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.
Answer:
Molar mass = 12.51 g/mol
Explanation:
V = 3L
P = 1.25 atm
T = 20 C = 20 + 273 = 293 K (Upon conversion to kelvin temperature)
n =?
m = 1.95g
Molar mass = ?
The variables are related by the follwing equation;
pV = nRT
where r = gas constant = 0.0821 L atm K−1
Solving for n, we have;
n = pV / RT
n = (1.25 * 3 ) / (0.0821 * 293)
n = 3.75 / 24.0553 = 0.1559 mol
The relationship between number of moles, n and molar mass is given as;
n = mass / molar mass
Molar mass = Mass / n = 1.95 / 0.1559
Molar mass = 12.51 g/mol
Protons, neutrons, and electrons
Answer:
ΔG = - 442.5 KJ/mol
Explanation:
Data Given
delta H = -472 kJ/mol
delta S = -108 J/mol K
So,
delta S = -0.108 J/mol K
delta Gº = ?
Solution:
The answer will be calculated by the following equation for the Gibbs free energy
G = H - TS
Where
G = Gibbs free energy
H = enthalpy of a system (heat
T = temperature
S = entropy
So the change in the Gibbs free energy at constant temperature can be written as
ΔG = ΔH - TΔS . . . . . . (1)
Where
ΔG = Change in Gibb’s free energy
ΔH = Change in enthalpy of a system
ΔS = Change in entropy
if system have standard temperature then
T = 273.15 K
Now,
put values in equation 1
ΔG = (-472 kJ/mol) - 273.15 K (-0.108 KJ/mol K)
ΔG = (-472 kJ/mol) - (-29.5 KJ/mol)
ΔG = -472 kJ/mol + 29.5 KJ/mol
ΔG = - 442.5 KJ/mol
