AAMC FL2 C/P
Boyle's Law
A principle that describes the relationship between the pressure and volume of a gas at constant temperature P1V1=P2V2
Doppler effect
An observed change in the frequency of a wave when the source or observer is moving
The lone pair of electrons in ammonia allows the molecule to: A. assume a planar structure. B. act as an oxidizing agent. C. act as a Lewis acid in water. D. act as a Lewis base in water.
The answer to this question is D because, by definition, a Lewis base is a substance that donates an electron pair in forming a covalent interaction.
Venturi effect
Reduction in pressure of a fluid resulting from the speed increase as fluids are forced to flow faster through narrow spaces.
Which statement about the cooperativity of RIα/C activation and RIα protein folding is supported by the data in figures 2 and 3? A. Both activation and folding are cooperative. B. Activation is cooperative, but folding is not. C. Folding is cooperative, but activation is not. D. Neither activation nor folding is cooperative.
The answer to this question is A because both curves have a sigmoidal shape, which is indicative of cooperative processes.
What is the role of the solid-state catalyst in the Haber process? A. It increases the amount of ammonia produced per unit time. B. It increases the total amount of ammonia produced. C. It decreases the amount of ammonia that decomposes per unit time. D. It decreases the total amount of ammonia produced.
The answer to this question is A because catalysts increase the rate of chemical reaction, which is the amount of product formed per unit time.
Based on the information in the passage, PDK1 catalyzes the addition of phosphate to what functional group? A. Hydroxyl B. Amine C. Carboxyl D. Phenyl
The answer to this question is A because reactions involving either Ser or Thr would involve the hydroxyl group in the side chain of these amino acids.
Protein secondary structure is characterized by the pattern of hydrogen bonds between: A. backbone amide protons and carbonyl oxygens. B. backbone amide protons and side chain carbonyl oxygens. C. side chain hydroxyl groups and backbone carbonyl oxygens. D. side chain amide protons and backbone carbonyl oxygens.
The answer to this question is A because secondary structure is represented by repeated patterns of hydrogen bonds between the backbone amide protons and carbonyl oxygen atoms.
It is possible to design a reactor where the SCY conductor and the nitrogen/ammonia electrode operate at different temperatures. Which combination of temperatures is expected to give the best results? A. SCY temperature higher than electrode temperature B. SCY temperature lower than electrode temperature C. SCY temperature the same as electrode temperature D. The temperature of the components does not make a difference.
The answer to this question is A because the proton conductivity of SCY increases with increasing temperature, while the favorability of reaction decreases with overall temperature. It is therefore beneficial to maintain the SCY conductor at a higher temperature.
Based on the data presented in figures 2 and 3, what is the most likely role of Y229 in protein stability and cAMP activation? A. Y229 is important for protein stability but not critical for cAMP activation. B. Y229 is important for cAMP activation but not critical for protein stability. C. Y229 is important for protein stability and critical for cAMP activation. D. Y229 is not important for protein stability and not critical for cAMP activation.
The answer to this question is A because the thermal melt shows that removal of Y229 decreases stability, therefore Y229 is important for stability. Removal of Y229 has little effect on protein activation, as the activation curve is similar to WT activation.
How much work did an 83-year-old female do while stretching the rubber band to the limit of her strength? A. 4 J B. 5 J C. 6 J D. 7 J
The answer to this question is A because the work done is W = 0.5 × kx2 where x = 0.20 m. So W = 0.5 × 200 (N/m) ×(0.2 m)2 = 4.0 J.
Which single bond present in nitroglycerin is most likely the shortest? A. C-H B. C-O C. C-C D. O-N
The answer to this question is A. All of the bonds listed are single bonds. Since hydrogen has a much smaller atomic radius than second period elements, the covalent bond between C and H is shorter than any of the other bonds listed. Hydrogen has the smallest atomic radius so the covalent bond it forms (it donates one electron and carbon donates on electron to bond) will be the shortest. Also: the bond enthalpies for all of those single bonds are in the passage. The strongest bond will have the highest bond enthalpy and the shortest bond distance.
