Princeton Review MCAT Demo Test Physics/Chem

Question 53: Based on the information in the passage, which of the following is true? a. compared to fatty acids, molecules of traiclyglycerol are relatively inert. b. a competitive inhibitor for ATGL would decrease DAG catabolism c. HSL is responsible for generating glycerol and MAG, but not DAG d. TAGs contain two stable ester linkages

Answer: A Glycerol: 3 carbon structure with 3 -OH groups. Glycerol backbone is found in all lipids known as triglycerides. Fatty acid: carboxylic acid consisting of a hydrocarbon chain and a terminal carboxyl group, especially any of those occurring as esters in fats and oils Ester linkage: triglyceride lipids are composed of 3 ester linkages. Esters can be formed when a carboxylic acid reacts with an alcohol group, this process also forms a water molecule.

Question 25: 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

Answer: A Nucleophilicity trends: across a row in the periodic table, there is increasing electronegativity which decreases lone pair availability. Higher electron density will increase the nucleophilicity. Down the row of the periodic table, increasing polarization of the nucleophile increases the nucleophilicity. The electron density of larger atoms is more readily distorted, or polarized, since the electrons are further from the nucleus. Very good nucleophils: I-, HS-, RS- good: Br-, HO-, RO-, NC-, N3- fair: NH3, Cl-, F-, RCO2- weak: H2O, ROH very week: RCO2H

Question 45: 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.

Answer: B *Electronegativity*: tendency of an atom to attract a pair of electrons towards itself. *Electron affinity*: the amount of energy released or spent when an electron is added to a neutral atom or molecule in the gaseous state to form a negative ion. *Ionization energy*: energy required to remove an electron from a neutral atom. *Nuclear shielding*: shielding effect, describes the decrease in attraction between an electron and the nucleus in any atom with more than one electron shell. The more electron shells there are, the greater the shielding effect experienced by the outermost electrons. *Effective nuclear charge*: the net positive charge experienced by valence electrons. Approximated by the equation: Zeff = Z - S (Z is atomic number / # protons in the nucleus, and S is number of shielding electrons / # of electrons between the nucleus and electron in question / # of non valence electrons) Acidity trend: increases across a row, and increase down a row.

Question 16: Lipase works to hydrolyze which of the following functional groups? a. amide b. ester c. ether d. hemiacetal

Answer: B Amide: derivative of carboxylic acids -- hydroxyl group replaced by amine or ammonia. Lone pair of electrons delocalized into the carbonyl, forming a partial double bond between N and carbonyl C. R-CO-NR2 Ester: Derived from an acid. At least one OH group is replayed by an -O- group. Usually derived from a carboxylic acid and an alcohol. R-CO-OR Ether: contain an ether group (oxygen atom connected to 2 alkyl or aryl groups). R-O-R' Hemiacetal: Results from the addition of an alcohol to an aldehyde or ketone. Single alcohol has been added to the carbonyl group. R-C-OR-OH-H Lipase: a pancreatic enzyme that catalyzes the breakdown of fats to fatty acids and glycerol or other alcohols.

Question 37: The dancer is wearing a necklace which consists of a chain and a pendant of mass, m. When the dancer is spinning at a constant speed, the pendant pulls away from her body, and the chain makes an angle, theta, with the dancer's body. If the pendant is spinning at a constant speed, v, and the chain has a tension, T, what is the radius of the circle created by the pendant? a. mv^2/T b. mv^2/Tsin(theta) c. mv^2/Tcos(theta) d. Tsin(theta)/mv^2

Answer: B Centripetal force: Fc = mv^2 / r = m x a_c Fc in this case would be T x sin(theta) T x sin(theta) is the x-component of the pendant away from the body Uniform circular motion: traveling in a circular path at a constant speed. R is radius of the path, the period T is the time it takes to complete a circle, speed is given by circumference over the period. v = 2πr/T a = 2πv/T centripetal acceleration: a = v^2/r

Question 39: 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 degree angle with her body. Which of the following is true for the circular motion of the basket of flowers? 1. the spinning causes a centripetal force, which pulls the basket out from the center of the circle. 2. the dancer's arm provides the centripetal force, which pulls the basket in towards the center of the circle. 3. 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, III

Answer: B Centripetal force: the force that acts on a body moving in a circular path and is directed toward the center around which the body is moving.

