stidy pack 2 chem phys

What is the approximate isoelectric point of the C-terminal amide-capped tripeptide Asp–Ala–Cys–NH₂?

(Note: pK_a of N-terminus = 9.8; pK_a of Asp side chain = 3.9; pK_a of Cys side chain = 8.4.) 

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A

9.8

Answer choice eliminated

B

6.9

C

6.1

D

3.9

Solution: The correct answer is C.

1. At this pH, the side chain of Asp has a charge of –1, the side chain of Cys has a charge of –1, and the N-terminus has a charge of +0.5, making the net charge about –1.5.

2. This isoelectric point calculation averaged the pK_a of the N-terminus and the pK_a of the Asp side chain instead of the Asp and Cys side chains.

3. The isoelectric point (pI) for two neutral side chains that ionize to a negative charge when there is only an unmodified N-terminus is calculated as follows: pI = (pK_a of the Asp side chain + pK_a of the Cys side chain)/2 = (3.9 + 8.4)/2 = 6.15 ≈ 6.1.

4. At this pH, the side chain of Asp would have a charge of –0.5, the N-terminus would have a charge of +1, and the side chain of Cys would have a charge of 0, making the net charge +0.5.

Simplified explanation:

For isoelectric point, pL = pka1+pka2/2

The two pKas are the range at which the zwitterion (neutral charge) exists. If you understand isoelectric focusing technique, then you understand why knowing the zwitterion range is important.

The C-terminus (COOH) is amide capped, so it doesn't contribute to this question and can be ignored.

At low pH, the NH3+ form of N-terminus always dominates, so the whole molecule has a +1 charge.

As pH increases, Asp deprotonates first (due to lower pKa), creating a -1 charge, this neutralizes the +1 charge from the NH3+ and now we have a Zwitterion that is neutral.

As pH increases further, Cys deprotonates (pKa 8.4) and now we have a -1 net charge.

So we now know that the Zwitterion (neutral) exists between pKa 3.9 to 8.4.

Plug those into the equation: pI = 3.9+8.4/2 = 6.1

C C-terminus is capped, so it is neutral, given by the question stem.

Asp has a (-) charge

Ala neutral

Cys neutral

N terminus has a (+) charge.

The N terminus (+) and Asp (-) cancel out, so it is a neutral chain. That is how you know to use the PI equation for a neutral chain, PI = (Pka 1 + Pka 2) / 2

What is the maximum number of full α-helical turns in a protein monomer composed of 4428 amino acid residues?

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Solution: The correct answer is B.

1. This result is based on 4.1 amino acid residues per helical turn, which is for a π-helix, not an α-helix.

2. An α-helix has 3.6 amino acid residues per full helical turn. 4428/3.6 = 1230 full α-helical turns.

3. This result is based on 3 amino acid residues per helical turn, which is for a 3₁₀-helix, not an α-helix.

4. This result is based on 2.7 amino acid residues per helical turn. There are no protein helices with this ratio.

3.5 AA per turn

For example if you have a negatively charged side chain at one position and a positively charged one four amino acids later, the two will align pretty well and be able to interact and stabilize the helix.

This sort of info can be used to predict where alpha helices may form in proteins where the structure is not solved, allowing scientists to quickly narrow their searches for proteins with particular structural features. It can also be used from an engineering perspective to try to design peptides/proteins with desired properties.

What is the approximate molecular weight in kDa of a protein composed of 90 amino acid residues?

_A

7.2 kDa

B

8.1 kDa

C

9.0 kDa

D

9.9 kDa

Solution: The correct answer is D.

1. This result is based on 80 Da per amino acid residue, not 110.

2. This result is based on 90 Da per amino acid residue, not 110.

3. This result is based on 100 Da per amino acid residue, not 110.

4. In a protein, an average amino acid residue is 110 Da in molecular weight. Thus, 90 × 110 = 9900 Da or 9.9 kDa.

Which type of turn in a protein has a hydrogen bond between the i and i + 5 end residues of the turn?

_

A

π

B

γ

C

β

D

α

Solution: The correct answer is A.