The concentration of enzyme for each experiment was 5.0 μM. What is kcat for the reaction at pH 4.5 with NO chloride added when Compound 3 is the substrate? A. 2.5 × 10-2 s-1 B. 1.3 × 102 s-1 C. 5.3 × 103 s-1 D. 7.0 × 105 s-1
The answer to this question is A. The fact that the rate of product formation did not vary over time for the first 5 minutes implies that the enzyme was saturated with substrate. Under these conditions, kcat = Vmax/[E] = (125 nM/s)/5.0 μM = 2.5 × 10-2 s-1.
Four organic compounds: 2-butanone, n-pentane, propanoic acid, and n-butanol, present as a mixture, are separated by column chromatography using silica gel with benzene as the eluent. What is the expected order of elution of these four organic compounds from first to last? A. n-Pentane → 2-butanone → n-butanol → propanoic acid B. n-Pentane → n-butanol → 2-butanone → propanoic acid C. Propanoic acid → n-butanol → 2-butanone → n-pentane D. Propanoic acid → 2-butanone → n-butanol → n-pentane
The answer to this question is A. The four compounds have comparable molecular weights, so the order of elution will depend on the polarity of the molecule. Since silica gel serves as the stationary phase for the experiment, increasing the polarity of the eluting molecule will increase its affinity for the stationary phase and increase the elution time (decreased Rf).
Generation of Compound 3 is the result of Compound 2 undergoing: A. oxidation. B. reduction. C. hydrolysis. D. carboxylation.
The answer to this question is B because a carbon in Compound 2 has a decreased oxidation number in Compound 3, therefore it is reduced. There are many clues to answering this question: Dehydrogenases are oxidoreductases and are involved in oxidation/reduction reactions, so the substrate is either being oxidized or reduced. NADH is being oxidized (it loses H2), so the substrate must be reduced (gaining H2).Remember that gaining hydrogen or losing oxygen is reduction. Gaining oxygen or losing hydrogen is oxidation.
Based on the relative energy of the absorbed electromagnetic radiation, which absorber, a peptide bond or an aromatic side chain, exhibits an electronic excited state that is closer in energy to the ground state? A. An aromatic side chain; the absorbed photon energy is higher. B. An aromatic side chain; the absorbed photon energy is lower. C. A peptide bond; the absorbed photon energy is higher. D. A peptide bond; the absorbed photon energy is lower.
The answer to this question is B because aromatic side chains absorb in the near UV region of the electromagnetic spectrum, which has longer wavelengths, and hence lower energy, than peptide bonds. Because the energy of the photon matches the energy gap between the ground and the excited state, this implies that the aromatic side chain has more closely spaced energy levels. From the passage we see that the aromatic side chain absorbs electromagnetic radiation in the 250-290nm range whereas the peptide bond absorbs in the 190-250nm range. Remember, wavelength is INVERSELY proportional to frequency, so the longer wavelength light absorbed by aromatic side chains will have a lower frequency. Given our equation and without any calculations we can state that aromatic side chains will absorb less energy and have an electron excitation state closer to the ground
In the overall electrochemical reaction: N2(g) + H2(g) NH3(g) A. nitrogen is oxidized at the anode, and hydrogen is reduced at the cathode. B. nitrogen is reduced at the cathode, and hydrogen is oxidized at the anode. C. nitrogen is reduced at the anode, and hydrogen is oxidized at the cathode. D. nitrogen is oxidized at the cathode, and hydrogen is reduced at the anode.