Question 51: What is the equilibrium constant for the binding of the fourth NAD+ to GAPDH? dissociation constants K': K'1 = 0.15 x 10^5 K'2 = 0.63 x 10^5 K'3 = 1.3 x 10^5 K'4 = 1.5 x 10^5 a. 1.5x10^-5 b. 6.7 x 10^-6 c. 1.5 x 10^5 d. 6.7 x 10^6

Answer: B Dissociation constant is a specific type of equilibrium constant that measures the propensity of a larger object to split up into its component ions. AxBy -> xA + yB Kd = [A]^x[B]^y / [AxBy] Binding would be inverse of dissociation constant.

Question 27: In the electrolysis of water shown below, a current of 2 amps is applied to 180 mL of H2O (l) for 6 hours and 42 minutes. How many grams of H2(g) are formed? (Faraday's constant = 96.500 C/mol) 2H2O (l) + 2e- -> H2(g) + 2 OH- (aq) a. 0.25 g b. 0.5 g c. 5.0 g d. 10.0 g

Answer: B Electrolysis -- pass an electric current through ionic solution. Ions are forced to undergo oxidation (at anode) or reduction (at cathode). Calculate the quantity of substance produced or consumed: -write balanced half-reactions involved -calculate # of moles of electrons that were transferred -calculate # of moles of substance that was produced/consumed at electrode -convert the moles of substance to desired units of measure amp x s = C Coulumb - unit of electric charge; defined as quantity of electricity transported in one second by a current of one ampere. 6 hrs, 42 min = 6x3600 + 42x60 = 24,120 s 24,120 s x 2 amp = 48,240 C Faraday's constant = 96,500 C/mol 48,240 C x 1/96,500 = 1/2 mole of e- = # of moles of e- transferred .5 mole of e- x (1 mole H2/2 mole e-) = 0.25 moles of H2 .025 moles x 2 g/mol = 0.5 g of H2

Question 34: Besides detoxification of drugs such as acetaminophen, the liver is involved in and regulates several different biochemical pathways. Which of the following is NOT a biochemical activity of the liver? a. regulation of carbohydrate metabolism such as glycogenolysis, glycogenesis, and gluconeogenesis b. production of lipases and bile for fat digestion c. deamination of amino acid and conversion of the resulting ammonia to urea d. lipid metabolism, including cholesterol and lipoprotein synthesis

Answer: B Production of lipases occurs in the pancreas. Bile is produced in the liver but stored in the gallbladder.

Question 38: The 10 kg dancer leaps into the air with an initial velocity of 5 m/s at angle of 45 degrees from the floor. How far will she travel in the air horizontally before she lands on the ground again? a. 1.25 * root(2) m b. 2.5 m c. 2.5 * root(2) m d. 5 m

Answer: B Projectile motion: horizontal velocity remains unchanged throughout the motion. vertical velocity changes linearly due to gravity v_x = v0 x cos(theta) v_y = v0 x sin(theta) - gt magnitude of velocity v = square root of (v_x^2 + v_y^2) displacement: x = v0 x t x cos(theta) y = v0 x t x sin(theta) - 1/2gt^2 t = 2v0 x sin(theta) / g = 2 x 5 x sin(45)/g = 10 x1/root(2)/10 = 1/root(2) s x = 5 x cos(45) x t x = 5 x (1/root(2)) x (1/root(2)) = 2.5 m

Question 55: A researcher interprets Figure 1 to mean that G0S2 mRNA is ubiquitously expressed, but G0S2 protein expression is more tissue specific. Does the data in Figure 1 support this conclusion? a. Yes: all cells contain the same amount of mRNA for G0S2 but only some tissues contain G0S2 protein. b. Yes: reverse transcriptase PCR gives information on what transcripts are present in a cell, and western blots give information on protein expression c. No: gene expression regulation occur at the transcription level; mRNA produced by a cell will be translated into a protein d. No: the experiment didn't work properly; the data obtained from PCR should match what is obtained from RT-PCR, and this is not the case.

