MCAT- Chemical and Physical Foundation (NextStepTest#4)

Which of the following is most likely the net charge on a valine molecule in the human body?

0 This question is asking us to remember what factors contribute to the net charge on a valine molecule. Being an amino acid, valine has an acidic carboxy group that will be deprotonated at physiological pH. It also has a basic amine group that will be protonated at physiological pH, as well as a neutral side chain. Thus, it will exist as a zwitterion with net charge 0, as shown below.

What is the normality of a 0.015 M solution of phosphoric acid?

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). NOTE: Whenever you are trying to find the normality and you are given molarity, simply multiply the number of hydrogens with molarity. Units for Normality is N.

What is the pH of a 0.010 M perchloric acid solution?

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. pH = -log[H+] = -log[10-2] = 2.

The frequency used in U/S imaging must be greater than:

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

How many grams of hydrogen gas are required to completely react with 32 g of oxygen to form hydrogen peroxide?

2g

Britton-Robinson Buffer Solution

A "universal" pH buffer used for the range pH 2 to pH 12. Universal buffers consist of mixtures of acids of diminishing strength (increasing pKa) so that the change in pH is approximately proportional to the amount of alkali added.

Buffers

A solution that resists changes in pH upon addition of acid or base. While buffers cannot protect against addition of large amounts of acid/base, they are highly effective at maintaining pH when small to moderate quantities are added. A buffer must contain either a weak acid and its conjugate base or a weak base and its conjugate acid.

Strong/Weak Acids and Bases

A strong acid or base dissociates completely in water, while a weak acid or base dissociates incompletely.

Which of the following situations would present Participant 3 with the greatest risk for TBI?

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. Head Injury Criterion (HIC) = F x ∆t For this question, the best way to approach it is to see if there is anything different and then try to find the answer--> most likely the answer and in the question, it was.

Epsilon-amino acid (ε-amino group)

An amino acid in which the amino group is located on the carbon atom at the position ε to the carboxy group.

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.

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.

Answer: 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. 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. The rest makes sense Memorize the graph. Isothermal vs Adiabatic Graph

Which of the following correctly represent the units of the Boltzmann constant?

Answer: J/K F = -kT ln (KV) --> Hemholtz equation k is Boltzmann's constant. Note: k = 1.38✕10-23 (m^2•kg)/(s2•K). Using this... N= 1kg*m/s^2 =(N)*m/K J=N*m Answer=J/K which is the Boltzmann constant's unit

Aromatic

Aromatic compounds are conjugated cyclic molecules with a planar structure that also satisfy an additional criterion known as Hückel's rule: having (4n + 2) π-electrons, where n is an integer.

Aspartame (N-L-alpha-aspartyl-L-phenylalanine 1-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):

Aspartame contains two amino acids as mentioned above in the formula; thus, it is a dipeptide Aspartame is the methyl ester of the dipeptide containing aspartic acid and phenylalanine

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.

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 (r=8m) NEED TO THIS QUESTION AGAIN!! #43

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. NOTE: If Fc=Fg, then Fn=0. 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. 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

The first step of the formation of the imidazolinone ring of sgBP is most likely accomplished by the:

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

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. UV radiation is higher-energy, higher-frequency EMR than visible light or IR radiation. The passage states that UV absorbance spectroscopy was performed in order to assess whether the MH preparation process had changed the structure of A. The fact that A's peak UV absorbance remains the same in aqueous and MH solutions suggest that the structure is identical, so choice B is correct. (Choice D) If significant structural changes had occurred in A, we would expect to see a shift in UV absorption.

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

C. Sodium Bicarbonate, NaHCO3 The prefix "amphi-" means "both." Therefore, an amphiprotic species is one that can act as both an acid or a base. Sodium bicarbonate (choice C) dissolves in aqueous solution to produce sodium ions and bicarbonate ions. The former ion is neither acidic nor basic, but the bicarbonate ion, HCO3-, can act as a Bronsted-Lowry acid by loss of a hydrogen ion and can act as a B-L base by accepting a hydrogen ion to form carbonic acid, H2CO3.

Which of the following is a list of straight-chain hydrocarbons, each of which could have one triple bond?

C8H14; C6H10; C2H2 The formula for a straight-chain alkane is CnH2n+2. A double bond will replace two of those hydrogens with an extra C-C bond, giving a formula of CnH2n. A triple bond will replace four of those hydrogens with two extra C-C bonds, giving a formula of CnH2n-2.

Oxidoreductases

Catalyze oxidation-reduction reactions where electrons are transferred.

Conservation of Energy

Conservation of energy is a fundamental law of nature. This principle states that energy can neither be created nor destroyed, just transferred from one form to another. In other words, the total final energy within a system is equal to the total initial energy.

