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d1i1m1o1n [39]
3 years ago
13

A series of enzymes catalyze the reactions in the metabolic pathway X → Y → Z → A. Product A binds to the enzyme that converts X

to Y at a position remote from its active site. This binding decreases the activity of the enzyme. With respect to the enzyme that converts X to Y, substance A functions as ________.
Chemistry
1 answer:
Gnom [1K]3 years ago
7 0

The question is incomplete as it does not have the option which are:

A. A coenzyme.

B. An allosteric inhibitor.

C. The substrate.

D. An intermediate.

E. A competitive inhibitor.

Answer:

An allosteric inhibitor.

Explanation:

An allosteric inhibitor is a molecule which binds to the site other than the active site but inhibits the action of the Enzyme and thus the product.

The allosteric inhibitor is considered a non-competitive inhibitor which after binding to the enzyme cause some conformational changes which further changes the structure of the active site and thus substrate no longer binds the enzyme. This molecule thus inhibits the production.

Thus, An allosteric inhibitor is correct.

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4 0
3 years ago
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If a buffer solution is 0.130 M in a weak acid (K_a = 1.7 x 10^-5) and 0.590 M in its conjugate base, what is the pH?
Serggg [28]
Use the Henderson-Hasselbach equation:
pH = pKa + log[base]/[acid]
pH = -log(1.7 x 10^-5) + log(0.590/0.130) = 5.43
8 0
3 years ago
Seawater has a ph of 8.1. what is the concentration of oh–?
lara31 [8.8K]
PH scale is used to determine how acidic or basic a solution is.
pH can be calculated as follows;
by knowing the ph we can calculate pOH
pH + pOH = 14
pOH = 14 - 8.1 
pOH = 5.9 
pOH is used to calculate the hydroxide ion concentration 
pOH = -log[OH⁻]
[OH⁻] = antilog(-pOH)
[OH⁻] = 1.26 x 10⁻⁶ M
therefore hydroxide ion concentration is 1.26 x 10⁻⁶ M
4 0
3 years ago
A and B are two imaginary substances. The molecules of substance A have weak electrical forces between them. The
GrogVix [38]
I believe the answer is A, if they have a stronger connection lower the vapor pressure so higher the boiling point
7 0
1 year ago
Write the complete ionic equations, spectator ions and net ionic equation for the following.
gizmo_the_mogwai [7]

Answer:

Explanation:

1) ZnBr₂ (aq) + AgNO₃ (aq)

Chemical equation:

 ZnBr₂ (aq) + AgNO₃ (aq)  →Zn(NO₃)₂(aq) + AgBr(s)

Balanced chemical equation:

ZnBr₂ (aq) + 2AgNO₃ (aq)  →Zn(NO₃)₂(aq) + 2AgBr(s)

Ionic equation:

Zn²⁺(aq) + Br₂²⁻ (aq) + 2Ag⁺ (aq)+ 2NO⁻₃ (aq)  → Zn²⁺(aq) +(NO₃)₂²⁻(aq) + 2AgBr(s)

Net ionic equation:

Br₂²⁻ (aq) + 2Ag⁺ (aq)   →    2AgBr(s)

The Zn²⁺((aq) and NO⁻₃ (aq) are spectator ions that's why these are not written in net ionic equation. The AgBr can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

2) HgCl₂ (aq) + KI (aq)  →

Chemical equation:

HgCl₂ (aq) + KI (aq)  → KCl + HgI₂

Balanced chemical equation:

HgCl₂ (aq) + 2KI (aq)  → 2KCl(aq) + HgI₂(s)

Ionic equation:

Hg²⁺(aq)  + Cl₂²⁻  (aq) + 2K⁺(aq) + 2I⁻ (aq)  →  HgI₂ (s) + 2K⁺(aq) + 2Cl⁻ (aq)

Net ionic equation:

Hg²⁺(aq)  + 2I⁻ (aq) →   HgI₂ (s)

The Cl⁻ ((aq)  and K⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The HgI₂ (s) can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

 

3) Ca(OH)₂ (aq) + Na₂SO₄ (aq)

Chemical equation:

Ca(OH)₂ (aq) + Na₂SO₄ (aq)  →   CaSO₄(s) + NaOH(aq)

Balanced chemical equation:

Ca(OH)₂ (aq) + Na₂SO₄ (aq)  →   CaSO₄(s) + 2NaOH(aq)

Ionic equation:

Ca²⁺(aq)  + OH₂²⁻  (aq) + 2Na⁺(aq) + SO₄²⁻ (aq)  →   CaSO₄(s) + 2Na⁺(aq) + 2OH⁻ (aq)

Net ionic equation:

Ca²⁺(aq)   + SO₄²⁻ (aq)  →   CaSO₄(s)

The OH⁻ ((aq)  and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The CaSO₄ can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

4 0
3 years ago
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