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katrin [286]
4 years ago
9

Some enzymes have one or more sulfhydryl (thiol) groups that are important to enzymatic activity but that can react upon standin

g in solution to form inactive disulfide bonds.
Thiol reagents, such as 1,4-dithiothreitol (DTT), are often added to the solutions of such proteins to protect them from this reaction and to reverse it when it occurs. (The reverse reaction works best at slightly alkaline pH.) Draw the product formed when DTT reacts with a protein disulfide bond to liberate the free thiol groups. Which of the following occurs in this reaction? A. The protein disulfide is oxidized.
B. The protein disulfide is reduced.
C. DTT is reduced.
D. DTT is oxidized.
Chemistry
1 answer:
Jlenok [28]4 years ago
3 0

Answer:

A. Protein disulfide is oxidized.

Explanation:

When thiol reagents are introduced with some protein solutions they react with molecules of disulfide and oxidize the protein. There occurs inter-conversion of thiol molecules into free disulfide molecules. The DTT reduces the disulfide molecules bonds of proteins and it starts to peptide.

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JulsSmile [24]

The correct answer of gibbs free energy is -232 KJ.

ΔG = -nFE° = -2*96485*1.20 = -232 (kJ)

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Which feature do you usually see when tectonics plates move apart?
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A. new ocean ridges and seafloor 
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Octane (C8H18) is a component of gasoline that burns
MatroZZZ [7]

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How many Liters of 0.50M HCl are needed to neutralize 0.050L of 0.101M Ba(OH)2?
Aleksandr-060686 [28]

Answer:

V_{HCl}=5.05x10^{-3}L

Explanation:

Hello,

In this case, since hydrochloric acid and barium hydroxide are in a 2:1 molar ratio, for the neutralization, the following moles equality must be obeyed:

2*n_{HCl}=n_{Ba(OH)_2}

In such a way, in terms of molarities and volumes, we can compute the required volume of hydrochloric acid as shown below:

2*M_{HCl}V_{HCl}=M_{Ba(OH)_2}V_{Ba(OH)_2}\\\\V_{HCl}=\frac{M_{Ba(OH)_2}V_{Ba(OH)_2}}{2M_{HCl}} =\frac{0.101M*0.050L}{2*0.50M} \\\\V_{HCl}=5.05x10^{-3}L

Besr regards.

8 0
3 years ago
Read 2 more answers
The labels have fallen off three bottles containing powdered samples of metals; one contains zinc, one lead, and the other plati
Alinara [238K]

Here we have to identify the metal powder by the given disposal.

The identification of Zinc can be done by 1 m nitric acid (HNO₃) and Ni(NO₃)₂ which will produce hydrogen gas by reaction and displacement reaction as shown below.

Zn + 2 HNO₃ = Zn(NO₃)₂ + H₂ (g)

Zn + Ni(NO₃)₂ = Zn(NO₃)₂ + Ni

The identification of lead can be done by the reaction with 1 m nitric acid (HNO₃) which produces lead nitrate.

The reaction is-

Pb + 4HNO₃ = Pb(NO₃)₂ + 2NO₂ + 2H₂O

The identification of platinum can be done by the reaction with all the given disposal as it will not react with any of the compound.

1. Identification of Zinc (Zn):

(a) Zn metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than zinc.

Zn + NaNO₃ = No reaction

(b) Zn metal will react with 1 m HNO₃ to form hydrogen gas. The reaction is:

Zn + 2 HNO₃ = Zn(NO₃)₂ + H₂ (g)

(c) Zn will react with nickel nitrate [Ni(NO₃)₂] because it may only cause displacement reaction the reduction potential of Zn²⁺/Zn (-0.76) is less than that of Ni²⁺/Ni (-0.23). Thus the reaction will be:

Zn + Ni(NO₃)₂ = Zn(NO₃)₂ + Ni

2. Identification of lead (Pb):

(a) Pb metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than Pb.

Pb + NaNO₃ = No reaction

(b) Pb reacts with HNO₃ to form lead nitrate. The reaction is:

Pb + 4HNO₃ = Pb(NO₃)₂ + 2NO₂ + 2H₂O

(c) The standard reduction potential of Pb²⁺/Pb is more than nickel Ni²⁺/Ni thus there will be no reaction between Pb and NI(NO₃)₂.

Pb + Ni(NO₃)₂ = No reaction.

3. Identification of platinum (Pt)

(a) Pt metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than Pt.

Pt + NaNO₃ = No reaction.

(b) The standard reduction potential of Pt²⁺/Pt is so high (+1.188) thus there will be no reaction with HNO₃.

Pt + HNO₃ = No reaction

(c) The standard reduction potential of Pt²⁺/Pt is more than nickel Ni²⁺/Ni thus there will be no reaction between Pt and Ni(NO₃)₂.

Pt +  Ni(NO₃)₂ = No reaction.    

8 0
3 years ago
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