MCAT CP AAMC material

Approximately how many moles of Al3+ are reduced when 0.1 faraday of charge passes through a cell during the production of Al? (Note: Assume there is excess Al3+ available and that Al3+ is reduced to Al metal only.) _ A 0.033 mol B 0.050 mol C 0.067 mol D 0.10 mol

A A faraday is equal to one mole of electric charge. Because each aluminum ion gains 3 electrons, 0.1 faraday of charge will reduce 0.1/3 moles of aluminum, or 0.033 moles of aluminum.

Which reaction leads to the formation of DHB whose structure is shown in the passage? A) Carboxylation of hydroquinone B) oxidation of hydroquinone C) reduction of benzoquinone D) hydroxylation of benzoquinone

A) carboxylation of hydroquinone. The answer to this question is A because the carboxylation of hydroquinone leads to the formation of 2,5-dihydroxybenzoic acid according to the reaction C6H4(OH)2 + CO2 → C6H3(CO2H)(OH)2.

Which of the following atoms has the largest atomic radius? A Sodium B Aluminum C Sulfur D Chlorine DO NOT THINK IONS JUST STRAIGHT UP READ PERIODIC TABLE.

ATOMIC RADIUS INCREASES <- AND DOWN The correct answer is A. Sodium, aluminum, sulfur, and chlorine are all in the third row of the periodic table. Atomic radius tends to decrease from left to right across a given row of the periodic table. This is because, as one moves across the row, the effective nuclear charge increases, drawing in the outermost electrons. Because sodium is the first element in row three, it is the largest. Answer choice A is the best choice. BUT ANIONS ARE BIGGER RADIUS AND CATIONS ARE SMALLER.

What is the frequency of the emitted gamma photons? (Note: Use Planck's constant h = 6.6 x 10-34 Js and the elementary charge e = 1.6 x 10-19 C.) if emitted 140Kev. A. 2.11 × 1035 Hz B. 3.38 × 1019 Hz C. 3.01 × 10-20 Hz D. 1.45 × 10-47 Hz

Answer is B. The photon energy is 140 keV = 140 x 10^3 x 1.6 x 10−19 J = 2.24 x 10−14 J. This numerical value is inconsistent with the photon frequency derived as the ratio of 2.24 x 10−14 J and h = 6.6 x 10−34 Js, according to the formula E = hf. The photon energy is 140 keV = 140 x 10^3 x 1.6 x 10−19 J = 2.24 x 10−14 J. The photon frequency is the ratio of 2.24 x 10−14 J and h = 6.6 x 10−34 Js, which is 3.38 x 1019 Hz. This numerical value is consistent with the inverse of the actual photon frequency. The photon energy is 140 keV = 140 x 10^3 x 1.6 x 10−19 J = 2.24 x 10−14 J. This numerical value is inconsistent with the photon frequency derived as the ratio of 2.24 x 10−14 J and h = 6.6 x 10−34 Js based on formula E = hf.

The following reaction occurs spontaneously.Cd(s) + 2 H+(aq) → Cd2+(aq) + H2(g) Which of the following has the highest electron affinity? _ A Cd(s) B H+(aq) C Cd2+(aq) D H2(g)

B The reaction equation shows the reduction of H+ by Cd. Because the H+accepts the electron readily from Cd, it can be determined that H+ has the highest electron affinity.

What mass of CDP (403 g mol−1) is in 10 mL of the buffered solution at the beginning of Experiment 1? buffer solution is 16mmol CDP A 6.4 × 10−4 g B 6.4 × 10−3 g C 6.4 × 10−2 g D 6.4 × 10−1 g

C. 6.4x10-2g of CDP is present

A student plans to add HCl to a solution containing Pb(NO3)2(aq). To determine how much Pb2+ will precipitate from solution when the HCl is added, the student needs to know which of the following? _ A Ka for HCl B Ka for HNO3 C Ksp for PbCl2 D Keq for the reaction Pb2+ + 2 e- Pb

C. To determine how much Pb2+will precipitate, the student must know the solubility of PbCl2, which is related to the solubility product constant Ksp of PbCl2. Thus, C is the best answer.

When a weak acid (HA) is titrated with sodium hydroxide in the presence of an indicator (HIn), the pH at which a color change is observed depends on the: _ A final concentration of HA. B final concentration of HIn. C initial concentration of HA. D pKa of HIn.

D The indicator will change color over a specific pH range. The range at which the color change takes place depends on the point at which HIn is converted to In-, and this depends on the pKa of the indicator which is answer D.

To a first approximation, the ionization constant of H2S is: A Less than zero. B Between zero and one but close to zero. C Between zero and one but close to one. D Greater than one.

Ionization constant is how easily something splits into ions. so how strong of an acid is H2S is really the question Solution: The correct answer is B. H2S was described as being a weak acid. This means that a small fraction of the solute molecule ionize and release protons into solution. The value of the ionization constant is therefore somewhere between 0 and 1, but much closer to 0. As an example, phosphoric acid, among the strongest of the weak acids, has an ionization constant that is approximately 1 x 10^-2, or 0.01. This is much closer to 0 than 1.

Which of the following shows the electron configuration of chlorine in NaCl? _ A 1s22s22p63s23p4 B 1s22s22p63s23p5 C 1s22s22p63s23p6 D 1s22s22p63s23p44s2

Na gives up electron to Cl soooo just do Cl- electron configuration and you got it Solution: The correct answer is C. In NaCl, chlorine exists as the chloride ion. A chloride ion has 18 electrons with the electron configuration 1s22s22p63s23p6. Thus, C is the best answer.

Consider the reaction below. in the image carbonyl + NaBH4 -> what? Which of the following observations about the infrared spectrum of the reaction mixture would indicate that the reaction shown above occurred?

NaBH4 is reducing agent so carbonyl -> hydroxyl group. The correct answer is B. This question asks the examinee to identify the structural differences between a primary alcohol and an aldehyde in the IR spectrum. Since the aldehyde function has a carbonyl group and there are no other carbonyl groups in either the starting material or product, then the loss of the carbonyl stretch in the IR spectrum indicates complete consumption of the starting aldehyde. Thus, B is the correct answer. Both the starting material and the product have O-H and aliphatic C-H bonds, so answers A and C are not diagnostic. Answer D is consistent with no reaction or an incomplete reaction since the aldehyde function is destroyed in this process.

The process taking place at the cathode was: A oxidation by a loss of electrons. B oxidation by a gain of electrons. C reduction by a loss of electrons. D reduction by a gain of electrons.

RED-CAT; AN-OX Solution: The correct answer is D. According to the passage, H2(g) was produced at the cathode. The reaction taking place at the cathode was 2H+(aq) + 2e- → H2(g), which means the H+ was gaining electrons and undergoing reduction. Thus, D is the best answer. note= if H2 (g) is created then 2H+ 2e- => H2(g)

What is the molar concentration of Na+(aq) in a solution that is prepared by mixing 10 mL of a 0.010 M NaHCO3(aq) solution with 10 mL of a 0.010 M Na2CO3(aq) solution? _ A 0.010 mole/L B 0.015 mole/L C 0.020 mole/L D 0.030 mole/L

Solution: The correct answer is Since 1 equivalent of NaHCO3 provides 1 equivalent of Na+, the molar concentration of Na+(aq) in 0.010 M NaHCO3(aq) solution is also 0.010 M = 0.010 mole/L. The molar concentration of Na+(aq) in 0.010 M Na2CO3(aq) solution is 0.020 mol/L since 1 equivalent of Na2CO3 provides 2 equivalents of Na+. When equal volumes of these two solutions are mixed, the resulting molar concentration is equal to their average, (0.010 mol/L + 0.020 mol/L)/2.Thus, B is the correct answer. *** I guess you have to average the amount of moles from two equations if you are mixing to make a 3rd solution

Experiment 1 was repeated with 0.40 g of calcium, and the gas that evolved was collected. The identity of the gas, and its approximate volume at 1.0 atm and 27°C were:(Note: R = 0.0821 L•atm/mol•K) A H2, 250 mL. B H2, 500 mL. C O2, 250 mL. D O2, 500 mL.

