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Hint: Glycolysis is the first step of cellular respiration.It is common to both aerobic and anaerobic respiration.
Complete answer:
To answer this question, we must understand the regulation of glycolysis:
Glycolysis is a ten-step reaction that produces energy from glucose in the absence of oxygen. The location of glycolysis is cytoplasm of the cell. It degrades the six-carbon (6-C) glucose molecule to three-carbon (3-C) pyruvate molecule.
In the first step, phosphorylation takes place. In this step, glucose is converted to glucose-6-phosphate by transfer of phosphate group taken from ATP. This reaction is catalyzed by hexokinase. This enzyme is inhibited by high levels of glucose-6-phosphate. This means high levels of G-6-P inhibit hexokinase. This causes the cell to store glucose.
In the third step of glycolysis, fructose-6-phosphate (F6P) is converted into fructose-1,6-bisphosphate (F1,6BP). This reaction is catalyzed by phosphofructokinase (PFK). Here the phosphate group from ATP to fructose-6-phosphate. This produces fructose-1,6-bisphosphate (F1,6BP). This enzyme is inhibited by high levels of ATP, high levels of citrate. High levels of adenosine monophosphate (AMP) and adenosine diphosphate (ADP) activate this enzyme. This is the slowest step and thus rate limiting step.
In the tenth and last step of glycolysis, phosphoenolpyruvate (PEP) is converted to pyruvate by the enzyme pyruvate kinase. High levels of ATP inhibit this enzyme. This leads to a negative feedback for the enzyme pyruvate kinase.
Hence all these three steps set the pace of glycolysis.
> Thus the correct answer is option b) - Fructose-6-phosphate to fructose-1,6-diphosphate.
Note: Although phosphofructokinase (PFK) catalyzed step is considered as the rate-limiting step yet all these steps regulate glycolysis and thus set the pace of it. Many sources include all the three steps as rate-limiting steps.
Complete answer:
To answer this question, we must understand the regulation of glycolysis:
Glycolysis is a ten-step reaction that produces energy from glucose in the absence of oxygen. The location of glycolysis is cytoplasm of the cell. It degrades the six-carbon (6-C) glucose molecule to three-carbon (3-C) pyruvate molecule.
In the first step, phosphorylation takes place. In this step, glucose is converted to glucose-6-phosphate by transfer of phosphate group taken from ATP. This reaction is catalyzed by hexokinase. This enzyme is inhibited by high levels of glucose-6-phosphate. This means high levels of G-6-P inhibit hexokinase. This causes the cell to store glucose.
In the third step of glycolysis, fructose-6-phosphate (F6P) is converted into fructose-1,6-bisphosphate (F1,6BP). This reaction is catalyzed by phosphofructokinase (PFK). Here the phosphate group from ATP to fructose-6-phosphate. This produces fructose-1,6-bisphosphate (F1,6BP). This enzyme is inhibited by high levels of ATP, high levels of citrate. High levels of adenosine monophosphate (AMP) and adenosine diphosphate (ADP) activate this enzyme. This is the slowest step and thus rate limiting step.
In the tenth and last step of glycolysis, phosphoenolpyruvate (PEP) is converted to pyruvate by the enzyme pyruvate kinase. High levels of ATP inhibit this enzyme. This leads to a negative feedback for the enzyme pyruvate kinase.
Hence all these three steps set the pace of glycolysis.
> Thus the correct answer is option b) - Fructose-6-phosphate to fructose-1,6-diphosphate.
Note: Although phosphofructokinase (PFK) catalyzed step is considered as the rate-limiting step yet all these steps regulate glycolysis and thus set the pace of it. Many sources include all the three steps as rate-limiting steps.
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