1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
gladu [14]
3 years ago
7

Polyglutamate also exhibits pH-dependent unfolding. This transition occurs at a pH of about 7; below that pH the polypeptide cha

in is predominantly a-helical. Why is this transition for polyglutamate promoted by lowering the pH while the opposite applies to polylysine
Chemistry
1 answer:
Novosadov [1.4K]3 years ago
5 0

Answer:

Explanation:

As the pH drops, the NH2groups on the lysine side chains become charged and helices can no longer form because of charge repulsion between these groups. This might occur below the pKa of lysine if more than 50% of the lysine residues is to be charged in order to ‘break’ the helix. Another possibility is that the pKa of lysine residues might be different when in polylysine as compared with the monomer (free amino acid) in solution.

One will expect other residues that are positively charged at neutral pH to have a similar profile; namely, arginine and possibly histidine. Both arginine and histidine are bulkier than lysine. Even if there were some rotation of their side chains, steric interference would probably be so severe as to prohibit the formation of an -helix. The transition is inverted because at a low pH glutamate will be neutral whereas at a high pH it will assume a net negative charge (through dissociation of the carboxyl groups on its side chains). One will easily speculate that a polypeptide chain containing both glutamate and lysine residues will be able to form an helix at relatively neutral pHs. Under these conditions, lysines will be mostly positively charged and glutamates will be mostly negatively charged. This will allow these residues to make ionic bonds and salt bridges to stabilize the helix. At very low pH, however, lysine will be mostly positively charged, but it will be near to neutral glutamate residues. At very high pH, the Glu will be negatively charged, but it will be near neutral.

You might be interested in
You have a balloon containing 1 L of air at STP in a vacuum chamber. What will the volume of the balloon be when you reduce the
Ipatiy [6.2K]
P1V1/T1=P2V2/T2   P2=1/2 P1 V1=1 T1=298K

1 P1/298= (1/2) P1V2/373  cross P1

1/298=1/2V2/373
1/298=1/V2 746
v2=746/298
V2=2.5L


5 0
3 years ago
Determine the molality of a solution of methanol dissolved in ethanol for which the mole fraction of methanol is 0.135. Give you
Alja [10]

<u>Answer:</u> The molality of the solution is 0.11 m

<u>Explanation:</u>

We are given:

Mole fraction of methanol = 0.135

This means that 0.135 moles of methanol is present in 1 mole of a solution

Moles of ethanol = 1 - 0.135 = 0.865 moles

To calculate the mass for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of ethanol = 0.865 moles

Molar mass of ethanol = 46 g/mol

0.865mol=\frac{\text{Mass of ethanol}}{46g/mol}\\\\\text{Mass of ethanol}=(0.865mol\times 46g/mol}=39.79g

To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

Where,

m_{solute} = Given mass of solute (methanol) = 0.135 g

M_{solute} = Molar mass of solute (methanol) = 32 g/mol

W_{solvent} = Mass of solvent (ethanol) = 39.79 g

Putting values in above equation, we get:

\text{Molality of methanol}=\frac{0.135\times 1000}{32\times 39.79}\\\\\text{Molality of methanol}=0.106m\approx 0.11m

Hence, the molality of the solution is 0.11 m

6 0
3 years ago
How can u separate saturated, unsaturated and supersaturated solution?​
Kruka [31]

Answer:

Saturated = The solution cannot dissolve any more solute at a given temperature

2) Unsaturated = solution can dissolve more solute at a given temperature.

3) Supersaturated = Solution which has more solute than its saturated solution

Explanation:

<h2><em><u>Mark </u></em><em><u>me </u></em><em><u>brainlist</u></em></h2>
4 0
2 years ago
The data showed that recently the alligator population has decreased. How could the decrease in the alligator
Marina CMI [18]

Answer:

Since the alligator population decreased, this means there are less predators for the gar. Less predators for the gar mean the gar population will increase.  More gar also means more competition for the diving duck in terms of animals that eat crabs, and thus the diving duck population will also decrease.

Explanation:

Hope this helped <3

4 0
2 years ago
One mole of a monatomic ideal gas is subjected to the following sequence of steps: a. Starting at 300 K and 10 atm, the gas expa
Verdich [7]

Answer:

a) Q = 0; W = 0; ΔU = 0; ΔH = 0; ΔS = 0.09 atm.L/K

b) Q = 1250 J; W = 0; ΔU = 1250 J; ΔH = 1250 J; ΔS = -0.0235 atm.L/K

c) Q = 3653.545 J; W = - 3653.545 J; ΔU = 0; ΔH = 0; ΔS = - 3653.545 J

d) Q = - 2080 J; W = 830 J; ΔU = - 1250 J; ΔH = - 2080 J; ΔS = - 5.984 J/K

Explanation:

a) If there is a vacuum, the work is zero, as it is a free expansion, the volume increases, the pressure decreases, the temperature is constant and the internal energy is constant.

