Biochemistry : Biochemistry

Study concepts, example questions & explanations for Biochemistry

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Example Questions

Example Question #41 : Glycolysis

What enzymes in the glycolysis pathway in the liver catalyze irreversible reactions?

Possible Answers:

Glucokinase, pyruvate kinase, enolase

Glucokinase, phosphoglycerate kinase, mutase

Glucokinase, phosphofructokinase-1, pyruvate kinase

Aldolase, phosphofructokinase-1, mutase

Isomerase, aldolase, mutase

Correct answer:

Glucokinase, phosphofructokinase-1, pyruvate kinase

Explanation:

In the liver, glucokinase irreversibly converts glucose in the cell to glucose-6-phosphate. Phosphofructose kinase-1 irreversibly converts fructose-6-phosphate to fructose-1,6-bisphosphate. Pyruvate kinase converts phosphoenolpyruvate to pyruvate. All the other enzymes listed catalyze reversible glycolysis reactions.

Example Question #41 : Glycolysis

Which of the following is not a possible fate of the glucose which is taken into cells?

Possible Answers:

Oxidation to pyruvate via gluconeogenesis

Storage as glycogen

Synthesis into the extracellular matrix structure

Oxidation to ribose-5-phosphate via pentose phosphate pathway

Oxidation to pyruvate via glycolysis

Correct answer:

Oxidation to pyruvate via gluconeogenesis

Explanation:

Gluconeogenesis does not convert glucose into pyruvate, rather it builds glucose from non-carbohydrate organic compounds. All of the other answer choices are possible fates of glucose once it enters the cell.

Example Question #42 : Glycolysis

How many steps are required to complete glycolysis?

Possible Answers:

Four

Ten

Eight

Nine

Correct answer:

Ten

Explanation:

Glycolysis is a process that takes place via ten reactions, involving the activity of multiple enzymes and occurs in the cytoplasm of the cell in two distinct phases: an energy consumption phase and an energy production phase. The first step in glycolysis is the conversion of glucose to glucose-6-phosphate through the consumption on one ATP molecule. Glucose is reacted upon by the enzyme hexokinase to carry out this step. Kinases are a group of enzymes that add phosphate groups by removing them from an ATP. In the second step, glucose-6-phosphate is then reacted upon by phosphoglucose isomerase. Isomerases are a group of enzymes that rearranged the structure of a molecule without changing the molecular formula. In this case the phosphoglucose isomerase rearranges the glucose-6-phosphate into fructose-6-phosphate. The third step involves another kinase: phosphofructokinase-1. This enzyme attached another phosphate group to fructose-6-phosphate, creating fructose-1,6-bisphosphate. In the fourth step, this molecule is then reacted upon by fructose bisphosphate aldolase. An aldolase is an enzyme that creates or breaks carbon-carbon bonds. This step results in the creation of two molecules: dihydroxyacetone phosphate and glyceraldehyde-3-phosphate. The fifth step involves another isomerase, triose phosphate isomerase, which converts the dihydroxyacetone into glyceraldehyde-3-phosphate (G3P).The sixth step involves the enzyme glyceraldehyde-3-phosphate dehydrogenase (GAPDH). GAPDH moves a hydrogen onto the electron acceptor . A phosphate group from inorganic phosphate instead of ATP replaces the hydride group that was taken from G3P. This creates the molecule 1,3 bisphosphoglycerate. The seventh reaction involves yet another kinase, phosphoglycerate kinase. Kinases can also take phosphate groups away. During this step, two phosphate groups are transferred from the 1,3 Bisphosphoglycerate molecules onto 2 molecules of ADP to create two molecule of ATP. The 1,3 Bisphosphoglycerate then becomes 3-phosphoglycerate. The eighth reaction involves the enzyme phosphoglycerate mutase which is essentially another isomerase. It converts 3-phosphoglycerate to 2-phosphoglycerate. The ninth reaction involves the enzyme enolase which produces a double bond by removing the hydroxyl group on 2-phosphoglycerate which results in phosphoenolpyruvate (PEP). The tenth and final reaction of glycolysis involves the enzyme pyruvate kinase. Just like the previous kinase reaction, this kinase is going to remove phosphate groups from the molecule to produce 2 molecules of ATP (one per molecule of PEP created from 1 molecule of glucose. The final product of glycolysis is pyruvate.

Example Question #42 : Glycolysis

Which of the following is correct regarding glycolysis?

Possible Answers:

The "committed" reaction is also the rate-limiting reaction

Hexokinase is found in the liver

Glucose is converted to lactose

Fructose-2,6-bisphosphate is an inhibitor of phosphofructokinase

Correct answer:

The "committed" reaction is also the rate-limiting reaction

Explanation:

Phosphofructokinase catalyzes the most regulated step of glycolysis and limits the reaction rate of glycolysis. Glucose is converted to pyruvate during glycolysis, not lactate, which is the case in some organisms (humans) through the process of fermentation. Glucokinase, not hexokinase is found in the liver, fructose-2,6-bisphosphate activates, not inhibits phosphofructokinase.

Example Question #43 : Glycolysis

How is fructose metabolism handled in the liver?

Possible Answers:

It is phosphorylated by hexokinase to fructose-6-phosphate, which can enter the glycolytic pathway.

