AAMC Chemical and Physical Foundations of Biological Systems

Which reaction leads to the formation of DHB whose structure is shown in the passage

Carboxylation 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 experimental feature of the MALDI-MS technique allows the separation of ions formed after the adduction of tissue molecules?

Velocity of ions depends on the ion mass to charge ratio, 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?

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.

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?

The answer to this question is 0 because the energy in the photochemical reaction ΔH_u is the difference between the laser pulse energy E_m and the heat detected ΔH_nr, so the reading of the energy meter is the energy that is neither ΔH_u nor ΔH_nr. Based on the energy conservation, this is equal to zero for a laser used to dissociate a particular chemical bond.

Based on the ray diagram and distances shown in Figure 1, the focal length of the lens is (12 cm and 4 cm):

The answer to this question is A because the thin-lens formula yields f = (12 cm × 4 cm)/(12 cm + 4 cm) = 3 cm.

What is the kinetic energy of a photoelectron produced in the energy meter of the PAC device when the frequency of an incident photon that is NOT absorbed in the solution is f = 5.0 × 10¹⁵ Hz? (Note: Use h = 4.1 × 10^–15 eV•s.)

the kinetic energy of a photoelectron is equal to hf – 3.4 eV = 20.5 eV – 3.4 eV = 17.1 eV.

Which extraction procedure will completely separate an amide from the by-product of the reaction between an amine and excess carboxylic acid anhydride?

Add 0.1 M NaOH to quench unreacted anhydride, The amide is in nboth layers. 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.

The time dependence of the potassium current through a cell membrane channel subject to a constant 80-mV depolarization voltage is shown. What is the minimum electrical resistance of the ion channel during the time interval shown?

the electrical resistance R of a conductor through which a current I passes when subject to a voltage V is given by Ohm’s law: R = V/I = 80 mV/(400 × 10^–12 A) = 200 MΩ. Minimum is when I is biggest on the graph.

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.

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.

Compound 1 is a member of which class of lipid molecules?

Phosphatides because Compound 1 is a phosphatidyl choline and as such is a phospholipid member of the phosphatides. phosphat + idyl + choline = phosphate + alkyl (fatty) chain + choline molecule. This means that phosphatidylcholine is a phospholipid, which is composed of a choline molecule (an amine) linked to a phosphate and then to a fatty acid chain. Sphingosine is a type of hydrocarbon chain that is associated with many phospholipids but doesn't necessarily form a "backbone" (in the case of phospholipids that role is played by choline+phosphate). And if the backbone is glycerol, the molecule is likely a mono/di/triglyceride based on the number of fatty acids attached to it.

Why did the liposomes fluoresce during size-exclusion chromatography?

Liposomes can be difficult to detect since they do not absorb visible light and many molecules absorb UV light so fluorescence dye must be trapped inside. The experimental design allowed fluorescent dye to be trapped inside during liposome formation, which allowed their detection by fluorescence spectroscopy.

What does the behavior of liposomes prepared from compounds 1 and 2 upon mixing indicate about the energetics of their transformations? Liposomes prepared from:

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

What mass of Compound 1 (MW = 800 g/mol) is contained in the solution used to prepare liposomes that elute at 20 mL by size-exclusion chromatography?

The solution concentration was 0.10 mM for the liposomes that elute at 20 mL. Since the total volume of solution used in their preparation was 1 mL, the mass of lipid can be calculated as: (0.10 × 10^–3 mol/L) × (1 × 10^–3 L) × (800 g/mol) = 8 × 10^–5 g = 80 µg.

Liposomes derived from Compound 2 are prepared at pH 8.5 from two different solution concentrations (0.10 mM and 0.20 mM) as described in the passage. What is the expected appearance of the size-exclusion chromatograph of the liposomal suspension that results after mixing equal volumes of these?

mixing the two suspensions will result in the same liposomal suspension that is created when the lipid concentration is 0.15 mM. This is the average of the two concentrations, since the volumes were identical.

Assume the hydrolysis of ATP proceeds with ΔG′° = –30 kJ/mol.

ATP + H₂O → ADP + P_i

Which expression gives the ratio of ADP to ATP at equilibrium, if the [P_i] = 1.0 M? (Note: Use RT = 2.5 kJ/mol.)

The free energy of the reaction ΔG′° is related to the equilibrium constant K_eq = [ADP][P_i]/[ATP] as ΔG′° = –RTln(K_eq). Applying [P_i] = 1 and using the expression for K_eq gives ΔG′° = –RTln([ADP]/[ATP]), so that [ADP]/[ATP] = e^(30/2.5) = e¹².