Solution: The correct answer is A. Calcium undergoes the following reaction with water: Ca(s) + 2H2O(l)→ Ca2+(aq) + 2OH-(aq)+ H2(g). The gas produced was H2. If 0.40 g calcium reacted, then the number of moles of calcium reacted was equal to (0.40 g)/(40.1 g/mol) = 0.01 mol. The amount of H2 formed was also 0.01 mol. At 1.0 atm and 27°C, the volume of 0.01 mol H2 = [(0.01 mol)(0.0821 L×atm/mol×K)(300 K)]/1.0 atm= 0.246 L = 246 mL. Thus, A is the best answer. HAVE TO KNOW HOW Ca will dissosciate. since it can only make H2 gas then we know that it would be acidic gas

Ba2+(aq) is an ion that is very toxic to mammals when taken internally. Which of the following compounds, mixed in water, would be the safest if accidentally swallowed? A BaSO4, Ksp = 1.1 × 10-10 B BaCO3, Ksp = 8.1 × 10-9 C BaSO3, Ksp = 8.0 × 10-7 D BaF2, Ksp = 1.7 × 10-6

Solution: The correct answer is A. The lower the value of Ksp is, the lower the concentrations of the cation and anion in an aqueous solution and the lower the solubility of the compound in water. If mixed with water and accidentally swallowed, the Ba salt with the lowest value of Ksp would be the safest. Thus, Ais the best answer. Low Ksp means low solubility so it remains a solid form!! we want the drug to stay solid so we need to choose the compound with the lowest solubility

Which of the bromomethanes is LEAST polar? WHEN SEE POLAR THINK DIPOLE MOMENT!!!! _ A CBr4 B CHBr3 C CH2Br2 D CH3Br

Solution: The correct answer is A. The polarity of each of the bromomethanes is determined by the vector sum of all of the individual bond polarities within a given molecule of the compound. Carbon is sp3hybridized and tetrahedral in all of the bromomethanes. Therefore, the vector sum of the bond polarities in CBr4 is zero, whereas the vector sum is greater than zero in the other compounds. Thus, answer choice A is the best answer.

Which of the following statements most accurately describes the solubility properties of fatty acid salts? A They are soluble in polar media only. B They are soluble in nonpolar media only. C They can partially dissolve in both polar and nonpolar media. D They are completely insoluble in both polar and nonpolar media.

Solution: The correct answer is C. A fatty acid salt contains a long hydrocarbon chain, which is soluble in nonpolar solvents. The salt also contains the charged group -CO2-Na+,which is soluble in polar solvents. Thus, C is the best answer. amphipathic = polar and non polar

According to Equation 1, the concentration of the polymer with respect to ​​ is: Eq 1 : nCDP-> (CP)n + nHPO4 According to Equation 1, the concentration of the polymer with respect to ​​ is: A ​​. B ​​. C ​​. D ​​.

Solution: The correct answer is C. According to the balanced coefficients in Equation 1, each time one molecule of the polymer ​​ is formed, there also are nhydrogen phosphate ions ​​ produced. Thus, the concentration of ​​ must be ​​times the concentration of the ​​. The correct response is C.

If Solution A contains Ag+, the anion component must be: The table 1. shows that when Ag is mixed with Cl a white precipitate forms, when mixed with F- no precipitate forms, when mixed with CrO4-2 a red precipitate forms, when mixed with S a black precipitate formed. Also says that solution A was completely soluble in water. A CrO42-. B Cl-. C F-. D S2-.

Solution: The correct answer is C. According to the passage, the ionic compound in Solution A was completely soluble in water. The information in Table 1 shows that the only Ag+salt studied that is soluble in water is AgF. Thus, C is the best answer.

Suppose that CH4(g) reacts completely with O2(g) to form CO2(g) and H2O(g) with a total pressure of 1.2 torr. What is the partial pressure of H2O(g)? _ A 0.4 torr B 0.6 torr C 0.8 torr D 1.2 torr

Solution: The correct answer is C. The balanced equation for the complete combustion of CH4(g) is shown below. CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) The pressure of the gaseous products is 1.2 torr. For every three product molecules, two are water. Therefore, the partial pressure of water is 2/3 the total pressure, because the total pressure is a function of the total number but not kind of molecules. Two-thirds of 1.2 torr is 0.8 torr. Thus, answer choice C is the best answer. Basically write the balanced equation... Ch4 + 2O2 = CO2 + 2H2O soooo we know that when 3 moles of gas are made 2/3 of is H2O gas. if total pressure is 1.2 torr then 1.2 (0.6)= 0.72

How much sodium hydroxide is needed to completely saponify a triacylglycerol? A A catalytic amount, because OH- is continuously being regenerated during saponification B One-third of an equivalent, because each OH- ion reacts to form three fatty acid salts C One equivalent, because each OH- ion reacts to produce one molecule of glycerol D Three equivalents, because one OH- ion is required to saponify each of the three fatty acid groups

Solution: The correct answer is D. One hydroxide ion is required to hydrolyze one ester linkage of a triacylglycerol molecule. Because there are three ester linkages in a triacylglycerol, three equivalents of sodium hydroxide will be needed to completely saponify the triacylglycerol. Thus, D is the best answer.

When aqueous solutions of the various anions and cations were mixed, precipitates formed because: A few aqueous solutions can contain more than one cation or anion. B the anions precipitated as solid metals. C the solubilities of cations were decreased by the other cations. D the solubility product of a compound was exceeded.

Solution: The correct answer is D. The amount of a substance that will dissolve in water is described by the Ksp. The Ksp for a substance, AaBb, equals [A]a[B]b. If the amount of the compound present is in excess of the Ksp, then a precipitate would form to maintain the Ksp. Thus, D is the best answer.

Maltose phosphorylase (MP) is an enzyme that catalyzes the hydrolysis of the (1→4) glycosidic bond in maltose. The substrates of MP are maltose and inorganic phosphate and catalysis produces glucose and glucose 1-phosphate. Several MP amino acid side chains stabilize bound phosphate prior to catalysis (Figure 1). Which statement accurately describes the properties of maltose? AIt is a reducing disaccharide in which only one anomeric carbon is involved in the glycosidic bond. BIt is a reducing disaccharide in which both anomeric carbons are involved in the glycosidic bond. CIt is a nonreducing disaccharide in which only one anomeric carbon is involved in the glycosidic bond. DIt is a nonreducing disaccharide in which both anomeric carbons are involved in the glycosidic bond.

The answer is A because maltose contains a (1→4) glycosidic bond, which means that the carbons involved are the C1 and C4 of the respective monosaccharides. The C1 is the anomeric carbon and a disaccharide that has a C1 carbon that is not involved in a glycosidic bond is said to have a hemiacetal end. This is the requirement to be a reducing sugar. It is a Knowledge of Scientific Concepts and Principles question because you must recall the properties of maltose.

A protein contains four disulfide bonds. In order to break these bonds researchers added a minimum of: _ A 2 moles of NADH for each mole of protein. B 4 moles of NADH for each mole of protein. C 2 moles of NAD+ for each mole of protein. D 4 moles of NAD+ for each mole of protein.

The answer is B because each mole of NADH can reduce a mole of disulfide bonds. Since the protein has four disulfide bonds, four moles of NADH are needed. This is a Scientific Reasoning and Problem Solving question because you must use the scientific model of NADH oxidation and disulfide reduction to make a prediction about molar equivalency.

Which protein has the highest electrophoretic mobility in SDS-PAGE under non-reducing conditions? (Note: There are no disulfide interactions unless stated in the table.)

The answer is C because in an SDS-PAGE gel that is run under non-reducing conditions, proteins 1, 3, and 4 will run as monomers. Protein 2 will run as a dimer because the disulfide bonds between Cys residues are not reduced. The running masses will then be: A = 32, B = 38, C = 25, D = 38. As the smallest one, Protein 3 will have the greatest electrophoretic mobility. This is a Reasoning about the Design and Execution of Research question because you must understand the experimental conditions used in SDS-PAGE and how addition or lack of a reductant affects the result of the analysis.

At pH 7, which of the following peptides will bind to an anion-exchange column and require the lowest concentration of NaCl for elution? _ A AVDEKMSTRGHKNPG B YPGRSMHEWDIKAQP C HIPAGEATEKALRGD D EAPDTSEGDLIPEVS

The answer is C because the anionic peptides bind to anion-exchange columns. The strength of the binding depends on the overall charge of the peptide. Of the options given, only C and D have a net negative charge. The net charge of peptide C is -1, whereas the net charge of peptide D is -5. Peptide C would elute at a lower salt concentration than Peptide D.

