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KATRIN_1 [288]
3 years ago
14

What is the fate of glucose 6‑phosphate, glycolytic intermediates, and pentose phosphate pathway intermediates in this cell? Gly

colytic intermediates can only enter the pentose phosphate pathway through conversion to pyruvate and subsequent gluconeogenesis. The oxidative pentose phosphate pathway reaction catalyzed by glucose 6‑phosphate dehydrogenase is slowed down. Most of the glucose 6‑phosphate enters the pentose phosphate pathway. One molecule of glyceraldehyde 3‑phosphate and two molecules of fructose 6‑phosphate are used to generate three molecules of ribose 5‑phosphate. Most of the glucose 6‑phosphate enters the glycolytic pathway and is converted to fructose 6‑phosphate and glyceraldehyde 3‑phosphate. Under the given conditions, all triose phosphates are converted to pyruvate by the glycolytic pathway.
Chemistry
1 answer:
madreJ [45]3 years ago
3 0

Answer:

The Phosphorylated  glucose(glucose +inorganic phosphate), with the energy supplied from ATP hydrolysis formed glucose 6- phosphate, which is later converted to 2 molecules of fructose 6-phosphate- this is phosphorylation.And  represented the fate of  glucose -6-phosphate.

The fructose 6-phosphate are converted to triose phosphate- which is a 2-molecules of 3C compound. The latter is oxidized by NAD→ NADH+ to form intermediates in the glycolytic pathways .

These intermediates are converted to ribose 5-phosphates in the presence of transketolase  and transaldolase enzymes.And they are finally   converted to pyruvate in the glycolytic pathway with the production of 2ATPs per molecule of glucose.

Basically the phosphate pathway reaction is very slow due to enzyme catalysis.

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Step 2:

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Determination of the limiting reactant. To do this, we need to know which of the reactant is excess.

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From the balanced equation above,

98g of H2SO4 required 80g of NaOH.

Therefore, Xg of H2SO4 will require 1.6g of NaOH i.e

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Now comparing the mass of H2SO4 that reacted ( i.e 1.96g) and the mass of H2SO4 given ( i.e 2.94g), we can see clearly that there are left over ( i.e 2.94 - 1.96 = 0.98g) of H2SO4. Therefore, H2SO4 is the excess reactant and NaOH is the limiting reactant.

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Determination of the mass of water produced from the reaction. This is illustrated below:

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80g of NaOH reacted to produced 36g of H2O.

Therefore, 1.6g of NaOH will react to produce = (1.6 x 36)/80 = 0.72g of H2O.

Therefore, the maximum mass of water (H2O) produced by the chemical reaction of aqueous sulfuric acid with solid sodium hydroxide is 0.72g

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