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DIA [1.3K]
3 years ago
7

Lipoamide is a covalently attached coenzyme that plays a key role in several decarboxylase reactions, including glycine decarbox

ylase, and is attached to a lysine residue in the enzyme.
Review the molecule again. Manipulate it in order to identify the specific lysine residue that serves as the attachment site for lipoamide in the enzyme glycine decarboxylase. You may want to view the hint if you need help.

Choose one:
A. Lysine 17
B. Lysine 8
C. Lysine 118
D. Lysine 63
Chemistry
1 answer:
Reika [66]3 years ago
3 0

Answer:

Lysine 63.

Explanation:

The coenzyme may be defined as the non protein part of an enzyme that plays an important role in the enzyme functioning. Any cation or small molecule like amino acids can acts as coenzyme.

Lipoamide is important coenzyme that plays an important role in the metabolic pathways by undergoes the decarboxylation reactions. The reduced lipoamide known as dihydrolipoamide is used for the covalent attachment at Lysine 63 residues.

Thus, the correct answer is option (D).

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Determine the [OH−] of a solution that is 0.115 M in CO32−. For carbonic acid (H2CO3), Ka1=4.3×10−7 and Ka2=5.6×10−11.
lianna [129]

Answer:

[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.

Explanation:

Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.

Analysis:

            H₂CO₃(aq)     ⇄     H⁺(aq)    +    HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷

C(i)          0.115M                      0                  0

ΔC              -x                        +x                  +x

C(eq)    0.115M - x                   x                    x

            ≅ 0.115M

Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M

= 4.3 x 10⁻⁷  => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.

In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion  concentration, the hydroxide ion concentration is then calculated from

[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.

________________________________________________________

NOTE: The 2.32 x 10⁻⁴M  value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.

4 0
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