C/P Exam Q's
Reaction 1 can also be run using peptide derivatives that contain Se in place of S. Compared to the original reaction, the replacement of S with Se results in: A) A faster reaction, since Se is a better nucleophile than S B) A slower reaction, since Se is a weaker nucleophile than S C) A faster reaction, since Se is a better electrophile than S D) A slower reaction, since Se is a better electrophile than S
A - Se and S both carry lone pairs that allow them to act as nucleophiles, not electrophiles, so choices C and D can be eliminated. Nucleophilicity increases going down the periodic table due to greater size and polarizability, therefore Se is a stronger nucleophile than S, eliminating choice B. The use of a stronger nucleophile would cause the reaction using Se to go faster than the one using S.
In vivo, PO43- molecules utilize proton-coupled transport mechanisms to enter the cell. Which statement explains why this is necessary? A) PO43- is hydrophilic. B) PO43- is hydrophobic. C) The cell membrane is hydrophilic. D) The secondary active transport mechanism conserves ATP.
A - The structure of phosphate shows it to be negatively charged and polar, which makes it hydrophilic. Thus, it cannot cross the hydrophobic interior of the membrane.
Which of the following types of electromagnetic radiation would have the shortest wavelength? A. Radiation that ejects an electron from an sp orbital B. Radiation that ejects an electron from an sp2 orbital C. Radiation that ejects an electron from an sp3 orbital D. Radiation that excites but does not eject an electron from an sp3 orbital
A - This question asks us about the wavelength of EMR that ejects an electron from an atom. Shorter-wavelength EMR (such as γ rays)carries much more energy than longer-wavelength EMR (such as radio waves). Therefore, we must look for the answer choice that involves the highest-energy EMR. The closer an electron is to the nucleus, the harder it is to eject. Because sp-hybridized orbitals have the most s character of all of the answer choices, they contain the electrons that are hardest to eject
Redox Reactions
A primary alcohol can be oxidized to an aldehyde by a mild oxidizing agent (such as PCC) or to a carboxylic acid by a strong oxidizing agent like NaCr2O7. A secondary alcohol will be oxidized to a ketone by either a mild or a strong oxidizing agent. A strong oxidizing agent will likewise oxidize an aldehyde to a carboxylic acid. A strong reducing agent, such as LiAlH4, can reduce a carboxylic acid directly to an alcohol, while weak reducing agents such as NaBH4 will not reduce carboxylic acids at all. A special agent, DIBAL, can reduce a carboxylic acid to an aldehyde when applied at a precise 1:1 ratio. Both mild and strong reducing agents can reduce aldehydes and ketones to primary and secondary alcohols, respectively.
Bronsted-Lowry Base
Accepts protons H+ (Conjugate Acid)
Dissociation Energy & Acidity
Acidity is inversely proportional to bond strength of H - a small dissociation energy = least energy to dissociate = H is easily abstracted
Which two organelles appear most similar when viewed under the electron microscope? A) Nucleus and lysosome B) Smooth ER and golgi apparatus C) Mitochondrion and plasma membrane D) Peroxisome and nucleus
B
Lipase works to hydrolyze which of the following functional groups? A) Amide B) Ester C) Ether D) Hemiacetal
B - According to the passage, lipase is an enzyme that breaks down fats, or triglycerides, which contain three ester functional groups. Amides are the functional groups present in proteins so eliminate choice A, hemiacetals are present in sugars (eliminate choice D), and ethers are not largely important functional groups in any biologically important molecules (eliminate choice C).
In the electron transport chain, which of the following is the component that is reduced but never oxidized? A. Ubiquinone B. Oxygen C. Nicotinamide adenine dinucleotide D. ADP
B - Almost all elements in the chain are at some point reduced and oxidized. This is the driving force behind the electron transport chain. However, at the end of the chain, oxygen is reduced to form H2O, which is the end fate of the electrons in the chain.
Water is a rare substance in that the solid is less dense than the liquid at the freezing point, resulting in a solid form that floats on top of the liquid. Which of the following best explains this phenomenon? A) The bent structure of the water molecule results in a molecular dipole that maximizes the close molecular packing structure in the solid state. B) The bent structure of the water molecule and ratio of covalently-bonded hydrogens to lone pairs of electrons on the oxygen atom maximizes the hydrogen bonding that occurs in the solid phase, producing a hexagonal structure with large empty spaces. C) The degree of ionization in the solid state is less than in the liquid phase. D) The London dispersion forces of water significantly decrease in the solid phase as compared with the liquid phase.
B - For most substances, the liquid becomes more dense as the average kinetic energy (temperature) decreases, and the solid is more dense than the liquid due to close-packing solid-state structures, resulting in the formation of the solid at the bottom of the liquid. However, as stated in the question, solid water (ice) is significantly less dense than the liquid form at 0°C (the melting/freezing point). Remember, the water molecule is bent (shown below), with a bond angle of approximately 104.5°. This, combined with the degree of hydrogen bonding that can occur between water molecules, yields a solid crystalline structure with relatively large amounts of empty space. As a result, solid water is less dense than its liquid form.
Electronegativity, electron affinity, and ionization energy all increase across a periodic table row and decrease down a periodic table group. Why does acidity not follow this trend? A) Acidity relates to the ability of the nuclear protons to attract hydrogen atoms and not valence electrons. B) Acidity relates to the stability of the conjugate base, and larger atoms form more stable conjugate bases even when they are less electronegative than smaller atoms. C) Acidity is not influenced by nuclear shielding and effective nuclear charge, while the other trends are. D) Acidity increases across a periodic table row like the other trends, but it increases down a group because the electronegativity of the elements approaches that of hydrogen.
B - Given H-X, acidity increases as dissociation into H+ and X- increases. As the electronegativity of X increases across a periodic table row, the dissociation and acidity increase just like the other trends in the question. As the size of X increases, the electrostatic force between H and X decreases. This also makes the dissociation more likely and therefore acidity increases going down the periodic table (i.e., the opposite direction of the other trends listed). Larger atomic anions are more stable conjugate bases because they distribute charge better. The size effect overcomes high values of electronegativity as illustrated by the fact that HI is a strong acid and HF is not. Choice A is eliminated because acidity relates to the ability of an atom to attract valence electrons away from a hydrogen ion, thereby making proton donation (loss of a hydrogen ion) easier. Choice C is eliminated because acidity does increase with increasing effective nuclear charge across the periodic table. Choice D is eliminated because the electronegativity of elements only approaches that of hydrogen when going down the columns on the right side of the periodic table. In addition, this choice does not adequately explain the role of atomic radius in determining acidity.
Which of the following compounds is likely the most efficient at hydrolyzing a triglyceride? A) HCl(aq) B) NaOH(aq) C) H2O(aq) D) Na2SO4(aq)
B - Hydroxide ions (OH-) are very efficient at hydrolyzing (also known as saponifying) triglycerides due to the high nucleophilic capacity of these ions (choice B is correct). It is possible for water to do this, but it would occur much more slowly since water is not as good of a nucleophile (choice C is incorrect).
