C/P Section Bank
Hydroxylation
addition of a hydroxyl group
Carboxylation
addition of carboxylic acid groups, usually to serve as calcium-binding sites
glycerol-3-phosphate
three carbon molecule that combines with fatty acids to form triglyceride
SN1 reaction
*unimolecular nucleophilic substitution reactions: 2 steps 1. Leaving group leaves forming a positively charged carbocation (rate limiting step) *The rate of rxn depends only on the concentration of the substrate rate = k [R-L] R-L is the alkyl group containing the leaving group *Anything that accelerates the formation of the carbocation increase the rate of rxn 2. Nucleophile attacks the carbocation (unstable) *results in substitution product - 50%/50% racemic micture 3 OH > 2 OH > 1 OH
SN2 reaction
-bimolecular nucleophilic substitution reactions - only 1 step (concerted reaction) -nucleophile attacks the compound at the same time as the leaving group leaves -Nucleophile actively displaces the leaving group in a backside attack for this to occur, nucleophile must be strong & substrate can't be sterically hindered -concentrations of substrate & nucleophile have role in determining the rate --> rate = k[Nu][R-L] -Position of the substituents around the substrate carbon is inverted 1 OH > 2 OH > 3 OH
thin lens equation
1/f = 1/di + 1/do
A combination of KM and kcat with which relative values is indicative of the lowest catalytic efficiency for an enzymatically catalyzed reaction? A. A high KM and a low kcat B. A high KM and a high kcat C. A low KM and a high kcat D. A low KM and a low kcat
A
What fraction of the starting amount of tritium remains after 13,500 days? A. 1/8 B. 1/3 C. 1/2 D. 3/4
A Passage states that the half life of tritium is 4500 days. 13500 is 3 half lives passed. (1/2^3)
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
A Refer to the beginning of the passage (hydrolysis of a terminal peptide bond) --> peptide bond formed between the carbonyl of one AA and amino of another AA --> R-C=O to NHR 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.
Dnmt3a was purified using which type of column chromatography? A. Affinity B. Anion-exchange C. Cation-exchange D. Size-exclusion
A The answer to the question is A because the use of histine tagging and a nickel column is a form of affinity chromatography.
What is the structure of the cytosine base after catalysis by Dnmt3a?
B Passage states "Dnmt3a, a DNA methyltransferase, modifies the C5 position of cytosine bases within CpG sites"
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
C 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.
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
A 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)
photoelectric effect
The emission of electrons from a material when light of certain frequencies shines on the surface of the material E = hf KE = Etot - W KE = hf - W Eo = hVo KE = 1/2mv^2 KE = hc/λ - Eo
reduction reaction
a chemical change in which electrons are gained, either by the removal of oxygen, the addition of hydrogen, or the addition of electrons
Based on the passage, an experimental feature that distinguishes the PAC technique from the classic calorimetry technique based on thermometers is that PAC: A. can be used on samples with specific heats larger than water's. B. enables the measurement of fast and localized heat transfer processes. C. is based on the second law of thermodynamics. D. is useful on samples in the solid phase only.
B In classic calorimetry, the material is combusted in a contained such that no gases can escape under water; the entire sample is heated. The temperature of the water rises and this change is measured and used to calculate the energy released. According to the passage, PAC uses a microphone to detect the heat produced by very fast heating processes in the specific volume where light is absorbed. All calorimetry is based on the second law of thermodynamics.
The ATP-dependent phosphorylation of a protein target is catalyzed by which class of enzyme? A. Oxidoreductase B. Transferase C. Hydrolase D. Ligase
B This is correct. This Biochemistry question falls under the content category "Principles of chemical thermodynamics and kinetics." The answer is B because kinases catalyze the transfer phosphate groups from ATP to target proteins and are classified as transferases. The item stem describes a kinase. This is a Knowledge of Scientific Concepts and Principles question because you must recall the classification of a kinase.
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+
B This is correct. 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.
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
B This is correct. 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.
Which of the following peptides is most likely to form a covalently bonded dimer? A. AVTSYWP B. GHICEPN C. NFMNELI D. ELIPWQN
B This is correct. This Biochemistry question falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer is B because the presence of cysteine indicates that the peptide could form a disulfide-linked dimer. This is a Scientific Reasoning and Problem Solving question because you must make a inference about covalent bond formation based on the scientific model of disulfide bonds.