The answer to this question is B because oxidation always occurs at the anode and reduction at the cathode of an electrochemical cell. Since nitrogen decreases in oxidation state during the reaction, it is reduced. Hydrogen, on the other hand, increases in oxidation state and is, therefore, oxidized. RED CAT AN OX Looking at Reaction 1:N2 starts at 0 ox state and goes to -3 so it's reduced. H2 starts at 0 ox state and goes to +1 so it's oxidized
A patient puts on a mask with lateral openings and inhales oxygen from a tank as shown. What phenomenon causes static air to be drawn into the mask when oxygen flows? A. Doppler effect B. Venturi effect C. Diffusion D. Dispersion
The answer to this question is B because oxygen pressure is the sum of the oxygen static pressure P and the oxygen flow pressure ρv2/2. In the area of the mask openings, Pair = P + ρv2/2, thus Pair > P. Air enters the mask because the static pressure of the air is larger than the static pressure of the oxygen in flow. This is the Venturi effect, and the mask is called the Venturi mask.
At 25°C, the formation of [Cu(NH3)4]2+ according to Equation 1 is most likely a: The value of the formation constant of [Cu(NH3)4]2+ is 5.6 1011 at 25°C. A. spontaneous process with positive ΔG°. B. spontaneous process with negative ΔG°. C. nonspontaneous process with positive ΔG°. D. nonspontaneous process with negative ΔG°.
The answer to this question is B because the equilibrium constant for the reaction is very large (much greater than 1). This necessarily means that ΔG° is negative and the reaction is spontaneous.
An inflatable cuff was used to temporarily stop blood flow in an upper arm artery. While releasing the pressure to deflate the cuff, a stethoscope was used to listen to blood flow in the forearm. The blood pressure reading was 130/85. Given this information, which of the following statements is LEAST likely to be true? A. 85 mmHg was the diastolic pressure. B. Blood flow was heard when the pressure of the cuff was greater than 130 mmHg. C. 130 mmHg was the systolic pressure. D. Blood flow was heard when the pressure of the cuff was 90 mmHg.
The answer to this question is B because the question indicates that the cuff was inflated to temporarily stop blood flow in the artery. The systolic pressure is determined from the first sound of blood flow that can be heard once the pressure exerted by the inflatable cuff falls below the pressure in the artery. The blood pressure reading was 130/85, which indicates that blood flow started again when the pressure was 130 mmHg. Therefore, blood flow was not heard when the pressure of the cuff was greater than 130 mmHg. A is not the correct response because the lower number of the blood pressure reading represents the diastolic pressure. C is not the correct response because the higher number of the blood pressure reading represents the systolic pressure. D is not the correct response because blood flow would be heard when the pressure of the cuff is 90 mmHg, as this pressure is higher than the diastolic pressure.
What are the structural features possessed by storage lipids? A. Two fatty acids ester-linked to a single glycerol plus a charged head group B. Three fatty acids ester-linked to a single glycerol C. Two fatty acids ester-linked to a single sphingosine plus a charged head group D. Three fatty acids ester-linked to a single sphingosine
The answer to this question is B because triacylglycerols are neutral storage lipids. They consist of three fatty acids ester-linked to a single glycerol.
In [Cu(NH3)4]2+, the subscript 4 indicates which of the following? A. The oxidation number of Cu only B. The coordination number of Cu2+ only C. Both the oxidation number of Cu and the coordination number of Cu2+ D. Neither the oxidation number of Cu nor the coordination number of Cu2+
The answer to this question is B. Because ammonia is neutral, the number 4 reflects only the number of ammonia molecules that bind to the central Cu2+cation and does not indicate anything about its oxidation number.
Consider the reaction shown in Equation 1 at equilibrium. Would the concentration of [Cu(NH3)4]2+ increase if the equilibrium were disturbed by adding hydrochloric acid? [Cu(H2O)4]2+(aq) + 4NH3(aq) [Cu(NH3)4]2+(aq) + 4H2O(l) A. Yes, because the equilibrium in Equation 1 would shift to the left B. No, because the equilibrium in Equation 1 would shift to the left C. Yes, because the equilibrium in Equation 1 would shift to the right D. No, because the equilibrium in Equation 1 would shift to the right
The answer to this question is B. Hydrochloric acid will protonate ammonia in a Brönsted acid-base reaction and reduce the amount of ammonia present. The disturbed equilibrium responds in a way to restore ammonia, but this causes the amount of [Cu(H2O)2(NH3)2]2+ to decrease. This means that the equilibrium shifts to the left.