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, which are intended to resist changes in pH, consist of a weak acid or weak base and its corresponding salt. The passage states that the buffer solution is acidic (pH = 3.8), so we can assume that a base is not present in the solution. Therefore, any Roman numeral (RN) that is a weak acid or a salt formed from a weak acid is a potential component of the buffer mentioned in the passage. In RN questions, we should begin by looking for the RN that appears most often in the answer choices and discount those choices that contain an answer that is obviously wrong. RN I and II are in 3 choices, so we can begin with either of those, and we can discount any answer that has III, a weak base. A polyprotic, weak acid fits the criteria for an acidic buffer solution (I). Additionally, Na2HPO4 (phosphoric acid) is a salt formed from a weak acid and would therefore be an appropriate choice as a buffer (II). II and III only Ammonia is a weak base in solution, so it is unlikely to be a component of the buffer solution from the passage because the passage states that the solution is acidic. Thus, C can be discounted.
Which of the following statements regarding the anode of a voltaic cell is false? A) The anode is the site of oxidation B) Current flows away from the anode C) The anode has a lower reduction potential than the cathode D) Electrons flow away from the anode
B - In all electrochemical cells, the anode is the site of oxidation (choice A is true). Additionally, electrons always flow from anode to cathode (choice D is true). Since current flows in the opposite direction of the flow of electrons, it flows from cathode to anode (choice B is false). In a voltaic cell, or galvanic cell, the cathode has a higher reduction potential than the anode (choice C is true).
When in vivo studies were performed on the three drug-polymer combinations, it was found patients in the polaxemer-188 group experienced the most stomach pains after administration. Given the results of the study, this is most likely due to: Gelucire - 0.6 g/mL Beta-Cyclodextrin 0.3 g/mL Polaxemer 188 - 0.1 g/mL A) Elevated drug concentration causing inflammation of the gastric mucosa. B) Tablet fragments causing inflammation of the gastric mucosa. C) Glipizide/polaxemer-188 matrices being the smallest of the 3 tested. D) Increased pH of the stomach contents.
B - Since the glipizide/polaxemer-188 pill displayed the lowest solubility, the pill must remain partially undissolved for a longer time than the other medications, and the resulting solids could cause the most inflammation.
Which of the following is NOT an assumption used for ideal gases? A) The molecules have no attraction or repulsion from one another B) The molecules undergo inelastic collisions C) Gases are made up of discrete, small particles called molecules D) The volume of the particles is small compared to the volume of the container
B - The molecules undergo elastic collisions, not inelastic collisions. Answer choices A, C, and D are assumptions that are made for ideal gases.
What is the primary role of the rough endoplasmic reticulum? A) Metabolize toxins B) Produce proteins to be sent through the secretory pathway C) Synthesize the Golgi apparatus D) Synthesize rRNA
B - The role of the rough endoplasmic reticulum is to produce proteins that are sent through the cell's secretory pathway. The smooth ER metabolizes toxins (choice A is incorrect). The Golgi apparatus is not itself synthesized by the rough ER (choice C is incorrect). rRNA is synthesized in the nucleolus (choice D is incorrect).
Boiling Point Elevation
Boiling point elevation directly follows from vapor pressure reduction. The lower the vapor pressure, the more energy that will be required to increase that vapor pressure to a level that matches the atmospheric pressure. In other words, the more solute particles present, the lower the vapor pressure, and the higher the boiling point. The extent of this elevation can be calculated using the formula ∆Tb = iKbm, where ∆Tb refers to the amount by which the boiling point is elevated, i refers to the ionization or van't Hoff factor, which refers to the number of ions each solute molecule dissociates into in solution, and m refers to solute molality.
A drive-reduction and cognitive theorist would argue that depression is most strongly correlated with a deficiency in which component of fulfillment? A) Socialization B) Stigma C) Arousal D) Self-efficacy
C - Drive-reduction theories suggest that depression stems from a reduction in the motivating forces of arousal. A cognitive theorist would argue that arousal is essential to sustaining most behaviors.
The preferred ion configuration of many elements on the periodic table is determined by: A) The electron configuration of the nearest noble gas; elements will gain or lose electrons until they have the same core electron configuration as this noble gas. B) The electronegativity of the element directly above it, within the same group; elements will gain or lose electrons until they have an electronegativity that matches the period immediately above them. C) The electron configuration of the nearest noble gas; elements will gain or lose electrons until they have the same valence electron configuration as this noble gas. D) The relative stability of the nearest d subshell; elements will gain or lose electrons until the outermost d subshell available is stable.
C - Elements will often adopt a configuration with a complete octet - the configuration found in the noble gases. This complete octet is preferred for the outermost valence electrons.
Which of the following could affect any of the dissociation constants for hemoglobin? I. Decreasing the temperature II. Adding NH3 III. Increasing the [O2] A) I only B) III only C) I and II only D) I, II, and III
C - K′ is constant for any equilibrium at a certain temperature, and changing the temperature altersK′. Therefore, Item I is correct and choice B can be eliminated. According to the passage, lowering the pH alters the dissociation equilibrium of hemoglobin, preferentially stabilizes its tense form, and lowers its affinity for O2. This indicates that the hemoglobin-oxygen dissociation equilibrium as a whole and its dissociation constant are pH dependent. Ammonia, NH3, is a weak base and will affect the pH of the system and alter K′. Item II is correct and choice A can be eliminated. Adding [O2] will temporarily disturb the system's equilibrium, but the system will return to equilibrium according to Le Châtelier's principle until Q = K′. K′ will remain constant with the addition of [O2] which makes Item III false and choice C the correct answer.
Which of the following properties explains why sound wave velocity increases as it travels from air to a liquid medium at the same temperature? A) Kinetic energy of the medium B) Volume of the medium C) Density of the medium D) Bulk modulus of the medium
D - Density has an inverse relationship to sound wave velocity. A density increase (e.g., from warm air to cold air) would cause velocity to decrease. In general, sound travels fastest in solids and slowest in gases. The equation for sound velocity is given by v = √(Κ/ρ), where K is a coefficient of stiffness termed the bulk modulus (or the modulus of bulk elasticity for gases) and ρ is the density.
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 - The question stem asks about attenuation, which is a weakening of the U/S signal. Sound energy is attenuated as it passes through the body because parts of the signal are reflected, scattered, absorbed, refracted or diffracted. Amplification does the opposite of attenuation; it makes the signal stronger. Since this is an "EXCEPT" question, amplification is our answer.
Concentration cells
Flow of electrons and the direction of current are opposite of one another
Gibbs free energy
Gibs free energy uses Kelvin; Room temp at standard conditions os 0C/273K; Under non standard conditions room temp is 25C/298K
Commonly Found Strong Bases
Potassium hydride: KH Lithium hydride: LiH Lithium hydroxide: LiOH Sodium hydroxide: NaOH Potassium hydroxide: KOH Rubidium hydroxide: RbOH Caesium hydroxide: CsOH Magnesium hydroxide: Mg(OH)2 Calcium hydroxide: Ca(OH)2 Strontium hydroxide: Sr(OH)2 Barium hydroxide: Ba(OH)2
Freezing Point Depression
This property stems from the idea that solute molecules disrupt the lattice structure of the frozen solvent, so more added solute corresponds to more "difficulty" freezing and a lower freezing point. The equation for freezing point depression is ∆Tf = iKfm.