Experiments were then performed under three different conditions using both a 30 nucleotide substrate (30mer) and a 509 nucleotide substrate (509mer): Condition 1: Dnmt3a was pre-incubated with a biotinylated DNA substrate after which, non-biotinylated DNA substrate was added. Condition 2: Dnmt3a was pre-incubated with non-biotinylated DNA substrate after which, biotinylated DNA substrate was added. Condition 3: Dnmt3a was pre-incubated with buffer after which, both biotinylated and non-biotinylated DNA substrate were added. Which statement about the interaction between Dnmt3a and the DNA substrates is supported by the data in Figure 1? A Biotinylation prevents DNA strands from binding to Dnmt3a. B The stability of the protein/DNA fibers formed on the 30mer and 509mer DNA strands was the same. C The stability of the protein/DNA fibers formed on the 509mer DNA strand was greater than those formed on the 30mer DNA strand. D Biotinylated DNA strands are methylated at a higher rate than non-biotinylated DNA strands.

The answer is C because when the long (509mer) substrate is preincubated with Dnmt3a, the protein appears to form fibers that dissociate slowly from the DNA. This is supported by the ~3-fold increase in methylation rate when biotinylated substrate is used in the pre-incubation period over the rate observed when there is no pre-incubation. When the non-biotinylated substrate formed fibers on the 509mer DNA strand and the dissociation rate was slowed, the methylation rate was decreased because binding sites were not available for the biotinylated substrate. Note that this behavior is not observed for the 30mer, because it is not long enough to support stable fiber formation. It is a Data-based and Statistical Reasoning question because you must use the results of the experiments to come to the correct conclusion. Answer A is not correct because methylation of biotinylated DNA strands is observed. This would not occur if it were unable to bind to Dnmt3a. Answer B is not correct because no difference in the methylation rate of the 30mer is seen regardless of whether the pre-incubation step involved biotinylated or non-biotinylated DNA. This means that there is a rapid binding equilibrium, and stable protein fibers did not form on the shorter DNA substrate. By comparison, the methylation rates for the 509mer were affected by the identity of the pre-incubation substrate, indicating formation of stable protein fibers on the longer substrate. Answer D is not correct because the methylation rate of the substrates is not what is noted in the figure. Methylation of the non-biotinylated substrate is invisible in this experiment because this substrate does not bind to the avidin plate. This experiment does not compare rates of substrate methylation, it compares rates of methylation for biotinylated substrate as a function of the identity of the pre-incubation substrate.

In a certain kinetics experiment, the enzymatically catalyzed hydrolysis of ATP proceeds at a constant rate of 2.0 µM•s-1. If the volume of solution is 1.0 mL, what is the total number of ATP molecules that hydrolyzed after 1 min? A1.2 × 10-7 mol B3.3 × 10-6 mol C3.3 × 10-5 mol D1.2 × 10-4 mol _

The answer to this question is A. The total number of molecules that were hydrolyzed can be calculated by multiplying the rate in µM•s-1 by the time (in seconds) and the volume of the solution (in L): 2.0 × 10-6 mol•L-1•s-1 × 60 s × 1.0 × 10-3 L = 1.2 × 10-7 mol.

Which experimental feature of the MALDI-MS technique allows the separation of ions formed after the adduction of tissue molecules? A) distance travelled by ions depends on the ion charge B) velocity of ions depends on the ion mass-to-charge ratio. C) time of travel is inversely proportional to the ion mass-to-charge ratio D) Electric field between the MALDI plate and the MS analyzer is uniform

The answer to this question is B because the passage states that all ions travel the same distance of 0.5 m to the MS detector within the uniform electric field E region, and that the velocity of the ions is inversely proportional to their mass-to-charge ratio (m/z). Thus, the fastest ions are those with smallest m/z ratio, and these ions arrive first at the MS detector, being separated from the slower ones.

Which laser is suitable for the MALDI technique after its frequency is doubled? A) Laser A: wavelength 826nm, power 1.2 B) Laser B: wavlength 714, power 1.2 C) wavelength 650nm, power 1.5 D) wavelength 532nm, power 1.5mW

The answer to this question is D because the wavelength must be either 266 nm or 325 nm, and by doubling the frequency of the laser whose wavelength is 532 nm, the resulting wavelength is 532 nm/2 = 266 nm, because electromagnetic radiation wavelength and frequency are inversely proportional to each other. Also, the power of the radiation must be 1.5 mW.

The boiling point of glycerine in comparison with that of isopropyl alcohol, (CH3)2CHOH, is: A more than 10oC higher. B less than 10oC higher. C less than 10oC lower. D more than 10oC lower.

The correct answer is A. Glycerine (also known as glycerol) contains three hydroxyl groups, each capable of participating in hydrogen bonds. Isopropyl alcohol contains only one hydroxyl group. Glycerine is expected to be much more extensively hydrogen bonded than isopropyl alcohol. As hydrogen bonding increases the boiling point of a liquid, glycerine has a significantly higher boiling point than does isopropyl alcohol. This situation is represented by answer choice A.

H2O is liquid at room temperature, whereas H2S, H2Se, and H2Te are all gases. Which of the following best explains why H2O is liquid at room temperature? _ A Hydrogen bonds form between H2O molecules. B Oxygen lacks d orbitals. C H2O has a lower molecular weight. D H2O is more volatile.

The correct answer is A. H2O is capable of forming intermolecular hydrogen bonds, but H2S, H2Se, and H2Te are not. Thus, A is the best answer. Its always about the Hbond come on now.

What is the best explanation for the fact that a solution of NaNO2(aq) is basic? A NO2- is hydrolyzed with the formation of OH -(aq) ions. B Na+ is hydrolyzed with the formation of OH -(aq) ions. C NaNO2(aq) decreases the Ka of HNO2(aq). D NaNO2(aq) increases the Ka of HNO2(aq).

The correct answer is A. The NO2- reacts with water, forming OH- ions. Thus, A is the best answer. ** the Ka of equations will not be affected by adding more reactants. important to know that NaNO2 is a salt and nitrite will react with water to be hydrolyzed forming OH ions which would increase the pH .

A compound was analyzed and found to contain 12.0 g carbon, 2.0 g hydrogen, and 16.0 g oxygen. What is the empirical formula for this compound? _ A CH2O B C6HO8 C C6H12O6 D C12H2O16

The correct answer is A. The empirical formula shows the lowest whole number ratio of moles of each element in a mole of the compound, or the corresponding number of atoms per molecule. The number of moles of each element is found by dividing its mass in grams by its atomic weight. From the periodic table, the atomic weights of C, H, and O are 12.0, 1.0, and 16.0, respectively. The number of moles of carbon in 12.0 g C is 12/12 = 1; of hydrogen in 2 g H = 2/1 = 2; and of oxygen in 16 g O = 16/16 = 1. Therefore, the compound contains a 1:2:1 molar ratio of C:H:O, giving the formula C1H2O1, or simply CH2O. This formula is the empirical formula because the subscripts cannot be further reduced by a division by two. Thus, answer choice A is the best answer.

When 2.0 mL of 0.1 M NaOH(aq) is added to 100 mL of a solution containing 0.1 M HClO(aq) and 0.1 M NaClO(aq), what type of change in the pH of the solution takes place? AA slight (<0.1 pH unit) increase BA slight (<0.1 pH unit) decrease CA significant (>1.0 pH unit) increase DA significant (>1.0 pH unit) decrease

The correct answer is A. The pH will increase because a strong base is added. The increase will be small because the base is being added to a buffer solution. Thus, Ais the best answer.

With respect to bonding and electrical conductivity, respectively, sulfur hexafluoride, SF6(g), would be described as: A covalent and a nonconductor. B ionic and a nonconductor. C covalent and a conductor. D ionic and a conductor.

The correct answer is A. The passage states that the vast majority of covalent compounds are comprised exclusively of nonmetallic elements, whereas binary ionic compounds are made up of a metal and a nonmetal. Because neither sulfur nor fluorine is a metallic element, sulfur hexafluoride is a covalent compound. The passage also states that aqueous solutions of covalent compounds do not conduct electricity. Sulfur hexafluoride is covalent and a nonconductor. Thus, A is the best answer.

