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For a thermodynamic mixture of isomeric products |
the relative mole ratio of products is directly related to the relative stability of these products |
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#2. What can be inferred from the results of scheme A about the relative thermodynamic stabilities of Compounds 4a and 4b |
Compound 4a is more stable than 4b The image shows that 4a is produced 60% and 4b 40%. –So 4a will be more stable among the two thermodynamic products |
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#3. Which of the following compounds is most likely to be a major product in step 2 if the aldol condensation is NOT followed by a dehydration reaction? |
Ans in image –coz if there is no dehydration, then the OH would be present instead of the double bond –product of aldol condensation is b-hydroxyketone |
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#4. Compound X is shown below.(image) If compound 2 is replaced with compound X, what will be the structure of the final product in steps 1 and 2? (Note: assume that the yields of steps 1 and 2 are not affected by substitution) |
Ans in image –coz this is the product of Michael addition followed by intramolecular aldol condensation (Robinson annulation) |
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#5. What is the total amount of charge and energy, respectively, that the capacitor will store if it is connected to the battery on the railcar? |
1.20C and 7.20J C=Q/V Energy stored by a capacitor = Ue = (1//2)CV^2 = (1/2)QV = (1/2)Q^2V Given V= 12V C = 100 x 10^-3 F Q = VC = (12)(100 x 10^-3) = 1.20 C Ue = (1/2)QV = (1/2)(1.20)(12) = 7.20J |
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#6. Suppose that the railcar passes by a horn that is emitting a sound with a frequency f. Which of the following describes f’ that the person on the railcar hears? |
f’ > f before passing the horn, f’ < f after passing it Due to doppler effect, the frequency that the person on the railcar hears before passing the horn is larger than the actual frequency of the sound emitted, while the person hears a frequency lower than the actual frequency after passing the horn –Before passing the horn, the object(rail car) is coming closer to the source (horn) so f’>f –After passing the horn, the object(railcar) is moving away from the source (horn) so f<f’ |
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Doppler effect |
The apparent frequency (f’) of the source is increased as the source approaches the observer, and is decreased as the sources leaves the observer f’ = ((V+_ Vo)/(V-+Vs))f |
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#7. If no braking occurs, a total of how much power would be required to keep the railcar moving at 40m/s? |
Power = work done/ time Work done = Fd v= d/t –> d =vt W= Fd –> Fvt P = Fv = (1000N)(40) = 40000 W = 40 kW There are lot of force values given in the passage but we use 1000N coz the power required must match the work done by the friction force that tends to slow down the railcar Rest of the values of forces involve braking so can’t use it since the question says that "If no braking occurs…." |
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#8. If a generator-brake system alone were engaged when the railcar was moving, which of the following graphs would most accurately represent the subsequent speed of the railcar? |
Ans Image–> Coz the speed of the car decreases in time such that the decelerating force declines linearly with speed, according to the passage. Because the decelerating force is directly proportional to the deceleration acceleration, it follows that the decelerating acceleration itself declines linearly with speed. Finally, because the decelerating acceleration is the rate of change of velocity, or the slope of the velocity vs. time graph, then the slope of the graph must decrease in time F=ma, m will be constant, since force is decreasing linearly, acceleration is decreasing. Slope of v, t graph = a, so in the answer choices the slope cannot stay the same because a is changing not constant, and it is decreasing overall since deceleration First to be clear, the passage states "Force decreases linearly with Speed" BUT the axes on the graphs are Velocity vs Time. Choice A would be if you had a CONSTANT decelerating force Choice D shows the expected result of a LINEARLY DECREASING decelerating force. It helps to think of it conceptually. At high speed the force of deceleration will also be high, so the velocity will drop quickly (steep slope). At low speed the force of deceleration will be small, so the rate at which velocity drops will be slow (shallow slope). |
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#9. The relative thermodynamic stability of isomeric organic compounds can be inferred from which of the following types of experimental data? |