Lead storage battery
When discharging, its two cells are a lead anode and a lead (IV) oxide cathode that are connected by sulfuric acid. When it is completely discharged, the battery has two identical lead sulfate (PbSO4) electrodes. To bring the battery back to its charged state, these electrodes dissociate to bring back the original anode (Pb) and cathode (PbO2).
ΔG = -RTln(K/Q) = RT ln(Q/
ΔG can be calculated with the equation: ΔG = -RTln(K/Q) = RT ln(Q/K). The question stem states that K=10 and Q = 29. Since Q is greater than K, Q/K must be greater than one, and ln(Q/K) must be positive. The values of R and T are always positive. ΔG will therefore yield a positive value (choices A and B are incorrect). The forward reaction must be nonspontaneous. As a result. the reaction will progress leftward to reach equilibrium (choice D is incorrect).
Which of the following is a weak acid in aqueous solution? A. HF B. HCl C. HBr D. HI
A - All but one of the hydrohalic acids completely dissociate. The one exception is hydrofluoric acid, which is a weak acid in aqueous solution due to its unusually high degree of covalent bonding between the fluorine and the hydrogen.
Which of the following elements is MOST likely to be a strong nucleophile? A) Hydroxide ion B) Water C) Ethanol D) Tert-butanol
A - Charge, electronegativity, H-bonding capacity, and steric bulk are important in determining how strong a nucleophile is. Here, hydroxide ions are negatively charged (OH-), strongly electronegative, and have no steric hindrance (choice A is correct). Water and ethanol are not very strong nucleophiles (choices B and C are incorrect). Tert-butanol has a lot of steric hindrance (choice D is incorrect).
Medical students may feel helpless when confronted with unethical behavior because they believe that reporting will have serious negative repercussions. As time goes by, they become more stressed and upset about the situation. What type(s) of stress are they likely feeling? I. Distress II. Eustress III. Neustress A) I only B) II only C) III only D) I and III only
A - Distress is a negative type of stress that builds over time and is bad for your body (I). It happens when you perceive a situation to be threatening to you some way (physically or emotionally) and your body becomes primed to respond to the threat; eustress is a positive type of stress that happens when you perceive a situation as challenging, but motivating. Eustress is usually enjoyable; neustress is a neutral type of stress. Neustress happens when you are exposed to something stressful, but it doesn't actively or directly affect you. For example, news about a natural disaster on the other side of the world may be very stressful, but your body doesn't perceive that stress as good or bad for you so you aren't affected
Lathyrism is known to target and degrade collagen in the lower limbs. Which pair of amino acids would best provide disulfide links to stabilize the folded form of collagen? A) C and C B) M and M C) M and C D) S and S
A - Disulfide linkages only form between the side chains of cysteine (C) residues. When not part of such a linkage, cysteine side chains contain a thiol group (S-H). Through the process of oxidative folding, two thiol groups can connect and form an S-S bond, as shown below. These linkages are an important part of protein tertiary and quaternary structure. Methionine does contain sulfur, but it cannot form disulfide bonds. This apparent contradiction can be explained by looking at the structure of methionine. The side chain of this amino acid includes a sulfur atom bound to carbon instead of hydrogen. As such, it cannot easily lose its proton and form a disulfide linkage through oxidation.
Which of the following accurately describes ethanoic acid? A) Weak acid B) Strong acid C) Titration indicator D) Conjugate base
A - Ethanoic acid, as the name suggests, is an acidic compound. The removal of a proton from ethanoic acid results in the formation of a conjugate base (choice C is incorrect). In this context, ethanoic acid is not used as a titration indicator (choice D is incorrect). Based on the information provided in reaction 3, ethanoic acid does not completely dissociate in solution and instead exists in an equilibrium state consisting of its undissociated reactants and dissociated products. Thus, it is a weak acid (choice B is incorrect).
At which nucleophilic site is serine most likely to be acylated by penicillin? A) Side chain OH B) Carbonyl carbon C) Carbonyl OH D) Alpha carbon
A - Good nucleophiles are electron-rich species. The likely nucleophile would also need to minimize steric hindrance to be acylated by penicillin. The side chain hydroxyl group fits both of these criteria. A carbonyl carbon would more properly be thought of as electrophilic. The question asks about a nucleophilic site. The carbonyl OH is a possible nucleophile, but its proximity to the carbonyl group makes it more sterically hindered than the side chain OH, reducing its ability to interact with penicillin. Also, the resonance stabilization conferred in this carboxylic acid makes its -OH group relatively less nucleophilic than a typical alcohol.
The class of medications described in the passage (opiods) are similar to endogenous hormones produced in what endocrine gland? A) Anterior pituitary B) Posterior pituitary C) Adrenal cortex D) Adrenal medulla
A - Opioid analgesics mimic the effect of a group of hormones called endorphins, which naturally inhibit pain by blocking the release of certain neurotransmitters at nociceptors. Endorphins are produced in the anterior pituitary gland, along with other hormones such as luteinizing hormone (LH), follicle stimulating hormone (FSH), and growth hormone (GH). The primary hormones produced by the posterior pituitary are vasopressin, involved in fluid reuptake in the kidney, and oxytocin, a hormone associated with emotional bonding.
What effect will the addition of a fluorine substituent have on carbocation stability? A) The fluorine group will be destabilizing because it is highly electronegative. B) The fluorine group will be stabilizing because it is highly electronegative. C) The fluorine group will be destabilizing because it has additional lone pair electrons. D) The fluorine group will be stabilizing because it has additional lone pair electrons.
A - The first part of each answer describes whether the fluorine group is stabilizing or destabilizing. From Table 1, we see that the fluorine group has an RSE of +2.9 and is therefore destabilizing (eliminate B and D). Next, we must explain this destabilizing influence. Carbocations are typically unstable due to their high concentration of positive charge. If a substituent can contribute some electron density to the carbocation, this positive charge will be "balanced" and the species stabilized. In contrast, electron-withdrawing substituents pull even more electron density away from the carbocation, decreasing its stability further. Since fluorine is destabilizing, and since it is highly electronegative, we can deduce that it must be electron-withdrawing.
Modern MRI machines use electricity to generate their magnetic fields inside a circular chamber instead of using permanent magnets. Which of the following would NOT increase the strength of the MRI field? A) Increased radius of the MRI chamber B) Increased power supplied to the MRI machine C) Decreased resistance of the MRI machine D) Increased current through the MRI machine
A - The magnetic field strength generated at the center of a circuit loop is given by B = µoI/2r, where B is the field strength, µo is the permeability of space (a constant), I is the electrical current through the circuit, and r is the radius of the loop. Field strength is directly proportional to current and inversely proportional to radius. Increasing the radius of the chamber will lower the strength of the field at the center of the chamber.
Which of the following must be true if the penicillin-serine reaction "locks" in the ester formation as described in the passage? A) The thermodynamic cost of breaking the penicillin-serine ester is larger than the thermodynamic release of amide formation. B) The thermodynamic release of forming an amide is equal to the thermodynamic cost of forming an ester. C) The thermodynamic cost of breaking the penicillin-serine ester is smaller than the thermodynamic release of amide formation. D) The thermodynamic release of forming the penicillin cyclic amide is larger than the thermodynamic release of breaking the penicillin-serine ester.
