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Viefleur [7K]
3 years ago
7

Fructose-2,6-bisphosphate is a regulator of both glycolysis and gluconeogenesis for the phosphofructokinase reaction of glycolys

is and the fructose-1,6-bisphosphatase reaction of gluconeogenesis. In turn, the concentration of fructose-2,6-bisphosphate is regulated by many hormones, second messengers, and enzymes.
How do the following affect glycolysis and gluconeogenesis?
Chemistry
1 answer:
azamat3 years ago
8 0

Answer: Glycolysis is stimulated by a high concentration of fructose-2,6-bisphosphate, and the gluconeogenesis is stimulated by a low concentration of fructose-2,6-bisphosphate.

Explanation: Fructose-2, 6-bisphosphate (F2, 6P) is an allosteric activator of the key enzyme in the glycolysis cycle, phosphofructokinase (PFK). F2, 6P also acts as an inhibitor of fructose bisphosphate phosphatase (FBPase) in gluconeogenesis. The concentration of F2, 6P is governed by the balance between its synthesis and breakdown, catalysed by phosphofructokinase-2 (PFK-2) and fructose-bisphosphatase-2 (FBPase-2), respectively. These enzymes are found in a dimeric protein and are controlled by a phosphorylation/dephosphorylation mechanisms. Phosphorylation of the dimeric protein results in an increased concentration of FBPase-2, leading to a decreased concentration of F2, 6P, thus activating the gluconeogenesis cycle. The concentration of PFK-2 is increased when the dephosphorylation of the dimeric protein takes place, leading to the increased concentration of F2, 6P, thus stimulating glycolysis cycle.

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The two types of energy changes that occur are heat and light changes.
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3 years ago
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If a metal rod has a density of 7.85 g/cm and a volume of 4 cm, then what is its mass?
Bond [772]

Answer:

<h3>The answer is 31.4 g</h3>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume of metal = 4 cm³

density = 7.85 g/cm³

The mass is

mass = 7.85 × 4

We have the final answer as

<h3>31.4 g</h3>

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8 0
3 years ago
The equilibrium constant for the reaction 2x(g)+y(g)=2z(g) is 2.25 . what would be the concentration of y at equilibrium with 2
Troyanec [42]

[\text{Y}] \approx0.337\;\text{mol}\cdot\text{dm}^{-3} at equilibrium.

<h3>Explanation</h3>

Concentration for each of the species:

  • [\text{X}] = \dfrac{n}{V} = 2\; \text{mol}\cdot \text{dm}^{-3};
  • [\text{Y}] = \dfrac{n}{V} = 0\; \text{mol}\cdot \text{dm}^{-3};
  • [\text{Z}] = \dfrac{n}{V} = 3\; \text{mol}\cdot \text{dm}^{-3}.

There was no Y to start with; its concentration could only have increased. Let the change in [\text{Y}] be +x \; \text{mol}\cdot \text{dm}^{-3}.

Make a \textbf{RICE} table.

Two moles of X will be produced and two moles of Z consumed for every one mole of Y produced. As a result, the <em>change</em> in [\text{X}] will be +2\;x \; \text{mol}\cdot \text{dm}^{-3} and the <em>change</em> in [\text{Z}] will be -2\;x \; \text{mol}\cdot \text{dm}^{-3}.

\begin{array}{l|ccccc}\textbf{R}\text{eaction}&2\; \text{X}\; (g) & + &\text{Y}\; (g) & \rightleftharpoons &2 \; \text{Z}\; (g)\\\textbf{I}\text{nitial Condition}\; (\text{mol}\cdot\text{dm}^{-3})& 2 & &0 & & 3 \\\textbf{C}\text{hange in Concentration}\; (\text{mol}\cdot\text{dm}^{-3})\;& +2\;x & &+x &&-2\;x\\\textbf{E}\text{quilibrium Condition}\; (\text{mol}\cdot\text{dm}^{-3})& & &&&\end{array}.

Add the value in the C row to the I row:

\begin{array}{l|ccccc}\textbf{R}\text{eaction}&2\; \text{X}\; (g) & + &\text{Y}\; (g) & \rightleftharpoons &2 \; \text{Z}\; (g)\\\textbf{I}\text{nitial Condition}\; (\text{mol}\cdot\text{dm}^{-3})& 2 & &0 & & 3 \\\textbf{C}\text{hange in Concentration}\; (\text{mol}\cdot\text{dm}^{-3})\;& +2\;x & &+x &&-2\;x\\\textbf{E}\text{quilibrium Condition}\; (\text{mol}\cdot\text{dm}^{-3})& 2 + 2\;x & &x&&3-2\;x\end{array}.

What's the equation of K_c for this reaction? Raise the concentration of each species to its coefficient. Products go to the numerator and reactants are on the denominator.

K_c = \dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]}.

K_c = 2.25. As a result,

\dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]} = \dfrac{(3-2x)^{2}}{(2+2x)^{2} \cdot x} = K_c = 2.25.

(3-2\;x)^{2}= 2.25 \cdot(2+2\;x)^{2} \cdot x\\4\;x^{2} - 12 \;x + 9 = 2.25 \;(4\;x^{3} + 8 \;x^{2} + 4 \;x)\\4\;x^{2} - 12\;x + 9 = 9 \;x^{3} + 18\;x^{2} + 9\;x\\9\;x^{3} + 14\;x^{2} + 21\;x - 9 = 0.

The degree of this polynomial is three. Plot the equation y = 9\;x^{3} + 14\;x^{2} + 21\;x - 9 on a graph and look for any zeros. There's only one zero at x \approx 0.337. All three concentrations end up greater than zero.

Hence the equilibrium concentration of Y: 0.337\;\text{mol}\cdot\text{dm}^{-3}.

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=20.0066 moles so round it up to 3 S.F

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To reach 01.kg/L concentration of salt in the tank will take around 18.9 min

Explanation:

Attached

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4 years ago
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