AAMC FL2 Flashcards

Solution: The correct answer is A.

1. In the studies, one can reason from the passage that the chymotrypsin-like activity of proteasome releases AMC and produces a fluorescent signal. The passage states that chymotrypsin cleaves peptide bonds after aromatic residues, which have the structure –(C=O)–NH–. Thus, the chymotrypsin-like proteasome should hydrolyze the corresponding amide bond connecting the C-terminus of the model tetrapeptide and the NH₂ group of AMC.
2. This oxygen atom cannot be part of an amide bond, which is the type of bond cleaved by the chymotrypsin-like activity of proteasome.
3. The amide bond cleaved by the chymotrypsin-like activity of proteasome has the structure –(C=O)–NH–, not –(C=O)–⁺O=CR₂.
4. The substrate for the proteasome is succinyl–LLVY–AMC. By peptide structure interpretation, only the C=O in tyrosine (Y) is being used to make a bond to AMC, and an amide bond is required. This carbon atom is C=O and cannot make an amide bond to the C-terminus of the model tetrapeptide; an amine nitrogen atom is needed.

Solution: The correct answer is D.

1. The value for this option was due to an incorrect conversion of 100 µM to 1 × 10^–7 M.
2. If 200 nM was used instead of 2 nM, this value would be calculated.
3. This is the value if 10 µM was used instead of 100 µM and converted to M.
4. The passage states that the proteasome concentration was 2 nM, and the substrate concentration was 100 µM. The proteasome concentration can be converted to molarity via (2 nmol•L^–1) × (1 mol/1 × 10⁹ nmol) = 2 × 10^–9 mol•L^–1, and the substrate is (100 µmol•L^–1) × (1 mol/1 × 10^–6 µmol) = 1 × 10^–4 mol•L^–1. The ratio of the substrate to the proteasome concentration is (1 × 10^–4 mol•L^–1) / (2 × 10^–9 mol•L^–1) = 5 × 10⁴.

1 nm = 1x10^-9 M

1 uM = 1x10^-6 M

Solution: The correct answer is A.

1. The least polar compound, capable on only London dispersion forces, will elute first because it has the weakest attraction to the polar stationary phase (n-pentane). The second compound to elute will be a slightly polar compound that can accept but not donate hydrogen bonds (2-butanone). The third compound to elute will be a more polar compound that can both accept and donate hydrogen bonds via 1 OH group (n-butanol). The fourth compound to elute will be a compound that can both accept and donate hydrogen bonds via 1 OH group and accept hydrogen bonds to C=O (propanoic acid).
2. n-Butanol and 2-butanone are reversed in their elution order. As 2-butanone is less polar than n-butanol, it should elute before n-butanol.
3. This is the reverse elution order. The order of elution on silica gel is least polar (first) to most polar (last), but this order is most polar (first) to least polar (last).
4. The elution positions of propanoic acid and n-pentane should be switched.

The order of polarity between carboxylic acid, ketone, aldehyde, and alcohol, from highest to lowest polarity, is typically as follows:

1. Carboxylic Acid: The -COOH group in carboxylic acids is highly polar due to both the carbonyl group (C=O) and the hydroxyl group (O-H) capable of forming hydrogen bonds. This gives carboxylic acids their high polarity.
2. Alcohol: Alcohols (-OH group) are also highly polar because of the ability of the hydroxyl group to form hydrogen bonds. However, they are slightly less polar than carboxylic acids because they lack the additional polar carbonyl group.
3. Aldehyde: Aldehydes have a polar carbonyl group (C=O) that makes them polar, but they do not have the ability to form hydrogen bonds like alcohols and carboxylic acids. Their polarity is less than that of alcohols and carboxylic acids.
4. Ketone: Ketones also contain a carbonyl group (C=O) and are polar. However, because they have two alkyl groups attached to the carbonyl, which are less polar than the -OH or -COOH groups, ketones are less polar than aldehydes.

In summary, the order of polarity is:
Carboxylic Acid > Alcohol > Aldehyde > Ketone.

Solution: The correct answer is D.

C. Because the daughter nuclei that result from the radioactive decay of the parent nuclei may or may not be also radioactive, this time might be infinite in case there are no radioactive daughter nuclei. However, the half-life is always a finite amount of time

D. The half-life of a radioactive material is defined as the time it takes for half of all the radioactive nuclei in a sample to decay into their daughter nuclei, which may or may not be radioactive.

Solution: The correct answer is D.

1. The magnitude of the force required to stand up needs not be increased beyond the weight of the person as long as the person stands up gradually.
2. The friction with the ground is used at all times as long as the soles of the feet are in contact with the ground.
3. The energy required to stand up is fixed, being equal to the weight of the person multiplied by the distance over which the center of mass of the person moves upwards.
4. As the person is attempting to stand, the only support comes from the feet on the ground. The person is in equilibrium only when the center of mass is directly above their feet. Otherwise, if the person did not lean forward or slide the feet under the chair, the person would fall backward due to the large torque created by the combination of the weight of the body (applied at the person’s center of mass) and the distance along the horizontal between the center of mass and the support point.

**

When a person sits on a chair, their center of mass is lower and their body is in a stable position. To stand up, they need to shift their center of mass over their feet in a way that maintains balance.

Leaning forward or sliding the feet under the chair helps the person bring their center of mass over their feet, ensuring they stay balanced as they push themselves upward. Without this movement, the person might lose balance and fall backward.

So, leaning forward or adjusting the feet is necessary to maintain equilibrium during the transition from sitting to standing.

Solution: The correct answer is B.

1. Because an aromatic ring absorbs photons in the longer wavelength, near UV region, and ΔE = hc/λ, the absorbed photon energy is lower, not higher.
2. The near UV region where an aromatic ring absorbs electromagnetic radiation is lower in energy than the far UV region where peptide bonds absorb electromagnetic radiation. Thus, the excited state of the electrons in the aromatic ring will be closer to the ground state energy.
3. Although it is true that the absorbed photon energy is higher for a peptide bond, that indicates that the electronic excited state that is farther from the energy of the ground state.
4. Because a peptide bond absorbs photons in the shorter wavelength, far UV region, and ΔE = hc/λ, the absorbed photon energy is higher, not lower.