Absorption of ultraviolet light by organic molecules always results in what process? A. Bond breaking B. Excitation of bound electrons C. Vibration of atoms in polar bonds D. Ejection of bound electrons
The answer to this question is B. The absorption of ultraviolet light by organic molecules always results in electronic excitation. Bond breaking can subsequently result, as can ionization or bond vibration, but none of these processes are guaranteed to result from the absorption of ultraviolet light. C. Different effects dominate at different frequencies. With IR, there isn't enough energy in the photon to excite electrons, so the energy goes into vibrations. With UV, there is always enough energy in the photon to cause an excitation of the electron, and that's where the energy goes, rather than vibration. D. This is an example of the photoelectric effect. You can absorb energy without causing emission.
What is the pH of a buffer solution that is 0.2 M in HCO3- and 2 M in H2CO3? (Note: The first pKa of carbonic acid is 6.37.) A. 4.37 B. 5.37 C. 6.37 D. 7.37
The answer to this question is B. The pH of the solution can be calculated using the Henderson-Hasselbach equation: pH = pKa + log([base]/[acid]). Plugging in the values provided in the question gives pH = 6.37 + log(0.2/2) = 5.37.
Why does NH3 displace H2O in the formation of [Cu(NH3)4]2+? NH3 contains more lone pairs of electrons than H2O. NH3 is a stronger Lewis base than H2O. NH3 donates a lone pair of electrons more readily than does H2O. A. I and II only B. I and III only C. II and III only D. I, II, and III
The answer to this question is C because Reaction 1 is a Lewis acid-base reaction. The fact that the reaction proceeds in the forward direction indicates that NH3 is a better Lewis base toward Cu2+ than H2O. This also means that NH3 donates a lone pair of electrons more readily than does H2O.
Which amino acid will contribute to the CD signal in the far UV region, but NOT the near UV region, when part of a fully folded protein? A. Trp B. Phe C. Ala D. Tyr
The answer to this question is C because all chiral nonaromatic amino acids will contribute solely to the CD signal in the far UV region. Aromatic amino acids absorb in the near region when a part of tertiary structures, according to the first paragraph. In other words, you are looking for the not aromatic amino acid because it will not affect the near UV region.
Which statement correctly describes the structure of the DNA double helix? A. Nitrogenous bases pair with other bases in the same purine or pyrimidine groups. B. The two DNA strands of the double helix are oriented in the same direction. C. The amount of guanine will equal the amount of cytosine in a DNA sequence. D. Sugar-phosphate backbones form the interior of the double helix.
The answer to this question is C because as guanine and cytosine form base pairs on opposite DNA strands, they will occur in equal amounts within a specific DNA sequence. A is incorrect because purines and pyrimidines do not form base pairs with other members within the same group. B is incorrect because the strands are antiparallel, meaning that they are oriented in opposite directions. D is incorrect because sugar-phosphate backbones form the exterior of the double helix.
What expression gives the amount of light energy (in J per photon) that is converted to other forms between the fluorescence excitation and emission events? A. (6.62 × 10-34) × (3.0 × 108) B. (6.62 × 10-34) × (3.0 × 108) × (360 × 10-9) C. (6.62 × 10-34) × (3.0 × 108) × [1 / (360 × 10-9) - 1 / (440 × 10-9)] D. (6.62 × 10-34) × (3.0 × 108) / (440 × 10-9)
The answer to this question is C because the equation of interest is E = hf = hc/λ, where h = 6.62 × 10 −34 J ∙ s and c = 3 × 10 8 m/s. Excitation occurs at λe = 360 nm, but fluorescence is observed at λf = 440 nm. This implies that an energy of E = (6.62 × 10 −34) × (3 × 10 8) × [1 / (360 × 10 −9) − 1 / (440 × 10 −9)] J per photon is converted to other forms between the excitation and fluorescence events.