Which of the following properties is associated with the existence of glycine as a dipolar ion in aqueous solution? _ A High dipole moment B High molecular weight C Low dielectric constant D Low solubility in water

The correct answer is A. The problem states that glycine exists "as a dipolar ion in aqueous solution" and asks which of four properties is associated with this fact. Polarity in neutral molecules results from an uneven distribution of electron density, which can arise from separation of unlike charges. This occurs in zwitterions and in ylides. In addition, molecules that contain strongly electron-withdrawing or electron-donating substituents are highly polar and possess correspondingly high dipole moments. Thus, answer choice A is the best answer.

To examine the dissociation rate of Dnmt3a from DNA, His6-tagged Dnmt3a was expressed in Escherichia coli and purified with nickel-nitrilotriacetic acid-agarose chromatography. Reducing SDS-PAGE confirmed the purity and showed a single band at 35 kDa. Experiments were then performed under three different conditions using both a 30 nucleotide substrate (30mer) and a 509 nucleotide substrate (509mer): Dnmt3a was purified using which type of column chromatography? A Affinity B Anion-exchange C Cation-exchange D Size-exclusion

The correct answer is A. This is correct. This a Biochemistry question that falls under the content category "Separation and purification methods." The answer to the question is A because the use of histine tagging and a nickel column is a form of affinity chromatography.

What fraction of the starting amount of tritium remains after 13,500 days? Half life = 4500 days A 1/8 B 1/3 C 1/2 D 3/4

The correct answer is A. This is correct. This is a Physics question that falls under the content category "Atoms, nuclear decay, electronic structure, and atomic chemical behavior." The answer to the question is A because after 13,500 days, the equivalent of three tritium half-lives have passed, which means that (1/2)^3, or 1/8th, of the original sample remains.

Which of the following compounds has the most ionic character? A KBr(s) B CsCl(s) C NaI(s) D RbBr(s)

The correct answer is B. Cs is not in table, but should be in 0.8 column so chlorine 3-.8 equals highest electronegative difference which is 2.2 The compound with the greatest difference in electronegativities between the metal and nonmetal has the most ionic character. The data in Table 1 show that electronegativities tend to decrease down a group of the periodic table. Cs would have an electronegativity of approximately 0.8 or lower and Cl has an electronegativity of 3.0. The difference between the electronegativities of these two elements is the greatest of the compounds listed. Thus, B is the best answer.

The advantage of the Doppler ultrasound technique over the standard ultrasound technique is that it also allows: A. distinguishing between fluids and tissue. B. measuring the blood flow. C. measuring the tissue density. D. measuring the heart wall thickness.

The correct answer is B. Both the Doppler ultrasound and the standard ultrasound techniques can distinguish between fluids and tissues. The Doppler effect allows the observer to distinguish between a stationary and a moving object that reflects ultrasound waves, in this instance the blood in flow. The Doppler ultrasound technique cannot provide information regarding the tissue density. The Doppler ultrasound technique cannot provide information regarding tissue thickness, such as the heart wall thickness.

Which of the compounds shown below can form hydrogen bonds with water? CH3-CH2-CH3Compound 1 CH3-CH2-NH2Compound 2 _ A Compound 1 only B Compound 2 only C Both compounds D Neither compound

The correct answer is B. In Compound 2, which is an amine, two hydrogen atoms are covalently bonded to the nitrogen atom, and nitrogen is an electronegative element. Compound 1, an alkane, does not contain any hydrogen atoms covalently bonded to an atom of an electronegative element. Thus, B is the best answer.

Which experimental technique was most likely used by the students to determine the rate of reaction? A Monitor the increase in absorbance of the solutions at 200 nm. B Monitor the increase in absorbance of the solutions at 360 nm. C Monitor the decrease in absorbance of the solutions at 200 nm. D Monitor the decrease in absorbance of the solutions at 360 nm. ***Two students performed an assay to determine the concentration of lactase in a commercial preparation used to make milk consumable by those who are lactose intolerant. Lactase catalyzes the cleavage of lactose (β-D-galactopyranosyl-1→4-D-glucose) into β-D-galactose and D-glucose.Rather than use lactose as a substrate, which is difficult to quantify spectroscopically, the students chose Compound 1, which produces β-D-galactose and yellow Compound 2 under the action of lactase (Reaction 1)

The correct answer is B. This is a General Chemistry question that falls under the content category "How light and sound interact with matter." The answer to this question is B since the experiment describes that the substrate was chosen based on the fact that it produced a yellow colored product, Compound 2. The complementary color to yellow is purple, which is 360 nm. It is a Reasoning about the Design and Execution of Research question because you are asked to reason about the appropriateness of particular research designs.

After substrate exposure, the reactions were initiated by addition of 3H-labeled methyl-donating substrate, which undergoes β- decay with a half life of 4500 days. Samples were loaded onto plates containing immobilized avidin, which strongly binds biotin, and were washed extensively prior to analysis by scintillation counting (Figure 1). In the pre-incubation experiments described in the passage, the wash step after the binding reaction is designed to remove which molecule(s) from the plate? Biotinylated substrate Non-biotinylated substrate Methylated cytosine A I only B II only C II and III only D I, II, and III

The correct answer is B. This is correct. This is a Biochemistry question that falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer to the question is B because the wash step is designed to remove the non-biotinylated substrated. The biotinylated substrate is bound to the avidin on the plate. The biotin/avidin association is so tight, that the wash step would not remove the bound substrate. The non-biotinylated substrate would not bind to the avidin plate, and it would be washed away in the wash step. The step after the wash is a scintillation count, which would be used to quantify the amount of 3H-methyl groups.

What is the structure of the cytosine base after catalysis by Dnmt3a? Draw cytosine normal and then draw where it would be methylated. "Dnmt3a, a DNA methyltransferase, modifies the C5 position of cytosine bases within CpG sites."

The correct answer is B. This is correct. This is a Biochemistry question that falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer to the question is B because this structure represents cytosine that has been methylated at the C5 position.

Alcohols generally require acid catalysis in order to undergo substitution by nucleophiles. The acid catalyst enhances the reaction by: _ A increasing the solvent polarity. B creating a better leaving group. C neutralizing basic impurities. D protecting the alcohol group.

The correct answer is B. This question asks the examinee to remember that the substitution reaction in question serves to replace the hydroxyl group and that hydroxide ion is one of the worst leaving groups in substitution reactions. Under acidic conditions, the hydroxyl group is protonated such that the leaving group is now water, a superior leaving group, rather than hydroxide ion. Thus, B is the best answer. Solvent polarity and basic impurities are not particularly relevant, and protecting the alcohol group defeats the purpose of the reaction.

Na2CO3 + HCl → CO2 + H2O + NaCl Consider the above unbalanced equation. For this reaction, how many mL of a 2 M solution of Na2CO3 are required to produce 11.2 L of CO2 at STP? A 125 mL B 250 mL C 375 mL D 500 mL

The correct answer is B. This question deals with stoichiometry, which examines the quantities of reactants and products associated with a chemical reaction. The first step in working stoichiometry problems is to balance the equation. For this reaction, the balanced equation is: Na2CO3 (aq) + 2 HCl (aq) → CO2 (g) + H2O (l) + 2 NaCl (aq) The problem seeks the volume of 2 M Na2CO3solution which, when reacted with HCl solution, will produce 11.2 L of CO2 gas at STP (standard temperature and pressure). It is an important fact that one mole of an ideal gas, at STP, will occupy a volume of 22.4 L. Therefore, 11.2 L of CO2 gas, at STP, must represent 0.5 mole of CO2. (Virtually all gases can be approximated as ideal gases at common temperatures and pressures.) According to the balanced chemical equation, one mole of CO2 is produced when one mole of Na2CO3 reacts. Therefore, the amount of Na2CO3 required to produce 0.5 mole of CO2 gas must also be 0.5 moles. We must find the volume of 2 M Na2CO3 solution which contains the required 0.5 moles of reactant: Working through the numbers for this problem, we find that the answer is 250 mL. Thus, answer choice B is the best answer. In problems of this sort, it is crucial to enter the proper units for each number. When the units in the equation cancel out to leave the result in the correct units, it is an important indicator that the problem has been set up correctly.

When glycerol reacts with three different fatty acids, how many stereogenic centers does the product triacylglycerol contain? A 0 B 1 C 2 D 3

The correct answer is B. When glycerol reacts with three different fatty acids, only carbon 2 in the resulting triacylglycerol is attached to four different groups. Thus, B is the best answer.