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
C Keep in mind that His (H) is a neutral amino acid at pH 7 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. This is a Reasoning about the Design and Execution of Research question because you must understand the design of an anion-exchange experiment and the role of elution in that design to determine the appropriateness of a technique.
Which experimental results confirm that Compound 1 is an uncompetitive inhibitor of a particular enzyme? In the presence of Compound 1: A. the KM increases and Vmax decreases. B. the KM decreases and Vmax increases. C. both the KM and Vmax decrease. D. both the KM and Vmax increase.
C The answer is C because the presence of an uncompetitive inhibitor results in a lower apparent KM and Vmax of the enzymatically catalyzed reaction. This is a Knowledge of Scientific Concepts and Principles question because you must recall the effect of an uncompetitive inhibitor.
To determine a protein's thermodynamic stability, chemical denaturation studies can be performed. Assuming that only the native and unfolded states can be observed under experimentally available conditions, what is the most likely shape of the curve for the dependence of the fraction of folded protein upon denaturant concentration? A. Hyperbolic B. Linear C. Sigmoidal D. Exponential
C The answer is C because the unfolding of proteins is a cooperative process. Cooperative processes are marked by sigmoidal curves.
Based on the results of the kinetics studies and the observed Tm, what is the best conclusion regarding the role of the metal centers in catalysis and enzyme conformational stability? A. ZnA is responsible for catalysis, and ZnBprovides conformational stability. B. ZnB is responsible for catalysis, and ZnAprovides conformational stability. C. Both ions are involved in catalysis, and both provide conformational stability. D. Both ions are involved in catalysis, but neither provides conformational stability.
C This is a Biochemistry question that falls under the content category "Principles of chemical thermodynamics and kinetics." The answer to this question is C because the data suggest that replacement of any of the residues has a significant effect on conformational stability (as evidenced by the lowering of Tm) as well as the catalytic rate (as evidenced by the lower values for kcat). This strongly suggests a significant role for both Zn ions in the catalytic sequence. It is a Data-based and Statistical Reasoning question because you are asked to analyze and interpret data presented in Table 1 to arrive at the conclusion.
Which cation is most likely to be found in place of Fe(II) in the square planar binding domain of hemoglobin? A. Mg2+ B. Li+ C. Co2+ D. Na+
C This is a General Chemistry question that falls under the content category "Atoms, nuclear decay, electronic structure, and atomic chemical behavior." The answer to this question is C because Co2+ is closely related to Fe2+ as a transition metal and can support a square planar coordination environment. It is a Scientific Reasoning and Problem Solving question because you are asked to bring together theory and evidence to draw a conclusion.
According to the data in Table 1, what is one of the values of the electromagnetic energy delivered during one pulse by the ionizing radiation? A. 2.0 µJ B. 3.5 µJ C. 7.5 µJ D. 8.0 µJ
C This is a Physics question that falls under the content category "Translational motion, forces, work, energy and equilibrium in living systems." The answer to this question is C because the energy delivered in one pulse is given by the product between the power and the pulse duration. According to data in Table 1, the energy of the 266 nm radiation is 7.5 µJ and that of the 325 nm is 4.4 µJ. It is a Data-based and Statistical Reasoning question because you must use, analyze and interpret data in a table in order to determine the required energy.
Why does the proposed mechanism involve deprotonation prior to reaction? Deprotonation as described in the passage causes the reacting: A. amine group to be more nucleophilic. B. carbonyl to be more electrophilic. C. water molecule to be more nucleophilic. D. zinc ion to be more electrophilic.
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. Deprotonation of the reacting water will make it more nucleophilic and reactive towards the electrophilic carbonyl atom of the amide functional 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.
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.
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.
proteolytic cleavage
Hydrolysis of a protein by a proteolytic enzyme, eg. trypsin, chymotrypsin Hydrolysis (substitution, elimination, fragmentation rxns)
What is the approximate molecular weight of active Dnmt3a? A. 35 kDa B. 70 kDa C. 105 kDa D. 140 kDa
Passage states that Dnmt3a is active as a homotetramer (4) Reducing SDS-PAGE was used - single band confirmed at 35kDa 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.
