FL4 C/P
Bufffer
A solution that resists changes in pH upon addition of acid or base. While it cannot protect against addition of large amounts of acid/base, it's effective at maintaining pH when small to moderate quantities are added.
Strong/weak acids and bases
A strong acid or base dissociates completely in water, while a weak acid or base dissociates incompletely. I.e., HBr and HF both dissociate in water to produce H+ ions, but at equilibrium, a solution of HBr will produce more ions in solution than HF, bc HBr is a strong acid while HF is a weak acid. Because strong acids and bases completely ionize, the [H+] and pH (or [OH−] and pOH) are easily calculated, since the concentration of protons or hydroxyl groups is equal to the molarity of the solution. Thus, a 0.4 M solution of KOH has a [OH−] of 0.4, and a 0.007 M solution of HI has a [H+] of 0.007. For solutions so dilute that the H+ or OH− normally present in pure water (1 × 10−7 M) is comparable or larger to the amount produced by acid/base ionization, the ions present due to water must be considered.
At what angle relative to the velocity of a red blood cell should the transducer be held during Doppler ultrasound to observe the largest Doppler shift? A. 0º B. 30º C. 45º D. 90º
A. 0º Rearranging the equation Vb = FdVs/2F0cos(θ) shows that the Doppler shift, FD, is proportional to cos(θ). Therefore, the Doppler shift will be largest at θ = 0°, where cos(θ) takes its maximum value of 1. Fd = (Vb(2F0 • cos(θ))/Vs) When a wave-emitting detection device is used, its accuracy is optimized if the waves emitted by the device travel directly parallel to the waves that are being measured (θ = 0). The deviation caused by non-parallel angles is known as the cosine effect, because the measured velocity will equal the true velocity × cos(θ).
In one trial of this experiment, significant impurities were detected in the extracted caffeine. Based on the results of the experiment, which of the following would be the expected melting point range for this extracted sample? A. 195°C - 220°C B. 245°C - 267°C C. 233°C - 236°C D. 190°C - 195°C
A. 195°C - 220°C A melting point range includes the temperature when the first crystal of a compound starts to melt and the temperature when the compound is entirely melted. For a pure compound, the melting point range is narrow. Therefore, melting point determination is sometimes used to identify an unknown pure compound. However, presence of impurities in a compound lowers and broadens the melting point range. According to the passage, the melting point of pure caffeine is 235°C, making choice A the best answer.
What is the pH of a 0.010 M perchloric acid solution? A. 2 B.4 C.7 D.12
A. 2 Perchloric acid is a strong acid that completely dissociates in aqueous solution, so the hydrogen ion concentration is 1.0 x 10-2 M. The pH = -log[H+] = -log[10-2] = 2.
In 250 mL of the MH solution with the most favorable solubility profile, how many moles of nicotinamide (MW = 122 g/mol) are present? A. 3.1 x 10-2 B. 3.1 x 10-1 C. 1.6 x 10-2 D. 1.6 x 10-1
A. 3.1 x 10-2 Table 1 shows there are 40 mg/mL of total hydrotrope in each MH solution. The solution with the most favorable solubility is the N:B:C ratio of 15:20:5. In this solution, the concentration of nicotinamide (N) is 15 mg/mL. Since there are 250 mL of MH, the amount of nicotinamide is 3750mg 250mL x 15mg = 3750mg (~4000mg = 4g) mole nicotinamide = 4 g nicotinamide x (1 mol nicotinamide / 122 g nicotinamide [round to 120g]) = 4 g x 1 mol / 120 g = 3.33 x 10-2 mol
According to the results of the experiment, the minimum risk for TBI during the pillow fight occurred during which time interval? A. 5.90 sec to 5.91 sec B. 5.91 sec to 5.92 sec C. 5.92 sec to 5.93 sec D. 5.95 sec to 5.96 sec
A. 5.90 sec to 5.91 sec According to the passage and Figure 2, higher TBI risk is associated with larger HIC. HIC is proportional to the average linear force, a product of mass and acceleration. Thus, the smaller the magnitude of acceleration, the lower the risk conferred to the individual. The rate of change of acceleration for any given time interval is the magnitude of the slope of the graph of acceleration vs. time for that interval. Using Figure 1, the rate of change in the interval between 5.90 sec and 5.91 sec is approximately zero, making it the lowest risk period for TBI.
The light blue appearance of S157T is most directly attributable to what other phenotypic change versus the wild type? A. Increased λmax B. Decreased λmax C. Decreased ɛ D. Increased ɛ
A. Increased λmax Table 1 shows three substitutions that lead to a change in the quality of sgBP's blue color. Specifically, as the wavelength of peak absorbance decreases from S157T to Q62M to S157C (611 nm, 608 nm, and 604 nm), the apparent color changes from dark blue to blue to light blue. The light blue appearance of S157T would seem to be due to it having a higher peak absorbance wavelength than the wild type. WRONG: C) While the molar absorptivity of S157T is lower than sgBP, there's no clear pattern in the color change reported in Table 1. Both S157T and S157C have lower molar absorptivity than sgBP, but S157T appears light blue, while S157C appears dark blue.
Which specific class of enzymes is primarily responsible for the release of free glycerol from stored triglycerides? A. Lipases B. Carboxylases C. Phosphorylases D. Kinases
A. Lipases Lipases are the enzymes that digest lipids (fats). Most dietary fats originally exist in the form of triglycerides. Since lipases typically catalyze hydrolysis reactions, they are a subclass of the hydrolases.
It is found that in the absence of molecular oxygen, the resulting imidazolinone-containing reactant is not fluorescent. According to Figure 1, what best explains this inability to fluoresce? A. The 5-membered ring is not conjugated with the aromatic phenol ring of tyrosine. B. The tyrosine side chain in the final chromophore remains deprotonated. C. The reactant lacks conjugation among any double bonds. D. The more thermodynamically favorable serine keto tautomer is formed.
A. The 5-membered ring is not conjugated with the aromatic phenol ring of tyrosine. In step 3 Figure 1, the Cα-Cβ bond of Tyr is oxidized to a double bond in a reaction that consumes molecular oxygen. This double bond places the 5-membered ring into an aromatic system in conjugation with the aromatic phenol ring of the tyrosine side chain. P2 states "imidazolinone structure, together with the tyrosine phenolic group, form a conjugated system of π-electrons that cause excitation and emission in ... GFP-like proteins." Without the molecular oxygen needed for this oxidation, such conjugation will not arise. Its absence could explain the failure of the molecule to fluoresce. WRONG: B. The protonation of the tyrosine side chain is not dependent on the presence of molecular oxygen. C. The molecule is not entirely without conjugation—the tyrosine side chain is aromatic, so conjugation does exist. However, the molecule lacks the specific conjugation between the imidazolinone structure and the tyrosine phenolic group required for excitation and emission in GFP fluorophores. D. Serine does not exist as a tautomer.
The first step of the formation of the imidazolinone ring of sgBP is most likely accomplished by the: A. attack of Gly64 amide nitrogen on the electrophilic Gln62 carbonyl carbon. B. attack of Gln62 amide nitrogen on the electrophilic Gly64 carbonyl carbon. C. attack of Tyr63 phenolic hydroxyl group on the electrophilic Gln62 carbonyl carbon. D. attack of Tyr63 phenolic hydroxyl group on the electrophilic Gly64 carbonyl carbon.
A. attack of Gly64 amide nitrogen on the electrophilic Gln62 carbonyl carbon. In the QYG chromophore of sgBP, serine is replaced by glutamine. According to the first paragraph of the passage, "CP chromophores form an imidazolinone structure through autocyclization and dehydration reactions, shown in steps 1 and 2 of Figure 1." Given this, choice A, the attack of the nucleophilic amide nitrogen of glycine 64 on the electrophilic carbonyl carbon of glutamine 62, is most likely to occur during the first step of the formation of the imidazolinone ring of sgBP. WRONG: B) Attack of the Gln62 amide nitrogen on the electrophilic Gly64 carbonyl carbon would form a sterically constrained three-membered ring, unlike the five-membered heterocycle formed by step 1. C. Steric constraints and resonance effects prohibit this reaction from occurring and would not form the product. D. Steric constraints prohibit this reaction from occurring and would not form the product.
Compared to the cytotoxic activity of ZSTK474 against the cell lines tested, the cytotoxicity of each target compound was: A. enhanced against at least one cell line tested. B. enhanced against all cell lines tested. C. reduced against all cell lines tested. D. reduced against at least one cell line tested.