Answer: B Reverse transcriptase PCR: technique used to detect RNA expression. Creation of complementary DNA transcripts from RNA. Traditional PCR is used to amplify target DNA sequences. Instead, RT-PCR is used to clone expressed genes by reverse transcribing the RNA of interest into its DNA complement through the use of reverse transcriptase.

Question 48: Two circular pipes both have radii of 3.5m. They both steadily eject water horizontally, and the water from each pours onto a measuring grid 5 m below. The stream from the first pipe lands on the measuring grid 20 cm away from a position directly below the pipe mouth, and the stream from the second pipe lands 5 cm away from a position directly below its pipe mouth. How long would it take the second pipe to pour out as much water as the first pours out in 1 minute? a. 2 min b. 4 min c. 8 min d. 16 min

Answer: B Volumetric flow rate: volume of fluid which passes per unit time, represented by Q. Q = v x A v = flow velocity A = cross sectional vector area Velocity is 4 times as much since the distance is 4 times greater. Therefore, it takes 4 times as much time to reach the same volume of fluid. (?) *Equation of continuity*: for an incompressible fluid flowing through a tube of varying cross-section, the mass flow rate is the same everywhere in the tube. Mass flow rate = rate at which mass flows through given point, so total mass divided by time interval. density1 x area1 x velocity1 = density2 x area2 x velocity2 (cross sectional area) *Bernoulli's equation*: Pressure, speed, and height (y) at 2 points in a steady-flowing, non-viscous, incompressible fluid are related by the equation: P1 + 0.5density x v1^2 + density x g x y1 = P2 + 0.5density x v2^2 + density x g x y2 Velocity is greater in the narrow section of the pipe. Where velocity is increased, pressure is lower.

Question 58: Two resistors R1 and R2 are connected in series to a battery with no internal resistance. The current through each is measured. A third resistor, R3, is then connected in parallel to R2. How does this affect the currents through R1 and R2? a. I1 and I2 both increase. b. I1 increases and I2 decreases c. I1 decreases and I2 increases d. I1 and I2 both decrease

Answer: B When resistors are connected in parallel, the supply current is equal to the sum of the currents through each resistor. Currents in the branches of a parallel circuit add up to the supply current. When resistors are connected in parallel, they have the same potential difference across them. 1/R = 1/R1 + 1/R2 +... When resistors are connected in series, the current is the same through each resistor. The total resistance of the circuit is found by adding up the resistance values of the individual resisters: R = R1 + R2 +...

"The chestipiece usually has a diaphragm and a bell, the latter for transmitting low frequency sounds (20 to 80 Hz) and the former for high frequency (80 to 200 Hz)." Question 5: What is the highest period of a sound wave that can be detected well by the diaphragm on a stethoscope? a. 0.05 seconds b. 0.0125 seconds c. 0.005 seconds d. the bell can detect all sound waves equally well, so the period is unlimited

Answer: B 0.0125 seconds Mistake: Calculation is wave period = 1/frequency Unit of wave period: seconds OR can calculate by: wave period = wavelength / velocity (not applicable in this situation)

Question 49: 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 2e- and one proton d. the carboxylic acid (pKa = 2.19) is mostly in its protonated form

Answer: C

Question 47: What are the units of the rate constant k for a second order reaction? a. 1/s b. 1/s^2 c. L/(mol x s) d. L / (mol x s^2)

Answer: C 0 order reactions: rate = k[A]^0 M/t = k k units: M/s 1st order reactions: rate = k[A] M/t = kM k units: = 1/s 2nd order reactions: rate = k[A]^2 M/t = kM^2 k units = 1/(sM) 3rd order reactions: rate = k[A]^3 M/t = kM^3 k units = 1/(sM^2) n order reactions: rate = k[A]^n M/t = kM^n k units = 1/(sM^n-1)

Question 31: Which of the following is NOT a mechanism of action for the metabolism of acetaminophen? a. oxidation b. glucuronidation c. acetylation d. sulfonation

Answer: C Acetylation - introducing an acetyl group into a compound (-OCH3) Glucuronidation - transfer of the glucuronic acid component of uridine diphosphate glucuronic acid to a substrate by any of several tyeps of UPD-glucuronosyltransferase.

