Biochemistry : Biochemistry

Study concepts, example questions & explanations for Biochemistry

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Example Questions

Example Question #232 : Catabolic Pathways And Metabolism

In what part of the cell does the pyruvate dehydrogenase complex function?

Possible Answers:

In the nucleus

Cytoplasm

The mitochondrial matrix

The mitochondrial intermembrane

Within the inner membrane of the mitochondria

Correct answer:

The mitochondrial matrix

Explanation:

The pyruvate dehydrogenase complex (PDC) is preparing pyruvate for the Krebs cycle by converting it to acetyl-CoA. Because the Krebs cycle functions within the mitochondrial matrix, the PDC is also taking place there.  This ensures quick and easy movement from the PDC into the Krebs cycle.

Example Question #41 : Citric Acid Cycle

Which process involved in cellular respiration produces the largest quantity of high energy electron carriers?

Possible Answers:

Glycolysis

ATP synthase

Pyruvate dehydrogenase complex

Electron transport chain

Krebs cycle

Correct answer:

Krebs cycle

Explanation:

The Krebs cycle produces the most high energy electron carriers of any process involved in cellular respiration. Per glucose molecule, the Krebs cycle produces  and .

Example Question #42 : Citric Acid Cycle

Which of the following molecules enter the Krebs cycle directly, following glycolysis?

Possible Answers:

Pyruvate dehydrogenase

Acetyl-CoA

Pyruvate

Ubiquinone

Glucose

Correct answer:

Acetyl-CoA

Explanation:

Ubiquinone is a part of the electron transport chain, and has little to do with the Krebs cycle. Glucose is broken down during glycolysis, and does not enter the Krebs cycle directly. Many students make the mistake of thinking that pyruvate enters the Krebs cycle, since it is produced in glycolysis, and the Krebs cycle follows glycolysis. However, pyruvate is first converted to acetyl-CoA by the pyruvate dehydrogenase complex in the mitochondrial matrix, and acetyl-CoA enters the Krebs cycle.

Example Question #15 : Reactants And Products Of The Citric Acid Cycle

What is the role of isocitrate dehydrogenase in the citric acid cycle of the mitochondria?

Possible Answers:

Isocitrate dehydrogenase converts citrate to alpha-ketoglutarate and is inhibited by 

Isocitrate dehydrogenase converts alpha-ketoglutarate to isocitrate and is activated by 

Isocitrate dehydrogenase converts isocitrate to alpha-ketoglutarate and is inhibited by 

Isocitrate dehydrogenase converts citrate to isocitrate and is inhibited by 

Isocitrate dehydrogenase converts citrate to isocitrate and is activated by

Correct answer:

Isocitrate dehydrogenase converts isocitrate to alpha-ketoglutarate and is inhibited by 

Explanation:

Isocitrate dehydrogenase activation leads to oxidative decarboxylation of isocitrate in a two step process producing alpha-ketoglutarate and . In the mitochondria, the reaction produces also a charged electron carrier molecule, , from . Isocitrate dehydrogenase, inhibited by  and activated by , is a major regulator enzyme of the citric cycle.

Example Question #961 : Biochemistry

Which of the following steps within the citric acid cycle results in the production of a molecule of carbon dioxide ?

Possible Answers:

The conversion of malate to oxaloacetate

The conversion of succinyl-CoA to succinate

The conversion of isocitrate to alpha-ketoglutarate 

The conversion of succinate to fumarate

The conversion of fumarate to malate

Correct answer:

The conversion of isocitrate to alpha-ketoglutarate 

Explanation:

The only step of the citric acid cycle listed that results in the production of  as a side product is the conversion of isocitrate to alpha-ketoglutarate. In this step, the enzyme, isocitrate dehydrogenase catalyzes the conversion of isocitrate to alpha-ketoglutarate, while also converting  to  and  as side products, and generating a molecule of  in the process (i.e. reducing the carbon count from 5 in isocitrate to 4 in alpha-ketoglutarate). 

The conversion of alpha-ketoglutarate to succinyl-CoA also produces a molecule of  as a side product. However, this step is not listed as an answer choice.

None of the other answer choices listed produce  as side products. 

Example Question #1 : Glycolysis Regulation

The enzyme phosphofructokinase is an important enzyme that plays a significant regulatory role in glycolysis. Which of the following conditions would be expected to result in increased activity of this enzyme?