1. A π-turn has a hydrogen bond between the i and i + 5 end residues of the turn.

2. A γ-turn has a hydrogen bond between the i and i + 2 end residues of the turn.

3. A β-turn has a hydrogen bond between the i and i + 3 end residues of the turn.

4. An α-turn has a hydrogen bond between the i and i + 4 end residues of the turn.

Gay Bitches Ate Pie

Gamma -> Beta -> Alpha -> Pi

i+2 -> i+3 -> i+4 -> i+5

his actually makes sense if you think about the fact that it's a distance between two points on different residues that have 3 residues between them and that the point at which the hydrogen bonds occurs does not include the entire span of the residues, hence the decimal! One refers to distance between hydrogen bonds (an approximate, measurable distance), the other refers to the relative positions of the residues themselves (a discrete, countable quantity).

i”th residue and the amino group of the “i + n”th residue. So for example, if some secondary structure has turns at i + 4, then the carboxyl group of amino acid 1 would hydrogen bonds with the amino group of amino acid 5. This bonding pattern allows the polypeptide to adopt a 3D shape which looks like our familiar alpha helix! (But this pattern applies for both right-handed and left-handed alpha helices). Similarly, the beta turns also have a specific pattern and all of those others that we don’t really talk about but occur in nature.

Hopefully that makes sense, but if not you should check out the picture of the alpha helix atA in Alpha looks like a 4

B for Beta has a 3 in it

γ looks mike someone holding up 2 fingers

Then "pi" and "five" semi-rhyme.

"i" and the amide of amino acid "i + number". The smaller the number, the tighter the turns are. 𝛾 turns are the tightest, Pi turns are the "loosest".

What is the percentage of the V_max attained for a single-substrate enzyme following Michaelis–Menten kinetics at [S]₀ = 4 × K_M?

_

A

83% V_max 

B

80% V_max 

C

75% V_max 

D

67% V_max 

Solution: The correct answer is B.

1. This rate is for [S]₀ = 5 × K_M.

2. For [S]₀ = 4 × K_M, V₀ = (V_max ×  4 × K_M)/(4 × K_M + K_M) = 0.8 × V_max  or 80% V_max.

3. This rate is for [S]₀ = 3 × K_M.

4. This rate is for [S]₀ = 2 × K_M.

Recall the Michaelis-Menten equation: V0 = (Vmax * [S])/(Km + [S]).

The main thing here is when they ask for the "percentage of Vmax", they're asking for V0 in terms of Vmax, since Vmax is always 100%, or the theoretical maximum reaction rate. This means they are asking for the term V0/Vmax.

Rearranging for this term, we get V0/Vmax = [S] / (Km + [S]).

Plugging in our S0 = 4*Km, we get V0/Vmax = 4*Km / (Km + 4*Km) = 4Km/5Km = 0.8

Thus, at t = 0, V0 = 0.8 Vmax, or 80% of Vmax.

Just plug it into the michaelis menten equation:

V = Vmax * (S/(S+Km))

V = Vmax * (4Km/(4Km+Km))

V = Vmax * (4Km/5Km)

V = Vmax *(4/5)

V = 80% Vmax

Researchers studied the occurrence of cardiovascular injury in patients exposed repeatedly to therapeutic electromagnetic radiation in the form of high-energy X-rays of frequency 1.5 × 10¹⁷ Hz. Researchers performed computed tomography (CT) scans using low-energy X-rays of wavelength 10 nm in air. The X-rays were collimated by passing through an opening of width D = 1.0 μm and length L = 10.0 μm made in a copper plate. The X-rays were absorbed selectively by the irradiated heart tissue based on the atomic number of the chemical species present in the scanned area. X-ray detectors placed in a circular pattern around the patient's heart collected the X-rays, which passed through the tissues. Based on the number of X-ray photons collected, each detector generated an electric current (typical intensity 15 μA) that was sent to a computer, which created a composite image from all detectors. Data revealed that calcium-containing plaques were formed in the left ventricle of patients subjected to an absorbed dose per unit time in excess of 0.04 Gy/s. The unit of 1 Gy represents one joule of radiation energy absorbed by one kilogram of absorbent material.

What is the gap between the atomic energy levels that correspond to the emission of therapeutic high-energy X-rays?

(Note: Use h = 6.6 × 10^–34 J•s.) 

A

5.1 × 10^–14 J

B

6.6 × 10^–15 J

C

8.1 × 10^–16 J

D

9.9 × 10^–17 J

Solution: The correct answer is D.