*

Higher frequencies/shorter wavelengths have more energy. Lower frequencies/longer wavelengths have less energy.

So the part that absorbs light with less energy will be less excited, i.e. closer to the ground state, than the part that absorbs higher energy light.

Solution: The correct answer is C.

**You add up the charges of the basic/acidic amino acids. Only basic and acidic amino acids are either protonated (+ charge) or deprotonated (- charge) at physiological pH.

1. This charge counts negative charges from the C-terminus and the glutamate (E) side chain and ignores the positive charges from the N-terminus and the lysine (K) side chain.
2. This charge ignores the contribution of +1 charge from lysine (K).
3. The sT-loop contains 1 positively charged side chain on K and 1 negatively charged side chain for E, making the net charge zero.
4. This charge ignores the contribution of –1 charge from glutamate (E).

Solution: The correct answer is D.

1. Replacing phosphothreonine with alanine (A) will replace a phosphate side chain with a methyl side chain, which does not have the appropriate size or charge.
2. Replacing phosphothreonine with glutamine (Q) will replace a phosphate side chain with an amide chain, which does not have the appropriate size or charge.
3. Replacing phosphothreonine with glycine (G) will replace a phosphate side chain with a hydrogen atom side chain, which does not have the appropriate size or charge.
4. The phosphorylated threonine (T) would most likely be mimicked by glutamate (E) in terms of size and charge. The carboxylate side chain of glutamate has a negative charge like phosphate and multiple oxygen atoms, also like phosphate.

** All sigmoidal curves suggest that its a cooperative process

Solution: The correct answer is A.

1. Because both curves have a sigmoidal shape, these are indicative of cooperative processes.
2. To NOT be cooperative, the folding curve would have to have a hyperbolic instead of a sigmoidal shape. Its sigmoidal shape makes it cooperative.
3. To NOT be cooperative, the activation curve would have to have a hyperbolic instead of a sigmoidal shape. Its sigmoidal shape makes it cooperative.
4. To NOT be cooperative, both the activation and folding curves would require a hyperbolic instead of a sigmoidal shape. Their sigmoidal shapes make them cooperative.

Solution: The correct answer is A.

1. A higher melting temperature is indicative of a more stable protein, as more energy is needed to unfold the protein. L203A has an approximate T _m of 50°C; therefore, it is the most stable and even more stable than the WT protein.
2. With a T _m of approximately 42°C, I204A is not the most stable and less stable than the WT protein.
3. With a T _m of approximately 34°C, Y229A is tied for the least stable variant with R241A.
4. With a T _m of approximately 34°C, R241A is tied for the least stable variant with Y229A.

Solution: The correct answer is A.

1. Based on the cAMP activation information given in the note in Figure 2, changing Y229 to A229 has no effect on kinetics. However, based on the data in Figure 3, Y229A is less stable than the WT protein, indicating that Y229 is essential for protein stability.
2. Based on the data in figures 2 and 3, the opposite is true. The Y229A variant has lower protein stability, but the same cAMP activation.
3. Although the data in Figure 3 confirms that Y229 is important for protein stability (Y22A has a much lower T _m than WT protein), the note in Figure 2 states that Y229A has the same cAMP activation as WT protein.
4. Although the note in Figure 2 confirms that replacing Y229 with alanine has no effect on cAMP activation, the data in Figure 3 shows that Y229A is much less stable than WT protein.

Solution: The correct answer is B.

1. The Doppler effect describes the relationship between the frequency of a wave emitted by a source and the frequency of the same wave as it is perceived by an observer due to the relative motion between the source and the observer.
2. The oxygen pressure is the sum of the oxygen static pressure P and the oxygen flow pressure ½ ρv ². In the area of the mask openings, P _air = P + ½ ρv ², thus P _air > P. Air enters the mask because the static pressure of the air is larger than the static pressure of the oxygen that flows in the mask. This is the Venturi effect, and the mask is called the Venturi mask.
3. Diffusion is caused by a difference in concentration, whereas the concentration of oxygen in the tank is the same as in the air; otherwise, the breathing process could be affected.
4. Dispersion is an optical effect that does not apply to the flow of gases.

Solution: The correct answer is C.

1. There is the same 50% pyrimidine percentage in AT as GC base pairs, so there is no binding advantage for pyrimidines between these base pairs.
2. There is the same 50% purine percentage in AT as GC base pairs, so there is no binding advantage from purines between these base pairs.
3. Duplex DNA with a lower (A + T):(G + C) ratio melts at a higher temperature than comparable length duplex DNA with a greater (A + T):(G + C) ratio because GC base pairs create stronger pi-stacking interactions in the duplex than AT base pairs.
4. Both AT and GC base pairs have the same number of phosphate groups.

Solution: The correct answer is D.

1. The lone pair of electrons in ammonia allows the molecule to adopt a trigonal pyramidal molecular geometry, not a planar structure based on VSEPR (Valence Shell Electron Pair Repulsion) theory.
2. Oxidizing agents accept electrons in redox reactions. Nitrogen cannot expand its octet and ammonia's lone pair of electrons typically form a covalent bond, preventing the central nitrogen from accepting more electrons.
3. The lone pair of electrons in ammonia does not act as a Lewis acid, an electron pair acceptor, in water. Ammonia’s lone pair of electrons generally form a bond with the central nitrogen atom, making it a Lewis base.
4. A Lewis base is a chemical species that donates a pair of electrons, which ammonia does to form a covalent bond with a hydrogen atom.

Solution: The correct answer is A.

1. The passage states that the SCY conductors are favored because their proton conductivities substantially increase with temperature. The last paragraph states that the “decomposition of ammonia, is insignificant below 300°C but increases dramatically thereafter.” This indicates that the proton conductivity of SCY increases with increasing temperature, while the favorability of reaction decreases with overall temperature. It is beneficial to maintain the SCY conductor at a higher temperature than the electrode temperature.
2. While the SCY can generate protons at a lower temperature, it will not produce the best results as the proton conductivities substantially increase with increasing temperatures.
3. The SCY conductor can be maintained at the same temperature as the electrode temperature, but it will not produce the best results as higher temperatures will increase the proton conductivities.
4. The passage states that higher temperatures will increase the proton conductivities with the SCY conductor, and the nitrogen/ammonia electrode will see the decomposition of ammonia at temperatures above 300°C.