A. enhanced against at least one cell line tested. Passage tells us that ZSTK474 is compound 4a. Know that the lower the amount of the drug it takes to kill 50% of the cells (i.e. smaller IC50 values), the greater the cytotoxic effects against the tested cell lines. Table 1 shows that the IC50 values (i.e., the cytotoxicity) of each target compound were smaller than compound 4a (ZSTK474) in at least one (5b enhanced relative to 4a), but not all, tested cell lines.
In comparison to piperazine, the inductive effect should cause the pKb of methyl-piperazine to be: A. lower, because of electron donation by the methyl substituent. B. lower, because of electron withdrawal by the methyl substituent. C. higher, because of electron donation by the methyl substituent. D. higher, because of electron withdrawal by the methyl substituent.
A. lower, because of electron donation by the methyl substituent. pKb measures basicity; smaller pKb corresponds to stronger base. Methyl-piperazine is a stronger base than piperazine because of the inductive effect generated by the electron-donating N-methyl group of methyl-piperazine. This effect increases the availability of electron charge density available to donate, thereby increasing the strength of N-methyl piperazine to act as a Lewis base.
If negative charge repulsion in C4S units contributes to the elastic modulus of articular cartilage, given passage information, which of the following is likely observed as a result of OA in articular cartilage? A.Decreased charge repulsion and decreased compressive strength B. Increased charge repulsion and increased compressive strength C. Decreased charge repulsion and increased compressive strength D. Increased charge repulsion and decreased compressive strength
A.Decreased charge repulsion and decreased compressive strength According to paragraph 2, "GAGs ... interact non-covalently with one another to exert outward pressure within the cartilage" and "Loss of C4S is believed to be responsible for changes in the biomechanical properties of cartilage seen in osteoarthritis (OA)." If negative charge repulsion in C4S units contributes to the elastic modulus of the cartilage, decreased C4S content due to OA would then decrease negative charge repulsion and change a mechanical property of the cartilage—namely, its elastic modulus, which is defined in the passage as being a "a measure of compressive strength." This is very nearly the compressive strength mentioned in the question stem. Overall, this is consistent with the positive correlation between elastic modulus and GAG content shown in Figure 2.
All of the following can be concluded about the heat engines tested EXCEPT: A.if the efficiency of engine 1 increases, the interior area of its heat cycle in Figure 2 will decrease. B.without a change in temperature, no net work could be extracted from the heat cycle. C.isothermal expansion is followed by an adiabatic expansion in Figure 2. D.isothermal compression is followed by an adiabatic compression in Figure 2.
A.if the efficiency of engine 1 increases, the interior area of its heat cycle in Figure 2 will decrease. The formula for the efficiency of any system is output work/input energy. Therefore, the efficiency of engine 1 is W/QH. Efficiency is increased by increasing W. Since work in Figure 2 is the interior area of the cycle, if W increases, the interior area should increase as well. This makes choice A, which states the opposite, correct for this EXCEPT question WRONG: B. In Figure 2, process 1 is at a higher temperature than process 2. If these two processes occur at the same temperature, process 1 will be on top of process 2. Subsequently, the interior area of the cycle will reduce to zero. Since the interior area is W, no net work could be extracted if TH and TC are equal. C. Process 1 is an isothermal expansion - that part of the graph is at a constant temperature of 900 K and volume increases from 1 m3 to 2 m3. Process 2 is an adiabatic expansion - no heat exchange (mentioned in P2 of the passage) and the volume increases from 2 m3 to 3 m3. Therefore, an isothermal expansion is followed by an adiabatic expansion. D. Process 3 is an isothermal compression - that part of the graph is at a constant temperature of 300 K and volume decreases from 3 m3 to 4 m3. Process 4 is an adiabatic compression - no heat exchange (mentioned in P2 of the passage) and the volume decreases from 4 m3 to 1 m3. Therefore, an isothermal compression is followed by an adiabatic compression in Figure 2.
Arrhenius
Acids (produce H+): substances that dissociate in a manner that increases the concentration of protons (H+ ions) in solution Bases (produce OH-): substances that dissociate in a manner that increases the concentration of hydroxide (OH-) ions
Alkanes and alkenes
Alkanes and alkenes are hydrocarbons; alkanes only have single C-C bonds, while alkenes have at least one C=C double bond. They interact with each other through weak London dispersion forces, have low melting and boiling points, and do not engage in meaningful acid-base chemistry.
Alkanes
Alkanes have only single bonds between carbons, and end with the suffix -ane. Straight-chain alkanes have the general formula CnH2n+2. They are considered to be saturated compounds, because they only contain single bonds. This means that they contain as many hydrogen atoms as is possible for a given carbon chain.
Hydrocarbons
Alkanes, alkenes, and alkynes are hydrocarbons—the simplest examples of organic molecules, made up only of carbon and hydrogen atoms.
Alkenes
Alkenes have at least one double carbon-carbon bond (C=C), and end with the suffix -ene; an alkene with two double bonds is a diene, and an alkene with three double bonds is a triene. Straight-chain alkenes with a single double bond have the general formula CnH2n. Molecules with at least one double bond (or a ring structure) are considered unsaturated.
Alkynes
Alkynes have at least one triple carbon-carbon bond and have the characteristic suffix -yne; they are not an important topic for the MCAT, but should be aware of what they are. Straight-chain alkynes with a single triple bond have the general formula CnH2n-2.
Nitrogen-containing functional groups
Amines (R-NH2, R-NHR', or R-NR'R") Imines (R=NH or R=NR') Enamines (C=C-NH2, C=C-NHR, or C=C-NRR') Nitrogen-containing compounds with medium melting/boiling points that can act as weak bases
Amphoteric
Amphoteric (or amphiprotic) compounds can act as either an acid or a base, depending on the other reactants present.
Brønsted-Lowry
An acid is a proton donor (like the Arrhenius definition) and a Brønsted-Lowry base is a proton acceptor, or any compound that can gain a bond to H+, such as NH3, which is not an Arrhenius base.
Doppler ultrasound
Analyze the speed of blood flow relative to the transducer. When sound waves are reflected back to the Doppler US device, blood (moving sound) and transducer (stationary observer) and Doppler equation can be applied to determine speed and direction
Aromatic compound
Aromatic compounds are conjugated cyclic molecules with a planar structure that also satisfy an additional criterion known as Hückel's rule. i.e., benzene Many other biologically relevant aromatic rings contain non-carbon molecules, aka heterocyclic rings. Antiaromatic molecules are highly unstable compounds with 4n π-electrons.
Which of the following is most likely the net charge on a valine molecule in the human body? A. -1 B. 0 C. +1 D. +2
B. 0 Valine is a non-polar AA so net charge at physiological pH will be 0. Carboxyl group will be deprotonated (COO-) and amino group will be protonated (NH3+). Thus, it will exist as a zwitterion with net charge 0.
Participant 1 rides a fourth roller coaster as shown below. What is the minimum ramp height H if the ride at the top of the loop maintains the minimum speed needed to stay on the track throughout the loop? (Note: Neglect friction.) A.10 m B.20 m C. 30 m D. 35 m
B. 20 m At the top of the loop, the gravitational and the normal forces (if any) point downward toward the center of the loop. Therefore, the net force causing centripetal acceleration is the sum of the gravitational force and the normal force. When the centripetal force is the minimum amount needed for the ride to stay on track, the normal force is zero. Note: At the top, the normal force can be zero but the gravitational force cannot. Let's designate the initial height where the ride starts as H. At the very beginning, the energy of the ride is the gravitational potential energy. Therefore, Ei = mgH. At the top of the loop, the ride includes both gravitational and potential energy. Therefore, Ef = mgh + ½ mv2 = mg (2r) + ½ mv2. The net acceleration at top (centripetal acceleration) = v2/r. Since we concluded that the normal force at the top is zero, the net acceleration at the top is the gravitational acceleration, g. Therefore, v2 = gr. Substituting this equation into the 'Ef' equation, we obtain the following: Ef = 2mgr + ½ mgr = (5/2) mgr To obey conservation of energy, Ei = Ef. Therefore: mgH = (5/2) mgr. We conclude that H is 5r/2. Since r is 8 m, H is 5(8 m)/2 = 20 m.
