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:
energy flow and nutrient cycles ( photosynthesis , food webs, decomposition webs) sediment transport and soil formation. the water cycle. reproduction/ regeneration mechanisms.
Fluorine has the highest. Fluorine's electronegativity is 4.0
Answer:
Explanation: The chemical reaction is written by writing down the chemical formulas of the reactants on the left hand side and the chemical formulas of products on the right hand side separated by a right arrow.
This is a single displacement reaction in which a more reactive element displaces the less reactive element from its salt solution. Thus sodium is more reactive than Mg and thus displaces it from 
.
The number of atoms of each element must be same on both sides of the reaction so as to follow the law of conservation of mass.
Thus the equation is balanced.
Answer:
Explanation:
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In this case, sine the solution of this problem require the application of the Raoult's law, assuming heptane is a nonvolatile solute, so we can write:
Thus, we first calculate the mole fraction of chloroform, by using the given masses and molar masses as shown below:
Therefore, the partial pressure of chloroform turns out to be:
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