< 0 because ΔG° = –RT(lnK).

Solution: The correct answer is A.

1. The passage notes that the formation of [Cu(NH₃)₄] ²⁺ takes place stepwise, with an ammonia molecule replacing one water molecule in each step. The second step would have an additional water molecule replaced by an ammonia molecule to make [Cu(H₂O)₂(NH₃)₂] ²⁺ through the following steps: Step 1) [Cu(H₂O)₄] ²⁺(aq) + NH₃(aq) ⇌ [Cu(H₂O)₃(NH₃)] ²⁺(aq) + H₂O(l) Step 2) [Cu(H₂O)₃(NH₃)] ²⁺(aq) + NH₃(aq) ⇌ [Cu(H₂O)₂(NH₃)₂] ²⁺(aq) + H₂O(l)
2. This is not an intermediate in forming the [Cu(NH₃)₄] ²⁺ complex because there are five ligands around copper, not four.
3. This is the product that forms after the third step in the reaction. Step 3) [Cu(H₂O)₂(NH₃)₂] ²⁺(aq) + NH₃(aq) ⇌ [Cu(H₂O)(NH₃)₃] ²⁺(aq) + H₂O(l)
4. This complex is not an intermediate in the formation of [Cu(NH₃)₄] ²⁺(aq) as water molecules are replaced by ammonia molecules in a stepwise manner. This option indicates that the ammonia molecule adds to the copper complex, and the water molecule is not displaced in the same step.

one donor atom while the other atom acts as an acceptor to receive the electron pair.

Solution: The correct answer is B.

1. If hydrochloric acid is added to the reaction, ammonia will be protonated and converted to ammonium, thus decreasing the ammonia concentration. This will cause the equilibrium to shift to the left to accommodate the disruption and make less [Cu(NH₃)₄] ²⁺, not more.
2. According to Le Chȃtelier’s principle, this disruption to the equilibrium causes the reaction to shift in response to produce more ammonia. The shift will cause the equilibrium to shift to the left, decreasing the amount of [Cu(NH₃)₄] ²⁺ formed.
3. The acid would react with the ammonia in the solution, and the equilibrium would shift to the left to increase the ammonia concentration, thus decreasing the amount of [Cu(NH₃)₄] ²⁺ formed.
4. Ammonia will be converted to ammonium and will be removed from the solution. Equilibrium would shift to the left, decreasing the amount of [Cu(NH₃)₄] ²⁺ formed.

Solution: The correct answer is B.

1. That value is the oxidation number of Cu, not the number of ligands bound to the copper.
2. The subscript 4 indicates the number of neutral ammonia ligands directly bonded to the central Cu²⁺ cation.
3. The subscript 4 does not represent both oxidation and coordination numbers of the copper ion. The oxidation number is determined by the overall charge of the complex, which is +2 in this case, while the coordination number is 4 due to the presence of neutral ammonia molecules in the complex.
4. Although the 4 does not correspond to the oxidation number of copper, which is +2, it does correspond to the number of ammonia ligands bound to the copper.

Solution: The correct answer is C.

1. Although NH₃ is a stronger Lewis base than H₂O, Option I is incorrect because H₂O has two lone pairs of electrons, whereas NH₃ has one pair of electrons.
2. Although NH₃ donates a lone pair of electrons more readily than does H₂O, Option I is incorrect because H₂O has two lone pairs of electrons, whereas NH₃ has one pair of electrons.
3. NH₃ is a stronger Lewis base than H₂O because the nitrogen atom is less electronegative than the oxygen atom. This results in the oxygen atom having a stronger attraction for its lone pairs of electrons, making them less available for donation and H₂O less Lewis basic than NH₃.
4. Only options II and III are correct. Option I is incorrect as NH₃ has fewer lone pairs of electrons than H₂O.

Solution: The correct answer is B.

1. The value for this option only occurs when the ratio of [base] /[acid] is 1:100.
2. To find the pH of a solution, use the Henderson-Hasselbalch equation: pH = pKa + log([base] /[acid] ). The pH = 6.37 + log(0.2/2) = 5.37 because the log(0.1) is -1.
3. The value for this option is obtained if the ratio of [base] /[acid] is incorrectly set to 1 and is used in the Henderson-Hasselbalch equation. Since the log(1) is equal to 0, that makes the pH of the solution equal to the pK _a.
4. The value for this option is obtained if the [base] /[acid] ratio is inverted to [acid] /[base] and used in the Henderson-Hasselbalch equation. This would make the pH = 6.37 + log(2/0.2) = 7.37 since the log (10) is 1.

The # after the exponent, as long as the front is 1x, is what the log is equal to...

log(1x10^-6) = -6

log(1x10^8) = 8

Solution: The correct answer is C.

1. Based on this value, the K _sp value would be 5.8 × 10^–8 instead of 4.9 × 10^–9.
2. Using this concentration, the K _sp value of CaCO₃ would be 2.4 × 10^–9 instead of 4.9 × 10^–9.
3. The dissolution of CaCO₃ is CaCO₃(s) ⇌ Ca²⁺(aq) + CO₃ ^2–(aq) and the solubility product constant expression is K _sp = [Ca²⁺] [CO₃ ^2–] . Equal amounts of Ca²⁺ and CO₃ ^2– are produced when CaCO₃ dissolves in solution, so the expression reduces to x² = 4.9 × 10^–9, which is equal to 49 × 10^–10. Taking the square root, x = 7.0 × 10^–5 M.
4. This value would indicate that CaCO₃ has a K _sp value of 2.4 × 10^–17 _, not 4.9 × 10^–9.

Solution: The correct answer is A.