Approximately half of hemoglobin A's oxygen binding sites are occupied at what partial pressure of O2? A. 20 mmHg B. 30 mmHg C. 50 mmHg D. 80 mmHg
B. 30 mmHg One-half of Hb's oxygen binding sites are filled at its P50 value. In Figure 2, Y for Hb A is near 50% at a pO2 of approximately 30 mmHg.
Following the procedure of the experiment, how much total caffeine could you expect to extract? A. 6.50 g B. 9.84 g C. 11.72 g D. 12.97 g
B. 9.84 g Caffeine is largely nonpolar, so that it will be more soluble in dichloromethane than in water. Therefore, it will dissolve more in the DCM layer. Using the partition coefficient equations and the caffeine properties provided by the passage, we must add up the amount of caffeine dissolved in each layer in both the first extraction and the second extraction. First extraction: According to paragraph 1, the solubility of caffeine at standard temperature is 0.14 g/mL in DCM and 0.02 g/mL in distilled water. Multiplying each number by 100 ml of DCM and water used in the extraction results in 14 g / 100 mL and 2 g / 100 mL, respectively. Ψ ≅ solubility of a solute in organic layer / solubility of a solute in aqueous layer: Ψ = (14 g / 100 mL)/ (2 g / 100 mL) = 7 A tablet weighing 10 g was used. If we assume that X grams dissolve in DCM, then (10-X) g dissolve in the aqueous layer. Therefore: Ψ = (X g / 100 mL of CH2Cl2) / ([10 - X] g/ 100 mL of H2O) = 7 X = 70 - 7X 8X = 70 X = 70/8 = 8.75 g dissolved in DCM If 8.75 grams dissolve in DCM, then 10 g - 8.75 g = 1.25 is dissolved in the aqueous layer, which is used in the second extraction. Ψ = (X g / 100 mL of CH2Cl2) / ([1.25-X] g / 100 mL of H2O) = 7 X = 8.75 - 7X X ≅ 1 g Total caffeine extracted ≅ 8.75 g + 1 g = 9.75 g Choice B is the closest answer. Notice that calculations result in weight slightly less than that of a tablet. Tablets use binders, so the actual amount of caffeine is slightly lower.
1) Which of the following reagents is/are likely to be used to form the Britton-Robinson buffer solution used in the experiment? I. A polyprotic, weak acid II. Na2HPO4 III. NH3 A. I only B. I and II only C. II and III only D. I, II, and III
B. I and II only Buffer solutions, intended to resist changes in pH, consist of a weak acid or weak base and its corresponding salt. Passage states the buffer solution is acidic (pH = 3.8). Can assume a base is not present in the solution. Look for answer choices containing a weak acid or a salt formed from a weak acid as it may be a potential component of the buffer. A polyprotic, weak acid fits the criteria for an acidic buffer solution (RN I). Na2HPO4 (phosphoric acid) is a salt formed from a weak acid and would be an appropriate choice as a buffer (II). NH3 can be discounted as it is a weak base (RN III).
The Arrhenius equation, k = Ae-Ea/(RT), expresses the relationship between the rate constant, k, and the temperature of a reaction. According to this equation, which of the following will increase the rate of a reaction? A. Decreasing the pre-exponential factor, A B. Increasing the temperature, T C. Increasing the activation energy, Ea D. Decreasing the temperature, T
B. Increasing the temperature, T While this question introduces the Arrhenius equation, you can answer it simply by remembering the principle in general chemistry that increasing temperature increases the rate of reaction. However, notice that increasing T causes the exponent of e to be less negative, so k increases.
Which of the following boundaries would give a brighter ultrasound image at the boundary but the poorest overall resolution? A. Fat/muscle B. Muscle/bone C. Muscle/blood D. Water/fat
B. Muscle/bone Passage states that if the ∆Z is very large, all of the ultrasound will be totally reflected at the boundary. The overall resolution will be poor because too much signal was reflected back, and there was not enough left to be able to penetrate further and continue imaging. If ∆Z is small, a small amount of signal will be reflected back, which would allow enough signal left to continue through for further imaging. Muscle and bone have the largest difference in impedance (5.7 - 1.7 = 4.0) of all the combinations given. *Table 1 in Q6
Three students in physics lab are given four capacitors, each with a capacitance of 4 μF, and are told to construct a circuit with the maximum possible total capacitance. Student 1 attaches all four capacitors in series. Student 2 attaches all four capacitors in parallel. Student 3 only inserts a single capacitor in his circuit. Which of the following is true? A. Student 1 and Student 3 will create circuits that have the same total capacitance. B. Student 2 will create the circuit with the highest capacitance. C. Student 2 and Student 3 will create circuits that have the same total capacitance. D. Student 3 will create the circuit with the highest capacitance.
B. Student 2 will create the circuit with the highest capacitance. Capacitors in parallel adds directly (sum of their individual capacitances), so student 2 will create a circuit w/ highest capacitance of 16μF. WRONG: A) capacitors in series are added reciprocally, so student 1 would create a circuit w/ 1μF capacitance. D) only using 1 capacitor, so the circuit would have 4 μF capacitance
The UV absorption of A in MH solutions and in aqueous solutions both peak at 314 nm, suggesting: A. A contains a C=O double bond. B. no significant structural changes in A occur during MH solution preparation. C. A in solution would be red. D. significant structural changes in A occur during MH solution preparation.
B. no significant structural changes in A occur during MH solution preparation. Passage states UV absorbance spectroscopy was used to see if the MH preparation process had changed the structure of A. Since A's peak UV absorbance remains the same in aqueous and and MH solutions solutions suggest the structure is identical. WRONG A) C=O double bond is detected by IR spectroscopy and NMR, not UV spectroscopy. C. Compounds that absorb light in the visible spectrum are colored, not UV spectrum. D. If significant structural changes had occurred in A, there would be a shift in UV absorption.
Use of CA4b is likely to increase the contrast obtained in CECT imaging because: A. cationic agents bind poorly to chondroiton sulfate in articular cartilage. B. the elements in CA4b have larger atomic numbers than the elements in cartilage. C. collagen and other components of articular cartilage are opaque to X-rays. D. articular cartilage is too thick to be imaged by traditional modalities.
B. the elements in CA4b have larger atomic numbers than the elements in cartilage. P2: "...CT imaging of cartilage samples treated with the novel cationic contrast agent CA4b, the technique, known as CECT..." P3: "X-ray attenuation, which is responsible for contrast in CECT imaging and is enhanced by the use of contrast agents, generally increases with the atomic number of atoms composing". Given this, it's reasonable to conclude that the contrast agent used in CECT (CA4b) may increase contrast in cartilage imaging because it is composed of elements with much greater atomic numbers than those found in cartilage.
The conversion between glucose and pyruvate strongly favors the formation of pyruvate, and yet the gluconeogenic pathway is able to utilize several shared enzymes with glycolysis to create glucose from pyruvate. It is able to do this primarily because: A. the formation of glucose from the gluconeogenic precursors is strongly favored by the increasing entropy of the reaction. B. the formation of glucose, fructose-6-phosphate, and PEP through gluconeogenesis-specific enzymes push the equilibrium of reactions catalyzed using shared enzymes to favor gluconeogenesis. C. the activity of the enzymes with two or more functions is reversed when those enzymes are altered by kinases. D. enzymes in gluconeogenesis bypass reversible, equilibrium-state steps common to both pathways.
B. the formation of glucose, fructose-6-phosphate, and PEP through gluconeogenesis-specific enzymes push the equilibrium of reactions catalyzed using shared enzymes to favor gluconeogenesis. Production of PEP, glucose, and fructose 6-phosphate by gluconeogenesis-specific enzymes that bypass irreversible steps of glycolysis push the equilibrium of reversible reactions in the direction of glucose production. WRONG A) While the universe as a whole demonstrates increasing entropy, the drivers of glycolysis and gluconeogenesis are reliant on changes in enthalpy through breaking and forming high-energy bonds, rather than entropic changes. C) While some enzymes that regulate the glycolytic and gluconeogenic pathways, such as PFK-2 and F2,6-BP, are bi-functional (the nature of their action depends upon their phosphorylation state), this is not true of the enzymes of glycolysis or gluconeogenesis. D) These steps are irreversible, not reversible, steps of glycolysis.
The area enclosed by the four steps shown in Figure 2 represents: A. the work done on the engine during one complete cycle. B. the work done by the engine during one complete cycle. C. the thermal energy change of the surroundings during one complete cycle. D. the heat transferred out of the engine during one complete cycle.
