GRE Subject Test: Biochemistry, Cell, and Molecular Biology : Help with Glycolysis

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Example Question #11 : Help With Glycolysis

How many molecules of pyruvate are produced from one molecule of glucose during glycolysis?

Possible Answers:

Correct answer:

Explanation:

The starting molecule in glycolysis is glucose, a six-carbon molecule while the ending molecule in glycolysis is pyruvate, a three-carbon molecule. During glycolysis glucose is split and its six carbons are used to make 2 molecules of three-carbon pyruvate because. Note that no carbon dioxide is released during glycolysis, but since aerobic metabolism (starting with the Krebs cycle) uses acetyl-CoA as a substrate, which is two carbons long, one molecule of carbon dioxide is released for each molecule of pyruvate produced during glycolysis.

Example Question #12 : Help With Glycolysis

Which of the following is true regarding glycolysis?

Possible Answers:

Glycolysis produces 4 net ATP

The inputs for glycolysis include glycogen and glucose

Glycolysis occurs in every cell

More than one of these are correct

Correct answer:

Glycolysis occurs in every cell

Explanation:

Glycolysis is the first step in producing ATP. Glycolysis is an anaerobic process that occurs in every cell. Certain cells, such as red blood cells, only rely on glycolysis for energy. In most of the other cells, glycolysis produces ATP and few intermediates that will be used in subsequent steps to generate more ATP; therefore, glycolysis occurs in every cell.

The major input for glycolysis is glucose. Glycogen, a storage form of glucose, needs to be broken down into individual glucose units before undergoing glycolysis. The net products of glycolysis are 2 NADH, 2 pyruvate molecules, and 2 ATP. There is a total of 4 ATP produced in glycolysis; however, two of the ATP molecules are consumed, leaving behind only 2 net ATP.

Example Question #13 : Help With Glycolysis

The first step of glycolysis hydrolyzes ATP to ADP and inorganic phosphate. What happens to the glucose molecule during this step?

Possible Answers:

The glucose is cleaved into two molecules of glyceraldehyde-3-phosphate

Glucose is cleaved into two molecules of pyruvate

The glucose is converted to fructose

The glucose is phosphorylated

The glucose is dephosphorylated

Correct answer:

The glucose is phosphorylated

Explanation:

The first step of glycolysis consumes a molecule of ATP, removing one of the phosphate groups to make ADP. This phosphate group is added to glucose to make Glucose-6-phosphate, therefore glucose is phosphorylated.

Example Question #14 : Help With Glycolysis

What enzyme converts glyceraldehyde-3-phosphate to 1, 3-bisphosphoglyceric acid during glycolysis? 

Possible Answers:

Phosphoglycerate kinase 

Glyceraldehyde-3-phosphate dehydrogenase

Pyruvate kinase 

Enolase 

Fructose-bisphosphate aldolase 

Correct answer:

Glyceraldehyde-3-phosphate dehydrogenase

Explanation:

The correct answer is glyceraldehyde-3-phosphate dehydrogenase. This enzyme sequesters a hydrogen from carbon 4 on glyceraldehyde 3-phosphate to reduce  to , and in its place, an inorganic phosphate molecule is transferred to this position. Fructose-bisphosphate aldolase converts Fructose-1, 6-bisphosphate to glyceraldehyde-3-phosphate and dihydroxyacetone phosphate. Pyruvate kinase converts phosphoenolpyruvate to pyruvate whereas enolase converts 2-phosphoglycerate to phosphoenolpyruvate. Phosphoglycerate kinase converts 1, 3-bisphosphoglycerate to 3-phosphoglycerate. 

Example Question #15 : Help With Glycolysis

Alcoholics often present with a deficiency in vitamin B1 (thiamine). What can you conclude about an alcoholic with thiamine deficiency?

I. He/she will not produce any pyruvate and NADH

II. There will be a buildup of lactic acid

III. He/she will not produce any acetyl-CoA from the breakdown of carbohydrates

Possible Answers:

III only

II and III

I and III

I only

Correct answer:

II and III

Explanation:

Thiamine is an important vitamin required for the conversion of pyruvate to acetyl-CoA. It is an important cofactor for the pyruvate dehydrogenase, an enzyme important for the conversion of pyruvate to acetyl-CoA. Pyruvate and NADH from glycolysis will continue to be produced; however, they cannot go any further without thiamine. This means that cells can’t undergo Krebs cycle and oxidative phosphorylation to produce ATP.

The buildup of pyruvate and NADH will cause the pyruvate molecules to undergo fermentation and produce lactic acid. It also will oxidize NADH, the product of which is essential for several cellular processes and needs to be regenerated.

Example Question #16 : Help With Glycolysis

Lactate dehydrogenase inhibitor is added to a mixture of cells. Which of the following molecules will build up inside a cell undergoing aerobic respiration?

Possible Answers:

NADH

Both of these molecules will build up

None of these molecules will build up

Pyruvate

Correct answer:

None of these molecules will build up

Explanation:

The question states that the cell undergoes aerobic respiration. This means that the products from anaerobic respiration (glycolysis) will go through Krebs cycle and electron transport chain (aerobic respiration) to generate ATP. Lactate dehydrogenase is an enzyme important for converting the pyruvate molecules (from glycolysis) to lactate and oxidizing NADH. This reaction occurs in anaerobic fermentation when there is tissue hypoxia (decrease in oxygen).

If this inhibitor was placed in a cell that is deprived of oxygen, then there would be a buildup of pyruvate and NADH; however, since the inhibitor is added to cells undergoing aerobic respiration there will be no buildup. The pyruvate and NADH will undergo aerobic respiration and generate ATP. Note that red blood cells (RBCs) are unique in that they only use anaerobic respiration for ATP; therefore, adding lactate dehydrogenase inhibitor to RBCs will lead to a buildup of pyruvate and NADH.

Example Question #17 : Help With Glycolysis

What is the rate-limiting enzyme for glycolysis?

Possible Answers:

Hexokinase

Pyruvate kinase

Pyruvate carboxylase

Phosphofructokinase 1

Correct answer:

Phosphofructokinase 1

Explanation:

Glycolysis has three irreversible enzymatic steps that help the substrate intermediates proceed in one direction through the glycolytic pathway: the enzymes are hexokinase, phosphofructokinase 1, and pyruvate kinase. Of these three enzymes, the most important enzyme that controls the rate of glycolysis is phosphofructokinase 1, or PFK-1. Pyruvate carboxylase is not used in glycolysis, but in gluconeogenesis.

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