1. The passage provides the average bond energies in kJ/mol of the C–H, C–O, C–C, and O–N single bonds. The strength of a bond is directly proportional to its bond energy, with shorter bonds being stronger than longer bonds. Based on the data provided in the passage, the C–H bond has the largest bond energy and, thus, the shortest bond length.
2. The C–O bond has the second highest bond energy, which makes it the second shortest bond length.
3. The C–C bond has the second lowest bond energy, which indicates that it is the second longest bond in the series.
4. The O–N bond length has the lowest bond energy indicating that it is the longest bond of the series.

Solution: The correct answer is D.

1. This ΔH°_f value is obtained by adding the ΔH°_f C₃H₅N₃O₉ and ΔH°_f CO₂ together and subtracting ΔH°_f H₂O, without multiplying each value by the coefficients in the balanced equation: [(–364.0) + (–393.5)] – (–241.8) = –515.7 kJ, which is approximately –500 kJ.
2. If the ΔH°_f for C₃H₅N₃O₉(l), CO₂(g) and H₂O(g) are added together instead of using Hess’s law, this value is obtained via: (–364.0 + –393.5 + –241.8) = –998 kJ, which is approximately –1000 kJ.
3. If incorrect coefficients are used when calculating ΔH°, a value near this option is obtained via: [6(–393.5) + 5(–241.8)] – [4(–364)] = –2107 kJ, which is approximately –2000 kJ.
4. The value of ΔH° can be calculated using the data provided in Table 1 using Hess’s Law, (the sum of ΔH°_f products) – (the sum of ΔH°_f reactants). The amount of nitroglycerin (in kJ) decomposed is calculated by: [12(–393.5) + 10(–241.8)] – [4(–364.0)] = –5684 kJ/4 moles; therefore, the value of 2 moles of nitroglycerin is –2842 kJ, which is approximately –3000 kJ.

Solution: The correct answer is D.

1. This calculated volume value is the result of using 1/6 of the correct number of moles to calculate the overall volume of N₂(g).
2. This is the value if 1/3 of the number of moles was used to calculate the total volume of N₂(g).
3. This value is obtained if the mole ratio is inverted to 2/3 instead of 3/2 when determining the correct amount of N₂(g).
4. According to the balanced reaction given in the passage, for every 4 moles of nitroglycerin consumed, 6 moles of N₂(g) are produced. Therefore, for every 1 mole of nitroglycerin, 1.5 moles of N₂(g) are produced. At STP, the molar volume of an ideal gas is 22.4 L•mol^–1, which means that 1.5 moles of N₂ will occupy 33.6 L.

Solution: The correct answer is C.

1. The C–H bond has a slight electronegativity difference between the two atoms, but it is not the least polar bond in nitroglycerin.
2. The O–C bond is the most polar bond because of the greatest difference in electronegativities between the two atoms.
3. The C–C bond is considered the least polar as there is no difference in electronegativity between the identical atoms.
4. The O–N is a polar bond because the difference in electronegativities is greater than not zero.

Solution: The correct answer is D.

1. This result is consistent with using a weight of 60 N instead of 540 N for the woman.
2. This result is consistent with climbing only 6 steps.
3. This is the power consumed by the person climbing 27 steps of height 7 cm.
4. The power consumed is P = ΔPE /time = mgh/Δt. From Table 1, there are 30 steps and Δt = 27 s. Then P = (54 kg × 10 m/s² × 30 steps × 0.15 m/step) / (27 s) = 90 W.

Solution: The correct answer is A.

1. The work done is defined as W = 0.5kx ², where x = 0.20 m for an 83-year-old female. Solving for W yields W = 0.5 ⋅ 200 (N/m) ⋅ (0.2 m)² = 4.0 J.
2. This implies the elastic constant is 160 N/m instead of 200 N/m.
3. This is the work done by a female in the 76−80 age group.
4. This is the work done by a female in the 71−75 age group.

Solution: The correct answer is D.

1. This is the algebraic difference between the minimum sound intensities heard by a 64-year-old male and a 74-year-old female.
2. If the ratio of the intensities is 40, then the decimal log of the ratio of their intensities is x = log 40 = 1.6. Consequently, the difference between the relative intensities of the two sound waves would be 1.6 B = 16 dB. This represents the approximate difference between the minimum sound intensities heard by a 67-year-old male and a 72-year-old female.
3. If the ratio of the intensities is 50, then the decimal log of the ratio of their intensities is x = log 50 = 1.7. Consequently, the difference between the relative intensities of the two sound waves would be 1.7 B = 17 dB. This represents the approximate difference between the minimum sound intensities heard by a 74-year-old male and a 77-year-old female.
4. The relative intensities of the two sound waves are 20 dB and 40 dB, respectively. Their difference is 20 dB, meaning that the decimal log of the ratio of their intensities is 2, which means that the ratio of their intensities is 10² = 100.

Here's how I did it:

Bf = Bi + 10 log (If / Ii)

The B will always be decibels, and the I will always be intensity. So it would be:

40 dB = 20 dB + 10 log (If / Ii)

---- subtract 20 -------

20 dB = 10 log (If / Ii)

------ divide by 10 -------

2 = log (If / Ii)

------ take 10 to the power on both sides so log gets cancelled out --------

10^2 = If / Ii

1. This implies the lenses have a positive focal length and the object is located very far away from the lenses, whereas the lenses have a negative focal length.
2. This implies the lenses have a positive focal length and the object is located close to the focal point, whereas the lenses have a negative focal length.
3. This implies the lenses have a positive focal length and the object is located between the focal point and the lens, whereas the lenses have a negative focal length.
4. The lenses have a negative focal length, which means they are diverging lenses. Such lenses form virtual and reduced images of objects situated at distances larger than the focal length.

real, reduced

virtual, reduced

• Nearsightedness (Myopia): Can see near objects clearly, distant objects are blurry. Corrected with concave lenses.
• Farsightedness (Hyperopia): Can see far objects clearly, near objects are blurry. Corrected with convex lenses.

Solution: The correct answer is C.

