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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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Is a theory testable, yes or no?
Paraphin [41]
Yes because a theory is based on results and the results are part of the experiment and it being tested. You have to test the experiment and get results so yes a theory is testable.
5 0
3 years ago
If a gas in a closed container is pressurized from 15.0 atmospheres to 16.0
Yuki888 [10]

Answer:

T₂ = 317.87 K

Explanation:

Given data:

Initial pressure = 15 atm

Final pressure = 16 atm

Initial temperature = 298 K

Final temperature = ?

Solution:

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

15 atm / 298K = 16 atm/T₂

T₂ = 16atm × 298 K / 15 atm

T₂ = 4768 atm. K / 15 atm

T₂ = 317.87 K

4 0
3 years ago
You have 50 ml of a complex mixture of weak acids that contains some HF (pKa = 3.18) and some HCN (pKa = 9.21). Which is larger,
bonufazy [111]

Answer:

\frac{[F^{-}]}{[HF]} is larger

Explanation:

pK_{a}=-logK_{a} , where K_{a} is the acid dissociation constant.

For a monoprotic acid e.g. HA, K_{a}=\frac{[H^{+}][A^{-}]}{[HA]} and \frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}

So, clearly, higher the K_{a} value , lower will the the pK_{a}

In this mixture, at equilibrium, [H^{+}] will be constant.

K_{a} of HF is grater than K_{a} of HCN

Hence, (\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})

So, \frac{[F^{-}]}{[HF]} is larger

5 0
3 years ago
For the reaction 2 s (s) + 3 o2 (g) → 2 so3 (g), how much so3 can be produced from 4 g o2 and excess s? 1. 0.25 mol so3 2. 0.13
igor_vitrenko [27]
The balanced equation for the above reaction is as follows;
2S + 3O₂ --> 2SO₃
Stoichiometry of O₂ to SO₃ is 3:2
O₂ is the limiting reactant and S is provided in excess. since O₂ is the limiting reactant, the whole amount is consumed in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of O₂ moles reacted- 4 g / 32 g/mol = 0.125 mol
3 mol of O₂ forms 2 mol of SO₃
therefore when 0.125 mol of O₂ reacts number of SO₃ moles - 2/3 x 0.125 mol
Number of SO₃ moles formed - 0.0833 mol
Answer is 4) 0.08 mol
8 0
3 years ago
An atom with the same number of protons and electrons has charge of
xz_007 [3.2K]

Answer:

Explanation:

0

All the charges of the protons are cancelled out by all the charges on the electrons.

6 0
2 years ago
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