AAMC FL 2 chem/phys Flashcards
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?
Remember E = hf and f = c/wavelength
but we need to find the excited wavelength (excited - emitted) to find this
Find the concentration given 2 nM and 100 micro molar
how to find sp^2 carbons
How many in this one?
have total 3 things bonded
3 atoms attached OR
1 atoms attached + 2 lone pairs OR
2 atoms attached + 1 lone pair
what is this?
histidine
how to find sp^3 carbons
have total 4 things bonded
4 atoms attached OR
3 atoms attached + 1 lone pair
2 atoms attached + 2 lone pairs
How to determine hybridization
what is the Kcat?
The kcat of an enzyme is its “turnover number”–the limiting number of reactions each active site can catalyze per second, per unit concentration of enzyme.
moles of product per second/moles of enzyme
if reaction rates were obtained in saturated conditions then it means the reaction reached
if reaction rates were obtained in saturated conditions then it means the reaction reached…. Vmax
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
B. Excitation of bound electrons
When ultraviolet light is absorbed by organic molecules, bound electrons (n, σ, and π) enter an excited state (σ* or π*) by certain rules of allowed excitation.
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
A.
The least polar compound, capable on only London dispersion forces, will elute first because it has the weakest attraction to the polar stationary phase (n-pentane). (bc the polar vs non polar wont attract)
The second compound to elute will be a slightly polar compound that can accept but not donate hydrogen bonds (2-butanone). The third compound to elute will be a more polar compound that can both accept and donate hydrogen bonds via 1 OH group (n-butanol). The fourth compound to elute will be a compound that can both accept and donate hydrogen bonds via 1 OH group and accept hydrogen bonds to C=O (propanoic acid).
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.
D.
“Half-life” is defined as the time it takes for half of a sample of radioactive material to decay. This means that after one half-life, half of the sample atoms’ nuclei will have decayed into their daughter nuclei, and half will remain in their original form, as stated in (D), which is correct.
The products of radioactive decay (daughter nuclei) may or may not be radioactive, making (C) incorrect.
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.
There is a distance between forces –> Torque problem
rotational equilibrium has torque balanced
tryptophan is what kind of amino acid
aromatic
Only aromatic amino acids contribute to the CD signal in the near UV region.
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
C
Only aromatic amino acids contribute to the CD signal in the near UV region. Alanine residues are NOT aromatic, but they contain peptide bonds, which absorb in the far UV region.
The near UV region where an aromatic ring. Far UV region where peptide bonds absorb electromagnetic radiation.
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.
An aromatic side chain; the absorbed photon energy is lower.
**The excited state of the electrons in the aromatic ring will be closer to the ground state energy
A synthetic peptide with the amino acid sequence KTFCGPEYLA was generated as a mimic of the T-loop.
What is the net charge of sT-loop at pH 7.2?
A.–2
B.–1
C.0
D.+1
C
The sT-loop contains 1 positively charged side chain on K and 1 negatively charged side chain for E, making the net charge zero
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
C.γ32P-ATP
proximal groups - alpha
terminal phosphate is transferred (gamma, y)
This experiment was repeated in the presence of a synthetic peptide that mimics the HM domain (sHM) of Ser/Thr kinases with the amino acid sequence FLGFTY. Phosphorylated sHM (spHM) was also used in place of sHM.
When used in place of spHM, which peptide would be most likely to achieve the same experimental results?
A.FLGFAY
B.FLGFQY
C.FLGFGY
D.FLGFEY
D.FLGFEY
second to last –> threonine
what reacts like a phosphorylated threonine (has a neg charge)
What amino acid has a neg charge?
E, glutamate
(answer D)
PDK1 phosphorylates specific Ser or Thr residues
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
PDK1 phosphorylates specific Ser or Thr residues, and the nucleophile accepting the phosphate in the side chains of both residues is –OH (hydroxyl).
A. Hydroxyl
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
A.Both activation and folding are cooperative.
Because both curves have a sigmoidal shape, these are indicative of cooperative processes.
In the case of Figure 3, cooperativity means that partial folding (or unfolding) induces a higher likelihood of that process continuing. Because both graphs show cooperativity, (A) is correct.
From the data presented in Figure 3, which RIα variant is the most stable?
A.L203A
B.I204A
C.Y229A
D.R241A
A.
A higher melting temperature is indicative of a more stable protein, as more energy is needed to unfold the protein. L203A has an approximate Tm of 50°C; therefore, it is the most stable and even more stable than the WT protein.
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.
Based on the cAMP activation information given in the note in Figure 2, changing Y229 to A229 has no effect on kinetics. However, based on the data in Figure 3, Y229A is less stable than the WT protein, indicating that Y229 is essential for protein stability.
Y229A variant has a similar activation curve to the WT protein, indicating that Y229 is not critical for cAMP activation, since substituting alanine in at that position has no effect.
A. Y229 is important for protein stability but not critical for cAMP activation.
The oxygen pressure is the sum of the oxygen static pressure P and the oxygen flow pressure ½ ρv2. In the area of the mask openings, Pair = P + ½ ρv2, thus Pair > P. Air enters the mask because the static pressure of the air is larger than the static pressure of the oxygen that flows in the mask. This is the Venturi effect, and the mask is called the Venturi mask.
In the overall electrochemical reaction:
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.