A 60-Ω resistor is connected in parallel with a 20-Ω resistor. What is the equivalent resistance of the combination? A. 80 Ω B. 40 Ω C. 15 Ω D. 3 Ω
The answer to this question is C because the equivalent resistance is given by the expression (1/(60) + 1/(20))-1 = 15 . (1/60) + (1/15) = (1/R) for parallel resistors.
In designing the experiment, the researchers used which type of 32P labeled ATP? A. α32P-ATP B. β32P-ATP C. γ32P-ATP D. δ32P-ATP
The answer to this question is C because the phosphoryl transfer from kinases comes from the γ-phosphate of ATP. Therefore, the experiment should require γ32P-ATP. ATP has 3 phosphate groups, the closest phosphate to the sugar is alpha, then beta and lastly gamma.
What causes duplex DNA with a certain (A + T):(G + C) ratio to melt at a higher temperature than comparable length duplex DNA with a greater (A + T):(G + C) ratio? A. Stronger van der Waals forces of pyrimidines B. Stronger van der Waals forces of purines C. Increased π- stacking strength D. Reduced electrostatic repulsion of phosphates
The answer to this question is C. G-C base pairs form stronger π-stacking interactions than A-T base pairs, thereby creating the most thermal stability. This disparity has often been used to explain the increased melting temperature of DNA rich in GC content.
What is the concentration of Ca2+(aq) in a saturated solution of CaCO3? (Note: The solubility product constant Ksp for CaCO3 is 4.9 × 10-9.) A. 2.4 × 10-4 M B. 4.9 × 10-5 M C. 7.0 × 10-5 M D. 4.9 × 10-9 M
The answer to this question is C. The solubility product constant expression for CaCO3 is Ksp = [Ca2+][CO32-]. Since equal quantities of Ca2+(aq) and CO32-(aq) are produced when CaCO3 dissolves, this expression reduces to 4.9 × 10-9 = x2, or 49 × 10-10 = x2. This can be solved directly by taking the square root of each side.
A person is sitting on a chair as shown. Why must the person either lean forward or slide their feet under the chair in order to stand up? A. To increase the force required to stand up B. To use the friction with the ground C. To reduce the energy required to stand up D. To keep the body in equilibrium while rising
The answer to this question is D because as the person is attempting to stand, the only support comes from the feet on the ground. The person is in equilibrium only when the center of mass is directly above their feet. Otherwise, if the person did not lean forward or slide the feet under the chair, the person would fall backward due to the large torque created by the combination of the weight of the body (applied at the person's center of mass) and the distance along the horizontal between the center of mass and the support point.
In industrial use, ammonia is continuously removed from the reaction mixture. This serves to drive Reaction 1 because of: A. Boyle's law. B. Charles's law. C. Heisenberg's principle. D. Le Châtelier's principle.
The answer to this question is D because removing a product as it forms causes a displacement from the equilibrium condition. The system will respond by shifting more reactants to the product side. This is an example of Le Châtelier's principle.
Which of the following best describes the bonds between Cu2+ and the nitrogen atoms of the ammonia molecules in [Cu(NH3)4]2+? A. Ionic B. Covalent C. Coordinate ionic D. Coordinate covalent
The answer to this question is D because the Lewis acid-base interaction between a metal cation and an electron pair donor is known as a coordinate covalent bond. Coordinate covalent is what transition metals form. "Coordinate" is when only one atom contributes both electrons in the bond. This would be the nitrogen atom in this case, with its lone pairs. Creates a transition metal/complex ion
Which of the following energy conversions best describes what takes place in a battery-powered resistive circuit when the current is flowing? A. Electric to thermal to chemical B. Chemical to thermal to electric C. Electric to chemical to thermal D. Chemical to electric to thermal
The answer to this question is D because the chemical energy of the battery elements is used as electrical energy to set the charge carriers in motion through the resistor, where they experience drag from the crystal lattice of the resistive conductor and dissipate their energy as heat from the resistor.