B. the work done by the engine during one complete cycle. Pressure-volume diagrams are a visualization tool for the study of heat engines. Since work is done only when the volume of the gas changes, the diagram gives a visual interpretation of the work done. Since the internal energy of an ideal gas depends upon its temperature, the PV diagram along with the temperatures calculated from the ideal gas law determine the changes in the internal energy of the gas so that the amount of heat added can be evaluated from the first law of thermodynamics. For a cyclic heat engine process, the PV diagram will be a closed loop. The area inside the loop is a representation of the amount of work done by the engine during a cycle.
Thalassemias are blood disorders in which the synthesis of either α-globin-like or β-globin-like chains are defective. Thalassemia involving non-expression of which protein chain is least likely to produce physical manifestations prior to birth? A. α globin B. β globin C. γ globin D. ζ globin
B. β globin Of the four proteins listed, Figure 1 shows that at all points prior to birth, at least two other globin chains are present at greater levels than β globin. Given this, β globin will be present in the smallest concentration, if at all, relative to the content of Hb proteins synthesized.
The enzyme glutathione reductase catalyzes the conversion of the molecule GSSG to GSH, which is essential in order for cells to counteract the effects of reactive oxygen species in the cytoplasm. Glutathione reductase cleaves a single disulfide bond within the GSSG molecule, resulting in two GSH molecules. Solely using this given information, how does the oxidation number change from a single GSSG molecule to a single GSH molecule? A. +2 to 0 B. +1 to 0 C. 0 to -1 D. 0 to -2
C. 0 to -1 In a disulfide S-S bond, the two sulfur atoms share their covalent electrons equally and therefore have an oxidation number of zero. When the disulfide bond gets reduced to two -SH sulfhydryl groups, S-H covalent electrons spend their time more around the sulfur in the GSH molecule, as opposed to the hydrogen. This is because sulfur is more electronegative than hydrogen. As a result, the sulfur in GSH has an oxidation state of -1. All else being equal, this suggests that a GSSG molecule has an oxidation state of 0, while a GSH molecule has an oxidation state of -1.
What is the normality of a 0.015 M solution of phosphoric acid assuming complete dissociation? A. 0.015 N B. 0.030 N C. 0.045 N D. 0.060 N
C. 0.045 N The chemical formula of phosphoric acid is H3PO4. Normality, in the context of acids, refers to the number of moles of protons per liter of solution (in other words, to the "molarity of protons"). Normality can be calculated by multiplying the molarity of the solution by the number of protons per molecule of acid (here, 3). (0.015 M solution) x (3 protons per molecule) = 0.045 N
The human frequency range is from 20 Hz to 20 kHz, where 8-20 kHz is high frequency. Ultrasound frequencies must be greater than which minimum threshold? A. 1 kHz. B. 10 kHz. C. 20 kHz. D. 40 MHz
C. 20 kHz. Ultrasound is defined as sound with a frequency above the human range of hearing ("ultra" = "above"). The human range of hearing is 20 Hz to 20 kHz, where high frequencies are 8-20 kHz, so the minimum threshold for ultrasound is 20 kHz.
An object made of silicon (specific heat = 710 J/kg°C) absorbs 3500 J of heat while increasing its temperature from 43°C to 53°C. What is the approximate mass of the object? A. 350 g B. 400 g C. 500 g D. 2050 g
C. 500 g The most efficient way to find this answer is by using the equation q (the heat in or out of a system) = mcΔT. Because this is a change in temperature, either centigrade or Kelvin temperatures can be used. Substituting in numbers, we get: 3500 J = m x 710 J/kg°C x (53°C - 43°C) Approximating values and rearranging gives: m = 3500 J / [700 J / kg°C] *10°C ≈ 3500/7000 kg = 0.5 kg = 500 g
How is Solubility Enhancement Ratio (SER) in Table 1 calculated? A. By dividing A's hydrophilic solubility by its hydrophobic solubility B. By multiplying A's hydrophilic solubility by its hydrophobic solubility C. By dividing the solubility of A in MH by its solubility in water D. By multiplying the solubility of A in MH by its solubility in water
C. By dividing the solubility of A in MH by its solubility in water Table 1 shows that the higher the solubility of A in a given solution is, the higher the SER. SER is a direct reflection of A's solubility in a given solution versus that of water. Using the solubility of A in water (0.165 mg/mL) from P3, we can see that SER is calculated by dividing the solubility of A in the mixed hydrotrope by the solubility of A in water. SER = (A solubility in MH/A solubility in water) i.e. table 1 row 1 for 10:20:10 mixture SER = [(8.6 mg/mL) / (0.165 mg/mL)] = 52.1
The λmax of S157C at pH 4.5 is greater than at pH 10. How does the proposed bonding of deprotonated C157 with Y63 at pH 10 account for this observation? A. Compared to hydrogen bonding at pH 4.5, the ion-dipole interaction at pH 10 destabilizes the sgBP chromoprotein barrel conformation. B. Compared to hydrogen bonding at pH 4.5, the ion-dipole interaction at pH 10 decreases the maximum normalized absorbance. C. Compared to hydrogen bonding at pH 4.5, the ion-dipole interaction at pH 10 promotes peak absorbance of higher energy radiation. D. Compared to hydrogen bonding at pH 4.5, the ion-dipole interaction at pH 10 increases the wavelength of maximum absorption.
C. Compared to hydrogen bonding at pH 4.5, the ion-dipole interaction at pH 10 promotes peak absorbance of higher energy radiation. P3 states that a H-bond forms "when both the phenolic side chain of Y63 and the hydroxyl group of Ser157 are protonated" and that "the strength of Y63-S157 bonding is thought to be important in ESPT, which determines the chromophore absorption spectrum." Figure 3 shows that at pH 10, the QYG chromophore tyrosine 63 is protonated and cysteine 157 is deprotonated. The two residues form a "protonation state-dependent ion-dipole" interaction that accounts for the "greater strength of ion-dipole bonds relative to hydrogen bonds." Figure 3 also shows that the side-chain pKa of cysteine is approximately 8.3. At pH 4.5, the side chain will exist in a mainly protonated form. This suggests that the longer maximum absorbance wavelength at pH 4.5 is because the Y63-C157 interaction predominantly consists of hydrogen bonding, and that the ion-dipole interaction which predominates at pH 10 promotes the maximum absorption of shorter-wavelength, higher-energy radiation. WRONG: A. There is no evidence in the passage to support this conclusion. B. Figure 2 shows that at pH 10, S157C has a higher maximum normalized absorbance (1) than at pH 4.5 (~0.5). D. Figure 2 shows that at pH 10, S157C has a wavelength of maximum absorption that is somewhat shorter than at pH 4.5.
The molecule below is: I. aromatic. II. non-aromatic. III. a heterocycle. A. I only B. II only C. I and III only D. II and III only
C. I and III only The molecule has 6 pi electrons, so according to Huckel's rule it is aromatic. It is also a heterocycle, a ring of atoms of more than one kind, so both I and III are correct.
Given the data presented in the passage, what is true regarding the effect of tested target compounds on the cytotoxicity of HCT116 cells? A. It is due to a mechanism of action other than PI3Kα kinase inhibition. B. It is consistent with the effect of treatment on the cytotoxicity of other tested cell lines. C. It cannot be predicted directly from target compound inhibition of PI3Kα kinase in vitro. D. It is negatively correlated with PI3Kα kinase inhibition by the target compound.
C. It cannot be predicted directly from target compound inhibition of PI3Kα kinase in vitro. The data shown in Tables 1 and 2 suggest that there is not an obvious correlation between the extent of PI3Kα inhibition by the target compounds and ZSTK474 and the extent of their cytotoxicity against HCT116C. Each Target compound (4b-c and 5a-b) has a greater cytotoxic effect against HCT116 cells than does ZSTK474 (compound 4a), but a smaller inhibitory effect. However, this does not represent, in general, a negative correlation. In Table 2, there is little relationship between the extent of PI3Kα inhibition and cytotoxicity when comparing the measures for any two compounds.
In addition to its X-ray-attenuating properties, researchers specifically chose CA4b from a host of other potential contrast agents. What is the most likely reason for its selection? A. Its irreversible binding of chondroitin sulfate. B. It mimics the properties of natural agents. C. Its charge character at neutral pH. D. It significantly increases the thickness of cartilage samples.