1. This is consistent with calculating the energy gap as (6.6 – 1.5) × 10^–14 J without scientific basis. 

2. This implies the X-rays' frequency is 1.0 × 10¹⁹ Hz.

3. This is consistent with calculating the energy gap as (6.6 + 1.5) × 10^–16 J without scientific basis. 

4. The gap between the atomic energy levels that correspond to the emission of high-energy X-rays used in the experiment is equal to the energy of an X-ray photon of frequency 1.5 × 10¹⁷ Hz, that is 6.6 × 10^–34 J⋅s × 1.5 × 10¹⁷ Hz = 9.9 × 10^–17 J. 

• he gap between the atomic energy levels that correspond to the emission of high-energy X-rays used in the experiment is calculated as the energy of an X-ray photon of a given frequency using the Planck-Einstein relation,

  E=hfcap E equals h f

  𝐸=ℎ𝑓

  .

• Using the given values:

• h(Planck's constant)

𝑓(frequency) =1.5×1017Hz1.5 cross 10 to the 17th power space Hz

=

6.6×10-34J⋅s×1.5×1017Hz=9.9×10-17J

6.6×10−34J⋅s×1.5×1017Hz=𝟗.𝟗×𝟏𝟎−𝟏𝟕J

. 

. 

What best approximates the typical number of electrons generated per second by the X-ray detectors that receive the transmitted low-energy X-rays?

(Note: Use elementary charge 1.6 × 10^–19 C.)

A

8 × 10¹²

B

9 × 10¹³

C

7 × 10¹⁴

D

5 × 10¹⁵

Solution: The corr

Solution: The correct answer is B.

1. This implies the typical current intensity is 8 × 10¹² × 1.6 × 10^–19 C/s = 1.3 μA instead of 15 μA.

2. The electric current of typical intensity 15 μA means there are 15 × 10^–6 C/s = 15 × 10^–6 electrons/(1.6 × 10^–19 s) = 9 × 10¹³ electrons/s.

3. This implies the typical current intensity is 7 × 10¹⁴ × 1.6 × 10^–19 C/s = 112 μA instead of 15 μA.

4. This implies the typical current intensity is 5 × 10¹⁵ × 1.6 × 10^–19 C/s = 8 mA instead of 15 μA.

Amp = Coulomb/sec

Convert the given value of 15 uA => 15 x 10^-6 C/sec

An elementary charge is the amount of charge carried per electron (elementary charge = C / e-) (e- = number of electrons)

Since we want the number of electrons, we can relate the C/sec and C/e- formulas

(C/sec) / (C/e-) => (C/sec) x (e-/C) ==> Coulomb cancels, thus we get # of electrons per second

Now we plug in values, (15 x 10^-6 C/sec) / (1.6 x 10^-19 C/e-) = 9 x 10^13 e-/sec

Hope this helps :)

hat is the chemical classification of the element detected in the left ventricle when the absorbed dose per unit time exceeded 0.04 Gy/s?

A

Alkali metal

B

Alkaline earth metal

C

Transition metal

D

Halogen

Solution: The correct answer is B.

1. Calcium is not an alkali metal because it is Group 2, not Group 1, of the periodic table.

2. The passage states that calcium plaques occurred in the left ventricle when the high-energy X-ray absorbed dose per unit time exceeded 0.04 Gy/s. Calcium is an alkaline earth metal because it is in Group 2 of the periodic table.

3. The passage does not state any of the transition metals (Groups 3–12 elements in the periodic table) as having occurred in the left ventricle when the high-energy X-ray absorbed dose per unit time exceeded 0.04 Gy/s. 

4. No halogen occurred in the left ventricle when the high-energy X-ray absorbed dose per unit time exceeded 0.04 Gy/s, according to the passage. Halogens are in Group 17 of the periodic table, whereas calcium, the element detected in the left ventricle, is in Group 2.

formed in the left ventricle of patients subjected to an absorbed dose per unit time in excess of 0.04 Gy/s. The unit of 1 Gy represents one joule of radiation energy absorbed by one kilogram of absorbent material.

Adapted from B. Messenger et al., "Coronary Calcium Scans and Radiation Exposure in the Multi-Ethnic Study of Atherosclerosis." Int. J. Cardiovasc. Imaging. ©2015 the Authors.

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Question

Which physical effect constitutes the basis on which the X-ray detectors for CT scan operate?