Question 40: Crown ethers (18-crown-6 shown below) are a family of compounds often used to aid in the crystallization of compounds bearing amine entities from cold, ether-type solvents. How do these compounds aid crystallization? a. crown ethers greatly decrease the polarity of the solvent, causing crystallization of the polar amine. b. crown ethers are not miscible in ether solvents and act as inhomogeneous sites for crystal formation. c. the crown ether saturates the hydrogen bonding capabilities of the amine, decreasing its interaction with the solvent d. the crown ether hydrogen bonds with solvent molecules, preventing them from interacting with amines in solution

Answer: C Crystallization process: two steps: nucleation and crystal growth Nucleation - step where solute molecules or atoms dispersed in the solvent start to gather into clusters and become stable Crystal growth - subsequent size increase of the nuclei Crown ethers: most crown ethers have 2 C atoms between the O atoms. Most useful property of crown ethers is ability to complex with cations. They are ethers, so soluble in organic solvents (carbon-based solvent) Crown ethers usually used to bring inorganic catalysts into the organic phase and to increase solubility of inorganic compounds in organic solvents to promote chemical reactions. Alcohol: intermolecular forces present affects boiling point. -OH group on alcohol leads to partially positive and negative charge -- attraction between two molecules -- H bonding (strongest type of intermolecular force) - this accounts for large boiling point. Ether: carbon and oxygen lead to partially negative and positive charges. There's is not enough difference in charge between and C and H, only a little bit of dipole dipole interaction, but not quite H bonding. diethyl ether: boiling point is higher than ethers. Ether molecules can't H-bond with each other, so that is not the intermolecular force. 2 molecules of diethyl ether interacting. London dispersion forces, with additive effect Crown ether: can interact with different ions. In the enter of the crown ether, it can fit a K+ ion perfectly for example (18-crown-6). partially negatively charged O atoms, negative charges hold the K+ ion in there. Outside of crown ether is non polar, which will dissolve in a non polar organic solvent, which would free up negative ions, increase nucleophilic strength of negative ions. Takes the cation leaving the anion to function as a better nucleophile. Can get different size crown ethers to take care of the different size cations.

Question 30: Based on information provided in the passage, which of the following statements describes a result of administration of acetaminophen to a patient with a history of chronic alcohol consumption? a. plasma NAPQI increases and GSH decreases b. plasma NAPQI increases and GSH increases c. Haptocyte NAPQI increases and GSH decreases d. Hepatocyte NAPQI increases and GSH increases

Answer: C Hepatocyte: cell of the main parenchymal tissue of the liver. These cells are involved in protein synthesis, protein storage, transformation of carbohydrates

Question 26: All of the following are products of the pentose phosphate pathway EXCEPT: a. glycolytic intermediates b. NADPH c. NADH d. Ribose-5-phosphate

Answer: C Pentose phosphate pathway is parallel to glycolysis. Generates NADPH and pentoses and ribose 5-phosphate (precursor for synthesis of nucleotides). Primary role is anabolic rather than catabolic. 2 distinct phases in pathway: 1) oxidative phase in which NADPH is generated 2) non-oxidative synthesis of 5-carbon sugars Most organisms, pathway takes place in the cytosol. Glucose-6-P NADP+ -> NADPH 6-phosphogluconolactone H2O -> H+ 6-Phosphogluconate NADP+ -> NADPH Ribose-5-phosphate Generates reducing equivalents (NADPH), ribose 5-phosphate (to synthesize nucleotides, nucleic acids), erythrose 4-phosphate (to synthesize aromatic amino acids)