The half-life of a radioactive material is: A. half the time it takes for all of the radioactive nuclei to decay into radioactive nuclei. B. half the time it takes for all of the radioactive nuclei to decay into their daughter nuclei. C. the time it takes for half of all the radioactive nuclei to decay into radioactive nuclei. D. the time it takes for half of all the radioactive nuclei to decay into their daughter nuclei.
The answer to this question is D because the half-life of a radioactive material is defined as the time it takes for half of all the radioactive nuclei to decay into their daughter nuclei, which may or may not also be radioactive.
What kind of image is formed by the lenses of the glasses worn by a 68-year-old male who sees an object 2 m away? A. Real and enlarged B. Real and reduced C. Virtual and enlarged D. Virtual and reduced
The answer to this question is D because the lenses have a negative focal length which means they are diverging lenses. Such lenses form virtual and reduced images of objects situated at distances larger than the focal length.
A glass rod is rubbed with a silk scarf producing a charge of +3.2 × 10-9 C on the rod. (Recall that the magnitude of the proton and electron charges is 1.6 × 10-19 C.) The glass rod has: A. 5.1 × 1011 protons added to it. B. 5.1 × 1011 electrons removed from it. C. 2.0 × 1010 protons added to it. D. 2.0 × 1010 electrons removed from it.
The answer to this question is D because the number of charges in excess can be computed as +3.2 × 10 -9 C/1.6 × 10 -19 C = +2.0 × 10 10. This means that the rod has an excess of positive charge, created by removing a number of +2.0 × 10 10 electrons from the material, as it is not possible to add protons in a manner described in this question. The formula for charge is q = neq is the amount of charge (in Coulombs)n is the number of electronse is the elementary charge (1.6 x 10-19 C)you are looking for the total number of electrons that were ripped away from it (since you can't move protons). therefore you would divide the total charge 3.2 x 10-9 by 1.6 x 10-19. To enable screen reader support, press Ctrl+Alt+Z To learn about keyboard shortcuts, press Ctrl+slash
When used in place of spHM FLGFTY, which peptide would be most likely to achieve the same experimental results? A. FLGFAY B. FLGFQY C. FLGFGY D. FLGFEY
The answer to this question is D because the phosphorylated threonine would most likely be mimicked by glutamic acid in terms of size and charge.
What is the average power consumed by a 64-year-old woman during the ascent of the 15-cm-high steps, if her mass is 54 kg? A. 10 W B. 20 W C. 40 W D. 90 W
The answer to this question is D because the power consumed is P=PE /time = mgh/t. From Table 1, there are 30 steps and t = 27 s. Then P = (54 kg × 10 m/s2 × 30 steps × 0.15 m/step) / (27s) = 90 W.
What is the ratio of the minimum sound intensities heard by a 64-year-old male and a 74-year-old female? A. 20 B. 40 C. 50 D. 100
The answer to this question is D because the relative intensities of the two sound waves are 20 dB and 40 dB, respectively. The difference is 20 dB, meaning that the decimal log of the ratio of their intensities is 2, which means that the ratio of their intensities is 100. So for a 64 y/o male, they had a intensity of 40 dB which means that dB=10Log(i-observed /i-source) equals 40 dB. For a 74 y/o female, they had an intensity of 20 dB which means that dB=Log(i-observed/ i-source). In order to determine the ratio, you need to do 40-20 which is 20dB difference between the two individuals and then solve for (i-observed/i-source). so 20/10 --> 2 ---> 10^2 gives you 100.