1. Given that 5% = 0.05 = 1/20 = 1/(2 ^x), solving for x yields ​​. This means the time it takes the ¹⁰C sample to decay is equal to 4.3 half-lives, that is 4.3 × 20 s = 86 s. The 75-year-old male takes 200 s to walk 5 laps, which is on average 40 s per lap. Therefore, he should walk about (86/40) × 1 lap = 2.1 laps until 5% of the sample is left.
2. Because 10% = 1/10 = 1/(2 ^x), solving for x yields ​​. This means the time it takes the ¹⁰C sample to decay is equal to 3.3 half-lives, that is 3.3 × 20 s = 66 s. The 75-year-old male takes 200 s to walk 5 laps, which is on average 40 s per lap. Therefore, he should walk about (66/40) × 1 lap = 1.6 laps until 10% of the sample is left.
3. The 75-year-old person takes 200 s to walk 5 laps, which is on average 40 s per lap. This time represents two half-lives of ¹⁰C, so the ratio of ¹⁰C sample left is 1/(2²) = 0.25, or 25%.
4. Using 75% = 0.75 = 1/1.33 = 1/(2 ^x), solving for x yields ​​. This means the time it takes the ¹⁰C sample to decay is equal to 0.4 half-lives, that is 0.4 × 20 s = 8 s. The 75-year-old male takes 200 s to walk 5 laps, which is on average 40 s per lap. Therefore, he should walk about (8/40) × 1 lap = one-fifth of a lap until 75% of the sample is left.

Solution: The correct answer is D.

1. A battery uses chemical energy to set the electric charges in motion; therefore, electric energy is not the source for chemical energy.
2. If chemical energy was converted to thermal energy, then only a small fraction could have been reconverted into electrical energy and this would require another specialized device besides the battery itself.
3. The primary energy of a battery is chemical, not electrical, because electrical energy requires that electric charges be kept separated in order to create an electrostatic field.
4. The chemical energy of the battery elements is used as electrical energy to set the charge carriers in motion through the resistor, where they experience drag from the crystal lattice of the resistive conductor and dissipate their energy as heat from the resistor.

Solution: The correct answer is A.

1. Secondary structure includes turns, helices, and beta sheets, all of which are established by hydrogen bonds between backbone amide protons and carbonyl oxygens.
2. These interactions may exist for tertiary structure, but they are not characteristic of secondary structure.
3. Although side chain hydroxyl groups can hydrogen bond to backbone carbonyl oxygens, this is not characteristic of secondary structure.
4. Side chain amide protons and backbone carbonyl oxygens may hydrogen bond in tertiary structure, but this is not characteristic of secondary structure.

1. During adulthood, adipocytes are unlikely to undergo mitosis. While there is a limited replicative capacity of adipocytes under certain conditions, adipocytes typically respond to changing conditions by altering their size.
2. Early in development, cardiac muscle cells are proliferative. However, as an organism reaches adulthood, mitotic capacity of such cells is decreased.
3. Gastrointestinal epithelial cells exhibit mitotic activity throughout adulthood. Specifically, this epithelium possesses stem cells that continuously proliferate. As such, gastrointestinal epithelial cells may have been used by the researchers to assess proliferative activity.
4. Neurons are post-mitotic cells that do not typically undergo cell division in adult mammals.

Solution: The correct answer is C.

1. Osmosis refers to the movement of water, not ions such as chlorine, from a low solute to a high solute solution.
2. Diffusion refers to the movement of a solute from an area of high concentration to an area of low concentration. However, when chlorine moves from pond water into the algae cell, it is moving from an area of low concentration to an area of high concentration.
3. Active transport mechanisms use energy in the form of ATP to transport solutes such as chlorine from an area of low concentration to an area of high concentration. In this instance, chlorine is being transported from the pond water, where it has a low concentration, to the cytoplasm of an algae cell, where it has a high concentration.
4. Facilitated diffusion utilizes transport proteins to promote the diffusion of a solute like chlorine from an area of high concentration to an area of low concentration. However, when moving from the pond water into an algae cell, chlorine is moving from an area of low concentration to an area of high concentration.

prolongation of the action potential.

Solution: The correct answer is B.

1. The passage does not contain any information about the quaternary structure of Wnt proteins.
2. The passage notes that Wnt proteins have an isoelectric point of 9 and therefore they are charged positively at physiological pH range.
3. Secretory proteins such as Wnt are synthesized on the rough endoplasmic reticulum, not the smooth endoplasmic reticulum.
4. As secretory proteins, Wnt proteins are folded in the rough endoplasmic reticulum, not the cytoplasm.

Solution: The correct answer is C.

1. The passage does not provide insight on the structure of β-catenin, such as its subunit composition.
2. The passage indicates that β-catenin can be phosphorylated. Phosphorylation, however, is not related to the number of disulfide bonds in a protein. Therefore, based on the information in the passage, one cannot conclude whether β-catenin has a high or low number of disulfide bonds.
3. The passage indicates that β-catenin can activate expression of Wnt target genes. To do this, Wnt target genes need to be transcribed. The figure also shows that β-catenin activates transcription factors for Wnt target genes. This suggests that β-catenin acts in the nucleus. Typically, proteins that act in the nucleus contain a nuclear localization sequence, which is one possible mechanism by which proteins can enter the nucleus. Thus, according to the passage, β-catenin could contain a nuclear localization sequence.
4. The passage indicates that β-catenin can be both phosphorylated, which primarily occurs at serine, threonine, and/or tyrosine residues, and ubiquitinated, which primarily occurs at lysine residues. To be modified, these residues must be surface exposed. However, all these residues are polar, not nonpolar.

Solution: The correct answer is A.

1. In the absence of Frizzled activation, β-catenin is phosphorylated and ubiquitinated. Ubiquitination marks proteins for degradation by a proteasome.
2. The Golgi apparatus is responsible for the packaging and transport of proteins to a location where they will be active. However, in the absence of Frizzled activation, β-catenin is ubiquitinated, meaning that it will be degraded.
3. The passage indicates that β-catenin is ubiquitinated. Thus, it will be degraded in the proteasome, not the lysosome.
4. β-catenin is ubiquitinated in the absence of Frizzled activation. Ubiquitinated proteins are actively degraded, not stored for later use.

hydrolyze amino acids.

Solution: The correct answer is C.