Conversion of pyruvate into glucose requires enzymes present in:

Conversion of pyruvate to glucose requires its initial conversion into oxaloacetate, in a reaction catalyzed by pyruvate carboxylase in the mitochondria. Oxaloacetate (OAA) is then decarboxylated and phosphorylated by cytosolic or mitochondrial forms of phosophoenolpyruvate carboxykinase (PEPCK). After transport of either OAA in the form of malate or PEP directly from the mitochondria, the remainder of gluconeogenesis takes place in the cytosol.

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 Attenuation 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.

Inductive effect

Electron donation or withdrawal through the sigma bonds of a molecule.

Which one of the engines in the study has the highest efficiency?

Engine 4 (Question #33) We need to calculate the efficiency for every engine in order to choose the one with the highest efficiency. We should know that for any engine, efficiency is defined as work output/work input. Therefore, for the engines in the study, the efficiency is calculated using the following formula: η (efficiency) = W/QH the first law of thermodynamics (conservation of energy), which tells us that QH = QC + W, using Figure 1 of the passage. In the above equation, we replace W with QH - QC: η = (QH - QC) /QH = 1 - (QC/QH)

Compared to the cytotoxic activity of ZSTK474 against the cell lines tested, the cytotoxicity of each target compound was:

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

Conjugated Systems

For the purposes of 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.

Tollens' test typically involves exposure of a carbohydrate to a solution of CuO in ammonia (NH3). Glucose yields a positive Tollens' test, but sucrose does not. Which of the following best explains this fact?

Glucose contains a hemiacetal group, while sucrose does not; this classifies glucose as a reducing sugar. Tollens' 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, sugars with hemiacetal groups can be oxidized and can thus function as reducing sugars.

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?

Hb A2 According to the passage, Hb A2 is a minor hemoglobin synthesized by adults. It contains a δ unit, which Figure 1 shows us is not present at all in fetuses of 20 weeks of gestation. As a result, it should be found in greater concentration in the maternal blood than in the fetal blood.

Hemiacetals and Acetals

Hemiacetals are compounds in which a terminal carbon atom is connected to (1) another carbon atom, (2) an -H atom, (3) an -OH group, and (4) an -OR group. Acetals are derivatives of hemiacetals in which the -OH group is replaced by an -OR' group. Additionally, when a glycosidic bond is formed between two monosaccharides (isolated sugar molecules) to form a disaccharide, a hemiacetal or hemiketal is converted into an acetal or ketal.

Antiaromatic

Highly unstable compounds with 4n π-electrons.

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

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. RN II is a salt formed from a weak acid (phosphoric acid), so it is also correct.

In comparison to piperazine, the inductive effect will cause the pKb of methyl-piperazine to be:

Lower, because of electron donation by the methyl substituent. The pKb is a measure of basicity, where a smaller pKb value corresponds to a stronger base. Methyl-piperazine is a stronger base than piperazine because of the inductive effect generated by the electron-donating N-methyl group of methyl-piperazine. This effect increases the availability of electron charge density available to donate, thereby increasing the strength of N-methyl piperazine to act as a Lewis base. Alkyl groups, such as the N-methyl group of methyl-piperazine, are generally electron-donating substituents.

The light blue appearance of S157T is most directly attributable to what other phenotypic change versus the wild type?

Increased wavelength max 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. (Using the table figure to solve the question)

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?

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.

Given the data presented in the passage, what is true regarding the effect of tested target compounds on the cytotoxicity of HCT116 cells?

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 PI3K 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?

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 ... CA4þ." 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 positive charge character of CA4b is central to its function in the CECT technique.

Which specific class of enzymes is primarily responsible for the release of free glycerol from stored triglycerides?

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 subculture of the hydrolyses.

When situated in the substrate affinity pocket of PI3Kα, the morpholine group oxygen of ZSTK47 is most likely to interact with the side chain of what amino acid residue? A. Aspartate B. Cysteine C. Lysine D. Tyrosine

Lycine From the passage: "one of the morpholine groups of ZSTK474 extends into the substrate affinity pocket of the enzyme, where its oxygen acts as a hydrogen bond acceptor for a primary amine of the enzyme." 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. A, B, D: None of these amino acids contain amino groups as part of their side chains.

According to the results of the experiments, what was the risk of concussion to Participant 3 when riding Coaster 2?

Missed this question because I did not read correctly and it cost me a point. When it asks for something that can be found in the passage, use the key words that was in the question.

Which of the following is NOT a functional group present in caffeine?

Primary Amine- they have two hydrogen bonds and only on R group. An imide is two acyl (O=C-R) groups bound to nitrogen. An amide contains the group C(O)NR2, where "R" can refer to an organic group or hydrogen atom.

Pyrrole

Pyrrole ring is nitrogen-based, not oxygen-based

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?

Question #23 Compared to hydrogen bonding at pH 4.5, ionic bonding at pH 10 promotes peak absorbance of higher energy radiation. The third paragraph of the passage states that a hydrogen 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 excited-state proton transfer (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 3B 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 predominately consists of hydrogen bonding, and that the ion-dipole bonding which predominates at pH 10 promotes the maximum absorption of shorter-wavelength, higher-energy radiation.

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?

