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andreyandreev [35.5K]
1 year ago
9

How is phosphorylation of glyceraldehyde 3-phosphate in the payment phase of glycolysis different from phosphorylation of glucos

e in the preparatory phase?
Chemistry
1 answer:
svet-max [94.6K]1 year ago
7 0

In the preparatory phase of glycolysis, two molecules of ATP are invested and the hexose chain is cleaved into two triose phosphates. During this, the phosphorylation of glucose and its conversion to glyceraldehyde-3-phosphate take place.  During this phase, the conversion of glyceraldehyde-3-phosphate to pyruvate and the coupled formation of ATP take place. Because Glucose is split to yield two molecules of D-Glyceraldehyde-3-phosphate, each step in the payoff phase occurs twice per molecule of glucose.

Glyceraldehyde 3-phosphate dehydrogenase Simultaneous oxidation and phosphorylation of G3P produce 1,3-bisphosphoglycerate (1,3-BPG) and nicotine adenine dinucleotide (NADH).

The divalent cation also affected the response of the enzyme from the endosperm and shoots to adenine nucleotides and inorganic pyrophosphate.

This phase is also called the glucose activation phase. In the preparatory phase of glycolysis, two molecules of ATP are invested and the hexose chain is cleaved into two triose phosphates. During this, the phosphorylation of glucose and its conversion to glyceraldehyde-3-phosphate take place. Steps 1, 2, 3, 4, and 5 together are called the preparatory phase.

For more information on phosphorylation click on the link below:

brainly.com/question/7465103

#SPJ4

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Please help, I really don’t understand this!!!
kap26 [50]

<u>Analysing the Question:</u>

We are given the balanced equation:

C₆H₁₂O₆ + 6O₂→ 6CO₂ + 6H₂O

from this equation, we can say that: <em>for every 1 mole of Glucose, we need 6 moles of Oxygen</em>

<u>Moles of Glucose used in the reaction:</u>

Molar mass of Glucose = 180 grams / mol

Given mass of Glucose = 1 gram

Mole of Glucose = Given mass / Molar mass

Moles of Glucose = 1 / 180 moles

<u>Mass of Oxygen required:</u>

We know that for every mole of Glucose, we need 6 moles of Oxygen

So, for 1/180 moles of Glucose, we need 6 / 180 = 1 / 30 moles of Oxygen

Mass of 1 / 30 moles of Oxygen:

Mass = Molar mass * number of moles

Mass of Oxygen = 32 * 1/30

Mass of Oxygen  = 32 / 30

Mass of Oxygen = 1.06 grams

5 0
3 years ago
How many moles are in 5.6x1025 molecules of calcium oxide?
Vlad1618 [11]

Answer:

<h2>93.02 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{5.6 \times  {10}^{25} }{6.02  \times  {10}^{23} }  \\  = 93.023...

We have the final answer as

<h3>93.02 moles</h3>

Hope this helps you

8 0
3 years ago
What is scientific explanation of the structure of atoms and how they interact with other atoms? (F) be element (g)compound (k)a
Marina86 [1]

Answer:k

Explanation:

Atomic theory would be the theory or explanation of the function of atoms and their relationship with each other.

7 0
3 years ago
Read 2 more answers
What was the result of heating the mixture? All BUT ONE choice is correct.
snow_tiger [21]

Answer:

w gang alright

Explanation:

ay its b alright

4 0
3 years ago
A mixture of compounds containing diethylamine, phenol, ammonia, and acetic acid is separated using liquid-liquid extraction as
umka2103 [35]

Answer:

I would expect to extract the acetic acid.

Explanation:

In the first step, since we are adding a concentrated acid,<u> it will react with the bases present in the mixture (diethylamine and ammonia) </u><u>forming salts</u><u>, </u><u>which are soluble in water</u>. Therefore, after draining the aqueous layer, we will have phenol and acetic acid left in the organic layer.

In the second step, we are adding a diluted base, so it will react with a strong acid. This compound is acetic acid, and its salt will be present in the aqueous layer. Phenol will be left on the organic layer.

7 0
3 years ago
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