Possible Answers:

High levels of NADH

High levels of citric acid cycle intermediates

Low levels of ATP

None of these

Low levels of ADP

Correct answer:

Low levels of ATP

Explanation:

We're told in the question stem that phosphofructokinase is an important regulatory enzyme for the glycolysis pathway. In this pathway, glucose is partially oxidized to provide energy for the cell. Therefore, when the cell has plenty of energy available to meet its metabolic needs, the activity of phosphofructokinase will be reduced in order to suppress glycolysis. Alternatively, when the cell is in need of energy, it will turn glycolysis on by increasing the activity of this enzyme. Thus, compounds that indicate the cell has a lot of energy available, such as ATP and NADH, would be expected to be allosteric inhibitors of this enzyme. So, if the cell has a low amount of ATP, the energy carrying molecule, we would expect the activity of this enzyme to be high. High levels of NADH and citric acid cycle intermediates signal that the cell has enough energy, therefore these would serve to reduce the activity of this enzyme. Low levels of ADP would signal that the cell most likely has a high amount of ATP and is thus in a state of energy surplus. Consequently, this scenario would likely reduce the activity of this enzyme.

Example Question #2 : Glycolysis Regulation

Which of the following choices will result in continuous glycolysis?

Possible Answers:

Loss of allosteric binding site for fructose-2,6-bisphosphate on phosphofructokinase-1 (PFK-1)

Loss of allosteric binding site for fructose-1,6-bisphosphate on pyruvate kinase

Loss of allosteric binding site for ATP on phosphofructokinase-1 (PFK-1)

ATP binding to the allosteric site on pyruvate kinase

Correct answer:

Loss of allosteric binding site for ATP on phosphofructokinase-1 (PFK-1)

Explanation:

When there are high levels of ATP in the blood, ATP itself can act as a signal for the inhibition of ATP production. phosphofructokinase-1 (PFK-1) and pyruvate kinase are major sites of glycolytic regulation. ATP can inhibit these enzymes by binding to their allosteric sites. If these allosteric binding sites are lost, ATP can never bind, and glycolysis will continue indefinitely. Conversely, glycolytic intermediates (like fructose-1,6-bisphosphate) and glycolytic activators (like fructose-2,6-bisphosphate) can act on the same enzymes to increase their activity. Loss of allosteric binding sites for fructose-1,6-bisphosphate and fructose-2,6-bisphosphate on pyruvate kinase and PFK-1, respectively, will result in the slowing down or inhibition of glycolysis.

Example Question #1 : Glycolysis Regulation

All of the following stimulate glycolysis except __________.

Possible Answers:

decreased pH level

high level of carbon dioxide

increased level of pyruvate

decresed level of ATP

increased level of AMP

Correct answer:

increased level of pyruvate

Explanation:

Glycolysis will be stimulated in situations that require the body to make more ATP.  When the pH is low, ATP is depleted, AMP is at high levels, and carbon dioxide is increased, the body is likely going to need more of an energy supply.  All of these are related to exercise - a situation in which more ATP will be required.  However, an increase in levels of pyruvate implies that glycolysis is actually backed up and should not be stimulated.

Example Question #1 : Glycolysis Regulation

Fructose-2,6-bisphosphate will strongly upregulate which of the following enzymes?

Possible Answers:

Phosphofructokinase-2 (PFK-2)

Phosphofructokinase-1 (PFK-1)

Fructose-2,6-bisphosphatase (FBPase-2)

Fructose-1,6-bisphosphatase (FBPase-1)

Correct answer:

Phosphofructokinase-1 (PFK-1)

Explanation:

While phosphofructokinase-2 is responsible for creating fructose-2,6-bisphosphate, this molecule will actually upregulate the enzymatic activity of phosphofructokinase-1 (PFK-1). As a result, fructose-2,6-bisphosphate is responsible for increasing the amount of glycolysis done in cells via activation of the glycolytic enzyme PFK-1. 

Example Question #2 : Glycolysis Regulation

Glycolysis is an energy producing process that breaks down glucose. 

In glycolysis, feedback regulation is seen in which of the following examples?

Possible Answers:

Fructose-1,6-bisphosphate inhibiting pyruvate kinase

AMP inhibiting PFK-1 (phosphofructokinase-1) 

Hexokinase inhibited by ATP 

Insulin inhibiting pyruvate kinase.

ATP inhibiting PFK-1 (phosphofructokinase-1) 

Correct answer:

ATP inhibiting PFK-1 (phosphofructokinase-1) 

Explanation:

Feedback inhibition is a process by which the products of a reaction or series of reactions slows, stops or inhibits one of the previous reactions in the process, thereby controlling the rate of reaction, and rate of formation of the products.

In glycolysis, one of the end products is energy in the form of ATP. ATP acts as an inhibitor of phosphofructokinase-1, one of the main rate limiting enzymes in glycolysis. 

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