Would deviations from the ideal gas law be observed for gaseous nitrogen at 180 GPa and room temperature? "In the absence of an energy barrier, the solid form of nitrogen is predicted to be stable above 65 GPa pressure (1 atm = 1 × 105 Pa), with the diatomic form stable below that pressure." A No, because both temperature and pressure must increase before such deviations are observed B No, because nitrogen molecules are symmetrical and do not interact with each other C Yes, because at this pressure, molecular volumes and intermolecular forces become significant D Yes, because at room temperature, molecular volumes and intermolecular forces become significant

The correct answer is C. According to the passage, the solid form of nitrogen is predicted to be stable above 65 GPa. At 180 GPa, deviations from the ideal gas law would be observed for gaseous nitrogen. The ideal gas law assumes that molecular volumes and intermolecular attractive forces are negligible. At a pressure at which solid nitrogen is predicted to be stable, neither of these assumptions would be valid. Thus, C is the best answer.

According to trends in electronegativity, which of the following pairs of atoms is most likely to form an ionic bond? A N and O B C and F C Ca and I D Si and Cl

The correct answer is C. An ionic bond is most likely to form between elements of very high and very low electronegativity. In practice this generally means elements at the far right and far left of the periodic table respectively. Of the pairs of elements offered as choices, only calcium and iodine are found at opposite sides of the periodic table. An ionic compound of formula CaI2 is likely to form between them. Answer choice C is the correct answer.

What is the sum of the protons, neutrons, and electrons in strontium-90? strontium atomic number is 38. A 90 B 126 C 128 D 218

The correct answer is C. The number of protons in the nucleus of an element is given by the atomic number. In a neutral atom, the number of electrons is equal to the number of protons. The number of neutrons can be found by subtracting the atomic number from the mass number. The atomic number of strontium (Sr) is 38; so the number of neutrons is 90 - 38 = 52. The sum of protons, neutrons, and electrons in strontium is 38 + 38 + 52 = 128. This value is given as answer choice C.

Which of the following statements is consistent with the incorrect conclusion that HCl is an ionic compound? A It is a gas at room temperature. B A 1 M solution freezes below 0°C. C A 1 M solution conducts electricity. D It is composed of two nonmetals.

The correct answer is C. The passage states that an aqueous solution of an ionic compound conducts electricity. Thus, Cis the best answer.

For what mechanistic reason does G1 of lactase first act as a Brønsted acid during catalysis? A G1 becomes a better nucleophile. B G2 becomes a better nucleophile. C Glucose becomes a better leaving group. D Galactose becomes a better leaving group. ****A glutamate residue (Glu-1) in the active site donates a proton to an oxygen atom. Another glutamate on the opposite axial side (Glu-2) acts as a nucleophile to liberate D-glucose and generate an α-D-galactopyranosyl-modified enzyme intermediate. Then, Glu-1 deprotonates water, and the resulting hydroxide ion acts as a nucleophile to liberate β-D-galactose and regenerate the enzyme.

The correct answer is C. This is an Organic Chemistry question that falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer to this question is C. Protonation of the oxygen atom in glucose makes this substance a better leaving group in much the same way that protonation of an alcohol facilitates substitution of an -OH group. It is a Scientific Reasoning and Problem Solving question because you are asked to bring together theory, evidence, and observations to draw a conclusion.

Based on the description provided, if lactose was hydrolyzed under the action of lactase in O-18 labeled water, in which location(s) would the label appear? A Neither the galactose nor the glucose products B The glucose product only C The galactose product only D Both the galactose and the glucose products **8Lactase catalyzes the cleavage of lactose (β-D-galactopyranosyl-1→4-D-glucose) into β-D-galactose and D-glucose. A glutamate residue (Glu-1) in the active site donates a proton to an oxygen atom.

The correct answer is C. This is an Organic Chemistry question that falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer to this question is C. The cleavage reaction described is a hydrolysis of the glycoside linkage in a disaccharide. In this case, the deprotonated water attacks the galactose and so this sugar will be labeled with O-18. The glucose is protonated and acts a leaving group without reacting with an oxygen atom provided by water. It is a Scientific Reasoning and Problem Solving question because you are asked to bring together theory, evidence, and observations to draw a conclusion.

What are the oxidizing and reducing agents, respectively, in the reaction below? 2HCl + H2O2 + MnO2 → O2 + MnCl2 + 2H2O _ A H2O2; HCl B H2O2; MnO2 C MnO2; HCl D MnO2; H2O2

The correct answer is D. In the reaction pictured, Mn is reduced from +4 to +2; therefore, MnO2 is the oxidizing agent. O is oxidized from -1 in H2O2 to 0 in O2; therefore, H2O2 is the reducing agent. Thus, D is the best answer.

During glycolysis, pyruvate CH3C(=O)CO2- is reduced to lactate CH3CH(OH)CO2- by nicotinamide adenine dinucleotide (NADH). What is the balanced reaction for this conversion? A. CH3C(=O)CO2- + 2NADH → CH3CH(OH)CO2- + 2NAD+ B. CH3C(=O)CO2- + 2NADH + 2H+ → CH3CH(OH)CO2- + 2NAD+ C. CH3C(=O)CO2- + NADH + 2H+ → CH3CH(OH)CO2- + NAD+ D. CH3C(=O)CO2- + NADH + H+ → CH3CH(OH)CO2- + NAD+

The correct answer is D. Only 1 NADH molecule is required as reactant, H+ is missing as reactant, and only 1 NAD+ forms. Only 1 NADH molecule is required as reactant, only 1 H+ is needed as reactant, and only 1 NAD+forms. Although the number of NADH as reactant and NAD+ as product is correct, there should only be 1 H+as reactant, not 2. There is the correct number of NADH and H+ as reactants and NAD+ as product.

What bonding accounts for the expected increase in energy density of solid nitrogen as compared to methanol? A Solid nitrogen contains covalent and ionic bonds; methanol contains only weak ionic bonds. B Solid nitrogen contains covalent and ionic bonds; methanol has covalent bonds within each molecule and weak van der Waal's interactions between molecules. C Solid nitrogen contains only covalent bonds; methanol contains only weak ionic bonds. D Solid nitrogen contains only covalent bonds; methanol has covalent bonds within each molecule and weak intermolecular interactions.

The correct answer is D. Solid nitrogen contains only covalent bonds. Methanol contains only covalent bonds within each molecule but is capable of intermolecular hydrogen bonding because of the -OH group. Thus, D is the best answer.

Which of the following atoms has the largest first ionization energy? A Potassium B Zinc C Gallium D Krypton

The correct answer is D. The first ionization energy is the energy required to remove an electron from the outer shell of an atom. For any given period, it is lowest when the removal of the electron results in a complete shell or subshell, and highest when the removal of the electron disrupts a complete shell or subshell. Krypton, a noble gas, has a complete outer shell of electrons and therefore has an extremely high ionization energy. Potassium, with only one electron in its outermost shell, has a low ionization energy. Choice D is the correct answer.

Which of the following pairs of compounds provides an example of ionic and covalent bonding, respectively? A I don't think it's A B NaCl(s) and NaI(s) C I don't think it's C D NaCl(s) and HBr(g)

The correct answer is D. The passage states that the vast majority of covalent compounds are comprised exclusively of nonmetallic elements, whereas binary ionic compounds are made up of a metal and a nonmetal. Sodium is a metal and chlorine is a nonmetal; consequently, NaCl is an ionic compound. Hydrogen and bromine are both nonmetals, which means HBr is a covalent compound. Thus, D is the best answer.

What is the KM of Compound 1 with respect to lactase? A 0.05 mM B 0.10 mM C 5 mM D 20 mM

The correct answer is D. This is a Biochemistry question that falls under the content category "Principles of chemical thermodynamics and kinetics." The answer to this question is D because the value of 1/[S] when 1/Vo is zero (the x-intercept of the Lineweaver-Burk plot) is -1/KM and so KM = -(-1/0.05 mM-1) = 20 mM. It is a Data-based and Statistical Reasoning question because you are asked to analyze and interpret data presented within the context of an experiment to arrive at the conclusion.

If [E]T was the concentration of lactase in the kinetics trials, what expression gives the concentration of lactase in the commercial preparation of this enzyme? A [E]T × 25 B [E]T × 50 C [E]T × 200 D [E]T × 500 The students prepared stock solutions of Compound 1 in four different concentrations in pH 7 buffer. A stock solution of the enzyme was prepared by diluting 0.100 mL of the commercial preparation to 25.0 mL in the buffer solution. Experiments were initiated by mixing 1.0 mL of each substrate solution with 1.0 mL of the enzyme solution. The initial rates Vo were measured for each trial. The students then plotted 1/Vo versus 1/[S] (Figure 1) to determine KM, Vmax, and [E]T for the four trials.