According to the passage, what type of bond is cleaved by lysozyme? A. Peptide B. Phosphate ester C. Glycoside D. Pyrophosphate
Passage states: "cleaving N-acetylglucosamine oligosaccharides (NAGx)" C Glycoside bonds link monosaccharides together in an oligosaccharide in a special acetal linkage called a glycoside bond
Power (energy transferred)
Power = energy/time (W), (J/s)
Km
Substrate concentration at 1/2 Vmax
peptide bond
The chemical bond that forms between the carboxyl group of one amino acid and the amino group of another amino acid
Isomerization
The rearrangement of atoms in a molecule to form isomers.
Which of the two products was detected during the experiment? A. Compound 2a; it has three aromatic rings. B. Compound 2a; it is a diphenol. C. Compound 2b; it has extensive delocalization of electrons. D. Compound 2b; it is a mixed phenol and quinone.
This is an Organic Chemistry question that falls under the content category "How light and sound interact with matter." The answer to this question is C because Compound 2b has a much larger set of conjugated double bonds than Compound 2a. In the passage it was stated that Compound 1 was chosen as a substrate since it gave rise to an equilibrium mixture of products, one of which was intensely colored. It is a Reasoning about the Design and Execution of Research question because you are asked to reason about the features of a research study.
Tm temperature
Tm is the temperature at which 50% of the molecules are denatured or the fraction folded is 0.5
pyrophosphate bond
an unstable phosphate bond, such as the one in ATP. When this bond is created (ADP-->ATP), energy is absorbed. When the bond is hydrolyzed (ATP-->ADP), energy is released. However, energy is rarely simply released; most of the time, the hydrolysis of ATP is coupled with some necessary energy-absorbing reaction which therefore gets its activation energy from the degradation of ATP.
speed of light = wavelength x frequency
c = λf
Hydrolases
catalyze cleavage with the addition of water Ex. Phosphatases
Ligases
catalyze the formation of bonds with the input of ATP and the removal of water
phosphate ester bond
holds the phosphate to the glycerol's carbon
size exclusion chromatography
relies on porous beads; larger molecules elute first because they are not trapped in small pores
Transferases
transfer functional groups from one substrate to another Ex. Kinase - uses ATP to transfer one phosphate to another
Pyrophosphate
PPI
Which statement accurately describes the properties of maltose? A. It is a reducing disaccharide in which only one anomeric carbon is involved in the glycosidic bond. B. It is a reducing disaccharide in which both anomeric carbons are involved in the glycosidic bond. C. It is a nonreducing disaccharide in which only one anomeric carbon is involved in the glycosidic bond. D. It is a nonreducing disaccharide in which both anomeric carbons are involved in the glycosidic bond.
A 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.
When experiments are performed on enzymes that display traditional Michaelis-Menten kinetics, what shape does the graph of V0versus substrate concentration [S] have? A. A hyperbolic dependence on [S] B. A linear dependence on [S] C. A sigmoidal dependence on [S] D. A parabolic dependence on [S]
A The answer is A because traditional Michaelis-Menten kinetics describes a hyperbolic dependence of V0 on substrate concentration.
Which extraction procedure will completely separate an amide from the by-product of the reaction between an amine and excess carboxylic acid anhydride? A. Add 0.1 M NaOH(aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the ether layer by evaporating the solvent. B. Add 0.1 M HCl(aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the ether layer by evaporating the solvent. C. Add 0.1 M NaOH(aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the aqueous layer by neutralizing with HCl(aq). D. Add 0.1 M HCl(aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the aqueous layer by neutralizing with NaOH(aq).
A The answer to this question is A because the by-product of the reaction will be an acidic carboxylic acid and the excess unreacted starting material will also be acidic. Extraction with aqueous base will hydrolyze and extract both of these into the aqueous layer, leaving the neutral amide in the ether layer.
What is the best explanation for the observed behavior of the E147D variant? The substitution results in: A. a repulsive interaction that reduces conformational stability and thus enzymatic activity. B. a change in net charge that changes the reactivity of the nearby Zn ion. C. increased steric hindrance of the active site and reduced access by the substrate. D. increased capacity of the site to act as a general base which eliminates the proposed reaction mechanism.
A This is a Biochemistry question that falls under the content category "Principles of chemical thermodynamics and kinetics." The answer to this question is A because the E147D variant is fairly conservative, with both side chains being negatively charged. Aspartate has a slightly shorter chain and thus will be further from the active site. Since the resulting enzyme shows very low conformational stability, the best explanation is that a repulsive interaction develops that reduces conformational stability which eliminates the possibility of catalysis. 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.