C. Its charge character at neutral pH. CA4b is a cationic contrast-enhancing agent that, in the CECT technique, can be used to assess the GAG content of cartilage. Additionally, it is written in the passage that "GAG content is positively correlated with contrast in CT imaging of cartilage samples treated with ... CA4b." This correlation is likely because the extent of binding by cationic (positively charged) CA4b, and, thus, the extent to which contrast enhancement occurs, depends upon the GAG content to which CA4b binds. For this reason, the positively-charged CA4b is central to its function as a contrast-enhancing agent in the CECT technique.
Which of the following correctly represent the units of the Boltzmann constant? A. kg/s B. N/K C. J/K D. K/J•s
C. J/K According to the last paragraph, the Boltzmann constant has units of (m2•kg)/(s2•K). joule units: kg•m2/s2. Dividing by K gives us units for the Boltzman constant.
When situated in the substrate affinity pocket of PI3Kα, the morpholine group oxygen of ZSTK474 is most likely to interact with the side chain of what amino acid residue? A. Aspartate B. Cysteine C. Lysine D. Tyrosine
C. Lysine P1 states the the oxygen, of the morpholine group, acts a H-bond acceptor for a primary amine of the enzyme, so look for answer choice that has an amino group that can act as a H-bond donor Of the choices given, only the ε-amino group of lysine could be the amino group, acting as a hydrogen bond donor, described in the passage
Based on results presented in the passage, researchers hoping to alter the appearance of sgBP while maintaining its function as a CP providing a colored appearance would most logically choose to mutate which sgBP residue? A. Gln62 B. Glu144 C. Ser157 D. His196
C. Ser157 Table 1 shows that only modification of Ser157 resulted in mutant sgBP chromophores that express a different color.
Which of the following compounds is an amphiprotic species? A. Acetic acid, HC2H3O2 B. Sodium acetate, NaC2H3O2 C. Sodium bicarbonate, NaHCO3 D. Sodium carbonate, Na2CO3
C. Sodium bicarbonate, NaHCO3 An amphiprotic species is one that can act as an acid or a base. Sodium bicarbonate dissolves in aqueous solution to produce a strong base (NaOH), which completely dissociates, and carbonic acid, a weak acid. The solution is therefore basic. The bicarbonate ion, HCO3-, has an acidic proton, so it can act as a Bronsted-Lowry acid by loss of a hydrogen ion forming carbonate, or it can act as a B-L base by accepting a hydrogen ion to form carbonic acid, H2CO3.
A certain metabolic process in the liver produces NADH as a part of the process. If this process is upregulated, which of the following effects associated with gluconeogenesis is most likely to follow? A. Intracellular levels of oxaloacetate will increase. B. Plasma glucose concentrations will increase significantly. C. The rate of gluconeogenesis in the liver will decrease. D. Plasma glucose concentrations will decrease significantly.
C. The rate of gluconeogenesis in the liver will decrease. The key to this question is that we do not know which process the question stem refers to; we only know that it produces NADH. We are given no reason to assume that this process is the same as that described in the question stem, in which "the conversion of lactate to pyruvate is coupled with the reduction of NAD+ to NADH." Since we cannot assume that this is the process being discussed, the only information available for our use is that this process is producing larger-than-usual amounts of NADH. On the MCAT, it is very important to think about the most direct effect of the situation at hand (here, increased NADH levels), instead of thinking about potential factors that could have caused this situation. ***In other words, assume that the situation in the question stem has already occurred, and we now need to think about what will happen next.**** If NADH levels have already been increased, then by Le Chatelier's Principle, further conversion of NAD+ to NADH will be suppressed. Therefore, the production of OAA and pyruvate (two important gluconeogenic substrates) will be decreased, since pyruvate production is coupled with the conversion of NAD+ to NADH, and since OAA is produced from pyruvate. Thus, the rate of glucose production via gluconeogenesis is expected to decrease.
Aspartame (L-α-aspartyl-L-phenylalanine methyl ester) is a very well-known artificial sweetener found in the large majority of non-sugar containing food products. This compound is classified as a(n): A. hexose. B. phenol. C. dipeptide. D. amino acid.
C. dipeptide. Aspartame contains two amino acids, as mentioned in the formula, so it is a dipeptide. Aspartame is the methyl ester of the dipeptide containing aspartic acid and phenylalanine
Hb Gower 1 is the most common hemoglobin produced during the first month following conception. Based on passage information, the most likely subunit composition of the Hb Gower 1 protein is: A. α2γ2. B. α2ε2. C. ζ2ε2. D. α2β2.
C. ζ2ε2. Figure 1 indicates that only ζ and ε are present in meaningful quantities for the first month following conception. Consequently, Hb Gower 1 is most likely to be composed of ζ and ε chains.
Which of the following is NOT a functional group present in caffeine? A.Imide B.Amide C.Primary amine D.Tertiary amine
C.Primary amine Imide - two acyl groups bound to nitrogen Amide - contains the group C(O)NR2, where "R" can refer to an organic group or hydrogen atom Tertiary amine - nitrogen atom bound to three substituents. Multiple tertiary amines are present in caffeine All of the functional groups are part of caffeine except for a primary amine, which consists of nitrogen bonded to one R group and two hydrogen atoms. This structure is not part of caffeine
Which of the following situations would present Participant 3 with the greatest risk for TBI? A. A linear acceleration of 10 m/s2 for 2 ms in a car crash B. A single punch to the head causing acceleration of 100 m/s2 for 3 ms C. Riding a roller coaster of 9 m/s2 sustained for 40 s D. Total body acceleration of 15 m/s2 for 1 ms
C.Riding a roller coaster of 9 m/s2 sustained for 40 s According to the second paragraph of the passage, G-forces alone do not predict TBI. Besides the force, the time duration over which the force is applied is important in determining the severity of a brain injury. The second paragraph also states that the body can endure a large G-force if it occurs over a very short time period. Using Equation 1 from the passage, we can calculate the value of HIC for each of these activities. The acceleration values are given for all the choices. To incorporate acceleration, we rewrite the equation into HIC = m*a*t, where "m" represents the mass of the brain (which, since the subject is the same, can be assumed to be equal for all cases). Therefore, the impulse will be: A: HIC = m (10 m/s2) (0.002 s) = (0.02 m) Ns B: HIC = m (100 m/s2) (0.003 s) = (0.3 m) Ns C: HIC = m (9 m/s2) (40 s) = (360 m) Ns D: HIC = m (15 m/s2) (0.001 s) = (0.015 m) Ns According to Figure 2, the risk of TBI increases with HIC, making the conditions in choice C have the highest probability of brain injury.
Chromatography
Chromatography is a broad set of separatory techniques based on relative affinity, or tendency for a compound to attract to a certain solvent or structure. It is used to separate, identify, and purify the components of a mixture. Four common MCAT-tested techniques are ion exchange, surface adsorption, partition, and size exclusion. There are many chromatography techniques, but all share the principle that the molecules in a mixture are applied onto a stationary phase (usually a solid), while a fluid known as the mobile phase (generally a solvent chosen to match the target molecules, e.g. polar or nonpolar) containing the molecules of interest travels through the stationary phase. Molecules of interest in the mobile phase will interact with the stationary phase with different levels of intensity. Molecules that interact more strongly with the stationary phase will take longer to pass through it, whereas molecules that interact more weakly with the stationary phase will pass through it more quickly. Common factors that shape these interactions include molecular characteristics related to adsorption, polarity- or charge-based affinity for the stationary/mobile phase, and differences in molecular weight. A chiral stationary phase can also be used to separate stereoisomers based on the principle that the various enantiomers of a compound may interact differently with such a stationary phase. The list below contains some common types of chromatography that may appear on the MCAT. All are based on the principles outlined above. As such, when studying a chromatography technique, it is less helpful to focus on the details of the setup than it is to determine which molecules will interact more closely with the stationary phase, leading them to be retained longer. Column chromatography Ion-exchange chromatography Size-exchange chromatography Paper chromatography Thin-layer chromatography (TLC) Gas chromatography (GC) High-pressure liquid chromatography (HPLC) Gel chromatography Affinity chromatography
Circuits
Circuits are combinations of one or more resistors and/or capacitors connected by a conductive wire to which a battery is attached. The voltage differential of the battery (or "electromotive force") pushes current through the circuit. Circuits can contain multiple resistors, which can be connected either in parallel or in series. Circuits can also contain capacitors, which store charge in two physically separated components.