D. Heats of combustion The relative thermodynamic stability of isomers can be determined based on the amount of heat produced when the compounds are combusted; less heat, greater stability Key word is here "isomeric," meaning they have the same molecular formula but different connectivities. Because they are isomeric, their boiling points will probably be very close to each other (or close enough that it would be very difficult to distinguish between individual molecules). Heats of combustion, however, differs significantly. For example, if you have 2-methylbutane vs pentane, because that methyl carbon is branched (as a tertiary carbon) will confer more stability than as a pentane molecule. |
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#10. Raadioactive tritium (3H) labeleed guanine has been used to measure the rate of biochemical processes that involve its building or incorporation Given that water is the solvent for this type of experiment, what is the best site for tritum labeling? |
I Because the best site for tritium labeling would not exchange the tritium ions for protons in water. All of the N-H sites (II-IV) would readily exchange tritium protons due to their lone pair-facilitating protonation and subsequent tritium exchange with water, but the C-H site (I), lacking a lone pair, would retain its tritium label |
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#14. The energy of activation for the reaction described in the passage is given by the energy of: |
The activated complex minus the energy of the reactants The activation energy for the reaction presents the minimum energy barrier necessary to be overcome by the reactants on the path to products B is free energy (comparing energy between the reactant and the final product) D is the energy difference between the reactant and the transition state which is the activation energy. |
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#16. What type of reaction is occurring between I2 and Zn? |
Oxidation-reduction because both Zn and I2 change oxidation states during the reaction |
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#18. What is the period of the voltage source that operates the plasma pencil? |
1000ns Because the period of a time-varying signal is the shortest repitition time. According to fig 2, each pulse is on for 500ns, then off for 500ns. Thus the time between consecutive pulse is 500ns + 500ns = 1000ns |
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#19. What is the maximum energy of the UV photons generated by this plasma pencil? (Note: speed of light is c = 3.0 x 10^8 m/s; Planck’s constant is h = 6.63 x 10^-34 J.s) |
E=hf = hc/lamda given: lamda = 200-300nm the question is asking for MAXIMUM energy, so you would want to minimize the denominator, so use 200nm for lamda E = (6.63 x 10^-34 J.s)( 3.0 x 10^8 m/s)/ (200 x 10^-9 m) = 1 x 10^-18 |
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#20. That the electric field is uniform between the electrodes means that the electric field lines: |
are equally spaced at both electrodes and between them –By definition, electric field lines are spaced in a uniform field |
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#24. What additional substance is necessary for reaction 2 to take place? |
FAD –Because among the choices listed, only FAD is a cofactor oxidant because an oxidizing agent is required for Reaction 2 to proceed. Its oxidized form is FAD and its reduced form is FADH2 |
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What is one characteristic of steroid |
FOUR FUSED RINGS |
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#26. The pH of a 1L phosphate buffer solution was measured as 7.6, but the experimental procedure calls for a pH 7.2 buffer. Which method will adjust the solution to the proper pH? (Note: pKa values for phosphoric acid are 2.2, 7.2, and 12.3.) |
Alter the ratio of monosodium/disodium phosphate added to favor the monosodium species because in order to lower the pH of a buffer, the proportion of acidic buffer component must be increased. Adding strong base, diluting with water, or adding a different salt will NOT lower the pH for MCAT purposes, diluting a solution does NOTHING to the pH because you are diluting A- and the HA the same amount. The ratio does not change in the HH equation. As for why C is correct, look no further than our good friend LeChatlier. The general reaction is NaH2PO3<-> Na2HPO3 + H+ The more the reaction favors the dissociation of the monosodium phosphate, the more acidic the solution. Altering the ratio to favor the monosodium/disodium phosphate to favor the monosodium species is a fancy way of saying increasing the monosodium concentration. This increases the concentration of reactants, which will push the reaction right (forming more H+). This, in turn, decreases the pH of the solution. Could you explain how you got the reaction formula? The formula is actually just a simple dissociation/deprotonation of a diprotoc acid. This is analogous to carbonic acid deprotonating to become bicarbonate (h2co3 <-> h+ + hco3-). In the case of this reaction, the negative hco3- would with Na+ in solution making it neutral NaHCO3. The more Na the molecule has, the more hydrogen’s are released. In the question above, rather than carbonic acid, it is dealing with dissociation of phosphorous acid. H3po3 <-> h2po3- + h+ <-> hpo3(2-) + 2H+ . The reaction continues but for our purposes we only care about the latter half of that equilibrium. Again, in this solution, the negative