A - The passage tells us that once penicillin forms an ester with serine, it prevents the further reactions by the enzyme because the serine residue is now locked into the ester with penicillin. For this to be the case, the ester must be stable. It must be so stable that breaking it apart would be too "expensive" - or too hard to do given the subsequent thermodynamic release that would occur from amide formation. If the cost to break open the penicillin-serine ester were small, then the molecules wouldn't be locked into that ester configuration.
A researcher performs an organic reaction in which a benzene with an alcohol substituent turns into a benzene with an aldehyde substituent. Which of the following techniques can be used to determine the progression of the reaction? A) Thin layer chromatography B) Dissolution in an organic solvent C) Protonation using an activated water D) Treatment with an oxidizing agent
A - Thin layer chromatography (TLC) separates molecules on the basis of polarity. Aldehyde is a nonpolar functional group, whereas alcohol is a polar functional group. Species with an alcoholic functional group would be attracted to the polar stationary phase, and thus produce a lower Rf than species with an aldehyde. This can be used to determine whether or not the product has formed (choice A is correct). B, C, and D are procedural steps you can complete in the lab, but they are not commonly used techniques for determining the progression of a reaction.
Boiling chips and vacuum distillation, respectively, are used in distillations to: A) Provide nucleation sites that give the liquid a place to start forming bubbles to prevent superheating; lower the boiling points of the substances to be distilled. B) Lower the boiling points of the substances to be distilled; work synergistically with the vacuum system to further lower the boiling points. C) Lower the boiling points of the substances to be distilled; provide nucleation sites that give the liquid a place to start forming bubbles to prevent superheating. D) Provide nucleation sites that give the liquid a place to start forming bubbles to prevent superheating; speed up the distillation process by vacuuming the first distillate out of the apparatus.
A - When transitioning from liquid to gas during boiling, the liquid needs nucleation sites, or places to start forming bubbles. This is typically achieved either by scratching the inside of the flask or by introducing boiling chips. Next, remember that boiling occurs when the Pvap of the substance in question equals the Patm. Typically, we boil substances by increasing the temperature, thereby increasing Pvap. Alternatively, however, we can lower boiling point by reducing Patm, which can be accomplished through the introduction of a vacuum. Vacuum distillation is often used when components have very high boiling points and would otherwise be difficult to distill.
Acid-Base Chemistry
A higher Ka is associated with stronger acidity. Adding electron-withdrawing functional groups (like fluorine atoms) provides inductive stabilization of the conjugate base by pulling electron density away from the negatively-charged position (shown below). This stabilization increases the acidity of the original compound. Inductive effects increase if the electron-withdrawing groups are closer to the acid and if more of them are present.
Which of the following values gives the number of parts per million (ppm) of NaCl in a 0.90% saline solution? A) 900 B) 9,000 C) 90,000 D) 900,000
B 1,000 parts/ 100 parts = 10,000 / 1% .90% (10,000/1%) = 9,000
In the research cited, if it was found that Hispanic patients were prescribed less pain medication than Caucasian patients due to Hispanic patients having disproportionally less insurance than Caucasian patients, then insurance coverage would be considered: A) A moderating variable. B) A mediating variable. C) A confounding variable. D) Both an independent and dependent variable.
B - A mediating variable is one which explains the association between two other variables through a causal relationship. Here, insurance coverage connects patient race and physician prescriptions. A moderator variable is one that influences the strength of a relationship between two other variables, and a mediator variable is one that explains the relationship between the two other variables. As an example, consider the relation between yearly income and frequency of testicular self-exams. Age might be a moderator variable, in that the relation between yearly income and testicular self-exam might be stronger for older men and less strong or nonexistent for younger men. Education might be a mediator variable in that it explains why there is a relation between self-exam and yearly income. This would mean when we remove the effect of education, the relation between the two variables weakens.
Which of the following statements is correct with regard to the role of creatine phosphokinase? A) It is involved in the committed step for creatine phosphate formation. B) It stabilizes the transition state in the formation of creatine from creatine phosphate. C) It lowers ΔG, making the reaction more spontaneous. D) It raises ΔG, making the reaction more spontaneous.
B - Creatine phosphokinase is the enzyme used to catalyze the reversible reaction between creatine and creatine phosphate. Catalysts increase the reaction rate by lowering the activation energy (stabilizing the transition state) of the reaction (choice B is correct). Note that this enzyme will stabilize the transition state between creatine and creatine-phosphate in BOTH directions. Because the specific reaction is reversible, it cannot be a "committed step" (choice A is wrong). Furthermore, catalysts do not affect ΔG and spontaneity (choices C and D are wrong).
A book rests horizontally on a table. The book experiences a gravitational force of mg due to the earth's gravity. According to Newton's third law: A) The book experiences a normal force of mg pushing up due to the table. B) The earth experiences a gravitational force of mg from the book. C) The table exerts a gravitational force of mg on the earth. D) The earth exerts a normal force up on the table equal to mg plus the weight of the table.
B - Newton's third law can be expressed as: FA on B = −FB on A Here, the force of the earth pulling on the book is equal to and opposite of the book pulling up on the earth. Got this right; but it was a tough question
What alteration to an HCN channel will reduce the transport of K+ in favor of Na+ transport? A) Increasing the diameter of the channel by adding amino acids B) Decreasing the diameter of the channel by removing amino acids C) Replacing the serine groups in the channel with alanine D) Replacing the alanine groups in the channel with serine
B - Paragraph 1 states that HCN channels are permeable to K+ and Na+ ions. Since Na+ is chemically similar to K+, one can assume the difference must be due to the size of the atom. The K+ ion is larger than the Na+ ion, so reducing the diameter of the channel can allow Na+ to enter while preventing K+ entry.
A researcher seeks to monitor the conversion of retinal to retinol using infrared spectroscopy. Which of the following will indicate the reaction is complete? A) The disappearance of peaks in the 3200-3500 cm-1 region B) The disappearance of a peak in the 1700-1750 cm-1 region C) The disappearance of a peak in the 1580-1640 cm-1 region D) The appearance of a peak in the 1700-1750 cm-1 region
B - Retinol differs from retinal in that it contains -OH groups, but does not contain a carbonyl group (C=O). The carbonyl stretching frequency falls in the range of 1700-1750 cm-1, while the O-H stretching frequency is expected to fall in the range of 3200-3500 cm-1. The reaction is thus complete when the signal in the 1700-1750 cm-1 range has completely disappeared. The 1580-1640 peak is indicative of C=C groups, which both molecules have and cannot be used to determine when retinal has been converted.
Which of the following best explains why, at a constant voltage, warmer solutions are ejected more quickly than cooler ones? A) The droplets have smaller volume at higher temperatures. B) The surface tension is lower for solutions at higher temperatures. C) The electrical forces grow smaller as temperature increases. D) The electrical forces grow larger as temperature increases.
B - Surface tension decreases as temperature rises, such that as the average kinetic energy of the fluid molecules increases, they are more likely to be able to break the intermolecular bonds holding the molecules together at the surface. Surface tension also decreases as the surface area of the fluid rises due to the molecules being stretched further apart, resulting in a thinner film of surface tension.
