) and the enol T mismatch is a G•T base pair that mimics a G•C base pair (3 hydrogen bonds)
  4. 3 in the wobble pair, 3 in the enol T mismatch. This answer choice incorrectly lists 3 hydrogen bonds in the wobble pair. The wobble pair is an A•T base pair (2 hydrogen bonds). Answer choice C is our best answer choice.

52) Similar to our previous question in this question set, this is a passage-related question and the author asks about a specific set of details from the passage. We can quickly look back at Figure 1 from the passage and look at the active site of DNA polymerase:

We want to find an answer choice that is likely to be found in the pocket of a catalytically active DNA polymerase. In other words, which metal could be substituted in for M2+ in the figure.

  1. Zn. Zinc forms a divalent cation: Zn2+. We’re looking for an answer choice that is NOT likely to be found in the pocket of a catalytically active DNA polymerase. Our correct answer will not form a divalent cation.
  2. Fe. Iron forms a divalent cation: Fe2+. We’re looking for an answer choice that is NOT likely to be found in the pocket of a catalytically active DNA polymerase. Our correct answer will not form a divalent cation.
  3. Mg. Magnesium forms a divalent cation: Mg2+. We’re looking for an answer choice that is NOT likely to be found in the pocket of a catalytically active DNA polymerase. Our correct answer will not form a divalent cation.
  4. Na. Sodium is the only answer choice listed that will not form a divalent cation, but rather will form a monovalent cation: Na+. This is going to be our best answer choice.

 

Exam 4 C/P Solutions: Passage 10

53) This is a passage-related question that is essentially like a pseudo-standalone question. The only vital information we will need from the passage to answer this question is to find the energy stored in the defibrillator’s capacitor. The author says, “Suppose a capacitor in a portable defibrillator needs to store 400 J.” Gravitational potential energy has the same units: kg x m2 / s2 or Joules (J). Formula for gravitational potential energy is given as mass (m) x gravitational acceleration (g) x height (h) or simplified as mgh. We can set the energy from the passage equal to gravitational potential energy:

400 J = (5 kg) (9.8 m/ss) (h)

Solving for h gives us a height of 8.1 meters. This was a math problem where we did minimal rounding or approximating. We solved for an exact height which is closest to answer choice B: 8 m.

54) This is a pseudo-standalone question that relies on relating current and time to solve for current. Note the voltage is given as a distractor:

We’re given current as 4.2 amp x hour, but we have to ensure we have the proper units for time:

1 hour x (60 minutes / 1 hour) x (60 seconds / 1 minute) = 3600 seconds. We can multiply:

4.2 A x 3600 s = 15,120 C

This was a math problem where we did minimal rounding or approximating. We solved for an exact charge which is closest to answer choice C: 15,000 C.

55) This is going to be similar to Question 54 in that is a pseudo-standalone question. Force is equal to mass x acceleration. We already know mass of the blood is 20 grams or 0.02 kg.

This 0.02 kg of blood enters the heart at 25 cm/s and leaves at 35 cm/s. That is a difference in 10 m/s in 0.1 seconds. The change in velocity over 0.1 seconds allows us to solve for acceleration:

(10 m/s) / 0.1 seconds = 1 m/s2

Multiplying force x acceleration gives us:

0.02 kg x 1 m/s2 = 0.02 N.

This was a math problem where we did minimal rounding or approximating. We solved for an exact force which is equal to answer choice A: 0.02 N.

56) This is a passage-based question that focuses on something we learned in Paragraph 2 of the passage where the author talks about defibrillators. The author tells us, “A portable defibrillator works by storing electric energy in a capacitor. Electrodes, sometimes called “paddles,” are placed on a patient’s chest, and current can be passed from one electrode to the other once a switch is flipped by a medical professional.”

All conductors contain electric charges. When exposed to potential difference, the positive charges in a conductor will migrate towards the negative end, the negative charges in the material will move towards the positive end of the potential difference. This flow of charge is electric current. We want to explain why the electric field inside the conductors that form the capacitor is zero. The effect of electric fields can be found by superimposing its E vectors, and in this case that effect ends up as a net zero. The charges are all functioning to essentially cancel out and make the electric field acting on them zero. That can happen once electrons have reached an equilibrium and the electric field produced in the conductor cancels with any external field. That’s a property of conductors.