In Half-Reaction 2, elemental hydrogen, in which hydrogen has an oxidation number of 0, loses electrons and is oxidized to form hydrogen cations (protons), which have an oxidation number of +1. In Half-Reaction 3, elemental nitrogen, in which nitrogen has an oxidation number of 0, gains electrons, and is reduced to form ammonia, in which nitrogen’s oxidation number is –3. Thus, the overall electrochemical reaction involves the reduction of nitrogen and the oxidation of hydrogen. Since reduction always occurs at the cathode and oxidation always occurs at the anode, (B) is the correct answer.
RedCat - reduction at cathode
oxidation at anode
In the stepwise formation of [Cu(NH3)4]2+ from [Cu(H2O)4]2+, which of the following ions would form in the second step?
A.[Cu(H2O)2(NH3)2]2+
B.[Cu(H2O)2(NH3)3]2+
C.[Cu(H2O)3(NH3)]2+
D.[Cu(H2O)3(NH3)2]2+
it goes from 4–> 3 –> 2 waters after the second step
If hydrochloric acid is added to the reaction, ammonia will be protonated and converted to ammonium, thus decreasing the ammonia concentration. This will cause the equilibrium to shift to the left to accommodate the disruption and make less [Cu(NH3)4] 2+, not more.
The equilibrium of Equation 1 would shift to the left, because a reduction in concentration of one of the reactants would reduce the rate of the forward reaction while leaving the rate of the reverse reaction unaffected. This, in turn, would cause the equilibrium concentration of the product, [Cu(H2O)2(NH3)2]2+, to decrease.
According to Le Chȃtelier’s principle, this disruption to the equilibrium causes the reaction to shift in response to produce more ammonia. The shift will cause the equilibrium to shift to the left, decreasing the amount of [Cu(NH3)4] 2+ formed.
Which of the following atoms will be expected to have the smallest second ionization energy?
A.Na
B.C
C.O
D.Ca
D. Ca
ionization energy increases up the table in the columns and across in the rows
Second ionization energy is the energy required to remove a second electron from an atom after the first one has already been removed. In general, it takes less energy to remove valence electrons than other electrons, because valence electrons are by definition in the atom’s outermost shell, and they are shielded from the nucleus by inner electrons. Thus, the second ionization energies of alkaline earth metals such as calcium are relatively low, and the loss of two electrons results in a stable, noble-gas-like electron configuration
Calcium, Ca, has an electron configuration of 1s2 2s2 2p6 3s2 3p6 4s2. Losing two electrons from calcium will require the lowest amount of energy because the ion’s resulting electron configuration is a stable noble gas configuration (1s2 2s2 2p6 3s2 3p6) with a full octet.
Which of the following species has the largest mass percent of oxygen?
A.H2O
B.CaCO3
C.CO2
D.HCO3–
percent mass formula = mass of element per mole / molar mass of compound
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.)
The dissolution of CaCO3 is CaCO3(s) ⇌ Ca2+(aq) + CO32–(aq) and the solubility product constant expression is Ksp = [Ca2+] [CO32–] .
solution combines ions
kip = [Ca2+] [CO32–]
let x equal each since they combine 1:1
Equal amounts of Ca2+ and CO32– are produced when CaCO3 dissolves in solution, so the expression reduces to x^2 = 4.9 × 10–9, which is equal to 49 × 10–10. Taking the square root, x = 7.0 × 10–5 M.
try converting 4.9 *10-9 to 49 * 10^-10 since this is easier to work with
pos ion C is attracted to the opposite NEG ion (right and bottom)
its repulsed by the other POS ion (left)
since ion C is equidistant from ions A and B, and all three ions carry the same magnitude charge, the upward component of the repulsive force and the downward component of the attractive force exactly counteract each other,
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.
Protons cannot be added or removed by rubbing two objects together
–> electrons change
Which single bond present in nitroglycerin is most likely the shortest?
A.C–H
B.C–O
C.C–C
D.O–N
The bond lengths of single bonds can be predicted by the atomic radii of the bonded atoms.
The smaller the atomic radii, the closer the nuclei can be, hence, the shorter the bond length.
Of the three atoms bonded to carbon in the answer choices, hydrogen is by far the smallest due to only having a single electron subshell, eliminating answers (B)
from the table C-H is the largest bond energy and the shortest bond length
delt H uses table information
- remember 2 moles product means we need to now check if eq is balanced
- now sum of enthalpy products - reactants
and use the number of atoms of each as the coefficient - solve
At STP, the volume of N2(g) produced by the complete decomposition of 1 mole of nitroglycerin would be closest to which of the following?
use the balance equation
According to reaction 1, the decomposition of four moles of nitroglycerin generates six moles of N2. Therefore, dividing the coefficients by four gives the decomposition of one mole of nitroglycerin, resulting in 6/4 = 1.5 moles of N2. At STP (0 °C, 1 atm), the volume of an ideal gas will be 22.4 L/mol. This gives the calculation:
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
Use the figures!!
The power consumed is P = ΔPE /time = mgh/Δt.
From Table 1, there are 30 steps and Δt = 27 s.
Then P = (54 kg × 10 m/s2 × 30 steps × 0.15 m/step) / (27 s) = 90 W.
How much work did an 83-year-old female do while stretching the rubber band to the limit of her strength?
What is the ratio of the minimum sound intensities heard by a 64-year-old male and a 74-year-old female?
male - 20 dB
female - 40 dB
The relative intensities of the two sound waves are 20 dB and 40 dB, respectively. Their 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 102 = 100.
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
neg focal length (table)
means this is diverging
SUV
small, upright, virtual
objects appear smaller
(all upright - virtual)
(all inverted - real)
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
several resistors in parallel always have a smaller effective resistance than would any of the individual resistors on their own.