In one performance, the dancer spins at a constant speed while holding a basket of flowers in her hand. While spinning, the dancer holds her arms out to the side, so her arms make a 90° angle with her body. Which of the following is true for the circular motion of the basket of flowers? I. The spinning causes a centripetal force, which pulls the basket out from the center of the circle. II. The dancer's arm provides the centripetal force, which pulls the basket in towards the center of the circle. III. The basket has a constant velocity which is equal to the tangential velocity of the dancer. A) I only B) II only C) II and III only D) I, II, and III
B - The centripetal force is by definition a force pointing into the center of the circle, so Item I is false, eliminating choices A and D. Since both remaining answer choices include Item II, Item II must be true and we can focus only on Item III. Remember that velocity is a vector, and since the basket is travelling in a circle, it is constantly changing direction. Therefore, the velocity is not constant and Item III is false, eliminating choice C. The correct answer is choice B. Note that Item II is in fact true: the dancer's arm is providing the centripetal force that pulls the basket into the circle.
Which of the following explains why the pitch of person's resonant voice harmonics rises when they inhale helium instead of air? A) The wavelength increases due to the change in temperature from the helium atoms. B) The change in timbre results in a higher perceived frequency of the voice. C) The frequency decreases due to the change in air density from helium atoms. D) The speed of the sound wave remains constant regardless of changes to the medium through which the sound wave propagates.
B - The pitch of a perceived sound is dependent on its frequency. Thus, when helium atoms are inhaled, the density of the air is being changed, which alters the timbre (an overall quality of voice that results from overlapping waves) to emphasize the higher-pitched harmonics. We perceive that as a high-pitched, "squeaky" voice. A decrease in frequency would result in a lower-pitched, deeper voice, not a high-pitched squeaky voice.
GAPDH converts an aldehyde into a carboxylic acid in order to change NAD+ to NADH. Which of the following is true regarding this reaction in humans? A) The aldehyde is reduced. B) NAD+ is oxidized. C) NAD+ gains 2 e- and one proton. D) The carboxylic acid (pKa = 2.19) is mostly in its protonated form.
C - An aldehyde (—CHO) gains an O when it is converted to a carboxylic acid (-COOH). By definition, gaining an O is oxidation which makes choice A incorrect. Choice B is incorrect because NAD+ gains elections in order to become a neutral molecule. Gaining electrons is reduction, not oxidation. Choice D is incorrect because according to the Henderson-Hasselbalch equation (pKa - pH = -log([A-] / [HA])), the conjugate base (deprotonated form) is favored if pH > pKa. Since the human body (with the exception of the gastrointestinal tract) has a pH higher than 2.19, the deprotonated form of the carboxylic acid is favored. Recalling that an H+ is equivalent to a proton, it is apparent that an H+ is added when converting NAD+ → NADH. Since NADH is a neutral molecule, the positive +1 charge from the proton and the +1 charge from the NAD+ must be balanced by two electrons.
What best describes the relationship between attribution theory and fundamental attribution error? A) Attribution theory states that attribute phenotypes are associated with inner psychological functioning; fundamental attribution error refers to a misdiagnosis of psychological state based on overreliance on an attribute phenotype. B) Attribution theory relates to an attempt by an individual to interpret actions by assigning causes to them; fundamental attribution error is when an individual interprets another's actions incorrectly by overemphasizing external events instead of internal characteristics. C) Attribution theory relates to an attempt by an individual to interpret actions by assigning causes to them; fundamental attribution error is when an individual interprets another's actions incorrectly by overemphasizing internal characteristics instead of external events. D) Attribution theory states that attribute genotypes are associated with inner psychological functioning; fundamental attribution error refers to a misdiagnosis of psychological state based on overreliance on an attribute genotype.
C - Attribution theory relates to the ways in which people attempt to explain various behaviors and events. Fundamental attribution error occurs when a person assigns too much weight to internal causes rather than external factors when trying to explain the cause of another person's behavior.
Early research into the mechanism of antibiotics theorized that it was implausible that the amide in penicillin would convert to an ester. Which of the following explains this doubt? A) Amides possess better leaving groups than esters. B) Esters possess poorer leaving groups than amides. C) Amides are less reactive than esters. D) Serine contains no good nucleophiles.
C - Carbonyl reactivity is determined by the strength of the leaving group attached to the carbonyl. Good leaving groups are usually weak bases. For the MCAT, you can simply remember the relative reactivities of these carboxylic acid derivatives, starting with the most reactive: acyl halides > anhydrides > esters and acids > amides. It is not thermodynamically favorable for a molecule to transform from a less reactive (more stable) form, like an amide, to a more reactive (less stable) form, like an ester.
Which of the following statements most accurately describes concentration cells? A) Nonspontaneous electrochemical cells that generate a current using a concentration gradient B) Spontaneous electrolytic cells that generate a current via a concentration gradient C) Galvanic cells that generate a current via a concentration gradient D) Batteries that store energy by creating concentration gradients
C - Concentration cells are spontaneous electrochemical cells that generate current using concentration gradients. Concentration gradients of charged particles are created and separated using a film of semipermeable membrane. The movement of these charged particles down a concentration gradient generates a current(choice A is incorrect). Concentration cells are spontaneous, as they rely on the spontaneous movement of particles down a concentration gradient (choice D is incorrect). Electrolytic cells are strictly nonspontaneous (choice B is incorrect).
What is the isoelectric point of glycine? A) 2.0 B) 4.5 C) 5.5 D) 9.0
C - Glycine, like all amino acids, has a carboxyl group (pKa = 2) and amino group (pKa = 9). To find the isoelectric point (pI), simply take the average of these two values. pI = (pKa, 1+ pKa, 2) / 2pI = (2 + 9)/2pI = 5.5
The flow rate of stomach content emptying is 100 cm3/sec. Patients who undergo gastric bypass surgery will increase this rate to almost 1600 cm3/sec. Assuming the flow of stomach contents approximates Poiseuille's Law, what change to their gastrointestinal connection would explain this, provided no other changes occur in the conditions of stomach content flow? A) New connection is 2 times longer B) New connection is 4 times longer C) New connection radius is 2 times larger D) The new connection radius is 4 times larger
C - If we model this GI connection as a pipe, we can quantify the flow using Poiseuille's Law, which states: Flow = ΔPπr4 / 8Lη ΔP is the pressure difference across the pipe, r is the radius of the pipe, L is the length of the pipe, and η is the viscosity of the fluid in the pipe. The question states that the flow increased from 100 to 1600 cm3/sec, a 16-fold change. Of all the answers provided, only doubling the radius would result in a 24 increase in the flow.
32P has a very short half-life of 14 days and decays into sulfur as shown below. 32 P --> 32 S + e- 15 16 This nuclear reaction is an example of: A) Alpha decay. B) Beta plus decay. C) Beta minus decay. D) Positron emission.
C - In β+ decay, or positron emission, the parent nucleus converts a proton into a neutron while emitting a positron (e+). In β- decay, the weak parent nucleus converts a neutron into a proton while emitting an electron (e-). The atomic mass stays constant while the atomic number increases by 1.
In modern radiotherapy, superconducting wire coils are used to create magnetic fields that focus particle beams in order to deliver the maximum dose of ionizing radiation to a tumor site while minimizing the destruction of healthy tissue. For which of the following types of radiation would this technique NOT be useful? A) Alpha particles B) Proton beams C) Gamma photons D) Beta particles
C - Magnetic fields exert forces on moving charged particles. Since photons have no charge, they are unaffected by magnetic fields (choice C is correct). All of the other particles named are positively or negatively charged.