Question 36: A 10 kg dancer raises her leg in the air in a developpe. Her 2 kg leg is 1 m long and can be assumed to have uniform mass throughout its length. What angle, theta, between the working leg and the standing leg will create the most torque in the dancer's hip? a. 30 degrees b. 45 degrees c. 90 degrees d. 135 degrees

Answer: C Torque = |r| x |F| x sin(theta) = 1m x 2 kg x sin (theta) - ?

Question 32: Which of the following offers the most likely explanation for the substantially higher peak plasma concentration of acetominophen following IV administration when compared to oral and rectal routes? a. the drug is poorly absorbed through mucous membranes in the oran 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

Answer: D

Question 54: Lipolysis is followed by beta-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

Answer: D *B-oxidation*: catabolic process by which fatty acid molecules are broken down. Occurs in the cytosol in prokaryotes and in the mitochondria in eukaryotes. Generates acetyl-CoA, which enters the citric acid cycle, and NADH and FADH2, which are co-enzymes used in the electron transport chain. Beta carbon of the fatty acid undergoes oxidation to a carbonyl group *Isomerase*: enzyme that catalyzes the conversion of a specified compound to an isomer. *Reductase*: enzyme that promotes the chemical reduction of a specified substance Both NAD and FAD play a role in cellular respiration. NADH is oxidized back to NAD+ and FADH2 is oxidized back to FAD.

Question 46: 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 both S and T c. Field lines point from S to T. d. Field lines point from T to S.

Answer: D Electric field lines point in the direction that a positive test charge would accelerate if placed upon the line. As such, lines are directed away from positively charged source charges and toward negatively charged source charges. Electric field: electric force per unit charge. Direction of field is the direction of the force it would exert on a positive test charge. Electric field is radially outward from a positive charge and radially toward a negative point charge.

Question 29: Which of the following best describes the structure of glucose shown below? a. pyranose form of D-glucose b. pyranose form of L-glucose c. furanose form of D-glucose d. furanose form of L-glucose

Answer: D Glucose has optical isomers L and D glucose. For Fischer projection with the most oxidized carbon at the top: -if OH on the bottom chiral center points to the right, it is referred to as D -if OH points to the left, it is referred to as L For Haworth projection: -if -CH2OH substituent is right of the anomeric carbon, it is D -if substituent is left to anomeric carbon, it is L Still don't understand this problem.

Question 24: Which of the following reaction conditions would be the least likely to cause inhibition of Reaction 1? a. Addition of Beta - mercaptoethanol (HSCH2CH2OH) b. pH 2 c. addition of iodoacetamide d. pH 8

Answer: D Iodoacetaminde: remember the structure (carbonyl with -NH2). Commonly used to bind covalently with thiol group of cysteine so that the protein cannot form disulfide bonds. Irreversible inhibitar of all cysteine peptidases. Reacts faster than iodoacetate. Iodoacetate / iodoacetic acid: (I with carboxylic acid) toxic compound. Reacts with cysteine residues in proteins, often used to modify SH groups to prevent the re-formation of disulfide bonds.

Question 21: The cyanine portion of the compound absorbs radiation with a wavelength of 560nm, 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.

Answer: D Mistake: E = hv (energy, Planck's constant, frequency) λ = c/ν (wavelength, speed of light, frequency) The greater the energy, then the larger the frequency, and the shorter the wavelength. So short wavelengths (560 nm) are more energetic than long wavelengths.

Question 12: 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+

Answer: D O2 > CoQ > FAD > NAD+ Mistake: I had reversed the order. O2 is the last electron acceptor in the chain for a reason. FAD is also introduced later in the ETC than NAD+ and generates less ATP.