The correct answer is D. This is a General Chemistry question that falls under the content category "Unique nature of water and its solutions." The answer to this question is D because the commercial preparation was first diluted by 1 → 250 to prepare the stock solution, and it was further diluted by 1 → 2 by mixing with the substrate stock solution to perform the kinetics experiments.

thermodynamic. product

The product of a reaction that is formed favorably at a higher temperature because thermal energy is available to form the transition state of the more stable product; has a larger overall difference in free energy between the products and reactants than the kinetic product. thermo is favored if allowed full time and high temp!!1

In order to study the kinetics of GalK catalysis, researchers coupled the phosphorylation of galactose with two other enzymatically catalyzed reactions. The coupled reactions were the pyruvate kinase catalyzed conversion of phosphoenolpyruvate to pyruvate and the lactate dehydrogenase catalyzed conversion of pyruvate to lactate. The reaction was monitored by the decrease in the concentration of NADH. The resulting kinetic data with respect to both ATP and galactose are shown in Table 1. The experiments were conducted using wild-type (WT) GalK and two variant forms of GalK where Asp45 was substituted with either Ala or Gly Which solution components are oxidized and reduced during the enzymatic assay described in the passage? A NADH is oxidized and pyruvate is reduced. B NAD+ is reduced and pyrvuate is oxidized. C Phosphoenolpyruvate is oxidized and NAD+is reduced. D Phosphoenolpyruvate is reduced and NADH is oxidized.

This Biochemistry question falls under the content category "Principles of chemical thermodynamics and kinetics." The answer is A because it can be assumed that NADH is being oxidized to NAD+ since the experiment quantifies the decrease in NADH. This must be coupled with a reduction reaction, which is the reduction of pyruvate to lactate. It is a Scientific Reasoning and Problem Solving question because you must use the model of NADH redox cycling to understand what will happen to pyruvate.

Which solution component will have the lowest concentration at the end of the kinetic assay described in the passage? A Lactate B ADP C ATP D NAD+ In order to study the kinetics of GalK catalysis, researchers coupled the phosphorylation of galactose with two other enzymatically catalyzed reactions. The coupled reactions were the pyruvate kinase catalyzed conversion of phosphoenolpyruvate to pyruvate and the lactate dehydrogenase catalyzed conversion of pyruvate to lactate. The reaction was monitored by the decrease in the concentration of NADH. The resulting kinetic data with respect to both ATP and galactose are shown in Table 1. The experiments were conducted using wild-type (WT) GalK and two variant forms of GalK where Asp45 was substituted with either Ala or Gly.

This Biochemistry question falls under the content category "Principles of chemical thermodynamics and kinetics." The answer is B because the GalK reaction produces ADP, but this ADP is immediately converted back to ATP by pyruvate kinase. Lactate and NAD+are also both end products of the assay. It is a Reasoning about the Design and Execution of Research question because you must understand the experimental setup to understand how the concentrations of the solution components will change.

Based on the data in Table 1, the highest catalytic efficiency results from which enzyme-substrate combination? A Wild-type with respect to galactose B Wild-type with respect to ATP C D45A with respect to galactose D D45A with respect to ATP

This Biochemistry question falls under the content category "Principles of chemical thermodynamics and kinetics." The answer is B because the catalytic efficiency is measured by the kcat/KM, which is the highest for the wild-type in with respect to ATP. The kcatis the highest for wild-type and the smallest KMfor wild-type occurs with ATP. The value of kcat/KM is 2.6 × 105 mol-1s-1 and the next highest value is 2.3 × 105 mol-1s-1 for D45G with respect to ATP. It is a Data-based and Statistical Reasoning question because you must use the data in the table to determine the trend in catalytic efficiency with the various substrates. EFFICIENCY = KCAT/KM!!!

The active site of MP contains a conserved glutamate residue (Glu487), which is believed to act as a general acid during catalysis, protonating the glycosidic bond. After this, it is likely that one of the phosphate oxygen atoms is responsible for the nucleophilic attack of the C1 carbon of the glycosidic bond. Which change in the protonation state of Glu487 is most likely responsible for the change in MP activity at either low or high pH? The change in activity at: A low pH is due to the protonation of Glu487. B low pH is due to the deprotonation of Glu487. C high pH is due to the protonation of Glu487. D high pH is due to the deprotonation of Glu487.

This Biochemistry question falls under the content category "Principles of chemical thermodynamics and kinetics." The answer is D because the pH-dependence of MP activity shows that activity decreases at both low pH and high pH. At low pH, protonation of functional groups is the most likely cause of decreased activity and at high pH deprotonation of functional groups is the most likely cause of decreased activity. The passage states that Glu487 protonates the substrate prior to nucleophilic attack. This means that Glu487 must be able to act as a Brønsted acid during catalysis, which implies that the pKa of this side chain must be significantly higher in the enzyme active site than it is in solution, otherwise it would exist in the deprotonated form in the active site. If the pH is raised sufficiently to cause deprotonation of Glu487, it will no longer be able to participate in catalysis.

What is the most likely reason for the decreased KMvalues observed in the D45G variant compared to the other two versions of GalK with respect to each substrate? A The substrate binding pocket is too crowded. B A key hydrogen bond between the enzyme and substrate is lost. C The enzyme is unfolded as a result of the substitution. D The Vmax is much lower, which means less substrate is needed to reach it.

This Biochemistry question falls under the content category "Principles of chemical thermodynamics and kinetics." The answer is D because, in this case, it seems most likely that the drop in KM is the result of the much lower Vmax (kcat•[Et]) since KM is just the substrate concentration necessary to reach 1/2 Vmax. It is a Data-based and Statistical Reasoning question because you must interpret a pattern in the kinetic data. Option A is incorrect. The substitution to a glycine would make the pocket less crowded, not more. Option B is incorrect. Neither alanine nor glycine makes hydrogen bonding interactions with the substrate through the side chain, so if this was the case, the low KM should be observed in both cases. Option C is incorrect. The enzyme still shows activity so it is not unfolded.

A protein with which properties will most likely have the largest negative net charge at pH 7? _ A A protein that binds to an anion-exchange column at pH 7 and requires a high concentration of NaCl for elution B A protein that binds to an anion-exchange column at pH 7 and requires a moderate concentration of NaCl for elution C A protein that binds to a cation-exchange column at pH 7 and requires a high concentration of NaCl for elution D A protein that binds to a cation-exchange column at pH 7 and requires a moderate concentration of NaCl for elution

This Biochemistry question falls under the content category "Separation and purification methods." The answer is A because a protein with a low pI would be negatively charged at pH 7. This protein, being anionic, would bind to an anion exchange column (eliminates choices C and D). Largest negative net charge implies the presence of a large quantity of negatively charged amino acids, allowing the protein to bind tightly to the column. A high concentration of NaCl would be required for elution (eliminates choice B). This is a Reasoning about the Design and Execution of Research question because you must understand how the design of ion-exchange chromatography experiments affects the separation of proteins with various charges.

Based on the results shown in Figure 2, what effect does mutation have on M1 mRNA? A The folded structures of wild-type and MBmutant M1 mRNA are similar, but the structure of HPmutant M1 mRNA is less compact. B The folded structures of wild-type and MBmutant M1 mRNA are similar, but the structure of HPmutant M1 mRNA is more compact. C Wild-type and MBmutant M1 mRNA have the same number of base pairs, but HPmutant M1 mRNA has fewer base pairs. D Wild-type and MBmutant M1 mRNA have the same number of base pairs, but HPmutant M1 mRNA has more base pairs.

This Biochemistry question falls under the content category "Separation and purification methods." The answer is B because the native gel shows that the size and shape of wild-type and MBmutant are similar, but HPmutant is more compact because it migrates further in the gel. The denaturing gel confirms that the number of nucleotides in each mRNA is the same. It is a Data-based and Statistical Reasoning question because you must use the results of the electrophoresis experiments to compare the folded structures of the biomolecules. more compact= smaller in size which means it travels farther in the gel.

