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1 year ago

The phosphorylation of fructose 6-phosphate to fructose-1,6-bisphosphate is the committed step in glycolysis because?.

1 answer:

7 0

The phosphorylation of fructose 6-phosphate to fructose-1,6-bisphosphate is the committed step in glycolysis because. it is the rate-limiting step

<h3>What is phosphorylation?</h3>

The first step in the metabolism of carbohydrates is frequently their phosphorylation. Because the phosphate group stops the molecules from migrating back across the transporter, phosphorylation enables cells to store carbohydrates. Glucose phosphorylation is a crucial step in the metabolism of sugar. In the first phase of glycolysis, D-glucose is converted to D-glucose-6-phosphate using the chemical equation D-glucose + ATP D-glucose-6-phosphate + ADP G° = 16.7 kJ/mol (° signifies measurement under standard conditions).

The rate-limiting stage in the liver's metabolism of glucose is the initial rate of phosphorylation of glucose (ATP-D-glucose 6-phosphotransferase) and non-specific hexokinase. Hepatic cells are freely permeable to glucose (ATP-D-hexose 6-phosphotransferase).

encouraging certain glucose transporters to translocate to the cell membrane.

To learn more about phosphorylation from the given link:

brainly.com/question/2138188

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You might be interested in

Find the equilibrium value of [CO] if Kc=14.5 : CO (g) + 2H2 (g) ↔ CH3OH (g) Equilibrium concentrations: [H2] = 0.322 M and [CH3

Annette [7]

Answer:

The equilibrium value of [CO] is 1.04 M

Explanation:

Chemical equilibrium is the state to which a spontaneously evolving chemical system, in which a reversible chemical reaction takes place. When this situation is reached, it is observed that the concentrations of substances, both reagents and reaction products, they remain constant over time. That is, the rate of reaction of reagents to products is the same as that of products to reagents.

Reagent concentrations and products in equilibrium are related by the equilibrium constant Kc. Being:

aA + bB ⇔ cC + dD

$Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }$

Then this constant Kces equals the multiplication of the concentrations of the products raised to their stoichiometric coefficients between the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

In this case:

$Kc=\frac{[CH_{3}OH ]}{[CO]*[H_{2} ]^{2} }$

You know:

Kc= 14.5
[H₂]= 0.322 M
[CH₃OH] =1.56 M

Replacing:

$14.5=\frac{1.56}{[CO]*0.322^{2} }$

Solving:

$[CO]=\frac{1.56}{14.5*0.322^{2} }$

[CO]= 1.04 M

The equilibrium value of [CO] is 1.04 M

8 0

3 years ago

A patient in need of treatment for the correction of a malocclusion would visit a/an

wariber [46]

A malocclusion is the t<span>he unusual alignment of the upper and lower teeth. The patient should visit an ORTHODONTIST to get his teeth corrected.</span>

4 0

3 years ago

What determines how much energy an area receives

Lemur [1.5K]

The more focused the rays are, the more energy an area receives, and the warmer it is. The lowest latitudes get the most energy from the Sun. The highest latitudes get the least. The difference in solar energy received at different latitudes drives atmospheric circulation.

7 0

3 years ago

How many moles of HCl would react with 37.1 mL of 0.138 M Sr(OH)2

Lena [83]

Answer is: 0.102 moles of HCl would react.

Balanced chemical reaction:

2HCl(aq) + Sr(OH)₂ → SrCl₂(aq) + 2H₂O(l).

V(Sr(OH)₂) = 37.1 mL ÷ 1000 mL/L.

V(Sr(OH)₂) = 0.0371 L; volume of the strontium hydroxide solution.

c(Sr(OH)₂) = 0.138 M; molarity of the strontium hydroxide solution.

n(Sr(OH)₂) = c(Sr(OH)₂) · V(Sr(OH)₂).

n(Sr(OH)₂) = 0.0371 L · 0.138 mol/L.

n(Sr(OH)₂) = 0.0051 mol; amount of the strontium hydroxide.

From balanced chemical reaction: n(Sr(OH)₂) : n(HCl) = 1 : 2.

n(HCl) = 2 · n(Sr(OH)₂).

n(HCl) = 2 · 0.0051 mol.

n(HCl) = 0.0102 mol; amount of the hydrochloric acid.

5 0

3 years ago

Choose all of the following mole ratios that are correct for the reaction given below. Be sure to choose all that apply.

Dennis_Churaev [7]

Answer:

2 moles NH3 = 1 mole of N2

3 moles of H2 = 1 mole of N2

3 moles of H2 = 2 moles of NH3

Explanation:

The balanced equation for the reaction is given below:

N2 + 3H2 —> 2NH3

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2 to produce 2 moles of NH3.

Thus, we can say that:

1 mole of N2 = 3 moles of H2

1 mole of N2 = 2 moles of NH3

3 moles of H2 = 2 moles of NH3

Thus, considering the options given above, the right answers to the question are:

2 moles NH3 = 1 mole of N2

3 moles of H2 = 1 mole of N2

3 moles of H2 = 2 moles of NH3

5 0

3 years ago

Read 2 more answers