anions bind Na+ making the compounds neutral. With this in mind, the reaction we use is Nah2po3 <-> Na2Hpo3 + H+ . You may ask why I got rid of the hydrogen on the left side and that is because this is the net equilibrium reaction. Above, you can see that two hydrogen’s are produced on the right and one on the left, they cancel leaving one left on the right. Now to lechatliers principle. What addition (or subtraction) would shift the rxn right? A few things can and I won’t name them all, but adding monosodium phosphate pushes the equilibrium left because you have more reactants than products. You can also take away disodium phosphate to shift the equilibrium right. This in turn produces more H+ in the solution, thereby lowering the pH. monosodium phosphate (NaH2PO4) has an additional H+ compared to the disodium phosphate (Na2HPO4), which has one less H+ since there are two Na+. This makes the monosodium the acid for this problem. You can solve this from this point in one of two ways: 1) Intuition: Monosodium phosphate is the acid in this case and will decrease the pH if more is added 2) pH = pKa + log (base/acid). If you increase acid, you have a smaller log value which decreases the pH |
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#27. When two amino acids are joined via a peptide bond, what is the mass of the byproduct of the reaction? |
18amu Peptide bond formation is the dehydration reaction so byproduct will be water |
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#28. Which experimental condition is NOT necessary to achieve reliable data for Michaelis-Menten enzyme kinetics?
A. |
The reaction is allowed to reach equilibrium before measurements are taken Because once the reaction reached equilibrium, it will be impossible and the kinetic data will look the same regardless of substrate concentration |
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#30. What is the general structure of hydroxamic acid? |
Ans–> image Becuase a hydroxamic acid was described in the passage as resulting from the acetylation (attachment RC=O) of the hydroxylamine nitrogen (NHOH) in compound 2 It cannot be choice D, because acetylation occurs at the carbonyl carbon of the acetyl group. Therefore, the nitrogen must be attached to the carbonyl group itself (as in choice C). |
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#37. Which atom is most likely involved in coordination of calcium ions found in HRP? |
D. Oxygen –and is D because an atom must be a Lewis base to coordinate to calcium ions, and oxygen is the only Lewis basic atom present in the side chains or backbones of the listed amino acids that has either a partially negative charge (in the peptide backbone or ser side chain) or negative charge (in Asp side chain) –the only other Lewis basic atom present is Nitrogen (distractor C), which has a partially positive charge in the peptide backbone due to resonance and is thus less likely to coordinate calcium ions this question requires: (1) Knowing that Lewis bases coordinate cations (2) Recognizing which side chains and backbones are Lewis bases (3) Understanding the effect of charge on the strength of the Lewis base interactions with cations |
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What coordinates with cations |
LEWIS BASES!! |
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#38. Which structure represents a component of the HRP cofactor? |
Ans image –> Because the cofactor is heme, which is a porphyrin. The basic unit of a porphyrin is the pyrrole ring -a 5 sided heterocycle containing one nitrogen atom |
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Resistivity = |
Inverse of conductivity |
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#40. Which two classes of enzymes are needed in the two-step conversion of cytosine to 5hmC? |
Transferase and oxireductase –Because the first step involves transfer of a methyl group to cytosine, and the next step involves the hydroxylation of that methyl group. Therefore, the two classes of enzymes needed are a transferase and an oxireductase. |
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#41. What is the correct expression for the deltaG’o for the transition observed in the experiments described in the passage? |
deltaG’o = -RTlnKeq Keq = [products]/[reactants] ΔG′° = -RTln([unfolded]/[native]) he equilibrium constant for DNA unfolding is Keq = [unfolded]/[native] as the unfolded DNA is considered to be a product. Thus, ΔG′° = −RTlnKeq = −RTln([unfolded]/[native]). |
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#42. Which statement about the unfolding cooperativity and pK of the oligonucleotides is consistent with the data in Figure 1? |
The oligonucleotides with the lowest pK displays the highest unfolding cooperativity –because the pK is the pH at which the fraction of folded DNA is 0.5. this occurs at the lowest value in 5hmC-WT. Cooperativity is measured as the slope of the unfolding transition. This is also highest in 5hmC-WT |
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What is pK? |
pH at which the fraction of folded DNA is 0.5 |
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#43. Based on the information in the passage and in figure 1, what effect does epigenetic modification have on iM pH-dependent denaturation |