Polyprotic acid
Compounds that have multiple H atoms that can participate in acid-base chemistry
Conservation of Energy
Conservation of energy is a fundamental law of nature: energy can neither be created nor destroyed, just transferred from one form to another. In other words, total final energy within a system is equal to the total initial energy. For problem-solving on the MCAT, it is important to determine whether we are only dealing with conservative forces or if non-conservative forces, such as friction, are incorporated.
Sulfur-containing functional groups
Contain the root "thio" and generally act similarly to the corresponding oxygen-containing groups
According to the experimental results, the elastic modulus of a sample of articular cartilage with mean CECT attenuation of 1500 HU is nearest: A. 0.1 MPa. B. 0.25 MPa. C. 0.3 MPa. D. 0.5 MPa.
D. 0.5 MPa. Pay close attention to the caption of Figure 2! CECT attenuation is represented by the gray line. The point on that line that corresponds to 1500 Hounsfield units corresponds to a GAG content of about 6% (as shown below). Next, we need to remember what the question is asking for: elastic modulus in units of MPa. If we look at the black line, which is a plot of E (in MPa) versus GAG content, we can see that a 6% GAG content corresponds to an E value of 0.5 MPa.
What is the pH of a 0.010 M sodium hydroxide solution at 25°C? A. 1 B. 2 C. 7 D. 12
D. 12 Should notice that choice D is the only option that is basic. To answer using math... NaOH (strong base) - completely dissociates in aqueous solution hydroxide ion concentration is also 0.010 M -log(1x10^-2) = pOH 2 pH + pOH = 14 (at 25ºC) pH = 14 - 2 = 12
How many grams of hydrogen gas are required to completely react with 32 g of oxygen gas to form hydrogen peroxide? A.0.5 g B.1.0 g C.1.5 g D. 2.0 g
D. 2.0 g The formation reaction for hydrogen peroxide is: H2 (g) + O2 (g) → H2O2 (l) Moving on to the stoichiometry calculation: 32 g O2 x (1 mol/32 g) x (1 H2/1 O2) x 2 g/mol = 2 g H2
Which of the following is a list of straight-chain hydrocarbons, each of which could have one triple bond? A. C4H10; C3H6; C2H2 B. C5H8; C3H4; C2H4 C. C5H10; C3H6; C2H4 D. C8H14; C6H10; C2H2
D. C8H14; C6H10; C2H2 A triple bond will replace four of those hydrogens with two extra C-C bonds - formula: CnH2n-2 Formula for a straight-chain alkane is CnH2n+2. A double bond will replace two of those hydrogens with an extra C-C bond - formula: CnH2n.
Which one of the engines in the study has the highest efficiency? A.Engine 1 B.Engine 2 C.Engine 3 D.Engine 4
D. Engine 4 Need to calculate the efficiency (η) for every engine to find the highest efficiency. For any engine, efficiency is defined as work output/work input. Efficiency is calculated using the following formula: η (efficiency) = W/QH Table 1 has QH and QC. Can rewrite the above equation by using 1st law of thermodynamics (conservation of energy), QH = QC + W, using Figure 1 of the passage replace W with QH - QC: η = (QH - QC) /QH = 1 - (QC/QH) Engine 1: η = 1 - (QC/QH) η = 1 - ((3.81 x 106 J) / (5.68 x 106 J)) This is approximately 1 - 4/6 = 1 - ⅔ = 1 - 0.67 Actual η = 0.33 Engine 2: η = 0.10 Engine 3: η = 0.56 Engine 4: η = 0.61 Thus, engine 4 has the highest efficiency (0.61).
Benedict's test typically involves exposure of a carbohydrate to a solution of CuO in ammonia (NH3). Glucose yields a positive Benedict's test, but sucrose does not. Which of the following best explains this fact? A. Sucrose contains a hemiacetal group, while glucose does not; this classifies sucrose as a reducing sugar. B. Sucrose contains a hemiacetal group, while glucose does not; this classifies glucose as a reducing sugar. C. Glucose contains a hemiacetal group, while sucrose does not; this classifies sucrose as a reducing sugar. D. Glucose contains a hemiacetal group, while sucrose does not; this classifies glucose as a reducing sugar.
D. Glucose contains a hemiacetal group, while sucrose does not; this classifies glucose as a reducing sugar. Benedict's test is intended to identify "reducing sugars," or sugars with the capacity to serve as reducing agents. Specifically, sugars with hemiacetal groups can undergo mutarotation, allowing them to be oxidized by CuO. The process of mutarotation requires ring opening, which occurs at a hemiacetal group. Thus, all monosaccharides are reducing sugars because they undergo mutarotation. Polysaccharides have a reducing end, but the glycosidic linkage between multiple sugar units prevents isomerization to the aldehyde from occurring. This means that sucrose gives a negative test. The hemiacetal group on glucose, shown below, marks it as a classic reducing sugar. Remember, a hemiacetal consists of a carbon atom directly attached to one -OR and one -OH group. The same carbon atom is also attached to a hydrogen atom and an R group.
Given passage information, which of the following compounds is most likely present in greater concentration in the maternal blood relative to the level at which it is found in the blood of a fetus at 20 weeks of gestation? A. 2,3-BPG B. O2 C. αγ dimers D. Hb A2
D. Hb A2 According to the passage, Hb A2 is a minor hemoglobin synthesized only by adults. As a result, it should be found in greater concentration in the maternal blood than in the fetal blood.
Which of the following statements correctly describe the methods used in the experiment? I. The retention factor in a TLC procedure depends on the solvent system, temperature, and the adsorbent. II. A polar compound will exhibit a smaller retention factor on a normal phase TLC plate than a less polar compound. III. Anhydrous methanol has greater eluting strength than pentane when used as solvents in a TLC procedure. A. I only B. I and II only C. II and III only D. I, II, and III
D. I, II, and III The retention factor is the distance the compound migrated divided by the distance of the solvent front. TLC depends on the differential affinity of a compound for the stationary vs. the mobile phase. Depending on how polar the solvent and the adsorbent are, a compound will move on a TLC plate at a certain rate (I). Temperature also affects the rate of movement. In normal phase TLC, a polar compound will be attracted to the adsorbent through electrostatic interactions and, therefore, will not move as far on a TLC plate, whereas a nonpolar compound will have more affinity for the mobile phase and will move further (II). Eluting strength depends on how strongly a compound adsorbs onto the adsorbent. Since typical adsorbents in normal phase TLC are highly polar, eluting strength increases with increasing solvent polarity. Methanol is more polar than pentane and therefore has a greater eluting strength (III).
2) In addition to a drug's solubility in hydrophilic media, which of the following is likely to be an important factor in determining its oral bioavailability? A. pKa B. pKb C. Solubility enhancement ratio D. Membrane permeability
D. Membrane permeability Asking what factor is likely to affect bioavailability, or the ability of a drug to be absorbed by the body and take effect. Passage states hydrophilic solubility is important, likely b/c the vast majority of biological solvents are hydrophilic. An orally-administered drug must be absorbed across the membranes of the gums, stomach, or small intestine to be effective. Therefore, membrane permeability must be important because if it is low, the drug will not cross these membranes into the bloodstream. While acidity or basicity (pKa/pKb) may have an effect on bioavailability, it is not as important as hydrophilic solubility or membrane permeability. The study only examined solubility, not bioavailability. Solubility enhancement ratio refers to hydrotropes, not the drug itself.
What amino acid in the Aequorea fluorophore would be designated as "X" in the X-Tyr-Gly GFP fluorophore? A. Glutamine B. Glycine C. Threonine D. Serine
D. Serine Three amino acids are shown in the fluorophore structure of Figure 1: serine, tyrosine and glycine. WRONG: A. Glutamine Glutamine corresponds to amino acid X in the QYG chromophore, not in the Aequorea.
A scientist wished to prepare a buffer for an experiment to be conducted at pH 9.7. Which of the following organic acids would be the best choice for this experiment? A. Acetic acid (pKa = 4.76) B. Carbonic acid (pKa = 6.35) C. Tricine (pKa = 8.05) D. Taurine (pKa = 9.06)
D. Taurine (pKa = 9.06) To construct the best possible buffer, choose the organic acid with the pKa closest to the pH at which the experiment will take place (9.7). An ideal buffer should have a pKa within 1 pH unit of the expected experimental conditions.
All of the following phenomena serve to attenuate the ultrasound signal as it passes through the body EXCEPT: A. absorption. B. refraction. C. scattering. D. amplification.
