Answer:
80 is the correct answer i think I'm not sure
The phosphorylation of fructose 6-phosphate to fructose-1,6-bisphosphate is the committed step in glycolysis because. it is the rate-limiting step
<h3>What is
phosphorylation?</h3>
The first step in the metabolism of carbohydrates is frequently their phosphorylation. Because the phosphate group stops the molecules from migrating back across the transporter, phosphorylation enables cells to store carbohydrates. Glucose phosphorylation is a crucial step in the metabolism of sugar. In the first phase of glycolysis, D-glucose is converted to D-glucose-6-phosphate using the chemical equation D-glucose + ATP D-glucose-6-phosphate + ADP G° = 16.7 kJ/mol (° signifies measurement under standard conditions).
The rate-limiting stage in the liver's metabolism of glucose is the initial rate of phosphorylation of glucose (ATP-D-glucose 6-phosphotransferase) and non-specific hexokinase. Hepatic cells are freely permeable to glucose (ATP-D-hexose 6-phosphotransferase).
encouraging certain glucose transporters to translocate to the cell membrane.
To learn more about phosphorylation from the given link:
brainly.com/question/2138188
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neutrons to the carbon-12 and carbon-14
The patient should be given 285.71 ml.
1000 ml contains 70 gr glucose.
x contains 20 gr glucose.
x=1000*20/70
Answer:
Pb(NO3)2
Cd(NO3)2
Na2SO4
Explanation:
In the first part, addition of HCl leads to the formation of PbCl2 which is poorly soluble in water. This is the first precipitate that is filtered off.
When the pH is adjusted to 1 and H2S is bubbled in, CdS is formed. This is the second precipitate that is filtered off.
After this precipitate has been filtered off and the pH is adjusted to 8, addition of H2S and (NH4)2HPO4 does not lead to the formation of any other precipitate.
The yellow flame colour indicates the presence of Na^+ which must come from the presence of Na2SO4.