Only hydroxymethylation results in significantly decreased stability because cytosine is more readily deprotonated –Because the drop on pK of the transition denotes a decrease in stability. Because DNA unfolding occurs as the pH increases, it can be inferred that this is due to cytosine deprotonation. Since 5hmC-WT has the lowest pK, hydroxymethylation decreases the stability by increasing acidity of cytosine |
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#45. The isoelectric focusing points pI for four proteins are shown in the following table At which buffer pH would two out of four of the proteins adhere to a cation-exhcange column? |
7.0 Because cation-exchange column only binds to positively charged proteins which only occurs when the pH is less than the pI. At pH 7.0, both proteins A and B would positively charged. |
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#46. A common column material used in size-exclusion chromatography is dextran, a polysaccharide of glucose. Which type of interaction occurs between proteins and the dextran column material? |
Hydrogen bonding because glucose has numerous hydroxyl groups that can hydrogen bond to the polar side chains that are typically exposed on a protein surface |
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#49. Which of the following properties of a 2.3 MHz ultrasound wave remains unchanged as it passes into human tissues? |
Frequency — frequency of a wave is not affected by the medium through which it propagates. Wave speed Amplitude Wavelength does |
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#50. Assume that in the study with the rat tissues, fluid flows at a speed of 0.30 mm/s through a typical capillary opening caused by a burst microbubble. Given this, which of the following is closest to the volume flow rate of fluid passing through the opening? A study was done on the tissues of rats treated with microbubbles burst by 2.3 MHz ultrasound. It was observed that the burst microbubbles made openings of ~2.5 × 104 μm2 through the capillary walls. |
7.5 × 10^6 μm3/s Volume flow rate = Av (2.5 x 10^4)(0.30) = 7.5 x 10^6 Q=AV Q=quantity of flow A=Area cross sectional V=velocity |
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Speed of sound = |
Wavelength x frequency |
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#52. Suppose that a blood vessel of cross-sectional area A carries microbubbles at a speed v into a capillary bed. If the capillary bed is made up of n capillaries, each with cross-sectional area a, with what speed will the blood flow in the capillary bed? |
if u= speed of flow in capillary bed Av = nau u=Av/na |
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A=B –> C+D Ksp =? |
Ksp = [A][B] |
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for spontaneous reactions |
delta G<0, Keq>1 |
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#57. Which of the following molecules have octahedral geometry? A. SF6 |
A. SF6 |
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#58. A tall tube is evacuated, and its stopcock closed. The open end of the tube is immersed into a container of water (densiy 10^3 kg/m^3) that is open to the atmosphere (pressure 10^5 N/m^2) When the stopcock opened, how far up the tube will the water rise? |
10m –Because the water will rise to a height such that the weight (mass multiplied by gravitational acceleration) of the water column equals the atmospheric pressure multiplied by the tube cross-sectional area A. Because mass is density times volume, it follows that 10^3 kg/m^3 x h x A x 10 = 10^5 x A, where h is the height sought. Solving for h yields h = 10^5/10^4 = 10m |
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#59. Which of the following is a second period element that is a covalent network solid in its standard? |
Carbon graphite –> covalent network solid |
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A change in frequency of sound, as a result of motion between the source of sound or the reflector of sound |
Dopper effect |
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how do we know what way the object is moving? |
if neg, the object is moving away from you if pos, the object is moving towards you |
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(PV = nRT) |
The Ideal Gas Law |
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pv/RT=n then n x 6.022×1023. The Avogadro number The Ideal Gas Law (PV = nRT) must be rearranged to solve for the number of moles of gas particles, n. The result is multiplied by Avogadro’s number NA to arrive at the required expression. |
Which expression gives the total number of gas particles (NOT moles, but individual particles) found in the lungs of an average human (V = 6.0 L) under normal physiological conditions (P = 1.0 atm, T = 37ºC)? |
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The Ka |
The stronger the acid the larger the what? |
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The Kb Base Dissociation Constant |
The stronger the base the larger the what? |
