Glycolysis Carbohydrate Intermediates - Biochemistry

Sucrose is the linking of glucose and fructose. Recall from the glycolytic pathway that fructose is further downstream than glucose, and therefore would allow for faster production of ATP.

First, we must realize that the first committed step is the first irreversible reaction of glycolysis that is unique to glycolysis (cannot lead to another process, such as the pentose phosphate pathway). This is the third step, in which fructose-6-phosphate is converted to fructose-1,6-bisphosphate (the correct answer).

Glucose is the beginning reactant of glycolysis, and pyruvate is the final product. Glucose-6-phosphate is the product of the first step of glycolysis overall, but not of the committed step.

Conversion of 1,3-bisphosphoglycerate to 3-phosphoglycerate is mediated by phosphoglycerate kinase. Conversion of phosphoenolpyruvate to pyruvate is mediated by pyruvate. In both these reactions adenosine diphosphate (ADP) is converted to ATP via substrate level phosphorylation. Conversion of 2-phosphoglycerate to phosphoenolpyruvate, mediated by enolase, does not produce ATP.

As glucose is introduced into the glycolytic pathway, it is first phosphorylated to create glucose-6-phosphate. That will then be converted to fructose-6-phosphate via phosphoglucose isomerase. That product will then be phosphorylated once more via phosphofructokinase-1 to create fructose-1,6-bisphosphate. Glucose-1,6-bisphosphate is never an observed intermediate in glycolysis.

In this question, we're shown a portion of glycolysis. We're asked to identify an intermediate in glycolysis based on the intermediate that comes before it and the one that comes after it.

To answer this, we'll need to know the pathway of glycolysis. The first intermediate shown here, glyceraldehyde-3-phosphate, is acted on by the enzyme glyceraldehyde-3-phosphate dehydrogenase. The product of this reaction is 1,3-bisphosphoglycerate, which is thus the correct answer. This intermediate is then acted on by the enzyme phosphoglycerate kinase to produce 3-phosphoglycerate.

Glucose-6-phosphate (G6P) is the first molecule of the pentose phosphate pathway where it is acted upon by glucose-6-phosphate dehydrogenase. G6P is the result of the hexokinase (first) reaction in glycolysis. What is key here is that the tissue in question is muscle. Because muscle cells lack the glucose-6-phosphatase necessary to produce free glucose from G6P, they cannot be said to perform gluconeogenesis. They do, however, perform glycogenesis through conversion of G6P to glucose-1-phosphate followed by conversion to uridine diphosphateglucose for addition to a growing molecule of glycogen.

Sucrose is the linking of glucose and fructose. Recall from the glycolytic pathway that fructose is further downstream than glucose, and therefore would allow for faster production of ATP.

First, we must realize that the first committed step is the first irreversible reaction of glycolysis that is unique to glycolysis (cannot lead to another process, such as the pentose phosphate pathway). This is the third step, in which fructose-6-phosphate is converted to fructose-1,6-bisphosphate (the correct answer).

Glucose is the beginning reactant of glycolysis, and pyruvate is the final product. Glucose-6-phosphate is the product of the first step of glycolysis overall, but not of the committed step.

Conversion of 1,3-bisphosphoglycerate to 3-phosphoglycerate is mediated by phosphoglycerate kinase. Conversion of phosphoenolpyruvate to pyruvate is mediated by pyruvate. In both these reactions adenosine diphosphate (ADP) is converted to ATP via substrate level phosphorylation. Conversion of 2-phosphoglycerate to phosphoenolpyruvate, mediated by enolase, does not produce ATP.

As glucose is introduced into the glycolytic pathway, it is first phosphorylated to create glucose-6-phosphate. That will then be converted to fructose-6-phosphate via phosphoglucose isomerase. That product will then be phosphorylated once more via phosphofructokinase-1 to create fructose-1,6-bisphosphate. Glucose-1,6-bisphosphate is never an observed intermediate in glycolysis.

In this question, we're shown a portion of glycolysis. We're asked to identify an intermediate in glycolysis based on the intermediate that comes before it and the one that comes after it.

To answer this, we'll need to know the pathway of glycolysis. The first intermediate shown here, glyceraldehyde-3-phosphate, is acted on by the enzyme glyceraldehyde-3-phosphate dehydrogenase. The product of this reaction is 1,3-bisphosphoglycerate, which is thus the correct answer. This intermediate is then acted on by the enzyme phosphoglycerate kinase to produce 3-phosphoglycerate.

First, we must realize that the first committed step is the first irreversible reaction of glycolysis that is unique to glycolysis (cannot lead to another process, such as the pentose phosphate pathway). This is the third step, in which fructose-6-phosphate is converted to fructose-1,6-bisphosphate (the correct answer).

Glucose is the beginning reactant of glycolysis, and pyruvate is the final product. Glucose-6-phosphate is the product of the first step of glycolysis overall, but not of the committed step.

Conversion of 1,3-bisphosphoglycerate to 3-phosphoglycerate is mediated by phosphoglycerate kinase. Conversion of phosphoenolpyruvate to pyruvate is mediated by pyruvate. In both these reactions adenosine diphosphate (ADP) is converted to ATP via substrate level phosphorylation. Conversion of 2-phosphoglycerate to phosphoenolpyruvate, mediated by enolase, does not produce ATP.

As glucose is introduced into the glycolytic pathway, it is first phosphorylated to create glucose-6-phosphate. That will then be converted to fructose-6-phosphate via phosphoglucose isomerase. That product will then be phosphorylated once more via phosphofructokinase-1 to create fructose-1,6-bisphosphate. Glucose-1,6-bisphosphate is never an observed intermediate in glycolysis.

Glucose-6-phosphate (G6P) is the first molecule of the pentose phosphate pathway where it is acted upon by glucose-6-phosphate dehydrogenase. G6P is the result of the hexokinase (first) reaction in glycolysis. What is key here is that the tissue in question is muscle. Because muscle cells lack the glucose-6-phosphatase necessary to produce free glucose from G6P, they cannot be said to perform gluconeogenesis. They do, however, perform glycogenesis through conversion of G6P to glucose-1-phosphate followed by conversion to uridine diphosphateglucose for addition to a growing molecule of glycogen.

Sucrose is the linking of glucose and fructose. Recall from the glycolytic pathway that fructose is further downstream than glucose, and therefore would allow for faster production of ATP.

In this question, we're shown a portion of glycolysis. We're asked to identify an intermediate in glycolysis based on the intermediate that comes before it and the one that comes after it.

To answer this, we'll need to know the pathway of glycolysis. The first intermediate shown here, glyceraldehyde-3-phosphate, is acted on by the enzyme glyceraldehyde-3-phosphate dehydrogenase. The product of this reaction is 1,3-bisphosphoglycerate, which is thus the correct answer. This intermediate is then acted on by the enzyme phosphoglycerate kinase to produce 3-phosphoglycerate.

Conversion of 1,3-bisphosphoglycerate to 3-phosphoglycerate is mediated by phosphoglycerate kinase. Conversion of phosphoenolpyruvate to pyruvate is mediated by pyruvate. In both these reactions adenosine diphosphate (ADP) is converted to ATP via substrate level phosphorylation. Conversion of 2-phosphoglycerate to phosphoenolpyruvate, mediated by enolase, does not produce ATP.

As glucose is introduced into the glycolytic pathway, it is first phosphorylated to create glucose-6-phosphate. That will then be converted to fructose-6-phosphate via phosphoglucose isomerase. That product will then be phosphorylated once more via phosphofructokinase-1 to create fructose-1,6-bisphosphate. Glucose-1,6-bisphosphate is never an observed intermediate in glycolysis.