What is the electronic configuration of the Co(II) center found in vitamin B12?

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. [Ar] 3d

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?

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^–6mol•L^–1•s^–1 × 60 s × 1.0 × 10^–3 L = 1.2 × 10^–7 mol.

Which of the two products was detected during the experiment?

Compound 2b because it has extensive delocalization of electrons

What is the value of K_M (in µM) for the WT enzyme?

1.31 because K_M is the concentration of substrate at which the kinetics experiment reaches half the maximum velocity and this can be read from the graph shown in Figure 1.

What was the most likely purpose of adding bovine serum albumin to the kinetics experiments in the passage? Bovine serum albumin:

Prevents the esterase from adhering to the walls of the vessel 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.

What mass of NaCl (MM = 58.5 g/mol) was present in 100 µL of the PBS solution?

can be obtained by multiplying the solution concentration, volume, and molar mass using the correct units to cancel: 58.5 g/mol × 0.1 × 10^–3 mol/L × 100 × 10^–6 L = 5.85 × 10^–7 g = 585 × 10^–9 g = 585 ng

Based on the results of the kinetics studies and the observed T_m, what is the best conclusion regarding the role of the metal centers in catalysis and enzyme conformational stability?

Both ions are involved in catalysis, and both provide conformational stability because the data suggest that replacement of any of the residues has a significant effect on conformational stability (as evidenced by the lowering of T_m) as well as the catalytic rate (as evidenced by the lower values for k_cat). This strongly suggests a significant role for both Zn ions in the catalytic sequence.

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.)

sp² → sp³ → sp²

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.

What is the approximate value of ΔS° for binding of NAG₃ to HEW at 27°C?

can be arrived at by using the data supplied in the passage and the relation ΔG = ΔH – TΔS. Solving for ΔS gives –(ΔG – ΔH)/T = –(–30,000 J + 50,000 J)/300 K = –67 J/K.

What quantity of NAG₃ was required to reach the equivalence point in the titration?

the graphs in Figure 1 both imply a 1:1 mole ratio of NAG₃ was added at the equivalence point and the solution contained

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 K_app = K₁. What change to the experimental conditions can result in measurable heat differences and K_app < K₁?

Increase the concentration of the second titration in compound 2. 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.

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

Saturating because the k_cat is used to describe the rate-limiting step of catalysis under saturating conditions of substrate.

At pH 7, which of the following peptides will bind to an anion-exchange column and require the lowest concentration of NaCl for elution?

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. so the one with less negative charge elutes at a lower concentration..

Which experimental technique was most likely used by the students to determine the rate of reaction?

Monitor the INCREASE in absorbance of the solutions at 360 nm 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.

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?

The galactose product only. 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.

For what mechanistic reason does G1 of lactase first act as a Brønsted acid during catalysis?

Glucose becomes a better leaving group. 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.

What method did the students use to calculate V_max?

Vmax is determined by the inverse of the y intercept of the Lineweaver Burk Plot sinSince the plot shown in Figure 1 is derived from the inverse of the Michaelis–Menton equation V_o = (V_max[S])/(K_M + [S]).

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?

[Et] x 500 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.

What is the structure of the cytosine base after catalysis by Dnmt3a?

The answer to the question is B because this structure represents cytosine that has been methylated at the C5 position. NH is the C1 and the carbonyl next to it is C2

Dnmt3a was purified using which type of column chromatography?

Affinity chromatography because the use of histine tagging and a nickel column is a form of affinity chromatography.

A description of the structures of four proteins is shown in the table.

Protein	Description
Protein 1	32 kDa monomer
Protein 2	Disulfide-linked homodimer comprised of 19 kDa monomers
Protein 3	Homotrimer comprised of 25 kDa monomers
Protein 4	Homodimer comprised of 38 kDa monomers

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.)

Protein 3 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.

A protein contains four disulfide bonds. In order to break these bonds researchers added a minimum of:

4 moles of NADH for each protein 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.

A protein with which properties will most likely have the largest negative net charge at pH 7?

A protein that binds to an anion exchange column at pH 7 and requires a high concentration of NaCl for elution. 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).

Based on the results shown in Figure 2, what effect does mutation have on M1 mRN

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.

Which solution components are oxidized and reduced during the enzymatic assay described in the passage?

NADH is oxidized and pyruvate is reduced, 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.

Which of the four DNA bases contains the largest number of hydrogen bond acceptors when involved in a Watson–Crick base pair?

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.

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 T_m of dsDNA in this experiment?

1. pH of solution

2. Ionic strength of solution

3. Length of DNA strands

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.