A

Doppler effect

B

Mass–energy conversion

C

Photoelectric effect

D

Venturi effect

Solution: The correct answer is C.

1. The Doppler effect manifests as a change in the observed frequency of a wave emitted by a source as a result of the relative motion between the source and the observed.

2. Mass–energy conversion refers to the effect that manifests as nuclear fission or fusion.

3. The passage states that detectors absorbed the transmitted X-rays and generated an electric current. This is the description of the photoelectric effect.

4. The Venturi effect refers to the measurement of volume flow rate or flow speed of a fluid based on the difference of static pressure in a tube with variable diameter

What is the effect of a temperature increase on the dimensions of the opening through which the X-rays for CT scan pass?

A

L increases more than D.

B

D increases more than L.

C

Both L and D increase by the same amount.

D

L increases by the same amount that D decreases.

Solution: The correct answer is A.

1. The linear increase in the length of the opening is ΔL = LαΔT and in the width is ΔD = DαΔT. Because L > D, then ΔL > ΔD. 

2. This implies the length is smaller than the width before the temperature increase, whereas the passage states that width D = 1.0 μm and length L = 10.0 μm. 

3. This implies the length is equal to the width before the temperature increase, whereas the passage states that width D = 1.0 μm and length L = 10.0 μm. 

4. Both L and D increase because of thermal expansion.

s just a matter of the L being greater than D. Both are subject to expansion, but since L is larger, the expansions is greater. If they were the same value, then your answer would have been correct.

epth expansion is the same as length expansion, it's just referring to a different measurement. Since L is a larger value than D, it will increase in length more when heated because both variables are directly proportional to the change in temperature.

What is the minimum radiation energy absorbed in two seconds by 20 grams of heart tissue that is associated with the formation of plaques? 

A

1.2 mJ

B

1.4 mJ

C

1.6 mJ

D

1.8 mJ

Solution: The correct answer is C.

1. This implies the exposure time is 26 s instead of 20 s.

2. This is the energy absorbed by 17 grams of tissue.

3. The minimum absorbed dose per unit time that causes plaques is 0.04 Gy/s = 4 × 10^–2 J/(kg⋅s). A 20 g mass tissue must absorb energy at a rate of 4 × 10^–2 J/(kg⋅s) × 20 × 10^–3 kg = 8 × 10^–4 J/s. The energy absorbed in 2 s is (8 × 10^–4 J/s) × 2 s = 1.6 mJ.

4. This implies the plaques were formed in the left ventricle when the absorbed dose per unit time exceeded 0.04 Gy/s.

What is the classification of neutral atoms of the chemical element found in plaques in the left ventricle based on its magnetic properties?

A

Diamagnetic

B

Ferromagnetic

C

Nonmagnetic

D

Paramagnetic

olution: The correct answer is D.

1. Calcium has the electronic structure [Ar] 4s². The presence of the empty d orbital allows an electron from the last occupied s orbital to acquire energy in the presence of a magnetic field and occupy the empty d orbital. This unpairs the last electrons and causes paramagnetic properties as opposed to diamagnetic.

2. Fe, Ni, and Co are the primary ferromagnetic materials of which none are found in the plaques.

3. Calcium has magnetic properties when exposed to an external magnetic field.

4. Calcium is the chemical element found in plaques in the left ventricle. Calcium has the electronic structure [Ar] 4s². The presence of the empty d orbital can cause an electron from the last occupied s orbital to acquire energy in the presence of a magnetic field and jump on the empty d orbital. This creates an energy state with unpaired electrons [Ar] 4s¹ 3d¹ that has paramagnetic properties.

paramagentism

Diamagnetic is characterized by filled quantum orbitals. A filled orbital ensures electrons have opposite 'spin' and therefore are relatively inert with respect to magnetics, and exposure to an applied magnetic field will have a repulsive effect because it cant respond to the field, but would like to maintain it's inert properties, so it repells. Somewhat analogous to polar and nonpolar interactions.

Paramagnetic has partially filled quantum orbitals which allows it to be influenced by an applied magnetic field, due to the augmenting effects of electron spin. This is weak though, because orbitals are just a marginal part of magnetism.

I’ve been reviewing and I’ve found what seem to be inconsistencies and I’m just hoping for some clarification. I understand that diamagnetic elements have all electrons paired and that paramagnetic elements have one or more unpaired electrons.