The liver does not have enzymes to metabolize five-carbon carbohydrates

It is phosphorylated by fructokinase to fructose-1-phosphate, which is then converted into precursors of glycolytic intermediates. 

An epimerase adds a carbon to fructose and induces a carbon skeleton rearrangement to form glucose, which can undergo glycolysis normally.

Correct answer:

It is phosphorylated by fructokinase to fructose-1-phosphate, which is then converted into precursors of glycolytic intermediates. 

Explanation:

An epimerase inverts the stereochemistry at a particular carbon, but it cannot add additional carbons. While hexokinase can phosphorylate fructose into a glycolytic intermediate in most tissue, the liver does not contain any hexokinase. In the liver, fructose is phosphorylated by fructokinase into fructose-1-phosphate. This is cleaved by an aldolase to form dihydroxyacetone phosphate (DHAP), a glycolytic intermediate, and glyceraldehyde, a precursor to glyceraldehyde-3-phosphate (G3P), another glycolytic intermediate. Glyceraldehyde can either be phosphorylated into G3P or be used to promote fat storage. 

Example Question #47 : Glycolysis

At the end of anaerobic glycolysis, what products are created?

Possible Answers:

Acetyl-CoA

1 Lactate and 1 pyruvate

2 Lactate and 2 ATP

1 Pyruvate 

2 Pyruvate, 2 NADH, and 2 ATP

Correct answer:

2 Pyruvate, 2 NADH, and 2 ATP

Explanation:

This is a slightly misleading question. Glycolysis is an anaerobic process. Its products are always 2 pyruvate, 2 NADH, and 2 ATP. The fate of the pyruvate depends on the absence and/or presence of oxygen, in which it will be reduced into lactate or cleaved into acetyl-CoA to enter the Krebs cycle, respectively.

Example Question #51 : Carbohydrate Metabolism

Glycolysis produces pyruvate. The fate of this pyruvate depends mainly on the availability of which of the following?

Possible Answers:

Phosphoric acid

Coenzyme A

Molecular oxygen

Calcium ions

Correct answer:

Molecular oxygen

Explanation:

If molecular oxygen is present, the products of glycolysis will continue and be fed into aerobic metabolism: the citric acid cycle and oxidative phosphorylation. If molecular oxygen is absent, glycolysis will lead to anaerobic metabolism/fermentation. Thus, the fate of pyruvate depends on the availability of oxygen.  does need to be recycled for glycolysis to proceed, but this can happen via the electron transport, which can only proceed if oxygen is present. Thus, while this is fundamentally correct, it is a downstream regulator. Coenzyme A is not a limiting factor in the fate of pyruvate, but it does get added after a molecule of carbon dioxide is released, producing acetyl-CoA during pyruvate dehydrogenation. Phosphoric acid and calcium ions are not relevant to this question.

Example Question #851 : Biochemistry

Which of these is not an end product of glucose metabolism by either aerobic or anaerobic means?

Possible Answers:

Fructose

Lactate

Carbon dioxide

Ethanol

Correct answer:

Fructose

Explanation:

Glucose can be catabolized by both aerobic and anaerobic means. When glucose undergoes oxidative phosphorylation (aerobic metabolism), the end products are carbon dioxide, water, and ATP. In the absence of oxygen, glucose can undergo either lactic acid or alcoholic fermentation. Lactate is a result of lactic acid fermentation, and ethanol and carbon dioxide are results of alcoholic fermentation. 

Example Question #3 : Other Glycolysis Concepts

Which of the following list the correct order of processes in the creation of ATP from glucose?

Possible Answers:

Glycolysis, pyruvate dehydrogenase complex, citric acid cycle, electron transport chain

Citric acid cycle, glycolysis, pyruvate dehydrogenase complex, electron transport chain

Electron transport chain, citric acid cycle, pyruvate dehydrogenase complex, glycolysis

Glycolysis, electron transport chain, pyruvate dehydrogenase complex, citric acid cycle

Glycolysis, citric acid cycle, pyruvate dehydrogenase complex, electron transport chain

Correct answer:

Glycolysis, pyruvate dehydrogenase complex, citric acid cycle, electron transport chain

Explanation:

Glucose goes through glycolysis to form pyruvate molecules which then proceed into the pyruvate dehydrogenase complex.  Acetyl-CoA is created from this, which can then move into the citric acid cycle.  High energy electron carries,  and , that were created in glycolysis and the citric acid cycle finally go through the electron transport chain to pump hydrogens through the mitochondrial membrane.  These hydrogens are used to generate ATP via the ATP synthase.

Example Question #852 : Biochemistry

If a cell is lacking in oxaloacetate, which of the following will occur?

Possible Answers:

The cell will die

The cell will overproduce ATP

Glycolysis will halt completely

There will be a buildup of pyruvate in the cell

Citrate will take over the role of oxaloacetate in the Kreb's Cycle

Correct answer:

There will be a buildup of pyruvate in the cell

Explanation:

If a cell is lacking in oxaloacetate, the Krebs cycle will be unable to continue.  Therefore, there will be no way for the electron transport chain to receive the high energy electrons it requires to create ATP.  And so, fermentation will take over in the cell for creation of ATP.  Glycolysis will not stop, but the end product pyruvate will build up because it will only be able to be used for fermentation, not in the pyruvate dehydrogenase complex.

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