Question#52- 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, it could quite probably do as the "non-covalent" interaction described by the passage as being responsible for an outward pressure in 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. Negative charge repulsion between and within GAGs is largely responsible for the compressive strength of articular cartilage.

The piezoelectric effect, when generating ultrasound waves, involves the conversion of:

Question: 7 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.

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 TLC plate than a less polar compound. III. Anhydrous methanol has a greater eluting strength than pentane when used as solvents in a TLC procedure.

RN 1- Retention factor (Rf) is the distance migrated by the compound divided by the total distance traveled by the solvent. RN 2- A polar compound will be attracted to the adsorbent (via dipole-dipole interactions) and, therefore, will move slower on a TLC plate, whereas a nonpolar compound will have more affinity for the mobile phase and will move faster. RN 3- Eluting strength depends on how strongly a compound adsorbs onto the adsorbent. Since typical adsorbents are highly polar, eluting strength increases with increasing solvent polarity. Methanol is more polar than pentane and therefore has a greater eluting strength.

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

Ser157 Table 1 shows that only modification of Ser157 resulted in mutant sgBP chromophores that express a different color. While many other mutations abolished the color of the chromophore altogether, such mutations would not maintain the function of the chromophore that, as stated in the first paragraph of the passage, is responsible for the colored appearance of CPs.

What amino acid in the Aequorea fluorophore would be designated as "X" in the X-Tyr-Gly GFP fluorophore?

Serine Three amino acids are shown in the flurophore structure of Figure 1: serine, tyrosine and glycine. You need to memorize the structure, function, and properties of the 20 amino acids.

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

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

Thalassemias

Thalassemia is a general term used for an autosomal recessive disorder where one of the globin genes is not expressed properly, leading to anemia. This can be the result of hemoglobin gene deletions, promoter mutations, nonsense and frameshift mutations, unstable mRNA, or unstable proteins. Specific terminology refers to the subunit that is not being made in adequate quantities, such as α-thallasemia or β-thallasemia (and so on). Commonly, the balance of the hemoglobin subunits can still be made properly, thus abnormal hemoglobin quaternary structures result from their associations that cannot function properly and have altered oxygen dissociation curves.

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?

The 5-membered ring is not conjugated with the aromatic phenol ring of tyrosine. In step 3 of Figure 1, the Cα-Cβ bond of tyrosine is oxidized to a double bond in a reaction which 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. Without the molecular oxygen needed for this oxidation, such conjugation will not arise. Its absence could explain the failure of the molecule to fluoresce.

Hemoglobin P50

The P50 is the oxygen tension at which hemoglobin is 50% saturated.

The heat cycle

The area enclosed by the four steps represents 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 work done. 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. Work is a general term in physics used to describe energy transfer. Its units are joules (J), and 1 J = 1 N•m or 1 kg•m2/s2. Work appears in several contexts in MCAT physics. W = PΔV= |F|∙d∙cos(θ)

Use of CA4b is likely to increase the contrast obtained in CECT imaging because:

The elements in CA4b have larger atomic numbers than the elements in cartilage. From the passage: "X-ray attenuation, which is responsible for contrast in CECT imaging and is enhanced by the use of contrast agents, generally increases with the atomic number of atoms composing" the target. 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.

Normal Force

The force perpendicular to a surface that prevents an object from falling through the surface

An object made of silicon (specific heat = 698 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?

The most efficient way to find this answer is by using dimensional analysis and the equation Q (the heat in or out of a system) = mcΔT. Q (J)=MCAT

Molarity

The number of moles of solute per liter of solution

Which of the following boundaries would give the lowest-quality image to a physician?

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

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

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 NAD+ levels will be unusually low, since the production of NADH naturally involves the depletion of NAD+. With low NAD+ levels, the future 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 oxaloacetate (OAA) is produced from pyruvate. Thus, the rate of glucose production via gluconeogenesis will decrease.

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?

The total capacitance of capacitors in parallel is simply the sum of their individual capacitances, so student 2 would create a circuit with 16 μF and the highest capacitance.

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?

This question asks us to infer which factor is most likely to affect bioavailability, or the ability of a drug to be absorbed by the body and take effect. The passage tells us that hydrophilic solubility is important, most likely because the vast majority of biological solvents are hydrophilic. In addition, 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.

According to the experimental results, the elastic modulus of a sample of articular cartilage with mean CECT attenuation of 1500 HU is nearest:

This question test your ability to interpret a graph with multiple DVs.

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)

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

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:

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, fructose 6-phosphate by gluconegenesis-sepcific enzymes that bypass irreversible steps of glycolysis push the equilibrium of reversible enzymes that function both in glycolysis and gluconegenesis in the direction of glucose production.

Adsorption

the process by which a solid holds molecules of a gas or liquid or solute as a thin film.

Electromagnetic (EM) waves

transverse waves that can propagate through vacuum, as well as through other media such as air and water. (E = hf = hc/λ)

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:

ζ2ε2. Answers were in the figure. Did not understand the question correctly and lost an easy point.