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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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4 years ago
The rate constant for the reaction 3A equals 4b is 6.00×10 how long will it take the concentration of a to drop from 0.75 to 0.2
Lelu [443]

This question is incomplete, the complete question is;

The rate constant for the reaction 3A equals 4B is 6.00 × 10⁻³ L.mol⁻¹min⁻¹.

how long will it take the concentration of A to drop from 0.75 to 0.25M ?

from the unit of the rate constant we know it is a second reaction order

OPTIONS

a) 2.2×10^−3 min

b) 5.5×10^−3 min

c) 180 min

d) 440 min

e) 5.0×10^2 min

Answer:

it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

Explanation:

Given that;

Rate constant K =  6.00 × 10⁻³ L.mol⁻¹min⁻¹

3A → 4B

given that it is a second reaction order;

k = 1/t [ 1/A - 1/A₀]

kt = [ 1/A - 1/A₀]

t = [ 1/A - 1/A₀] / k

K is the rate constant(6.00 × 10⁻³)

A₀ is initial concentration( 0.75 )

A is final concentration(0.25)

t is time required = ?

so we substitute our values into the equation

t = [ (1/0.25) - (1/0.75)] / (6.00 × 10⁻³)

t = 2.6666 / (6.00 × 10⁻³)

t = 444.34 ≈ 440 min     {significant figures}

Therefore it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

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3 years ago
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Explanation:

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