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kondaur [170]
3 years ago
14

6. Lysozyme is an enzyme that hydrolyzes sugar linkages in the bacterial cell wall and was first discovered by Alexander Fleming

. It is widely used throughout the food industry as a preservative to prevent food spoilage. To improve its stability, disulfides were introduced into lysozyme. Two variants were made, one with a disulfide between residues 3 and 97, and a second with a disulfide between residues 21 and 142. Both have increased stability, with ΔTm = +7 °C (Cys3-Cys97) and ΔTm = +10 °C (Cys21-Cys142). Explain why the thermal stability increases. Assuming both disulfides have the same bond strength, explain why the Cys21-Cys142 has a higher Tm than the Cys3-Cys97 variant. Hint: ΔTm is the temperature at which half the protein is in its unfolded state. Think in terms of the entropy difference between the native and mutants and what that does to the free energy state of the protein (6 points).
Chemistry
1 answer:
balu736 [363]3 years ago
7 0

Answer:

See explaination

Explanation:

The Cys3-cys97 and cys21-cys142 disulfides restrict the unfolded state of lysozyme enzyme to a class of more compact structures with a less exposed hydrophobic surface, compared to the unfolded states of reduced/non-crosslinked lysozyme. there are 2 major factors which lead to the stabilization of lysozyme due to disulfide bonds-

1- increase in the loop size due to the formation of disulfide bonds that leads to an increase in the even entropic effect.

2- the region formed should be flexible. the strain energy due to the formation of the disulfide bond is lower.

cys21-cys142 has a higher Tm than the cys3-cys97 because it involves flexible parts of the molecule. 21 and 142 residues are located on opposite sides of the active-site cleft where significant hinge-bending motion is seen. this introduces minimal strain in the protein.

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A buffer solution contains 0.306 M C6H5NH3Br and 0.418 M C6H5NH2 (aniline). Determine the pH change when 0.124 mol HCl is added
Ulleksa [173]

<u>Answer:</u> The pH change of the buffer is 0.30

<u>Explanation:</u>

To calculate the pH of basic buffer, we use the equation given by Henderson Hasselbalch:

pOH=pK_b+\log(\frac{[\text{conjugate acid}]}{[\text{base}]})

pOH=pK_b+\log(\frac{[C_6H_5NH_3^+]}{[C_6H_5NH_2]})        .....(1)

We are given:

pK_b = negative logarithm of base dissociation constant of aniline  = 9.13

[C_6H_5NH_3^+]=0.306M

[C_6H_5NH_2]=0.418M

pOH = ?

Putting values in equation 1, we get:

pOH=9.13+\log(\frac{0.306}{0.418})\\\\pOH=8.99

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

pH+pOH=14\\pH_{initial}=14-8.99=5.01

To calculate the molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles hydrochloric acid solution = 0.124 mol

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of HCl}=\frac{0.124}{1L}\\\\\text{Molarity of HCl}=0.124M

The chemical reaction for aniline and HCl follows the equation:

                   C_6H_5NH_2+HCl\rightarrow C_6H_5NH_3^++Cl^-

<u>Initial:</u>           0.418        0.124           0.306

<u>Final:</u>             0.294          -                0.430

Calculating the pOH by using using equation 1:

pK_b = negative logarithm of base dissociation constant of aniline  = 9.13

[C_6H_5NH_3^+]=0.430M

[C_6H_5NH_2]=0.294M

pOH = ?

Putting values in equation 1, we get:

pOH=9.13+\log(\frac{0.430}{0.294})\\\\pOH=9.29

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

pH+pOH=14\\pH_{final}=14-9.29=4.71

Calculating the pH change of the solution:

\Delta pH=pH_{initial}-pH_{final}\\\\\Delta pH=5.01-4.71=0.30

Hence, the pH change of the buffer is 0.30

8 0
3 years ago
A hydrogen-like ion is an ion containing only one electron. The energies of the electron in a hydrogen-like ion are given by the
yan [13]

Answer:

The ionization energy (in kJ/mol) of the helium ion is 21,004.73 kJ/mol .

Explanation:

E_n = -(2.18 10-18 J)\times \frac{Z^2}{n^2}

Z = atomic mass

n = principal quantum number

Energy of the electron in n=1,

E_1= -(2.18 10^{-18} J)\times \frac{4^2}{1^2}=-3.488\times 10^{-17} J

Energy of the electron in n = ∞

E_{\infty}= -(2.18 10^{-18} J)\times \frac{2^2}{\infty ^2}=0 J

Ionization energy of the He^+ ion:

I.E=E_{infty}-E_1=0-(-3.488\times 10^{-17} J)=3.488\times 10^{-17} J

I.E=3.488\times 10^{-20} kJ

To convert in into kj/mol multiply it with N_A=6.022\times 10^{23} mol^{-1}

I.E=3.488\times 10^{-20} kJ\times 6.022\times 10^{23} mol^{-1}=21,004.73kJ/mol

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2 years ago
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2 years ago
Each of the following values was read on an instrument of measuring device. In each case the last digit was estimated. Tell what
Drupady [299]

Answer:

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Explanation:

6 0
2 years ago
What is the ph of a buffer prepared by adding 0.809 mol of the weak acid ha to 0.608 mol of naa in 2.00 l of solution? The disso
Anarel [89]

The pH of the buffer is 6.1236.

Explanation:

The strength of any acid solution can be obtained by determining their pH. Even the buffer solution strength of the weak acid can be determined using pH. As the dissociation constant is given, we can determine the pKa value as the negative log of dissociation constant value.

pKa=-log[H] = - log [ 5.66 * 10^{-7}]\\ \\pka = 7 - log (5.66)=7-0.753=6.247\\\\pka = 6.247

The pH of the buffer can be known as

pH = pK_{a} + log[\frac{[A-]}{[HA]}}]

The concentration of [A^{-}] = Moles of [A]/Total volume = 0.608/2 = 0.304 M\\

Similarly, the concentration of [HA] = \frac{Moles of HA}{Total volume} = \frac{0.809}{2} = 0.404

Then the pH of the buffer will be

pH = 6.247 + log [ 0.304/0.404]

pH = 6.247 + log 0.304 - log 0.404=6.247-0.517+0.3936=6.1236

So, the pH of the buffer is 6.1236.

5 0
3 years ago
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