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iodine. because the number of iodine atoms provided in the reaction is less than the 2:1 ratio with zinc atoms that is required for a stoichiometric reaction. Limiting reagents are based on the reactants. You can cross out any answers that consists of products, and iodide is one of our products so this wouldn’t be an option |
Iodine and zinc are the reactants in an experiment to find the stoichiometry of a reactionA 2:1 ratio of moles of iodine atoms to zinc atoms in the product is consistently obtained. Gray, granular zinc metal (2.0 g) and purple iodine crystals (2.0 g) are added to an Erlenmeyer flask containing 25 mL of methanol (CH3OH). The limiting reagent in the experiments described in the passage is: |
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3.6 × 10-4 g The stock solution consisted of 72 g of glucose in 1.0 L of solution (72 μg/μL). Therefore, 2 μL contains 144 μg, or 1.4 × 10-4 g of the solute. |
What is the total mass of D-glucose dissolved in a 2-μL aliquot of the solution used for this experiment? |
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an oxidoreductase. because Reaction 2 is an oxidation of glucose, and thus the enzyme that catalyzes this reaction is classified as an oxidoreductase. |
The enzyme used in the blood glucose meter described in the passage is classified as: |
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60 μg The passage tells us that the meter strip measures the amount of reaction-2. It vaguely states, a two-electron oxidation, and if you miss this part, then you’re in trouble. So, from this your supposed to logically put together that for everyone one D-glucose you use, you get 2 moles of electrons. This is the stoichiometry bridge that links the 0.67umol in the stem question and the 180ug/mole. (0.67umole)(1mole of glucose/2mol of electrons)(180g glucose/1mol glucose)= 60ug |
The glucose meter measures the current produced during Reaction 2. If 0.67 μmol of electrons were measured, what mass of glucose was present in the sample? (Note: The molar mass of glucose is 180 g/mol = 180 μg/μmol.) two-electron oxidation |
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110Da |
The average molecular weight of an amino acid |
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H+ + NADPH + O2 → NADP+ + H2O2 because a careful examination of Reaction 1 and the passage indicates that the nonproductive reaction is between NADPH and O2, with NADP+ and H2O2 as the products and FAD as a regenerated cofactor (not in the balanced equation). |
What is the balanced equation for the nonproductive reaction when lysine is the substrate? Effect of pH on the rate constant kFAD for conversion of FADH-OOH to FAD. Reactions were performed in the presence of 15 μM amine and were initiated by the addition of 2 μM of SidA. When N6-trimethyllysine is used, regeneration of FAD occurs at the same rate as when no amino acid is present. The researchers confirmed that N6-trimethyllysine does occupy the active site during the process by examining the kinetics of hydroxylation of Compound 1 in its presence. |
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Acetylation |
refers to the process of introducing an acetyl group (resulting in an acetoxy group) into a compound, namely the substitution of an acetyl group for an active hydrogen atom |
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2.0 SidA is an enzyme that hydroxylates molecules. The kinetics of SidA with different substrates is shown in figure 1. KFAD is sort of like Kcat, which is a proxy for how fast catalysis is occuring. The higher the KFAD, the faster the reaction is occuring b/c vmax = kFaD [substrate] The question is asking at pH = 7.5 what is the relative rate of the reaction when comparing when the enzymes substrate is compound 1 and when the enzymes substrate is lysine. Then, I look at the figure and compare kFAD for compound 1 and lysine at pH 7.5 and notice it is about 2x higher and so I choose option D. |
At pH 7.5, at what relative rate does the enzyme hydroxylate Compound 1 compared to nonproductively producing H2O2 as a result of interacting with lysine? (Note: Assume equal availability of both substances.) |
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1.0 x 10-14 |
[H3O+][OH-] = |
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10- 4 because, at the highest pH tested (10) the [H3O+] is 10-10 M. This means that [OH-] is 10-4 M, because [H3O+] × [OH-] = 10-14 for aqueous solutions at 25°C. |
What is the concentration of hydroxide ion for the solutions with the highest pH that was studied? the highest pH studied was 10 |
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Oxygen |
Which atom is most likely involved in the coordination of calcium ions found in HRP? |
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Lewis Base because an atom must be a Lewis base to coordinate to calcium ions, and oxygen is the only Lewis basic atom present in the side chains or backbones of the listed amino acids that has either a partially negative charge (in the peptide backbone or Ser side chain) or a negative charge (in the Asp side chain). The only other Lewis basic atom present is nitrogen (distractor C), which has a partially positive chain in the peptide backbone due to resonance, and is thus less likely to coordinate calcium ions. It is a Scientific Reasoning and Problem Solving question because its solution requires (1) knowing that Lewis bases coordinate with cations, (2) recognizing which side chains and backbones are Lewis bases, and (3) understanding the effect of charge on the strength of Lewis base interactions with cations. |