What is the electronic configuration of the Co(II) center found in vitamin B12? A. [Ar] 3d7 B. [Ar] 4s2 3d5 C. [Ar] 4s2 3d7 D. [Ar] 4s2 4d5
A This is a General Chemistry question that falls under the content category "Atoms, nuclear decay, electronic structure, and atomic chemical behavior." The answer to this question is A. Co(II) is a dication and is formed from the atomic element by the loss of two 4s electrons. As a consequence, only seven 3d electrons remain in the valence shell. It is a Scientific Reasoning and Problem Solving question because you are asked to bring together theory and evidence to draw a conclusion.
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? A. 1.2 × 10-7 mol B. 3.3 × 10-6 mol C. 3.3 × 10-5 mol D. 1.2 × 10-4 mol
A This is a General Chemistry question that falls under the content category "Principles of chemical thermodynamics and kinetics." 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. It is a Scientific Reasoning and Problem Solving question because you are asked to determine and use a scientific formula to solve a problem.
An experimental setup designed to measure the resistance of an unknown resistor R using two known resistors R1 and R2, the variable resistor R3, a voltage source, and a voltmeter is shown. Which relationship gives the value of R when R3is adjusted so that the voltmeter reading is zero? A. R = R3 × R1/R2 B. R = R3 + R2 - R1 C. R = R3 - R2 + R1 D. R = R3 × R2/R1
A This is a Physics question that falls under the content category "Electrochemistry and electrical circuits and their elements." The answer to this question is A, because when the voltmeter reading is zero, the voltage across R is equal to the voltage across R1 and from Ohm's law, IR = I1R1, where I and I1 are the currents through the resistors. Moreover, IR3 = I1R2. By taking the ratio of these two equations, it follows that R/R3 = R1/R2, which is equivalent to R = R3 × R1/R2. It is a Reasoning about the Design and Execution of Research question because you must reason about the features of this experimental setup and the association between the variables that enable the use of this particular design for its stated purpose.
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.
A This is correct. 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.
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
A pI > pH: + pI < pH: - 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.
Oxidation
A chemical change in which a substance combines with oxygen, as when iron oxidizes, forming rust
series circuit
A circuit in which all parts are connected end to end to provide a single path of current.
parallel circuit
A closed electrical circuit in which the current is divided into two or more paths and then returns via a common path to complete the circuit.
Triacylglycerols (triglycerides)
A glycerol molecule esterified to three fatty acid molecules; the most common form of fat storage in the body Overall nonpolar and hydrophobic Can be seen in oily droplets
phosphonic acid
H3PO4
What does the behavior of liposomes prepared from compounds 1 and 2 upon mixing indicate about the energetics of their transformations? Liposomes prepared from: A. both Compound 1 and Compound 2 are under kinetic control. B. Compound 1 are under kinetic control, but those prepared from Compound 2 are under thermodynamic control. C. Compound 1 are under thermodynamic control, but those prepared from Compound 2 are under kinetic control. D. both Compound 1 and Compound 2 are under thermodynamic control.
B Kinetic - whatever is formed as fast as possible Thermodynamic - whatever is most stable This is a General Chemistry question that falls under the content category "Principles of chemical thermodynamics and kinetics." The answer to this question is B. The mixing experiments demonstrate that liposomes formed from Compound 1 cannot attain their thermodynamically preferred state and are therefore under kinetic control. Mixing liposomes of different sizes of Compound 2, on the other hand, results in the formation of new liposomes which are of an intermediate size indicating that they rapidly attain their thermodynamically preferred state. It is a Scientific Reasoning and Problem Solving problem because you are asked to bring together theory, evidence, and observations to draw a conclusion.
What was the most likely purpose of adding bovine serum albumin to the kinetics experiments in the passage? Bovine serum albumin: A. acts as a co-catalyst for the reaction. B. prevents the esterase from adhering to the walls of the vessel. C. helps buffer the solution from changes in pH. D. is a non-specific target for protease contaminants.
B Passage states that BSA "mobilizes proteins and lipids" 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 B since the passage specifically states that albumin is a protein that mobilizes proteins and lipids in serum. In the context of a kinetics experiment it is perfectly logical to assume that albumin is added to maintain homogeneity and prevent the enzyme from adhering to walls and other surfaces which would inhibit its activity. It is a Reasoning about the Design and Execution of Research question because you are asked to reason about the features of a research study.
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.
