C-P AAMC FL#2

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#1. What atom is the site of covalent attachment of AMC to the model tetrapeptide used in the studies?

I 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

#2. 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?

ans in image Because E = hf= hc/lamda Excitation occurs are LAMDAe = 360nm, but fluoroscence is observed at LAMDAf = 440nm. This implies that an energy of E =hc/[(360 x 10^-9)-(440 x 10^-9)] E = h x c x [(1/360 x 10^-9 ) – (1/440 x 10^-9)] J per photon converted to other forms between the excitation and fluorescene events

#9. Absorption of UV light by organic molecules always results in what process?

Excitation of bound electrons

#10. 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?

n-pentane –>2-butanone –> n-butanol –>propanoic acid 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)

#11. The half life of a radioactive material is:

The time it takes for half of all the radioactive nuclei to decay into their daughter nuclei –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 be radioactive

#12. A person is sitting on the chair as shown.

Why must the person either lean forward or slide their feet under the chair in order to stand up

To keep the body in equilibrium while rising 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 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

#14. 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?

Ala All chiral nonaromatic amino acids will contribute solely to the CD signal in the far UV region ITS IN THE PASSAGE!! READ AND TRUST YOUR GUT!!

#15. 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?

An aromatic side chain; the absorbed photon energy is lower –GROUND STATE = LOWEST ENERGY STATE! –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 Here, an "electron excitation state closer to the ground state" is just a fancy way of saying lower energy. To figure out which (aromatic side chain or peptide bond) will have lower energy let’s use the equation E = hf, which tells us the energy of electromagnetic radiation is proportional to its frequency. 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 state.

#16. A solution of poly-L-lysine in water (pH 7.5) provided CD spectrum (A). Upon adding enoguh trifluoroethanol (TFE) to generate a solution that was 50% TFE, the CD sepctrum changed to (B).

Which secondary structural changes most likely took place in the peptide as a result of adding TFE? The secondary structure of poly-L-lysine changed from:
A. random coil to alpha-helix
B. alpha-helix to random coil
C. beta-sheet to random coil
D. alpha-helix to beta-sheet

Alpha-helix to beta-helix After consulting Figure 1, it is evident that the predominant secondary structural feature found in poly-L-lysine dissolved in water (at pH 7.5) is the alpha-helix. After adding TFE, the secondary structure of poly-L-lysine changes to predominantly beta-sheet –LOOK AT THE FREAKIN GRAPH IN THE PASSAGE!!! Random coil -A protein that completely lacks secondary structure is a random coil. In random coil, the only fixed relationship between amino acids is that between adjacent residues through the peptide bond.

Random coil

Segments of proteins, and polypeptides that lack secondary structure, are often assumed to exhibit a random-coil conformation in which the only fixed relationship is the joining of adjacent amino acid residues by a peptide bond –so they are primary structures

#18. In designing the experiment, the researchers used which type of 32P labeled ATP?

Y32P-ATP Because the phosphoryl transfer from kinases comes from the Y-phosphate of ATP. Therefore, the experiment should require Y32P-ATP

#19. When used in place of spHM, which peptide would be most likely to achieve the same experimental results?

FLGFEY –Because the phosphorylated threonine would most likely be mimicked by glutamic acid in terms of size and charge

#23. Based on the data presented in figured 2 and 3, what is the most likely role if Y229 in protein stability and cAMP activation?

Y229 is important protein stability but not critical for cAMP activation –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 do we know there was a REMOVAL of Y229? –In this passage, it is a bit tricky to tell that there was a removal of Y229A as opposed to an addition unless you know what to look for. However, it all comes down to terminology and notation. First, they refer to Y229A as a variant, which in genetics terms usually refers to a mutation. Therefore, just from that, we know that there is a change from wild-type. Additionally, when you have notation like "Y229A" where there is a letter followed by a number then another letter, this means that the letter corresponding to the first amino acid is replaced with the amino acid corresponding to the second letter. For example, if I were to write "K78A", this would mean lysine (K) was replaced by alanine (A) in that mutation. Therefore, when we look at Y229A, this is a mutation resulting in the replacement of tyrosine (Y) with Alanine (A). Because of this, we know that there was a removal of a tyrosine (aka Y229) and that this adversely affected the protein stability.

#25. A patient puts on a mask with lateral openings and inhales from a tank as shown.

What phenomenon causes static ait to be drawn into the mask when oxygen flows?

Venturi effect –Because oxygen pressure is the sum of the oxygen static pressure P and the oxygen flow pressure rV^2/2. In the area of the mask openings, Pair=P+rV^2/2, thus Pair>P. air enters the mask because the static pressure of the air is larger than the static pressure if the oxygen in flow. This is the Venturi effect, and the mask is called the Venturi mask

#27. Which property of a substance is best used to estimate its relative vapor pressure?

Boiling point

#28. What are the structural features possessed by storage lipids?

Three fatty acids ester-linked to a single glycerol

For electorchemical rxns remember:

RED CAT –> reduction at cathode AN OX –> oxidation at anode

#31. The lone pair of electrons in ammonia allows the molecule to :

Act as a LEWIS BASE in water

#33. 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?