Based on the data in Table 1, what is the purification yield from the culture supernatant after the final ion-exchange chromatography step? A 0.1% B 20% C 40% D 60%

This Biochemistry question falls under the content category "Separation and purification methods." The answer is B because, while the specific activity represents a measure of solution purity, the activity units themselves provide the best measurement of yield. The total number of activity units in the initial culture was 3000 mg × 0.1 units/mg = 300 units. After the final step, the activity units that remain are 3 mg × 20 units/mg = 60 units. This represents a 60/300 = 0.2 or 20% yield.

Substituting residues in a peptide with which amino acid will most likely result in a peptide with an increased pI? _ A Lys B Glu C Gln D Val

This Biochemistry question falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer is A because Lys is a basic residue, meaning it would contribute to a high pI. Glu is an acidic residue, meaning it would contribute to a low pI. Gln and Val are both neutral. They would not contribute to the pI. This is a Knowledge of Scientific Concepts and Principles question because you must recall acid and base properties of amino acids.

Which nucleoside has the largest molecular weight? _ A Adenosine B Guanosine C Deoxyadenosine D Deoxyguanosine

This Biochemistry question falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer is B because DNA nucleosides have a lower molecular weight than RNA due to the lack of the 2′ hydroxyl group. Guanine has a higher molecular weight than adenine. This is a Knowledge of Scientific Concepts and Principles question because you must recall structure of nucleic acids.

Based on the data in Figure 3, what is the melting temperature Tm of wild-type M1 mRNA in a solution containing 100 mM KCl? A 45°C B 53°C C 64°C D 80°C

This Biochemistry question falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer is B because the Tm is the temperature at which 50% of the molecules are denatured or the fraction folded is 0.5. This occurs at approximately 53°C in 100 mM KCl. It is a Data-based and Statistical Reasoning question because you must interpret the data from the graph of thermal denaturation to determine melting temperature.

Which of the four DNA bases contains the largest number of hydrogen bond acceptors when involved in a Watson-Crick base pair? A) A B) C C) G D) T

This Biochemistry question falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer is B because the hydrogen bond acceptors are N and O. Adenine contains 1 donor and 1 acceptor, thymine contains 1 donor and 1 acceptor, guanine contains 2 donors and 1 acceptor, and cytosine contains 1 donor and 2 acceptors. This is a Knowledge of Scientific Concepts and Principles question because you must recall structure of a Watson-Crick base pair.

The side chain of which amino acid can form a bond that is similar to a peptide bond? A Gly B Phe C Lys D Ala

This Biochemistry question falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer is C because the lysine side chain can form isopeptide bonds by reacting with a carboxylic acid group, which is the same way that peptide bonds are formed. This is a Scientific Reasoning and Problem Solving question because you must apply the scientific model of a peptide bond to that of amino acid side chains.

Based on the data in Figure 3, the presence of which ion is most effective at stabilizing M1 mRNA? A K+ B Cl- C Mg2+ D Na+

This Biochemistry question falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer is C because three of the experimental conditions in Figure 3 result in equal stability that is higher than the condition representing only 100 mM KCl in solution. Introduction of Mg2+ to the solution increases the stability to the same level regardless of whether K+ and Cl- concentrations are higher or lower. The 1 M NaCl also results in the same stability, but the 1 M concentration is much higher than the 10 mM of Mg+ used. Therefore, Mg2+ shows the largest effect. It is a Data-based and Statistical Reasoning question because you must use the data in the figure to reach the correct answer.

Thermal denaturation experiments can be used to follow the transition of double-stranded DNA into single-stranded DNA. Which of the following parameters affects the Tm of dsDNA in this experiment? I)pH of solution II)Ionic strength of solution III)Length of DNA strands A I and II only B I and III only C II and III only D I, II, and III

This Biochemistry question falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer is D because all of these parameters would affect the thermodynamic stability of the DNA double helix. Significant drops in pH would result in protonation of hydrogen-bond acceptors, leading to a loss in base-pairing interactions. The presence of positive ions in solution (particularly Mg2+) leads to stabilization of the DNA fold via shielding of the repulsion between phosphate groups within the DNA backbone. The length of DNA strands would also play a role. Longer DNA strands are held together by more hydrogen bonds, meaning that more energy is required to denature the double-stranded DNA. This is a Reasoning about the Design and Execution of Research question because you must understand the consequences of changing variables on the results an experiment.

The first committed step of galactose catabolism is catalyzed by galactokinase (GalK), which transfers a phosphate group from ATP to galactose. Analysis of the substrate binding pocket shows that various residues engage in hydrogen bonding interactions with galactose. Arg36, Glu42, Asp45, and Gly180 represent four of these residues. Which amino acid in the substrate binding pocket is most likely to interact with the substrate through its backbone as opposed to its side chain? A Arg36 B Glu42 C Asp45 D Gly180

This Biochemistry question falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer is D because the interactions with the substrate are hydrogen bonds. Of the amino acids listed, only glycine is unable to form hydrogen bonds with side chain atoms, which means it is the most likely to use backbone atoms. It is a Scientific Reasoning and Problem solving question because you must use the scientific model of amino acid interactions to make a claim about substrate binding.

Which change to the sequence of wild-type M1 mRNA is most likely to increase its stability? A Changing position 132 to a G and position 151 to a T B Changing position 134 to an A and position 149 to a U C Changing position 129 to an A and position 154 to a U D Changing position 133 to a G and position 150 to a C

This Biochemistry question falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer is D because the stability of both DNA and RNA is largely dependent on the number of GC base pairs contained with the folded structure. Therefore, any change that converts and existing AU base pair to a GC base pair is most likely to increase the stability of the folded mRNA molecule. It is a Scientific Reasoning and Problem Solving question because you must use the scientific model of nucleic acid stability to make a claim.

Positions 129-154 of segment 7 of the M1 mRNA contain a total of how many purines and pyrimidines? A 17 purines and 9 pyrimidines B 9 purines and 17 pyrimidines C 14 purines and 12 pyrimidines D 12 purines and 14 pyrimidines

This Biology question falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer is A because the purines are A and G and the pyrimidines are C and U. Therefore, there are 17 purines and 9 pyrimidines. It is a Knowledge of Scientific Concepts and Principles question because you must know the difference between purines and pyrimidines. PURE AS GOLD: purine is Adenine and Guanine

What approximate volume of reaction solution was used in the experiments described in the passage if 0.5 mg of MgCl2 were added to the reaction mixture? A 500 μL B 750 μL C 2 mL D 50 mL

This General Chemistry question falls under the content category "Unique nature of water and its solutions." The answer is A because the concentration of MgCl2 was 10 mM, which means there were 10-2 moles of MgCl2per liter of solution. MgCl2 has a molar mass of 95.3 g/mol (approximately 100 g/mol). If 0.5 mg were added, that equates to 5 × 10-6 moles. And since 10-2 moles were present per liter, the number of liters is 5 × 10-4, or 500 μL. It is a Scientific Reasoning and Problem Solving question because you must apply the appropriate formula to solve a problem.

The relationship between the steroid hormone, estrogen, and the peptide hormone, insulin, is being investigated. In order to quantify levels of each of these hormones, tissue samples were homogenized and then placed in a mixture of 2:1 hexane/water. What is the expected result from this extraction method? _ A Estrogen would be in the hexane phase; insulin would be in the aqueous phase. B Both estrogen and insulin would be in the aqueous phase. C Insulin would be in the hexane phase; estrogen would be in the aqueous phase. D Both estrogen and insulin would be in the hexane phase.

This Organic Chemistry question falls under the content category "Separation and purification methods." The answer is A because the structural features of peptides and steroids imply that insulin is hydrophilic and estrogen is hydrophobic. The hydrophilic insulin peptide would be expected to be found in the aqueous phase, while the hydrophobic estrogen hormone would segregate into the hexane phase. This is a Reasoning about the Design and Execution of Research question because you must understand how the design of an extraction experiment would affect two biomolecules.

What is the most accurate description of the natural substrate of GalK? A It is a five-carbon aldose. B It is a five-carbon ketose. C It is a six-carbon aldose. D It is a six-carbon ketose.

This Organic Chemistry question falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer is C because galactose is a C-4 epimer of glucose, which means that it is a six-carbon aldose. It is a Knowledge of Scientific Concepts and Principles question because you must recall the structure and classification of galactose.