B 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
What is the ratio between the maximum and the minimum sound intensities that produce this particular loudness? A. 10^6 B. 10^5 C. 10^4 D. 10^3
B The answer to this question is B because the maximum sound intensity level is 80 dB and the minimum level is 30 dB, which correspond to intensities Imax = Io × 10^8 and Imin = Io × 10^3, based on the definition of the decibel units. Their ratio is then Imax/Imin = 10^5
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.
B 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 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.
B 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.
Why did the liposomes fluoresce during size-exclusion chromatography? A. The macromolecule had extensive conjugation. B. Fluorescent dye was trapped inside. C. Intermolecular interactions lower the energy of the excited state. D. Light reflects from the surface of the sphere.
B 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 B. Liposomes can be difficult to detect since they do not absorb visible light and many molecules absorb UV light. The experimental design allowed fluorescent dye to be trapped inside during liposome formation, which allowed their detection by fluorescence spectroscopy. It is a Reasoning about the Design and Execution of Research question because you are asked to reason about the features of a research study that suggests relationships between the variables.
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
B 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. It is a Data-based and Statistical Reasoning question because you are asked to analyze and interpret data presented in Figure 1 to arrive at the conclusion.
According to the passage and the data in Table I, what feature must distinguish laser type A from laser type C in order to be suitable for the cleavage of the bonds specified? Laser type A must: A. be better focused laser type C. B. have a higher frequency than laser type C. C. have a longer wavelength than laser type C. D. emit fewer photons per unit time than laser type C.
B This is a Physics question that falls under the content category "How light and sound interact with matter." The answer to this question is B because the photon energy must be larger than the bond energy in order to break the bond. Because the photon energy is directly proportional to the radiation frequency, laser type A must have a higher frequency than laser type C. It is a Data-based and Statistical Reasoning question because you must reason use data to explain relationships between variables or to make predictions.
The largest contribution to the stabilization of glycerol binding in GK is most likely due to which type of interaction? A. Hydrophobic interactions B. Aromatic interactions C. Hydrogen bonding interactions D. Disulfide bonding interactions
C This is correct. This Biochemistry question falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer is C because glycerol contains three hydroxyl groups that are capable of acting as hydrogen bond donors or acceptors. The residues that stabilize glycerol binding also contain functional groups that are capable of forming hydrogen bonds. Therefore, it is most likely that hydrogen bonding interactions are primarily responsible for glycerol stabilization. Glycerol cannot form disulfide bonds (lack of sulfhydryl group) and does not have an aromatic structure. While hydrophobic interactions may play a small role, the polarity of the substrate and amino acids within the cleft indicate that this is unlikely to be a dominant interaction. This is a Scientific Reasoning and Problem Solving question because you must use the scientific model of intermolecular interactions to predict how a protein binding site would stabilize glycerol.
The coefficient of viscosity of blood is directly proportional to the frequency of interactions of erythrocytes with the vessel walls. As blood enters small vessels, many erythrocytes tend to flow in the center of the vessel. Given this information, which graphic best describes the effect of vessel diameter on the coefficient of viscosity?
C coefficient of viscosity = frequency of RBC interactions with VESSEL WALLS small vessels - a lot in the center (less interaction with vessel walls) - low coefficient of viscosity
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
D 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.
What is the most likely reason for the decreased KM values 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.
D 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.
What is the reading of the energy meter in Figure 1 when an appropriate laser is used in PAC to dissociate a particular chemical bond? A. Em B. ΔHu C. ΔHnr D. 0
D The answer to this question is D because the energy in the photochemical reaction ΔHu is the difference between the laser pulse energy Em and the heat detected ΔHnr, so the reading of the energy meter is the energy that is neither ΔHu nor ΔHnr. Based on the energy conservation, this is equal to zero for a laser used to dissociate a particular chemical bond. https://www.reddit.com/r/Mcat/comments/4cmu1g/cp_section_bank_8/
Why did the researchers choose "A" as the replacement residue for each of the single site variants used in the study? Use of "A": A. reduces the net charge on the bimetallic center. B. increases strength of substrate binding to the active site. C. increases the conformational rigidity of the active site and the enzyme. D. reduces the interaction of the side chain with other active site components.
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. The researchers wanted to reduce or eliminate the interactions of each side chain with the active site and the best way to accomplish this while having a minimal impact on the overall structure of the enzyme is to use alanine. Alanine possesses a small hydrophobic side chain (methyl group) but participates in α helices and β sheets extensively. It is a Reasoning about the Design and Execution of Research question because you are asked to reason about the features of a research study that suggests relationships between the variables.
