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Andre45 [30]
3 years ago
9

Knowing that the protease shows highest activity between pH 4-6, what could be the identity of the base residue acting as a in t

he reaction? Explain and write the expected mechanism for the reaction. b. Assuming that the optimal pH for the reaction strictly depends exclusively on the protonation state of the active site residues at which pH do you expect maximal enzymatic activity? Explain your rationale. c. What is the fraction of deprotonated Cys and protonated B at the calculated optimum pH? d. How these fractions are changing at 1 pH unit below the optimal pH? Which fraction of the enzyme will be active at this pH?
Chemistry
1 answer:
olga_2 [115]3 years ago
8 0

hey there!:

A) Knowing theatre the protease is showing the highest activity at pH 4-6, implies that the amino acid that amino acid that it is acting in is an amino acid with a basic side chain. Therefore, the residues can be any one of the three basic amino acids being histidine, arginine or lysine , having basic side chains at neutral pH.

b) The mechanism of reaction of cysteine proteases is as follows:

 First step in the reaction is the deprotonation of a thiol in the cysteine proteases's active site by an adjacent amino acid with a basic side chain, which might be a histidine residue. This is followed by a nucleophilic attack by the anionic sulfur of the deprotonated cysteine on the substrate carbonyl carbon.

Here, a part of the substrate is released with an amine terminus, restoring the His into a deprotonated form, thus forming a thioester intermediate, forming a link between the carboxy-terminal of the substrate and cysteine, resulting in thiol formation. Thus the name thiol proteases. The thioester bond is then hydrolyzed into a carboxylic acid moiety while again forming the free enzyme.

C) cysteine proteases have a pka of 8-9 but when they are deprotonated by a His residue, their pka would come down to 6-8, which would be their optimal pH for functioning. This is because there is a deprotonation of the thiol group , later restoring the HIS deprotonated form and then formation of a thioester bond. This thioester bond when hydrolysed will a carboxylate moeity , which is responsible for bringing the pH down towards a more acidic side.  

d) at the optimal pH , the fraction of deprotonated cysteine and protonated B will be equal which will change with the change in pH.

Hope this helps!

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The estimated heat of vaporization of diethyl ether using the Chen's rule is A. 29.7 KJ/mol B. 33.5 KJ/mol C. 26.4 KJ/mol D. 36.
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Answer:

C. 26.4 kJ/mol

Explanation:

The Chen's rule for the calculation of heat of vaporization is shown below:

\Delta H_v=RT_b\left [ \frac{3.974\left ( \frac{T_b}{T_c} \right )-3.958+1.555lnP_c}{1.07-\left ( \frac{T_b}{T_c} \right )} \right ]

Where,

\Delta H_v is the Heat of vaoprization (J/mol)

T_b is the normal boiling point of the gas (K)

T_c is the Critical temperature of the gas (K)

P_c is the Critical pressure of the gas (bar)

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For diethyl ether:

T_b=307.4\ K

T_c=466.7\ K

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Applying the above equation to find heat of vaporization as:

\Delta H_v=8.314\times307.4 \left [ \frac{3.974\left ( \frac{307.4}{466.7} \right )-3.958+1.555ln36.4}{1.07-\left ( \frac{307.4}{466.7} \right )} \right ]

\Delta H_v=26400 J/mol

The conversion of J into kJ is shown below:

1 J = 10⁻³ kJ

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\Delta H_v=26.4 kJ/mol

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Explanation:

Hi there!

a) The concentration of the drug in the bottle is 294 mg/ 30.0 ml = 9.8 mg/ml

b) The drug has to be administrated at a dose of 0.0210 mg/ kg body mass. Then, the total mass of drug that there should be in the injection for a person of 70 kg will be:

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<span>Good Morning!
 
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