SCY temperature higher than electrode temp –Because the proton conductivity of SCY increases with increasing temp, while the favorability of reaction decreases with overall temp. It is therefore beneficial to maintain the SCY conductor at a higher temp READ THE FREAKIN PASSAGE ITS ALL IN THERE

#35. In the stepwise formation of [Cu(NH3)4]2+ from [Cu(H2O)4]2+, which of the following would form in the second step?

[Cu(H2O)2(NH3)2]2+ In the passage, the reaction was stated to proceed in four sequential steps, with each step involving the replacement of one water molecule by ammonia.

Coordinate covalent bond

A coordinate covalent bond, also known as a dative bond or coordinate bond is a kind of 2-center, 2-electron covalent bond in which the two electrons derive from the same atom. The bonding of metal ions to ligands involves this kind of interaction.

#37. Consider the reaction shown in equation 1 at equilibrium. Would the concentration of [Cu(NH3)4]2+ increase if the equilibrium were distributed by adding hydrochloric acid?

No, because the equilibrium in Equation 1 would shift to the left HCl will protonate ammonia in Bronsted 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

#38. In [Cu(NH3)4]2+, the subscript 4 indicates which of the following?

The coordination number of Cu2+ only 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

#42. What is the pH of a buffer solution that is 0.2M in HCO3- and 2M in H2CO3? (Note: The first pKa of carbonic acid is 6.37)

5.37 HANDERSON HASSELBALCH EQN pH = pKa + log ([HCO3-]/[H2CO3]) = 6.37 + log (0.1) = 6.37 + log (10)^-1 = 6.37-1 = 5.37

Some basic stuff about logs

log (10) = 1 log(100) = 2 log(1000) = 3 log (0.1) = -1 log(0.01) = -2 log(0.001) = -3

Some basic Ksp stuff

When taking blood pressure remember that

Blood flow will not be heard when the pressure in the cuff is greater than systolic pressure systolic/diastolic

Bonds between identical atoms are likely to be

LEAST polar

#52. What is the average power consumed by a 64-year-old woman during the ascent of the 15cm high steps, if her mass is 54kg?

Power = W/t W=Fd F=mg W=mgd P=nmgd/t Also n=30 =(30)(54)(10)(.15m)/(27) =90 W

#53. How much work did an 83 y/o female do while stretching band to the limit of her strength?

W = (1/2)Kx^2 = (1/2)(200)(0.2)^2 = 4J

#54. What is the ratio of the minimum sound intensities heard by a 64 y/o and a 74 y/o male

Ans: 100 64 y/o female -> lowest sound intensity = 20dB 74 y/o male -> lowest sound intensity = 40dB The difference is 20dB, meaning that the decimal log of the ratio of their intensities is 2, which means that the ration of their intensities is 100

#55. What kind of image is formed by the lenses of the glasses worn by a 68y/o male who sees an object 2cm away

Virtual and reduced 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

Converging mirrors have

POSITIVE focal length

Diverging mirrors have

NEGATIVE focal length

Real images are always

inverted and can be cast on screen

Virtual images are always

erect (noninverted) and cannot be cast on screen

Optical instruments

Eye =lens focuses real image on retina Glasses –Diverging (Concave) lens for near sightedness –Converging (Convex) lens for farightedness Magnifying glass –Virtual, erect, larger image formed when p<f for converging lens

#56. What is the approximate percentage of 10C sample left after the time it took 75y/o male to walk one lap around the gym? (Note: The half-life of 10C is 20 sec)

25% Because 75 y/o person takes 200s to walk 5 laps, which is on average 40s per lap. The time represents two half lives of 10C, so only 1/(2)^2 = 0.25 or 25% of 10C is left

#57. Which of the following energy conversions best describes what takes place in a battery-powered resistive circuit when the current is flowing?

Chemical to electrical to thermal

#58. Protein secondary structure is characterized by the pattern of hydrogen bonds between:

Backbone amide protons and carbonyl oxygens Because secondary structure is represented by repeated patterns of hydrogen bonds between the backbone amide protons and carbonyl oxygen atoms

#40. Which of the following atoms will be expected to have the SMALLEST second IE?
A. Na
B. C
C. O
D. Ca

D. Ca –you picked the answer with the highest IE!! READ! –Metals have lower IEs than nonmetals as long as the ionization event involves a valence e-. Since Na is an alkali metal, it has only one valence e- and has a large second IE. Ca is an alkaline earth metal and has two valence e-s. It will therefore have the smallest second IE of the four atoms listed.

#47. A glass rod is rubbed with a silk scarf producing a charge of +3.2 x 10^-9 C on the rod (recall that the magnitude of the proton and electron charges is 1.6 x 10^-19 C). The glass rod has:
A. 5.1 x 10^11 protons added
B. 5.1 x 10^11 electrons removed from it
C. 2 x 10^10 protons added to it
D. 2 x 10^10 electrons removed from it

D. 2 x 10^10 electrons removed from it the # of charges: (+3.2 x 10^-9 C) /(1.6 x 10^-19 C) = +2 x 10^10 C –this means that the rod has an excess of positive charge created by removing a number of +2 x 10^10 C electrons from the material as it is not possible to add protons in a manner described in this question

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