∴ n = 1 mole

∴ PV = RTn....ideal gas

∴ P1 = 10 atm

∴ R = 0.082 atm.L/K.mol

∴ T = 300 K = T2

∴ V2 = 3*V1

⇒ W = 0.....expands freely into vacuum

⇒ ΔU = Q = 0....first law

⇒ ΔS = -  nR Ln(P2/P1).....ideal gas

∴ V1*P1/T1 = V2*P2/T2

∴ T1 = T2 = 300 K

⇒ P2 = V1*P1 / V2 = V1*P1 / 3V1 = 10 atm/3 = 3.33 atm

⇒ ΔS = - (1mol)*(0.082 atm.L/K.mol) Ln ( 3.33/10)

⇒ ΔS = 0.09 atm.L/K

∴ ΔH = ΔU + (P2V2 - P1V1) = 0 + 0 = 0

b) heated reversibly at constant volume:

⇒ W = 0 ...at constant volume

∴ T2 = 400 K; T1 = 300 K

∴ V1 = V2

⇒ Q = ΔU = CvΔT....first law

∴ Cv = 12.5 J/K.mol.....monoatomic ideal gas

∴ ΔT = 400 - 300 = 100 K

⇒ Q = ΔU = 12.5 J/mol.K * 100K = 1250 J/mol * 1 mol = 1250 J

∴ ΔH = ΔU + PΔV = ΔU + 0 = 1250 J

∴ ΔS = - nR Ln (P2/P1)

∴ P2/T2 = P1/T1...constant volume

∴ P1 = 3.33 atm

⇒ P2 = P1*T2 / T1 = (3.33 atm)*(400K) / (300K) = 4.44 atm

⇒ ΔS = - (1mol)*(0.082atm.L/K.mol) Ln (4.44/3.33)

⇒ ΔS = - 0.0235 atm.L/K

c) reversibly expanded at constant temperature:

∴ T1 = T2 = 400K

∴ V2 = 3*V1

∴ ΔU = 0...constant temperature

⇒ Q = - W....fisrt law

∴ W = - ∫ PdV..... reversibly expansion

∴ P = nRT/V... ideal gas

⇒ W = - nRT ∫ dV/V

⇒ W = - nRT Ln (V2/V1)

⇒ W = - (1mol)*(8.314 J/K.mol) Ln (3)

⇒ W = - 9.134 J/K *400K = - 3653.545 J

⇒ Q = - W = 3653.545 J

⇒ ΔH = ΔU + P1V1 - P2V2 = 0 + nRT1 - nRT2 = 0 + 0 = 0

∴ ΔS = - nR Ln(P2/P1)

∴ P1 = 4.44 atm

⇒ P2 = V1*P1*T2/ V2*T1 = V1*(4.44atm)*(400K) / (3.V1)*(400K)

⇒ P2 = 4.44atm/3 = 1.48 atm

⇒ ΔS = - (1mol)*(8.314 J/mol.K) Ln (1.48/4.44)

⇒ ΔS = -9.134J/K * 400K = - 3653.545 J

d) reversibly cooled at constant pressure:

∴ T2 = 300 K;  T1 = 400 K

∴ P2 = P1

⇒ Q = ΔH = CpΔT

∴ Cp = 20.8 J/K.mol

∴ ΔT = 300 - 400 = - 100 K

⇒ Q = ΔH = 20.8 J/mol.K * ( -100K) = - 2080 J/mol * 1mol = - 2080 J

⇒ ΔU = nCvΔT = (1mol)*(12.5 J/mol.K)*( - 100K) = -1250 J

⇒ W = ΔU - Q = ΔU - ΔH = -1250 J - ( - 2080 J ) = 830 J

∴ ΔS = ∫ δQ/T = ∫ nCpdT/T

⇒ ΔS = nCp Ln (T2/T1)

⇒ ΔS = (1mol)*(20.8 J/mol.K) Ln (300/400) = - 5.984 J/K

7 0
3 years ago
Other questions:
  • Which force causes the rock materials to move from A to B A.Electricity
    11·1 answer
  • A material will float on the surface of a liquid if the material has a density less than that of the liquid. Given that the dens
    5·1 answer
  • Please answer number 76,77 and 78.
    15·1 answer
  • The equilibrium constant for the gas phase reaction
    9·1 answer
  • How does carbon 14 work
    13·1 answer
  • Serena is mixing a material into a beaker filled with a liquid. She notices that the material seems to disappear into the liquid
    5·1 answer
  • Which of the following is an sxample of using physical capital to save time and money
    15·1 answer
  • . Which one is formed by the combination of many tissues?​
    6·2 answers
  • How many moles are equal to 3.0x10^23 atoms of germanium? PLS ANSWER WITH WORK!
    7·1 answer
  • Chase had determined the boiling point of an unknown liquid to be 45.2 ⁰C when the correct boiling point is 44.32 ⁰C. What is th
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!