Which of the following is not a strong acid? A) Hydrochloric acid B) Hydroiodic acid C) Hydrofluoric acid D) Sulfuric acid
C - Most halogens form strong acids. The exception to this is fluorine, since hydrofluoric acid is not considered a strong acid (choice C is correct). The strong acids you should know for the MCAT include hydroiodic acid, hydrobromic acid, hydrochloric acid, perchloric acid, sulfuric acid, and nitric acid (choices A, B, and D are incorrect).
Which of the following waves would have a wavelength larger than any standard electrospun fiber? A) Red light B) Blue light C) Microwaves D) X-rays
C - Since only one of these options can be correct, we are simply looking for the choice with the largest wavelength. Visible light (ROYGBIV) ranges from approximately 7 x 10-7 m to 4 x 10-7 m in wavelength. X-rays are smaller than all visible light, with wavelengths around 1 x 10-11 m. Finally, microwaves are larger than visible light waves, with wavelengths of 1 x 10-3 m. As the longest-wavelength option, microwaves must be the correct answer.
Nitrogen primarily exists in the atmosphere as a diatomic gas. Which of the following is true about this form of nitrogen? A) The presence of a lone pair of electrons on each nitrogen atom in the molecule allows it to act as a strong Lewis base. B) The triple bond of electrons creates a region of high electron density that allows N2 to be very reactive as a nucleophile. C) Diatomic nitrogen gas is relatively inert and can be used as the atmosphere in laboratory conditions to prevent unwanted side reactions. D) Atmospheric nitrogen reacts spontaneously with carbon dioxide, which keeps atmospheric CO2 levels at a relatively low 0.04% (on a molar basis) of the atmosphere.
C - The MCAT will expect you to be familiar with N2 as a very inert gas. It makes up approximately 80% of the air you breathe, yet has no significant chemical reactions with your lungs - or with anything other than nitrogen-fixing plants. This information implies that nitrogen is very inert (unreactive). As such, it would serve as a good artificial atmosphere when working with reagents that might react with oxygen or other gases.
All of the following are products of the pentose phosphate pathway EXCEPT: A) Glycolytic intermediates B) NADPH C) NADH D) Ribose-5-phosphate
C - The products of the pentose phosphate pathway include fructose-6-phosphate and glyceraldehyde-3-phosphate (both glycolytic intermediates, choice A is wrong), NADPH (choice B is wrong), and ribose-5-phosphate (choice D is wrong). NADH is not a product of the pentose phosphate pathway, but is generated in several other pathways (e.g., glycolysis, PDC, Krebs cycle, and fatty acid oxidation, choice C is correct).
A ray of white light moves through the air and strikes the surface of water in a beaker. The index of refraction of the water is 1.33 and the angle of incidence is 30º. All of the following are true EXCEPT: I. the angle of reflection is 30º. II. the angle of refraction is 30º. III. total internal reflection will result, depending on the critical angle. A) I only B) I and III only C) II and III only D) I, II, and III
C - The question asks "all of the following are true EXCEPT,". The angle of incidence always equals the angle of reflection (I). This immediately lets us narrow it down to choice C. Light entering a more dense medium will bend towards the normal. With an angle of incidence of 30º, the angle of refraction must be less than 30º (II). Total internal reflection can only result when a ray of light begins in a higher-index material and reaches a boundary with a lower-index one (e.g. starting in water and moving towards air). Here, the light ray started in air (n = 1) and moved into water (n ∼ 1.3), making total internal reflection impossible (III).
Which of the following plays a large role in establishing resting membrane potential? A) Sodium voltage-gated channels B) Potassium voltage-gated channels C) Sodium potassium ATPase D) DNA polymerase II
C - The sodium potassium ATPase is involved in establishing resting membrane potential by pumping sodium ions outside of the cell and potassium ions into the cell (choice C is correct). Sodium voltage-gated channels are involved in membrane depolarization (choice A is incorrect). Potassium voltage-gated channels are involved in membrane repolarization (choice B is incorrect). DNA polymerase II is not involved in establishing resting membrane potential or in action potentials (choice D is incorrect).
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 - 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.
Electrolytic cells
Carry out nonspontaneous reactions and have a negative electromotive force. In other words, a source of electrical energy is required to perform the chemical half-reactions. This process is referred to as electrolysis, where the chemical species that undergo decomposition are known as electrolytes. Electrolytic cells are in direct contrast to voltaic cells. The anode of an electrolytic cell is positive, whereas the cathode is negative. In an electrolytic cell, an external voltage source makes the anode have a higher reduction potential than the cathode.
Competitive Inhibition
Competitive inhibition is when the inhibitor directly competes with the substrate for the active site on the enzyme. This increases Km since it now takes more substrate to ensure half of the active sites are occupied. If enough substrate is supplied, it will outcompete the inhibitor; therefore, Vmax does not change, although more substrate is needed to reach it.
Degrees of Unsaturation
Counting the degrees of unsaturation by the formula in this case would be too time consuming (and would give the wrong answer). Instead, from looking at the portion of the molecule that changes in the transformation (the spirolactone moiety), the total number of multiple bonds increases by one, but the number of rings decreases by one. Since rings and multiple bonds are both sources of degrees of unsaturation, the overall value remains unchanged. It is important to note that the removal of two protic hydrogens does not result in increasing the degree of unsaturation if the charges are localized on individual atoms. In cases like these, the charges count as formal-hydrogens on their respective oxygens in ;the degree of unsaturation calculation, and do not contribute to the formation of π-bonds or rings.
Which statement correctly describes both the 3s and 3p energy levels in the lowest-energy electron configuration of the phosphide anion P3− ? A) Both the 3s and 3p levels are half-filled. B) The 3s level is filled and the 3p level is half-filled. C) The 3s level is half-filled and the 3p level is filled. D) Both the 3s and 3p levels are filled.
D - An elemental atom of phosphorus has a total of 15 electrons and 15 protons (no net charge) with an electron configuration of 1s22s22p63s23p3. Therefore, a phosphide anion is the result of an elemental phosphorus atom gaining 3 additional electrons within the half-filled 3p subshell. As a result, the phosphide anion has a total of 18 electrons with a configuration of 1s22s22p63s23p6. This gives the phosphide anion a −3 charge and filled 3s and 3p energy levels.
Which of the following ranks the redox-active species of the electron transport chain in order of decreasing electron affinity? A) O2 > FAD > CoQ > NAD+ B) NAD+ > FAD > CoQ > O2 C) O2 > CoQ > NAD+ > FAD D) O2 > CoQ > FAD > NAD+
D - Because O2 is the terminal species in the electron transport chain (ETC), it must have the highest electron affinity and thus be first in the list (eliminating choice B). Since NADH donates its electrons to the first electron carrier in the ETC, the product of its oxidation, NAD+, would have the lowest electron affinity and would be last in the list (eliminating choice C). FADH2 enters the ETC by donating its electrons directly to ubiquinone (CoQ) so the oxidized FAD will have a lower electron affinity than CoQ (eliminating choice A).