1. Protein X can form a channel across the mitochondrial inner membrane, but this will not directly interfere with the functioning of the citric acid cycle, as the citric acid cycle occurs within the mitochondrial matrix.
2. Formation of a channel by Protein X across the mitochondrial inner membrane will not directly affect the functioning of the electron transport chain, as the electron transport chain can bypass the channel.
3. The gradual release of dye from liposome when Protein X is added indicates that the protein can form a channel, thus making the mitochondrial inner membrane permeable, which causes the dissipation proton gradient across the membrane.
4. Insertion of a channel across the mitochondrial inner membrane will most likely increase, not decrease, the pH of intermembrane space by the dissipation of the proton gradient across the membrane.

1. Histone acetylation is a post-translational modification and therefore it can be detected by western blots.
2. Southern blots are not an appropriate method to study post-translational modifications such as protein acetylation. They are primarily utilized to study DNA molecules.
3. Northern blots are used to study RNAs, not protein modifications.
4. RT-PCR is a method to analyze changes in protein expression at transcriptional levels and are not appropriate for studies of protein modification.

Prions;

Prions, however, lack nucleic acid such as DNA or RNA.

endosome

Proteases

Solution: The correct answer is A.

1. Compared to sGP, EGP contains an inserted nucleotide, which will produce a frameshift mutation. Frameshift mutations often alter the codon in which they were introduced, along with downstream codons. This will retain the same amino acid sequence from the amino terminus up to the point of the mutation. By contrast, the amino acid sequence from the point of the mutation through the carboxy-terminus will be altered, although the stop codon is maintained.
2. EGP contains an inserted nucleotide in its open reading frame. This will alter both the codon in which the nucleotide was inserted, as well as all downstream codons. However, downstream codons encode the carboxy-terminal sequence of the protein. Therefore, it is unlikely that EGP and sGP will possess the same carboxy-terminal sequences.
3. EGP contains an inserted nucleotide in its open reading frame. However, an inserted nucleotide is not associated with an additional amino acid, but instead produces a frameshift mutation. In a frameshift mutation, not only is the codon in which the nucleotide has been inserted altered, but all downstream codons are also altered.
4. Compared to sGP, EGP contains an inserted nucleotide. This insertion will produce a frameshift mutation, in which sGP and EGP will likely have an altered amino acid in both the codon in which the nucleotide was added, as well as all downstream codons. Removal of a single amino acid is not likely to be caused by this change.

Solution: The correct answer is B.

1. Regions 1 and 2 correspond to the variable and constant regions of the light chain of the antibody, respectively. However, only variable regions can target specific antigens. Instead, region 3 represents the variable region of the heavy chain, which can also be used to specifically target the HER2 antigen.
2. Regions 1 and 3 correspond to the variable regions of the antibody light and heavy chains, respectively. The variable regions of an antibody are responsible for detecting specific antigens, such as HER2.
3. Regions 3 and 4 correspond to the variable and constant regions of the antibody heavy chain. While the variable region of the heavy chain is involved in the recognition of specific antigens, such as HER2, the constant region is not. Instead, the variable region of the light chain is also involved in the detection of specific antigens.
4. Region 4 represents the constant region of the antibody heavy chain. However, constant regions of an antibody are not responsible for the detection of specific antigens.

Solution: The correct answer is D.

1. During the cross-bridge cycle, ADP is bound to myosin, not actin.
2. ADP is released from the myosin head during the “power stroke” of the cross-bridge cycle. This is when the myosin head bends, pulling on an actin filament. However, this occurs prior to ATP binding to myosin, and will not be directly affected by lack of ATP.
3. During the cross-bridge cycle, ATP binds to myosin, not actin. When this occurs, this allows for myosin to detach from actin, not the other way around.
4. During the cross-bridge cycle, ATP is needed to allow myosin detachment from an actin filament. Consequently, when ATP is unavailable, the myosin head remains attached to an actin filament. This maintains the muscle in a contracted state.

Solution: The correct answer is A.

1. The passage states, “[The baboons’] task was to tap on the option that matched the target color. When a correct response was produced, the subject received a banana chip.” A fixed ratio schedule of reinforcement provides the reinforcer after a consistent number of target behaviors. In the color-matching task, baboons received the reinforcer (the banana chip) after each target behavior (providing the correct response).
2. A variable ratio schedule of reinforcement provides the reinforcer after a varying number of target responses, which varies around a predetermined average. This is not the schedule of reinforcement described by the passage.
3. A fixed interval schedule of reinforcement provides the reinforcer when the target behavior is performed for the first time after a consistent interval of time has elapsed. This is not the schedule of reinforcement described by the passage.
4. A variable interval schedule of reinforcement provides the reinforcer when the target behavior is performed for the first time after a varying interval of time has elapsed. This is not the schedule of reinforcement described by the passage.

Yes, reinforcement schedules refer to when the reinforcer is provided following a specific behavior. These schedules dictate how and when reinforcement (the reward) is given, which affects how the behavior is learned and maintained. There are two main types of reinforcement schedules: continuous reinforcement and partial (or intermittent) reinforcement. Each of these has different subtypes based on the timing and frequency of reinforcement.

1. Continuous Reinforcement (CRF):

• When it's provided: A reinforcer is given every time the behavior occurs.
• Effect on behavior: This is typically used when first teaching a behavior because it leads to rapid learning. However, it can be less effective in maintaining the behavior over time once the reinforcement stops.

2. Partial (Intermittent) Reinforcement:

This type involves reinforcement being given only some of the time the behavior is performed. There are four main types based on timing and frequency:

a. Fixed-Ratio (FR) Schedule:

• When it's provided: Reinforcement is given after a set number of responses.
• Example: A worker is paid after completing 10 units of work. The reinforcer (payment) comes after a fixed number of responses (10 units).
• Effect on behavior: Produces a high rate of response, but there may be pauses after reinforcement (called a "post-reinforcement pause").

b. Variable-Ratio (VR) Schedule:

• When it's provided: Reinforcement is given after a variable number of responses, on average.
• Example: Slot machines in casinos. You don't know exactly when the next win will come, but over time, it's around every X number of plays.
• Effect on behavior: Leads to a high and steady rate of responding, because the person can’t predict when the reinforcement will occur.

c. Fixed-Interval (FI) Schedule:

• When it's provided: Reinforcement is given after a set amount of time has passed, and the behavior occurs.
• Example: A person receives a paycheck every two weeks. The reinforcer (paycheck) is delivered after a fixed interval of time, regardless of the number of responses in that time.
• Effect on behavior: Leads to a pattern of behavior where responses increase as the time for reinforcement approaches (called the "scalloping effect").

d. Variable-Interval (VI) Schedule:

• When it's provided: Reinforcement is given after a variable amount of time has passed.
• Example: Checking for emails, where you don't know exactly when you'll receive one, but on average, emails come at random intervals.
• Effect on behavior: Produces a moderate but steady rate of responding. It's harder to predict when the next reinforcement will happen.