a compound or ionic species that can donate an electron pair to an acceptor compound. |
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porphyrin ring-pyrrole ring, a five-sided heterocycle containing one nitrogen atom. Iron Prox His- his directly attached to Iron |
heme |
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Resistance |
R – A material’s opposition to the flow of electric current, measured in ohms |
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conductance |
it is the inverse of resistance, and is a measure of how easy it is to pass a current through a material |
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What is the resistivity of the best-performing PANI described in the passage? 200 Ω•cm because resistivity is the inverse of the conductivity, which is 1/5.0 × 10-3 (Ω∙cm)-1 = 200 Ω∙cm |
What is the resistivity of the best-performing PANI described in the passage? Polyanionic molecules can be used as templates in the HRP-catalyzed synthesis of PANI to ensure only the desired polymeric structure is obtained. The negative charge of the sulfonate (pKa = 0.7) in polystyrene sulfonate (PSS) interacts with the incoming aniline molecules (pKa = 4.6) and forces them into a conformation that results in a linear polymer. The best-performing PANI had a maximum conductivity of 5.0 × 10-3 (Ω∙cm)-1. |
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Deoxycytidine because the WT sequence does not contain deoxyguanosine. |
The complement to the WT sequence does NOT contain which nucleoside? The wild-type (WT) sequence was 5′-TTCCCTACCCTCCCCACCCTAA-3′ |
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ionization energy ↑ |
periodic table and to the right |
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Fluorine fluorine is the smallest halogen and appears at the top of the column of Group 17. With fewer electrons to shield the valence electron from the nucleus, fluorine will exhibit the highest first ionization energy. |
Which halogen has the highest first ionization energy? |
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d ≈ t2 because distance is proportional to the square of time, as seen from the graph that shows an arc of parabola where at time 6 units the distance is 36 units (slightly less than 40 units), and at time 10 units the distance is 100 units. |
Which of the following best describes the functional relationship between distance d and time t in the figure? |
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F = Flow F=∆P/R |
flow rate |
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0.65 mm. The frequency is the ultrasound so 2.3 MHz |
A study was done on the tissues of rats treated with microbubbles burst by 2.3 MHz ultrasound. It was observed that the burst microbubbles made openings of ~2.5 × 104 μm2 through the capillary walls. Given that the speed of sound in the rat tissues was 1500 m/s, the wavelength of the ultrasound wave used in the study was closest to: |
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m/s speed |
units for speed, frequency, and wavelength |
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according to the continuity equation, if u is the speed of flow in the capillary bed, then A ´ v = n ´ a ´ u, so u = A ´ v/(n ´ a). |
Suppose that a blood vessel of cross-sectional area A carries microbubbles at a speed v into a capillary bed. If the capillary bed is made up of n capillaries, each with cross-sectional area a, with what speed will the blood flow in the capillary bed? |
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Continuity equation |
Equation following the law that the mass flow rate of fluid must remain constant from one cross-section of a tube to anothe |
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The alkaline earth metals |
Calcium and magnesium belong to what group on the periodic table? |
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greater than 1. because a spontaneous reaction is one that exhibits ΔG < 0. Since ΔG = -RTln(Keq), this means that Keq must be > 1. |
If the reaction shown in Equation 1 is spontaneous, the value of K must be: |
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SF6 |
Which of the following molecules has an octahedral geometry? A. SF6 B. CBr4 C. XeF4 D. BrF3 |
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Carbon |
Which of the following is a second period element that is a covalent network solid in its standard state? A. Carbon B. Phosphorous C. Oxygen D. Iodine |
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Only have the reactancts in the equation and never have the soild of liqiud |
for Ksp |
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kw/ka |
kb= |
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the 2 is the limiting reagent |
2:1 ratio means |
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