  1. All of the electrons in the conductor are bound to atoms, and thus there is no way for an external electric field to penetrate atoms with no net charge. This answer choice is extreme, first of all. We typically stay away from answers that begin with “all.” Furthermore, conductors also have free electrons that can travel and orient themselves on the surface of the conductor.
  2. Free electrons in the conductor arrange themselves on the surface so that the electric field they produce inside the conductor exactly cancels any external electric field. This answer choice is consistent with what happens with conductors. We can imagine free electrons arrange themselves on the surface like we mentioned in our breakdown of the question. The charges are all functioning to essentially cancel out and make the electric field acting on them zero. That can happen once electrons have reached an equilibrium and the electric field produced in the conductor cancels with any external field. This is going to be our best answer choice.
  3. Free electrons in the conductor arrange themselves on the surface and throughout the interior so that the electric field they produce inside the conductor exactly cancels any external electric field. Free electrons are present on the surface, but not throughout the interior. The reason we get the electric field to be zero is because of the free electrons on the surface and produce an electric field that cancels the external field. Adding another electric field changes that net zero and balance.
  4. All electrons in the conductor, both free and bound, arrange themselves on the surface so that the electric field they produce inside the conductor exactly cancels any external electric field. Reasoning here is similar to answer choice A. We avoid answer choices that begin with “all.” Furthermore, bound electrons are not part of the rearrangement on the surface. Conductors have free electrons that can travel and orient themselves on the surface of the conductor. We can stick with answer choice B as our best answer choice.

 

Exam 4 C/P Solutions: Questions 57-59

57) This is a standalone question that simply involves calculating mean arterial pressure. The test maker was actually nice to us here and provides us with the corresponding equation for mean arterial pressure.

Blood pressure is the pressure exerted by blood on the walls of a blood vessel that helps to push blood through the body. Systolic pressure measures the amount of pressure that blood exerts on vessels while the heart is beating. The optimal systolic blood pressure is 120 mmHg. Diastolic pressure measures the pressure in the vessels between heartbeats. The optimal diastolic blood pressure is 80 mmHg.

In our question stem we’re given systolic/diastolic pressure of 135/90 which we can plug into the equation for mean arterial pressure:

(2 × 90 mmHg) + 135 mmHg
3

 

MAP =

MAP = 315 / 3 = 105 mmHg

This was a math problem where we did minimal rounding or approximating. We solved for an exact MAP which corresponds to answer choice C: 105 mmHg.

58) This is a standalone question that relies on our knowledge of action potential generation. I always recommend you review action potentials any chance you get as you review:

This is a comparison question so we can go through each of our four answer choices one-by-one. We’re looking for the event that occurs first.

  1. Voltage-gated sodium channels open at the axon hillock. If the threshold of excitation is reached at the axon hillock, voltage-gated Na+ channels open, and the membrane depolarizes. We can see this in our visual above, but we can’t comment just yet on if this event happens first until we go through our additional answer choices.
  2. Hyperpolarization stimulates the opening of ligand-gated potassium channels. At the peak action potential, K+ channels open and K+ begins to leave the cell. This is in response to depolarization. The membrane becomes hyperpolarized as K+ ions continue to leave the cell. The hyperpolarized membrane is in a refractory period and cannot fire. Potassium channels are voltage-gated. This answer choice is factually incorrect, so we can stick with answer choice A as our best option for now.
  3. Graded potentials propagate along the axon. We don’t want to get tricked here as location is key. Answer choice A mentioned voltage-gated sodium channels opening at the axon hillock. This choice mentions propagating along the axon which cannot happen until answer choice A happens.
  4. Calcium influx stimulates vesicle fusion and release of neurotransmitter. Once the threshold potential is reached, the neuron completely depolarizes. Depolarization causes the opening of voltage-gated calcium channels, and the influx of calcium into the cell triggers neurotransmitter release. This happens after action potentials arrive at axon terminals or, in other words, later than answer choice A. Answer choice A is our best answer.

59) This is a standalone question that asks us to calculate the energy needed to heat 2 kg of water 50o celsius (no phase changes). Quick glance at our answer choices shows our answers are in kJ. We can use Q=mcΔT here and solve for energy. As we go through our calculations, be sure to pay attention to units. We want to make sure we end up with kJ in our final answer.
Q = (mass) (specific heat) (change in temperature) = (2.0 kg) (4.2 kJ/°C•kg) (50 oC) = 420 kJ

Note the kilograms and degrees Celsius cancel out. We’re left with 420 as our answer in kJ.

This was a math problem where we did minimal rounding or approximating. We solved for an exact energy which corresponds to answer choice C: 420 kJ

Alternatively, if you did not know to use Q=mcΔT right away, you can use the units given to make sure the proper ones cancel out. Ideally, you remember all the necessary equations on exam day, but if you’re unsure about how to get to an answer, use the units as clues. Know that the test-maker might add distractors and give you values and units you might not use, so solving the problem correctly should always be your first goal!



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