Dnmt3a, a DNA methyltransferase, modifies the C5 position of cytosine bases within CpG sites. Dnmt3a is active as a homotetramer and binds to DNA in a cooperative manner, forming stable protein/DNA fibers. However, it was unknown how formation of these fibers affected the dissociation rate of Dnmt3a from DNA. To examine the dissociation rate of Dnmt3a from DNA, His6-tagged Dnmt3a was expressed in Escherichia coli and purified with nickel-nitrilotriacetic acid-agarose chromatography. Reducing SDS-PAGE confirmed the purity and showed a single band at 35 kDa. What is the approximate molecular weight of active Dnmt3a? A 35 kDa B 70 kDa C 105 kDa D 140 kDa

This is a Biochemistry item that falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer to the question is D because the enzyme is active as a homotetramer, which means that the molecular weight given by SDS-PAGE is four times less than the active molecular weight. This is due to the fact that the SDS-PAGE conditions denature the protein and eliminate quaternary structure.

Assuming binding is fast relative to subsequent catalytic steps, what relative effect does substituting the various active site residues have on catalysis versus substrate binding? The various substitutions affect: Asubstrate binding more than catalytic turnover. Bcatalytic turnover more than substrate binding. Csubstrate binding, but not catalytic turnover. Dcatalytic turnover, but not substrate binding.

This is a Biochemistry question that falls under the content category "Principles of chemical thermodynamics and kinetics." The answer to this question is B because the substitutions reduce kcat by three orders of magnitude (× 103), but reduce substrate binding (KM) by as much as a factor 6. It is a Data-based and Statistical Reasoning question because you are asked to analyze and interpret data presented within the context of an experiment to arrive at the conclusion. ***Km is binding affinity. Kcat is catalytic efficiency.

[One product, NAG3 has a high affinity for hen egg white (HEW) lysozyme: affinity constant KNAG3 = 1.3 × 105 M-1, ΔH° = -50 kJ/mol, ΔG° = -30 kJ/mol. Structural studies indicate that NAG3 makes hydrogen bonds with D59, W62, W63, D101, N103, and A107 in a cleft in HEW lysozyme that is adjacent to the catalytically active D52 and E35 residues.] Based on the structural studies, when NAG3 interacts with amino acids in HEW, it can hydrogen bond to the side chains of the amino acids at all of the positions described EXCEPT for the amino acid at which position? A 62 B 101 C 103 D 107

This is a Biochemistry question that falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer to this question is D. NAG3 interacts with the amino acids at positions 59, 62, 63, 101, 103, and 107. The side chains of the amino acids at all of the positions contain sites for potential hydrogen bonding with the exception of alanine which is at position 107.

What quantity of NAG3 was required to reach the equivalence point in the titration? A 25 nmol B 100 nmol C 25 µmol D 100 µmol

This is a General Chemistry question that falls under the content category "Unique nature of water and its solutions." The answer to this question is B because the graphs in Figure 1 both imply a 1:1 mole ratio of NAG3 was added at the equivalence point and the solution contained 1.0 mL of 0.10 mM = (1 × 10-3 L)(0.10 × 10-3 mol/L) = 1.0 × 10-7 mol = 100 nmol.

The biomolecules M are adducted by free protons from DHB and form [M+H]+ and [M+2H]2+ ions. The ions leave the MALDI plate due to the effect of the 4.5 kV electric voltage between the MALDI plate and the MS detector, and travel a distance of 0.5 m to the MS detector within the uniform electric field E region. The velocity of the ions is inversely proportional to their mass-to-charge ratio (m/z). Proteins can be "fingerprinted" using MALDI if they are subjected to proteolytic cleavage before analysis. What is the magnitude of the electric field used in the MALDI-MS imaging device described in the passage? A) 1 kV/m B) 3 kV/m C) 6kV/m D)9 kV/m

This is a Physics question that falls under the content category "Electrochemistry and electrical circuits and their elements." The answer to this question is D because the intensity of a uniform electric field E is related to the voltage V and distance d over which it is applied as E = V/d = 4.5 kV/(0.5 m) = 9 kV/m.

The reacting substrate carbon atom in the mechanism described in the passage undergoes which of the following hybridization state changes during the reaction? (Note: the middle hybridization state refers to an intermediate.) A sp2 → sp3 → sp2 B sp3 → sp2 → sp2 C sp3 → sp2 → sp3 D sp3 → sp3 → sp3

This is an Organic Chemistry question that falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer to this question is A. As described in the passage, deprotonation of water is the preliminary step in the reaction sequence. This knowledge and knowledge of the reaction pathways that lead to amide hydrolysis are enough to conclude that the sequence of events involves nucleophilic attack of coordinated hydroxide on the sp2-hybridized carbonyl carbon atom to generate an sp3-hybridized and tetrahedral intermediate which subsequently eliminates an amine (after proton transfer) to generate an sp2-hybridized carbon atom. It is a Scientific Reasoning and Problem Solving question because you are asked to bring together theory, evidence, and observations to draw a conclusion.

A sparingly soluble metal hydroxide, M(OH)2 has a molar solubility of S mol/L at 25°C. Its Kspvalue is: _ A S2. B 2S2. C 2S3. D 4S3.

This one is tough bro IDK. Solution: The correct answer is D. The Ksp for a substance, AaBb, equals [A]a[B]b.The Ksp for M(OH)2 = [M][OH-]2. If the solubility of M(OH)2 is S mol/L, then [M] = S mol/L and [OH-] = 2S mol/L. The Ksp = S(2S)2 = S(4S2) = 4S3. Thus, D is the best answer. so the Ksp of AaBb= [A]^a [B]^b so the Ksp of M(OH)2 would be [s][2s]^2 or we could think of it as [s][2s][2s] to get 4S^3.

What is the energy content in kcal of one peanut, if the temperature of 1 kg of water in a calorimeter increases by 50oC upon the combustion of 10 peanuts? (Note: For water, the heat of fusion is 1.4 kcal/mol, the specific heat is 4.185 J/g•oC or 1 cal/g•oC, and the density is 1.0 g/mL at 15oC. One kg equals 2.2 pounds.) A 0.5 kcal B 1 kcal C 5 kcal D 10 kcal

dimensional analysis should get to 50kcal for 10 peanuts and so one peanut is 5kcal.

succrose

non reducing sugar glucose + fructose

The pKa for the dissociation of ​​ to ​​ is 6.7. What is the initial ratio of A- to HA​​ in the buffer solution of Experiment 1? if the pH of buffer is 8.7!!! A 1:1 B 2:1 C 100:1 D 200:1

pH=pKA + log (A-/HA) if pH = 8.7 and pKa= 6.7 we need (log (A-/HA) to equal 2. log(100)=2 so ratio must be 100A- to 1 HA

lactose

reducing sugar glucose + galactose

Maltose

reducing sugar made of glucose + glucose

When performing experiments to measure the kcat of an enzyme, the substrate concentration should be: _

saturating. NOT Km. The answer is D because the kcat is used to describe the rate-limiting step of catalysis under saturating conditions of substrate

Ignoring stereochemistry, how many different tripeptides may exist that contain the same three amino acids as the molecule shown below? molecule shows peptide sequence AVM A1 B3 C6 D9

since it is a tripeptide, there are 3 codons. ignoring stereochemistry you would need to rearrange the peptides in different ways to see how many combos you could get. this is done by using n! or number of positions (factorial). 3! is = 3*2*1= 6 so answer is C. The correct answer is C. The formula for the number of possible peptides that contain one each of n amino acids is n! (n factorial). For n = 3 (a tripeptide), n! = 3! = 3 × 2 × 1 = 6 = Answer C. Alternatively, for a tripeptide ABC, the following combinations are possible: ABC, ACB, BAC, BCA, CAB, and CBA, or six = Answer C.

Which of the following properties is most useful in explaining the trend in the reactivities in Experiment 1? Exp 1: Pea-sized samples of five active metals were placed in deionized water, and observations were recorded in Table 1. A Electronegativity B Ionization potential C Electron affinity D Polarizability

since looking at reactivity of metals think IONS. The correct answer is B. The metals that reacted in Experiment 1 underwent oxidation. The energy required to remove an electron from an atom is the ionization potential. The reactivity of a metal depends on its ionization potential. Among the metals listed in Table 1, potassium has the smallest ionization potential and magnesium has the largest ionization potential. Therefore, potassium reacted the most vigorously and there was no obvious reaction with magnesium. Thus, B is the best answer.