D. amplification. Amplification opposite of attenuation; it makes the signal stronger. Attenuation - weakening of the U/S signal Sound energy is attenuated as it passes through the body bc parts of the signal are reflected, scattered, absorbed, refracted or diffracted.
Conversion of pyruvate into glucose requires PEPCK, an enzyme present in: A. the interstitial fluid only. B. the mitochondria only. C. the cytosol only. D. both the mitochondria and the cytosol.
D. both the mitochondria and the cytosol. Conversion of pyruvate to glucose requires its initial conversion into oxaloacetate, in a reaction catalyzed by pyruvate carboxylase in the mitochondria. OAA is then decarboxylated and phosphorylated by cytosolic or mitochondrial forms of PEPCK. After transport of either OAA in the form of malate or PEP directly from the mitochondria, the remainder of gluconeogenesis takes place in the cytosol.
The piezoelectric effect, when generating ultrasound waves, produces a charge imbalance through the compression or expansion of a material. This involves the conversion of: A. chemical energy to mechanical energy. B. electrical energy to potential energy. C. kinetic energy to electrical energy. D. electrical energy to mechanical energy.
D. electrical energy to mechanical energy. The piezoelectric effect begins with voltage generating a current through the crystal (electrical energy) and culminates in the crystal vibrating (mechanical energy). Mechanical energy is made up of kinetic and potential energy, as it is associated with the motion and position of an object.
Resistors
Devices that apply resistance to a circuit to manage current Circuits can contain multiple resistors, which can be connected either in parallel or in series. For a circuit with resistors R1, R2, ... Rn wired in series: Itotal = I1 = I2 = ... = In Vtotal = V1 + V2 + ... + Vn Rtotal = R1 + R2 + ... + Rn. If the resistors are wired in parallel: Itotal = I1 + I2 + ... + In Vtotal = V1 = V2 = ... = Vn 1/Rtotal = 1/R1 + 1/R2 + ... + 1/Rn
Electromagnetic (EM) waves
EM waves are transverse waves that can propagate through vacuum, as well as through other media such as air and water. EM waves have both electrical and magnetic components, with amplitudes perpendicular to each other and to the direction in which the wave is propagating, as shown below. EM waves propagate through space at the speed of light (c = 3 × 108 m/s NEED TO KNOW).
Ultrasound
Emits high frequency sound waves 1-18mgHz and then measures the amount of time taken for the emitted sound waves to reflect back and be detected by the sensor. Variations in time are used to predict the depths of internal structures in the body and produce an image of these structures in real time
EM radiation
Every form of EM radiation (radio, microwaves, infrafred, visible light, UV radiation, XR, gamma rays) oscillates in a periodic fashion with peaks and valleys, displaying a characteristic amplitude, wavelength, and frequency. These properties define the direction, energy, and intensity of the radiation. The standard unit for all EM radiation is the magnitude of the wavelength (in a vacuum), usually reported in "nm" for the visible light portion of the spectrum, and is measured by the distance between two successive peaks.
First order
First-order reactions are exemplified by radioactive decay and SN1 reactions that are dependent on carbocation formation.
Gluconeogenesis
Gluconeogenesis is a process that the body uses to create glucose from pyruvate. Occurs primarily in the liver, lesser extent in the adrenal cortex, and its goal is to ensure an adequate supply of glucose (which can then be converted into energy, or stored as glycogen) throughout the tissues of the body. It is upregulated by glucagon and by the presence of surplus pyruvate/acetyl-CoA. Gluconeogenesis is not the reverse glycolysis, although these two pathways share some of the same enzymes and steps (although they occur in reverse). However, they also differ at some crucial stages. Glycolysis contains some steps that are highly exergonic and essentially irreversible under biological conditions, so gluconeogenesis needs to bypass those steps. Additionally, glycolysis and gluconeogenesis need to be separated to prevent a futile cycle in which glucose is broken down to pyruvate and then pyruvate is built back up into glucose. In particular, the final stage of glycolysis (PEP → pyruvate) must be bypassed by gluconeogenesis. Thus, why gluconeogenesis has a two-step pathway split up between the mitochondria and cytosol, in which pyruvate carboxylase converts pyruvate to OAA in the mitochondria by adding a COO- group. OAA is briefly converted to malate for transport out of the mitochondria, where it is then converted immediately back to OAA. At this point, in the cytosol, PEPCK converts OAA to PEP. Additionally, the early stages of glycolysis (where phosphate groups are added to glucose) must be bypassed by gluconeogenesis. These are irreversible steps in glycolysis that involve the investment of ATP. Gluconeogenesis cannot simply reverse these steps, because doing so would mean creating ATP, which is the job of ATP synthase in the electron transport chain. Instead, gluconeogenesis bypasses these steps using enzymes that catalyze a simple hydrolysis reaction, splitting off a Pi from the carbohydrate.
pH and pOH
How acidic or basic a solution is can be expressed in terms of pH or pOH, which are defined as follows: pH = −log [H+] and pOH = −log [OH−]. For example, a solution with an H+ concentration of 10−4 M will have a pH of 4, and a solution with an OH− concentration of 10−9 M will have a pOH of 9. pH and pOH values can be estimated given a certain concentration using the following shortcut: p(N × 10−M) = (M−1).(10−N), such that a solution with an H+ concentration of 4 × 10−8 will have a pH = (8−1).(10−4) = 7.6. pH and pOH are related to each other through the equation pH + pOH = pKw, where Kw is the autoionization constant of water (Kw = [H3O+][OH−] = 1 × 10−14 at 25°C). Thus, at 25°C, pH + pOH = 14. Thus, a solution with a low pH will automatically have a high pOH, and vice versa. pH is most commonly used, but it is important to be able to interconvert pH and pOH values if needed. Acidic solutions have a high H+ concentration and a low pH. Basic solutions have a low H+ concentration and a high pH.
Hydrolases
Hydrolases catalyze reactions that involve cleavage of a molecule using water (hydrolysis). This case usually involves the transfer of functional groups to water. Hydrolases include amylases, proteases/peptidases, lipases, and phosphatases.
Non-conservative forces
If we need to account for nonconservative forces such as friction, we can do so using the equation: Etotal - Enonconservative = Efinal
Column chromatography
In column chromatography, the stationary phase is a vertical column packed with an adsorbent with carefully chosen properties. This adsorbent can attract sample molecules based on charge, size, or affinity for specific ligands.
Plasma membrane
In eukaryotic cells, composed of amphipathic phospholipids that have a polar head and a nonpolar tail. This structure contributes to the formation of a bilayer membrane: polar phosphate heads face the intracellular and extracellular environments (which are both aqueous solutions), and nonpolar tails remain inside the membrane. In addition to phospholipids, the presence of cholesterol and lipid rafts w/in the plasma membrane contribute to the fluidity of the membrane at lower temperatures and to its stability at higher temperatures.
Isomerases
Isomerases catalyze reactions that transfer functional groups within a molecule so that a new isomer is formed to allow for structural or geometric changes within a molecule.
Ligases
Ligases are used in catalysis where two substrates are stitched together (i.e., ligated) via the formation of C-C, C-S, C-N or C-O bonds while giving off a water (condensation) molecule.
Lyases
Lyases catalyze reactions where functional groups are added to break bonds in molecules or they can be used to form new double bonds or rings by the removal of functional group(s). Decarboxylases are examples of lyases.
Frequency (Hz) of a wave
Number of complete wavelengths that pass a given point per second. Longer wavelengths correspond lower-frequency radiation, and shorter wavelengths correspond to higher-frequency radiation. We can relate the velocity of a wave to its frequency and wavelength using the equation v = λf.
Oxidoreductases
Oxidoreductases catalyze oxidation-reduction reactions where electrons are transferred. In metabolism, these electrons are usually in the form of hydride ions or hydrogen atoms. When a substrate is being oxidized, it is the hydrogen donor. Examples include reductases, oxidases, and dehydrogenases.
Peptide bonds
Peptide bonds link together AA's - involves the formation of an amide via the condensation of the -COOH group of one amino acid with the -NH2 group of another (technically a condensation reaction, as it results in the production of a water molecule) Peptide bonds are stable under intercellular conditions, due in part to resonance stabilization; in addition to the form with a C=O double bond and a C-NH2 single bond, a resonance structure with a C-O− single bond and a C=N+H2 double bond exists. A consequence of this resonance structure is that peptide bonds are planar and do not rotate freely, which helps contribute to the structural stability of three-dimensional polypeptide structures. Peptide bonds are broken by hydrolysis (reverse of the condensation reaction through which a peptide bond is formed). The hydrolysis of peptide bonds is energetically favorable but is extremely slow under physiological conditions. For this reason, the breaking of peptide bonds (i.e., the destruction of the primary structure of a protein) is generally accomplished by specific enzymes in living cells.