The side chain of tryptophan will give rise to the largest CD signal in the near UV region when: A. present as a free amino acid. B. part of an α-helix. C. part of a β-sheet. D. part of a fully folded protein.
The answer to this question is D because tryptophan has an aromatic side chain that will give rise to a significant CD signal in the near UV region if it is found in a fully folded protein. The first sentence of the second paragraph states that the largest CD signal in the near UV region results from tertiary structures, meaning that if the amino acid is in a folded protein it will lead to the largest CD signal.
Which of the following atoms will be expected to have the smallest second ionization energy? A. Na B. C C. O D. Ca
The answer to this question is D. Metals have lower ionization energies than non-metals as long as the ionization event involves a valence electron. Since Na is an alkali metal, it has only one valence electron and has a large second ionization energy. Ca is an alkaline earth metal and has two valence electrons. It will therefore have the smallest second ionization energy of the four atoms listed, which include Na and two non-metals. Ionization Energy=energy required to remove electron. Think about it: With Ca, removing one electron gets us to within one e of the noble gas configuration. Therefore, removing this second electron (2nd ionization energy) shouldn't take much energy because getting to noble gas configuration is heavily favored!
Compared to the concentration of the proteasome, the concentration of the substrate is larger by what factor? A. 5 × 101 B. 5 × 102 C. 5 × 103 D. 5 × 104
The answer to this question is D. The proteasome was present at a concentration of 2 × 10-9 M, while the substrate was present at 100 × 10-6 M. The ratio of these two numbers is 5 × 104. Put the given concentrations into a ratio. The question asks by how much the substrate concentration is bigger, so put the substrate concentration on top. This is [S]/[E] = (100 microM)/(2 nanoM).Simplify the ratio by dividing. Since math and keeping track of units isn't my forte, I personally like to rewrite all units in SI units with scientific notation, like this: (100 x 10-6 M)/(2 x 10-9 M). Then I divide the numbers (100/2 = 50), and then I divide the powers of ten (10-6 / 10-9 = 103). So the answer is 50 x 103 M, which equals one of the answer choices, 5 x 104 M.
Based on the passage, the magnitude of ΔH° (in kJ) for the decomposition of 2 moles of nitroglycerin at 25°C is closest to which of the following? A. 500 B. 1000 C. 2000 D. 3000
The answer to this question is D. The value of ΔH° can be calculated using the data provided in Table 1 and applying Hess's Law. Two moles of nitroglycerin produce 6 moles of CO2(g) and 5 moles of H2O(g). The value of ΔH° for this amount of nitroglycerin combusted is 2(364.0) - 6(393.5) - 5(241.8) = -2842 kJ/mol.
What atom is the site of covalent attachment of AMC to the model tetrapeptide used in the studies? A. I B. II C. III D. IV
This question is A because AMC is attached to the peptide on the carboxyl side. This suggests that an amide linkage involving the N atom in AMC is used to covalently attach the fluorophore to the peptide. The passage says AMC is removed from a peptide via peptide bond hydrolysis. Thus, AMC must form a peptide bond with the peptide. Peptide bonds are between amine and carboxylic acid groups to form an amide. There is only one amine in AMC. It also helps to know that chymotrypsin cleaves on the carboxyl side of an aromatic amino acid, meaning the functional group on AMC that binds is the amine.
Bronsted-Lowry base
a molecule or ion that is a proton acceptor
Bronsted-Lowry acid
a molecule or ion that is a proton donor
Lewis acid
an atom, ion, or molecule that accepts an electron pair to form a covalent bond
Lewis base
an atom, ion, or molecule that donates an electron pair to form a covalent bond
Heisenberg uncertainty principle
states that it is impossible to determine simultaneously both the position and velocity of an electron or any other particle
Charles' Law
the law that states that for a fixed amount of gas at a constant pressure, the volume of the gas increases as the temperature of the gas increases and the volume of the gas decreases as the temperature of the gas decreases V1/T1 = V2/T2