Summary of Reinforcement Schedules:

• Continuous: Reinforcer is given every time the behavior occurs.
• Partial: Reinforcer is given only some of the time, and it can be based on the number of responses or the amount of time passed. The subtypes are:
  • Fixed-Ratio (FR): Reinforcer after a fixed number of responses.
  • Variable-Ratio (VR): Reinforcer after a variable number of responses.
  • Fixed-Interval (FI): Reinforcer after a fixed amount of time.
  • Variable-Interval (VI): Reinforcer after a variable amount of time.

Each of these schedules affects how quickly the behavior is learned, how resistant the behavior is to extinction, and the rate at which the behavior is performed. Let me know if you'd like more details or examples!

Solution: The correct answer is B.

1. Immigration in the United States was not a major factor in the increase of the population because the share of the immigrants in the population declined significantly in the post-WWII era until the 1970s.
2. Demographers define the baby boom generation as those individuals born between approximately 1946 and 1964. The projected increase in the share of the population over the age of 65 primarily stems from the baby boomers, the post-World War II generation in the United States and Canada. Birth rates were relatively high for almost two decades after World War II because of the fast economic growth experienced in the society. Therefore, the baby boom generation is the main sociohistorical factor that explains the projection in the passage about the increasing share of the population over 65 years of age.
3. The sexual revolution of the 1960s and 1970s was a cultural movement that challenged the traditional gender roles and understanding of sexuality. This movement is connected to feminist and LGBTQ movements and is not associated with an increase in the population of the baby boom generation.
4. The Great Depression experienced in the 1930s led to a significant decline in living standards and is not associated with the population boom experienced in the post-WWII era.

Conflict theory

difference in health outcomes by social status

early life events influence health outcomes in later life.

social backgrounds (such as race/ethnicity, gender, and class) produce differential outcomes for an individual.

1. Conversion disorder is characterized by impairments to voluntary motor or sensory function which are not due to a recognized neurological or medical condition. The symptoms described do not support a conversion disorder diagnosis.

1. Dissociative amnesia is a dissociative disorder where individuals cannot recall important autobiographical information, usually related to a trauma or stressor. This is most consistent to the symptoms described.

rewarded or given opportunities based on their abilities, skills, and achievements rather than their background, wealth, or social status

Solution: The correct answer is A.

1. According to the passage, the United States tends to have less social mobility than its peer countries. This empirical finding from comparative research had led some scholars to question assumptions about opportunity in American society. Such questioning of opportunity would be relevant to the concept of meritocracy, which assumes that opportunity is based on a combination of talent and effort.
2. Socialization refers to the learning of socially acceptable values and behaviors. The passage discussion does not refer to socialization.
3. Social identity refers to how someone defines themselves based on their social background. Social identity is not discussed in the passage in regard to social mobility.
4. Cultural capital refers to how someone’s cultural background might provide certain advantages to individuals to distinguish themselves from the rest of the society. Cultural capital is not discussed in the passage in regard to social mobility.

Solution: The correct answer is C.

1. Absolute poverty is an economic condition in which individuals cannot meet their basic needs. Absolute poverty is not discussed in the passage.
2. Marginal poverty stems from unstable employment conditions for an individual in which they cannot achieve minimum standards of living. Employment conditions are not discussed in the passage.
3. Relative poverty refers to social disadvantage by income or wealth as compared to the social advantages linked to income or wealth in a society. The passage discussion about social structural factors pertains to the median income in the United States and high levels of inequality, which is most consistent with the phenomenon of relative poverty.
4. Structural poverty is related to a lack of economic opportunities for individuals to leave poverty. A structural lack of economic opportunities is not discussed in the passage.

** Assimilated (adjusted) vs immigrant groups are two different people....

Solution: The correct answer is C.

1. The passage suggests that more assimilated groups would have less social support; however, lower levels of support are associated with worse health outcomes.
2. The passage suggests that more assimilated groups would have worse health outcomes, but this would not be due to a higher level of social support.
3. The passage suggests that strong social support in local immigrant communities may partly explain the relatively good health of individuals from some immigrant groups in the United States (when compared to U.S.- born individuals with otherwise similar demographic characteristics). Immigrant groups that are more assimilated (or as they become assimilated) tend to have worse health outcomes (or lose their previous health advantages) than less assimilated immigrant groups. Thus, the correct response reflects the hypothesis that more assimilated groups are likely to have less social support over time. Reducing that protective factor could lead to worse health overall.
4. The passage suggests that more assimilated groups would not have a higher level of social support.

Solution: The correct answer is D.

1. The passage states, “Although psychoactive drugs vary in terms of their risk of dependence, one of the factors associated with substance use disorders (SUD) in adolescents is the strong desire to ingest a drug (or other substance).” Habituation refers to reduced responsiveness to a repeating stimulus, which is not relevant to the symptom described by the passage.
2. Tolerance refers to when increasing amounts of a substance is required to receive the desired effect from it. This is not relevant to the symptom described by the passage.
3. Withdrawal refers to the symptoms experienced when a person stops using or reduces the dose of a substance. This is not relevant to the symptom described by the passage.
4. Craving refers to a strong desire to ingest a substance. Craving is the symptom described in the passage.

Solution: The correct answer is B.