The cyanine portion of the compound absorbs radiation with a wavelength of 560 nm, yet emits at 600 nm. Which of the following is the most plausible reason to explain the discrepancy? A) The molecule emits a more energetic photon than it absorbs because it decays to a new, lower energy ground state. B) The molecule emits a more energetic photon than it absorbs because it gains energy from the surrounding solvent in its excited state. C) The molecule emits a less energetic photon than it absorbs because the excited state of the molecule is more highly charged than the ground state. D) The molecule emits a less energetic photon than it absorbs because the excited molecule can lose energy through vibrational processes prior to photonic emission back to its ground state.
D - Choices A and B can be eliminated as shorter wavelengths equate to photons of larger energies (E = hc/λ). Choice C is incorrect as the excited state and ground state have the same overall charge (there is no electron transfer). Choice D is correct. In an idealized picture, a molecule absorbs a photon and emits the same photon when returning to the ground state; however, this is not universally the case. Once in the excited state the molecule can relax a bit of its energy as vibrational movement (heat), before losing what is left of its energy as an emitted photon. In this scenario, the emitted photon will always have a longer wavelength than the absorbed photon.
An object is placed in front of a concave mirror between the focal point and the mirror. If the object is then moved closer to the mirror, the image would be: A) Larger and farther from the mirror. B) Larger and closer to the mirror. C) Smaller and farther from the mirror. D) Smaller and closer to the mirror.
D - Conceptually, an object placed directly at the focal point would produce no image (i.e., an image infinitely far away). Therefore it makes sense that moving away from the focal point would produce an image closer to the mirror. This would eliminate choices A and B. Also, when an object is placed between the focal point and the mirror, the image is virtual and larger than the object. As the object moves closer to the mirror, the image would get smaller and eventually become the same size as the object when the object is touching the mirror. The answer is therefore choice D.
From one point in space, Point S, to another, Point T, electric potential increases continuously from 100 V to 200 V. Which of the following must be true of the electric field near these points? A) Field lines point away from both S and T. B) Field lines point towards bothS and T. C) Field lines point from S to T. D) Field lines point from T to S.
D - First, positive charges undergo a force from higher electric potential to lower electric potential. This happens because all things undergo forces from higher potential energy to lower, and from ΔPE= qΔφ, a positive charge means that q is positive, so ΔPE is negative (the positive charge is going to lower potential energy) when Δφ is negative (the positive charge is going to lower electric potential). Second, electric field lines point in the direction that a positive charge would undergo a force (by definition). Thus, electric field lines would point in the direction of lower electric potential, from 200 V to 100 V (that is, from T to S). Choice C is backwards and can be eliminated. The question stem doesn't describe the rest of the surroundings, so we have no way of knowing whether choices A or B would be mostly true; neither one would be completely true, because at least one field line (coming from T to S) points away from T, eliminating choice B, and at least one field line (coming from T to S) points towards S, eliminating choice A.
Which of the following offers the most likely explanation for the substantially higher peak plasma concentration of acetaminophen following IV administration when compared to oral and rectal routes? A) The drug is poorly absorbed through mucous membranes in the oral cavity and rectum. B) A substantial portion of the drug is excreted in the feces before it can be absorbed. C) Much of the drug is broken down in the digestive tract prior to absorption. D) The liver breaks down a large portion of the drug before it reaches the systemic circulation after oral or rectal administration.
D - IV administration allows a drug to bypass the liver (the primary organ responsible for drug metabolism, choice C is wrong) in order to reach the systemic circulation at the highest possible level. While it is true that absorption is poor through the oral and rectal mucosa, the drug is not being aborbed there. Both swallowing the drug (oral administration) and rectal administration introduce acetaminophen into the GI tract, and with a rise in plasma acetaminophen levels only a half hour after oral or rectal administration, acetaminophen is being readily absorbed by intestinal mucosa (choice A is true but does not explain the difference in peak plasma concentrations). With an onset of action of less than a half hour after oral administration, acetaminophen is readily absorbed by the oral mucosa (choice A is wrong). While some of the drug will be excreted unabsorbed, this is not the primary reason for the difference between the peak plasma concentrations (choice B is wrong).
Which of the following accurately describes the differences between a voltaic cell and an electrolytic cell? A) A voltaic cell is nonspontaneous, while an electrolytic cell is spontaneous B) In a voltaic cell, the anode is the site of oxidation, whereas in an electrolytic cell, it is the site of reduction C) In a voltaic cell, the cathode is negative, whereas in an electrolytic cell, it is positive D) In a voltaic cell, the anode is negative, whereas in an electrolytic cell, it is positive
D - In a voltaic cell, the anode is negatively charged, and the cathode is positively charged. In an electrolytic cell, the anode is positively charged, and the cathode is negatively charged (choice C is incorrect). A voltaic cell is spontaneous, while an electrolytic cell is nonspontaneous (choice A is incorrect). Regardless of the type of electrochemical cell, the anode is the site of oxidation (choice B is incorrect).
Lipolysis is followed by β-oxidation, in order to ensure the cell can harvest ATP from lipid molecules. Which of the following is true of this process? A) An isomerase and a reductase are required for complete oxidation of a monounsaturated fatty acid B) Lipolysis of a DAG will generate twice the amount of fatty acids and glycerol, compared to lipolysis of a MAG C) Unlike cell respiration, fatty acid catabolism starts in the mitochondria and finishes in the cytosol D) The cell requires more than twice the number of NAD+ electron carriers compared to FAD electron carriers, in order to harvest ATP from fatty acids
D - It is true that cells require more than twice the number of NAD+ electron carriers compared to FAD electron carriers, in order to harvest ATP from fatty acids (choice D is correct). As an example, catabolism of a ten-carbon saturated fatty acid will require four rounds of β-oxidation. This will generate four FADH2, four NADH, and five acetyl-CoA. The acetyl-CoAs will enter the Krebs cycle and will generate 15 more NADH and 5 more FADH2 (as well as 5 GTP). In total then, 19 NADH are generated (requiring 19 NAD+) and 9 FADH2 are generated (requiring 9 FAD). β-oxidation of a monounsaturated fatty acid requires only an isomerase, not a reductase (choice A is wrong). Lipolysis of a DAG will generate two fatty acids and one molecule of glycerol. Lipolysis of a MAG will generate one fatty acid and one molecule of glycerol, so the number of fatty acids generated is doubled, but not the amount of glycerol generated (choice B is wrong). Fatty acid catabolism starts in the cytosol, and is mostly performed in the mitochondrial matrix (choice C is wrong).
Abnormal functioning of which brain region(s) plays a role in the development of depression? I. Frontal lobe II. Limbic system structures III. Hypothalamus A) I only B) II only C) I and II only D) I, II, and III
D - The frontal lobe is involved in humans' ability to project future consequences of current actions (I). Limbic system structures regulate emotion and memory (II), and the hypothalamus coordinates many hormones, some of which are involved in mood regulation (III).
If a person drank a large quantity of hypersaline ocean water, the person could die because absorption of salt into the blood will cause it to become: A) Hypotonic compared with the cytosol of the body's cells, causing osmosis of water into the cells. B) Hypertonic compared with the cytosol of the body's cells, causing osmosis of water into the cells. C) Hypotonic compared with the cytosol of the body's cells, causing osmosis of water out of the cells. D) Hypertonic compared with the cytosol of the body's cells, causing osmosis of water out of the cells.