Bilayer membrane
Permeable to very small uncharged molecules (i.e. common blood gases) and lipid-soluble molecules (i.e. steroid hormones), that can freely diffuse through the cell membrane. Not permeable to larger, hydrophilic molecules, such as glucose. Such molecules require specialized transporters, which can be regulated to a much greater extent than diffusion processes. The lipid bilayer structure that results from entropically-favorable interactions between phospholipids ultimately allows the body to perform the physiologically essential task of closely regulating what goes into and out of the cell.
Zero-order
Physiologically, zero-order reactions are exemplified by enzyme-catalyzed reactions in which the enzyme is saturated—that is, when concentrations of the reactant far exceed the available active sites on enzymes. In such a situation, the catalysis is the rate-limiting step and the concentration of the reactant is irrelevant.
Bicarbonate buffer system
Plays a major role in maintaining the pH of blood within the narrow range of ~7.35-7.45 that is compatible with health. (CO2 + H2O ⇌ H2CO3 ⇌ HCO3- + H+) It also encodes a key relationship for understanding the importance of respiration in the context of human metabolism: through Le Châtelier's principle, increasing the amount of CO2 in the blood (as occurs as a result of cellular respiration) pushes the bicarbonate reaction to the right, increasing the amount of H+ present in the blood and thereby reducing its pH. Thus, it is essential for the body to be able to breathe out CO2 and breathe in O2.
Recrystallization
Recrystallization is used to purify a solid product that contains impurities. This process involves the dissolution of the solid in a solvent and subsequent heating. The solid then dissolves and is cooled, causing it to solidify (crystallize) again. As the lattice structures of solids tend to exclude impurities, each subsequent recrystallization results in a progressively purer compound.
Second order
Second-order reactions physically involve collisions between two reactant molecules, as in SN2 reactions. In multi-step reactions, the step with the slowest rate determines the overall rate of the reaction, so it is known as the rate-limiting step.
Separation techniques
Separation techniques used to prepare purified substances for analysis or reaction. The best technique for a given scenario depends on the phases of the substances being separated. If all are liquids, one may be able to utilize distillation, which aims to separate liquids by utilizing the difference between their boiling points. The liquids are initially held in the same round-bottom flask, termed the distilling flask. This flask is positioned above a heat source, typically a Bunsen burner. The top of the flask is connected to a column, which leads to a downward-sloping glass condenser over a receiving flask. The condenser is held within a glass casing through which cold water is pumped. As the round-bottom flask is heated, the liquid with the lower boiling point will begin to vaporize, and its vapor will travel up the column and re-condense to fall into the receiving flask. The eventual result is a receiving flask that is rich in the lower-BP liquid, while the distilling flask will still contain the liquid(s) with the higher BP. If boiling points are very high, a vacuum may also be used to lower atmospheric pressure, which lowers the boiling points of all substances involved.
pKa/pKb
Since K represents the ratio of products to reactants at equilibrium, the larger the dissociation constant for a species (Ka for acids, Kb for bases), the stronger the species. K values for weak acids and bases K values are usually converted to pK values (pK = −logK), much in the same way as [H+] values are converted to pH. Most organic acids and bases and species of biological origin are weak, so unfamiliar acids or bases are best assumed to be weak.
Conjugation
Special case of resonance that occurs when 3+ adjacent p-orbitals are aligned with each other, forming not just a π bond, but a π system. Electrons can delocalize throughout that π system. For the MCAT, conjugation can be associated with structures containing alternating single and double bonds in carbon chains. An important characteristic of compounds with conjugated systems is that they absorb ultraviolet (UV) light, and can therefore be well visualized using UV spectroscopy.
Doppler effect
The Doppler effect describes how the observed frequency of a sound emitted from an object can change if the object and/or the observer is in motion. If both the source and observer are moving relative to each other, the Doppler equation, shown below, is capable of accounting for the velocity of both the observer and the source: f' = f(v + vobserver)/(v - vsource) Sign convention is important. The vobserver term is positive if the observer is moving toward the source, while vsource is positive if the source is moving away from the observer.
Capacitor
The charge stored by a capacitor is a function of its capacitance and voltage (Q = VC). When capacitors are connected in series, their capacitance adds reciprocally, like how resistance adds for resistors in parallel (1/Rtotal = 1/R1 + 1/R2 + ... + 1/Rn). When capacitors are connected in parallel, their capacitance adds directly, like resistors in series (Rtotal = R1 + R2 + ... + Rn).
Energy of EM Wave
The energy of an EM wave can be related to its frequency and wavelength (E = hf = hc/λ), h is the Planck's constant. Wavelength of light is inversely proportional to frequency. An increase in frequency produces a proportional decrease in wavelength, with a corresponding increase in the energy of the photons that compose the light. Upon entering a new medium (such as glass or air from space), the velocity and wavelength of light change (although the velocity of light in a vacuum is higher than in any other medium), while the frequency remains unaltered.
Functional groups
The functional groups of a molecule predict its physical properties and reactivity
Rate law
The rate law of a general reaction of the form aA + bB → cC + dD is: rate = k[A]x[B]y. Reaction rate units of M/s or mol/L∙s. [A] and [B] are concentrations of the reactants in units of M or mol/L. The order of a reaction is defined by the sum of the exponents (x + y) in the rate law. Rate laws can be determined experimentally using the method of initial rates, where multiple trials are run with variations in the concentration of individual reactants. The initial rate is then measured, with the goal of identifying how changes in the concentration of a reactant affect the rate. Rate constant k can be calculated from any experimental trial after the exponents of the rate law have been established. It should be noted, though, that while the units of M (mol/L) vary according to the overall order of the reaction, the units of k must always involve inverse seconds. Although the rate constant must be experimentally determined, it varies according to the activation energy (Ea) and temperature. Decreasing Ea through a catalyst/enzyme and increasing the temperature will increase the reaction rate.
Reaction rate
The rate of a reaction—how fast it happens—is expressed as a function of the rate constant k and the concentration of some or all of the reactants.
Ohm's Law
The three quantities of current, voltage, and resistance are linked together in Ohm's law, which is essential knowledge: V = IR. Power, which is conceptually equivalent to work over time, can be expressed as P = IV
Transferases
Transferases catalyze transfer of a chemical group from one molecule (donor) to another (acceptor). Most of the time, the donor is a cofactor that is charged with the group about to be transferred. Examples include kinases and phosphorylases.
Degree of unsaturation
Used to predict the structural features of an unsaturated compound by indicating how many double-bond equivalents are present in the compound. It can be defined as (2C + 2 + N - H - X)/2 C is the number of carbons, N is the number of nitrogens, H is the number of hydrogens, and X is the number of halogens.
Lewis
acid: electron acceptor / base: electron donor Thus, a substance like BF3 can be considered a Lewis acid when it reacts with NH3. (mnemonic: acids are acceptors of electrons.)
oxygen containing functional groups
alcohols (RC-OH), aldehydes (RC(=O)H), ketones (RC(=O)R'), and carboxylic acids (R(C=O)OH) Due to hydrogen bonding, alcohols and carboxylic acids have higher melting/boiling points than aldehydes and ketones, and can function as organic weak acids. Carbonyl (C=O) carbons have a significant partial positive charge and therefore often act as electrophiles. The -OH group of carboxylic acids can be replaced by other functional groups to form carboxylic acid derivatives, the most notable (in increasing order of reactivity): amides (R(C=O)NR'R''), esters (R(C=O)OR'), acid anhydrides (R(C=O)O(C=O)R'), and acid halides (R(C=O)X)
Hückel's rule
having (4n + 2) π-electrons, where n is an integer
Henderson-Hasselbalch equation
pH = pKa + log[A-]/[HA] [HA] refers to a generic weak acid and [A-] refers to its conjugate weak base The basic idea of a buffer is that if we add a small to moderate amount of a strong acid or base to the buffered solution, it will entirely protonate or deprotonate some of the weak acid/base. However, since the Henderson-Hasselbalch equation refers to the log of the [A-]/[HA] ratio, doing so will only minimally affect the pH.