1. Stimulants, such as cocaine and amphetamines, have a relatively high risk of dependence (defined as an ever-increasing tolerance for the substance and the experience of withdrawal when substance use is reduced or ceased).
2. Relative to the other types of psychoactive drugs, hallucinogens have the lowest risk of dependence.
3. Alcohol, which functions as a sedative on the central nervous system, has a relatively high risk of dependence.
4. Given their impact on the central nervous system, sedatives have a relatively high risk of dependence.

Solution: The correct answer is A.

1. Conflict theory focuses on social inequalities. Stratification indicates that certain communities experience unequal access to resources and opportunities in a society. Therefore, disparities in access to SUD treatment among different social groups as discussed in the passage would be most consistent with a conflict perspective.
2. Racialization refers to the identification of certain behaviors with a racial category regardless of the consent of the members of that community. Racialization is not relevant to disparities in access to SUD treatments.
3. Socialization refers to the learning of social values and expected behaviors. This factor would not be relevant to disparities in access to SUD treatments.
4. Gentrification refers to the displacement of low-income residents in a neighborhood due to increasing property value. This phenomenon would not be directly relevant to disparities in access to SUD treatments.

Solution: The correct answer is D.

1. A judgement based on incorrect assumptions would be best described as bias. However, discrimination has to be based on differential treatment.
2. The physician’s bias would not directly indicate discriminatory behavior; therefore, this explanation would not be relevant to the behavioral aspect of discrimination.
3. A negative evaluation could affect the interaction, but such an interaction is not discussed in the question prompt.
4. The hypothetical physician displays a biased attitude, and thus a prejudice potentially based on a stereotype. However, no action or behavior is specifically identified with the scenario in the question. Without a description of differential treatment or behavior, discrimination is not identified.

Solution: The correct answer is C.

1. Social loafing refers to the fact that people are more productive alone than in a group. Research also suggests that individuals are less critical and less creative in groups. Thus, research indicates participants in groups will be less likely to arrive at the improved alternatives than participants working alone.
2. Groupthink occurs when members of a group prioritize agreement over critical thinking when coming to a decision. Due to groupthink, participants in groups are less likely to critically evaluate alternatives than those working alone.
3. Research on social loafing and related phenomenon support that participants working alone will generate more alternatives than those working in groups.
4. Given social loafing, participants in groups are likely to produce fewer alternatives than participants working alone.

situational pressures hinder groups from critically evaluating relevant information.

Solution: The correct answer is B.

1. Self-serving bias refers to the tendency to attribute one’s own successes to internal, stable attributes and failures to situational factors. This is not relevant to groupthink.
2. Confirmation bias is the tendency to put more weight on information that confirms one’s pre-existing attitudes. As with a group affected by groupthink, an individual’s confirmation bias causes the person to seek, and attend to, only information that confirms their existing point of view and to ignore disconfirming evidence.
3. Hindsight bias is the tendency to overestimate one’s ability to predict an outcome after it has already occurred. This is not relevant to groupthink.
4. Response bias is the tendency for research participants to respond inaccurately or falsely to self-report questions. This is not relevant to groupthink.

Solution: The correct answer is D.

1. Analytical intelligence is the ability to succeed at problem-solving tasks, such as using deductive reasoning. It is not directly relevant to the ability to delay gratification.
2. Creative intelligence is the ability to generate multiple, novel solutions to a problem. It is not directly relevant to the ability to delay gratification.
3. Interpersonal intelligence is the ability to understand other people’s emotions, connect with others, and manage relationships. It is not directly relevant to the ability to delay gratification.
4. Emotional intelligence refers to the ability to perceive, express, understand, and manage one’s emotions. Emotionally intelligent people are self-aware and can delay gratification in pursuit of long-term rewards, rather than being overtaken by immediate impulses.

Solution: The correct answer is D.

1. This option offers an explanation at an individual level which is not consistent with a functionalist approach.
2. This option refers to social inequalities. Functionalism does not focus on social inequalities. focuses on conflict theory
3. This option describes dynamics at an individual level which is not consistent with a functionalist perspective.
4. Functionalism makes a distinction between manifest, or intended, and latent, or unintended, functions of social practices which sustain social stability. The option describes a latent, or unintended, function. Because the expected function of yoga is not specifically smoking cessation, its utility as a cessation therapy is a latent function of the social activity.

how people create and interpret meaning through their interactions with others.

Sure! Here are a few examples of symbolic interactionism in action:

1. Language and Communication:
   • When we greet someone with a "hello" or shake hands, those actions hold symbolic meaning. In one culture, a handshake might represent respect, while in another, it could be a formal greeting. The meaning we attach to these actions depends on the social context and our shared understanding.
2. Self-Identity:
   • If someone is constantly praised for being a great student, they might start seeing themselves as "smart" or "capable." Their sense of identity is shaped by the way others view and interact with them. Over time, this influences how they see themselves and behave in future situations.
3. Social Roles:
   • Think of the roles of "parent" and "child." A parent might provide care and discipline, while a child might be expected to obey and learn. The behavior of both individuals is shaped by the symbolic meanings attached to their roles in society. If the parent starts treating the child as an equal, their relationship and interactions might change based on how they redefine their roles.
4. Fashion and Appearance:
   • People often make judgments based on how others dress or present themselves. A person wearing a suit might be perceived as professional, while someone in casual clothing might be seen as more laid-back. These assumptions are based on the symbols we attach to certain clothes or appearances.

In each of these cases, the symbols (like language, roles, or appearance) carry meanings that we learn through social interactions, and these meanings shape our thoughts, behaviors, and relationships.

Solution: The correct answer is B.

1. Impression management refers to direct attempts by an individual to control how they are perceived by others. This is not the best description of what is occurring in the prompt, because typically observational research participants are not directly trying to control the researchers’ perceptions.
2. The Hawthorne effect, named after an observational study of workers in the Hawthorne factory, refers to the tendency for observational research participant behavior to change when they know they are being observed. This best describes the phenomenon described in the prompt.
3. Self-fulfilling prophecy is the tendency to behave in ways that confirm expectations. Although observational research participants may be aware that they are being observed, it is unlikely that they would be aware of the specific expectations or hypotheses of the researchers.
4. The Thomas theorem argues that individual beliefs have real consequences. This phenomenon is not relevant to the study discussed in the question prompt.