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.
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Answer:
The volume that the sample of oxygen would occupy at 25 ° C if the pressure were reduced to 760.0 torr is 40.2 L
Explanation:
Boyle's law establishes the relationship between the pressure and the volume of a gas when the temperature is constant, so that the pressure of a gas in a closed container is inversely proportional to the volume of the container. That is, if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.
Boyle's law is expressed mathematically as:
Pressure * Volume = constant
or P * V = k
Considering an initial state 1 and a final state 2, it is true:
P1* V1= P2*V2
In this case:
- P1= 20.1 L
- V1= 1520 torr
- P2= 760 torr
- V2= ?
Replacing:
20.1 L* 1520 torr= 760 torr* V2
Solving:
V2= 40.2 L
<em><u>The volume that the sample of oxygen would occupy at 25 ° C if the pressure were reduced to 760.0 torr is 40.2 L</u></em>
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