Assume the hydrolysis of ATP proceeds with ΔG′° = -30 kJ/mol. ATP + H2O → ADP + Pi Which expression gives the ratio of ADP to ATP at equilibrium, if the [Pi] = 1.0 M? (Note: Use RT = 2.5 kJ/mol.) A. e2 B. e3 C. e6 D. e12
D This is a General Chemistry question that falls under the content category "Principles of chemical thermodynamics and kinetics." The answer to this question is D. The free energy of the reaction ΔG′° is related to the equilibrium constant Keq = [ADP][Pi]/[ATP] as ΔG′° = -RTln(Keq). Applying [Pi] = 1 and using the expression for Keq gives ΔG′° = -RTln([ADP]/[ATP]), so that [ADP]/[ATP] = e(30/2.5) = e12. It is a Scientific Reasoning and Problem Solving question because you are asked to determine and use a scientific formula to solve a problem.
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
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. It is a Scientific Reasoning and Problem Solving question because you are asked to bring together theory, evidence, and observations to draw a conclusion.
An ITC experiment is conducted by injecting Compound 1 into a solution of Protein A giving large measured heats and a high affinity constant K1. A CITC experiment under identical conditions but in the presence of 10 mM Compound 2 gives the same heat curve and Kapp = K1. What change to the experimental conditions can result in measurable heat differences and Kapp < K1? A. Dilute all of the solutions. B. Increase the concentration of Compound 1 in both titrations. C. Increase the concentration of Protein A in both titrations. D. Increase the concentration of Compound 2 in the second titration.
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 since the experiment depends on Compound 2 successfully competing for the binding sites available on Protein A and this is the only response that will improve the chances of this happening. It is a Reasoning about the Design and Execution of Research question because you are asked to reason about the features of a research study that suggests relationships between the variables.
The idea that enzymes undergo conformational changes upon substrate binding in order to bring specific functional groups into the correct catalytic orientation is described as the: A. transitionstate model. B. active-site model. C. lock-and-key model. D. induced-fit model.
D This is correct. This Biochemistry question falls under the content category "Principles of chemical thermodynamics and kinetics." The answer is D because the induced-fit model of enzyme catalysis states that the active site of the enzyme changes shape upon substrate binging and only the proper substrate causes a change that initiates catalysis. This is a Knowledge of Scientific Concepts and Principles question because you must recall the concept of substrate binding models.
When performing experiments to measure the kcat of an enzyme, the substrate concentration should be: A. limiting. B. equal to 1/2 KM. C. equal to KM. D. saturating.
D This is correct. This Biochemistry question falls under the content category "Principles of chemical thermodynamics and kinetics." The answer is D because the kcat is used to describe the rate-limiting step of catalysis under saturating conditions of substrate. This is a Reasoning about the Design and Execution of Research question because you must understand the design of an experiment used to measure the maximum velocity of an enzymatic reaction.
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.
D This is correct. 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 Glu487protonates 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. This is a Data-based and Statistical Reasoning question because you must interpret patterns in data to make a prediction about the change in enzyme activity.
How many bands would be observable if GK was subjected to native PAGE and SDS-PAGE analysis? (Note: Assume GK maintains a single state in solution.) A. Two bands in both the native PAGE and the SDS-PAGE B. One band in the native PAGE and 2 bands in the SDS-PAGE C. Two bands in the native PAGE and 1 band in the SDS-PAGE D. One band in both the native PAGE and the SDS-PAGE
D This is correct. This Biochemistry question falls under the content category "Separation and purification methods." The answer is D because GK is a homodimer, which means that both preservation and disruption of the quaternary structure results in a single gel band. This is because the subunits of the dimer are the same molecular weight. Therefore, both electrophoresis experiments should result in a single gel band. This is a Reasoning about the Design and Execution of Research question because you must understand the differences in the experimental design of two types of electrophoresis and how those differences affect the protein.
Magnitude of electric field (uniform)
E = V/d
Energy of a photon
E = hf = hc/λ Eo = hVo c = λv
second law of thermodynamics
Every energy transfer or transformation increases the entropy of the universe. The total entropy of an isolated system will never decrease over time. When energy is changed from one form to another, some useful energy is always degraded into lower-quality energy (usually heat).