Question 28: 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

Answer: D Virtual image forms when light rays diverge after reflecting off the mirror. The image is where the rays intersect, which would be behind the mirror. When object is located in front of focal point, image is upright and enlarged. Convex lens: the image is virtual, upright, reduced in size, located behind the convex mirror. As the object moves closer to the mirror, the image is closer to the mirror and the image size is increased. Concave lens: When object is located beyond center of curvature, image is located between center of curvature and the focal point. Inverted image, reduced size, real image. When object is located at center of curvature, image is also located at center of curvature. Image is inverted, dimensions are equal to object, real image. When object is located between center of curvature and focal point, the image will be located beyond the center of curvature. Image will be inverted, larger than object dimensions, real image. When object is located at focal point, no image is formed. Light rays reflect parallel. When object is located in front of focal point, the image will be located behind the mirror. Upright image, magnified image, virtual image.

Question 10: 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

Answer: Gamma particles Gamma particles are most powerful, need denser material to block. Alpha particle, is like a helium molecule, with two positive charges. Beta particle can be electron or positron, so can have negative or positive charge.

Question 50: Which of the following could affect any of the dissociation constants for hemoglobin? I. Decreasing the temperature II. Adding NH3 III Increasing the [O2]

Answer: I and II only Factors that affect affinity and dissociation constant are temperature and presence of a catalyst.

Question 15: Which of the following contributes to the fact that the small intestine cannot passively absorb bile salts? I. Taurine and glycine have low pKa values. II. Bile salts are hydrophobic III. Pancreatic secretions increase intestinal pH

Answer: I and III only Low pKa values: the lower the value, the more easily it gives up a proton. Primary bile acids: those that are synthesized in the liver Secondary bile acids: removal of glycine and taurine groups forms the secondary bile acids Bile acids are conjugated with taurine or glycine in the liver, and these are called bile salts. Amphipathic molecules: both hydrophobic and hydrophilic regions, form micelles. Intraluminal pH is rapidly changed from highly acid in the stomach to about pH 6 in the duodenum. The pH gradually increases in the small intestine from 6 to about pH 7.4. Pancreatic juice composed of 2 secretory products: digestive enzymes and bicarbonate. Bicarbonate is secreted from the epithelial cells lining small pancreatic ducts. Bicarbonate: O2COH; intermediate (deprotonated) form of carbonic acid. Serves a crucial role in physiological pH buffering system. Alkaline.

Is water an organic or inorganic solvent?

Answer: inorganic It has no carbon atom. It is polar and usually organic compounds are non-polar. Organic compounds dissolve more in organic solvent as compared to water.

Question 7: If I represents the intensity of the sound wave from an earpiece, A represents the area of the eardrum to which the sound is delivered, and t represents the time spent listening to the sound, which of the following expressions gives the energy received by the eardrum? a. 1/(IAt) b. IAt c. I/(At) d. At / I

IAt Mistake: You need to multiple the I intensity with the A area of the eardrum, not divide, to capture the entire area of the eardrum.

Absolute configuration question:

Note that for carbohydrates in their standard Fisher projection, if the -OH is to the right, the carbon is R and if the -OH is to the left, then the configuration is S.

What is allosteric regulation?

Regulation of an enzyme by binding an effector molecule at a site other than the enzyme's active site. Site where the effector binds is termed the allosteric site.

Question 2: If the conditions for Point 9 were replicated inside a glass container, what would an observer see? a. liquid b. vapor c. liquid and vapor d. supercritical fluid

Supercritical fluid Was unsure what supercritical fluid was. It is when the temperature and pressure of a substance is outside the critical point.

What is the committed step (in enzymology)?

The committed step is an effectively irreversible enzymatic reaction that occurs at a branch point during the biosynthesis of some molecules. After this step, the molecules are committed to that pathway and will ultimately end up in the pathway's final product.

What is western blot?

Used to detect specific proteins in a sample of tissue. Used to separate and identify proteins. Proteins is separated based on molecular weight through gel electrophoresis. The results are transferred to a membrane producing a band for each protein. Often used as a follow-up test to confirm the presence of an antibody and to help diagnose a condition.