D - The ingested salt will be absorbed into the blood, producing a relatively concentrated, or hypertonic, solution. Remember, NaCl does not readily diffuse through the phospholipid bilayer. However, water can diffuse via osmosis, resulting in a flow of water out of the cell in an attempt to reestablish isotonic solutions on both sides of the membrane. π = iMRT, where π represents osmotic pressure, M refers to molarity, R is the ideal gas constant, and T is the temperature in Kelvin.
In oxidative phosphorylation, what is the final step? A. O2 is oxidized to produce H2O. B. NADH is oxidized to form NAD+. C. H+ is hydrolyzed to form H2O. D. ADP is phosphorylated to produce ATP.
D - This question is testing your knowledge of oxidative phosphorylation. The end purpose of oxidative phosphorylation is to phosphorylate ADP into ATP. We should know for the MCAT that the very final step is ATP synthase allowing protons to move down the electrochemical gradient while forming ATP.
Which of the following phase changes are exothermic processes? I. Liquid to gas II. Liquid to solid III. Gas to solid A. I only B. II only C. I and III only D. II and III only
D - This question tests your understanding of reaction enthalpies. Reactions that proceed from solid to liquid to gas are endothermic reactions, since gases have more heat energy than liquids and liquids have more heat energy than solids. On the other hand, reactions that proceed from gas to liquid to solid release heat, making them exothermic reactions (II and III).
Subjects in the study who experience withdrawal from the opioids prescribed for injury are likely to exhibit all of the following upon physical examination EXCEPT: A) Increased perspiration. B) Tremors. C) Increased anxiety. D) Pupillary constriction.
D - Withdrawal from substances tends to produce the opposite effect of symptoms occurred while using the substances. Pupillary constriction is a physiological response to opioid use and thus would not be expected as a symptom during opioid withdrawal.
Bronsted-Lowry Acid
Donates protons H+ (Conjugate Base)
Commonly Found Strong Acids
Hydroiodic acid: HI Hydrobromic acid: HBr Hydrochloric acid: HCl Perchloric acid: HClO4 Sulfuric acid: H2SO4 Nitric acid: HNO3
Hydrolases
Hydrolases catalyze reactions that involve cleavage of a molecule using water (hydrolysis). This cases usually involves the transfer of functional groups to water. Hydrolases include amylases, proteases/peptidases, lipases, and phosphatases.
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. 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.
Noncompetitive Inhibition
Noncompetitive inhibition is when the inhibitor does not compete with the substrate for the active site, but reduces enzyme activity by binding to another site (the allosteric site) on the enzyme. In noncompetitive inhibition, the inhibitor can combine with either the enzyme or the enzyme-substrate complex. In pure noncompetitive inhibition, the value of Vmax is decreased. Since these inhibitors do not compete with the substrate, their activity is unaffected by substrate concentration. Since the inhibitor always affects a consistent proportion of the available enzyme, Vmax is reduced. However, Km remains the same since if Vmax, is reduced, Vmax/2 is reduced proportionally, and the amount of substrate required to reach this new, reduced Vmax/2 is the same as the original Km.
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.
Voltaic cells (galvanic cells)
Perform spontaneous reactions and have a positive electromotive force. Thus, the anode within a voltaic cell is negative, while the cathode is positive. In a voltaic cell, the cathode has a higher reduction potential than the anode.
Normality
Refers to the number of equivalents of acid or base that are used in a solution. For an acid with a single dissociating proton such as HCl, this is equal to the concentration (in moles) of the solution. For a multiprotic acid with multiple dissociating protons, this is roughly equal to the number of acidic protons times the concentration of the solution.
Ka Values
Since strong acids dissociate completely, strong acids typically have a larger Ka value. Weak acids have a Ka less than 1.0. Similarly, a larger Kb value corresponds with a strong base and weak bases have a Kb less than 1.0; The pKa is the pH at which half of the species in a solution are protonated, and the other half are deprotonated. If the pH is higher than the pKa, the species will be mostly deprotonated. Strong acids tend to have low pKa values, and strong bases tend to have low pKb values.
Conjugates
Strong acids have weak conjugate bases and strong bases have weak conjugate acids. This is because a strong acid will not form a species that readily accepts a proton. If it did, then it would not be a strong acid. The same logic applies to strong bases and their weak conjugate acids. As you may expect, this also means that weak acids tend to form strong conjugate bases and weak bases tend to form strong conjugate acids.
Enthalpy of Formation
The standard enthalpy of formation for an element in its naturally occurring form at standard conditions (298 K, 1 atm) is 0 kJ/mol (choice B is correct).
Decay
There are four primary types of decay: alpha decay, beta decay (β+ and β-), gamma decay, and electron capture. In alpha decay, an alpha particle, containing two protons, two neutrons, and a +2 charge, is emitted. In beta-minus decay, a neutron is converted into a proton in the nucleus, and a β- particle (an electron) is ejected to maintain charge balance. In beta-plus decay, a proton is converted into a neutron, and a β+ particle (a positron) is emitted to preserve charge. Gamma decay involves the emission of a gamma ray, which is a high-energy photon, from an excited nucleus. Finally, in electron capture, a nucleus "grabs" an electron, which changes a proton into a neutron.
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.
Uncompetitive Inhibition
Uncompetitive inhibition is when the inhibitor binds to only the enzyme-substrate complex, and inactivates it. This causes the number of active enzyme-substrate complexes to decrease, thereby decreasing Vmax. Km also decreases to exactly the same degree as Vmax. Since both kinetic factors decrease to the same extent, the slope of the Lineweaver-Burk plot of enzyme activity will be the same as for the uninhibited enzyme. The effects of an uncompetitive inhibitor cannot be overcome by increasing substrate concentration.
Vapor Pressure
Vapor pressure refers to the pressure of the vapor phase that exists (to some degree) immediately above the surface of any liquid. A higher vapor pressure indicates that a larger number of solvent particles were able to escape the liquid and enter the gas phase. When vapor pressure is equal to the atmospheric pressure exerted on the liquid's surface, the liquid will boil. Knowing this, the fact that adding more solute particles causes a reduction in vapor pressure is actually logical. Imagine the surface of an aqueous solution with a lot of solutes in it. Some of the solutes will be at the surface of the solution, so the space they take up is unavailable for the liquid-gas phase transition that is at the core of vapor pressure. This causes vapor pressure reduction.
Nickel-cadmium battery
When this battery discharges, the anode is composed of cadmium, and the cathode is composed of nickel (III) oxide-hydroxide. Both electrodes are in contact with a potassium hydroxide (KOH) solution. When fully discharged, the anode consists of cadmium hydroxide, and the cathode consists of nickel (II) hydroxide. To charge the nickel-cadmium battery, the electrodes are dissociated to return to the original compositions of the anode (Cd) and cathode (NiO(OH)).
Electromotive Force
When two half-reactions are found in a single electrochemical cell, the one with a higher E°reduction is the one that can be spontaneously reduced in the overall reaction. The one with a lower E°reduction is the one that can be spontaneously oxidized in the overall reaction.
