GRE Subject Test: Biochemistry, Cell, and Molecular Biology : Biochemistry

Study concepts, example questions & explanations for GRE Subject Test: Biochemistry, Cell, and Molecular Biology

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Example Questions

Example Question #3 : Enzyme Regulation

On a Lineweaver-Burk plot, an inhibited enzyme is shown to have a less negative x-intercept than the uninhibited enzyme, but the y-intercept remains the same. The type of inhibition displayed is __________ and the inhibited reaction has a __________ value.

Possible Answers:

competitive . . . larger

competitive . . . smaller

non-competitive . . . larger

non-competitive . . . smaller

Correct answer:

competitive . . . larger

Explanation:

The x-intercept on a Lineweaver-Burk plot tells us the negative reciprocal of .

Because the x-intercept is less negative, this tells us that the inhibited reaction has a larger . Having a different x-intercept but the same y-intercept is characteristic of competitive inhibition. The inhibitor and the substrate are competing for the same binding site. 

Example Question #1 : Enzyme Regulation

Which of the following choices describes a way to graphically determine the type of inhibition being displayed by an inhibitor?

I. Plot initial reaction rate versus the concentration of substrate for the uninhibited enzyme, and then compare to the inhibited enzyme

II. Plot the inverse of the initial reaction rate versus the inverse of the substrate concentration for the uninhibited enzyme, and then compare to the inhibited enzyme

III. Plot the concentration of the inhibitor versus the concentration of substrate

Possible Answers:

II only

I, II, and III

I and II

I only

Correct answer:

I and II

Explanation:

Plotting the concentration of the inhibitor versus the concentration of the substrate will not give you any useful information because the reaction rate is essential in determining the type of inhibitor present.

Plotting initial reaction rate versus substrate concentration, or plotting the inverses, describes the graphical representation of Michaelis-Menten kinetics and a Lineweaver-Burk plot, respectively. Both of these are excellent methods to visually determine the type of inhibition displayed. On the graph, the line representing the inhibited enzyme will shift in predictable fashions depending on the type of inhibition. 

Example Question #4 : Enzyme Regulation

You have an enzyme solution and you add an inhibitor molecule and observe a marked decrease in enzyme activity. You increase the substrate concentration but this does not lead to any observable increase in enzyme activity. What can you conclude about your inhibitor?

Possible Answers:

That it is a competitive inhibitor

That it is a noncompetitive inhibitor

That it is a kinase

That it binds the enzyme's active site

That is it an inorganic inhibitor

Correct answer:

That it is a noncompetitive inhibitor

Explanation:

Noncompetitive inhibitors bind to enzymes away from the active site (allosteric) and distort it, reducing its affinity for substrate. Since they do not directly compete with substrate for enzyme binding, increasing the substrate concentration in the presence of a noncompetitive inhibitor will have no affect. While enzyme inhibitors include both organic and inorganic molecules, there is not enough information in the question stem to conclude the chemical classification of the inhibitor.

Example Question #2 : Enzyme Regulation

How do competitive inhibitors affect enzyme efficiency?

Possible Answers:

Lower the Michaelis constant

Raise the maximum rate of the enzymatic reaction

Lower the maximum rate of the enzymatic reaction

Raise the Michaelis constant

Correct answer:

Raise the Michaelis constant

Explanation:

Competitive inhibitors can be overpowered by introducing excess substrate, so they do not affect the maximum rate of the enzyme. They do, however, make it so that more substrate is required in order to get the enzyme working at half of its maximum rate. As a result, competitive inhibitors act by raising the Michaelis constant of enzymes.

Example Question #21 : Enzymes

How does a noncomeptitive inhibitor affect an enzyme?

Possible Answers:

Raises the maximum rate of the enzymatic reaction

Lowers the maximum rate of the enzymatic reaction

Raises the Michaelis constant of the enzyme

Lowers the Michaelis constant of the enzyme

Correct answer:

Lowers the maximum rate of the enzymatic reaction

Explanation:

A noncompetitive inhibitor acts to decrease how fast the enzyme can act on substrates. It accomplishes this by lowering the maximum rate at which it can create products. Noncompetitive inhibitors do not alter the enzyme's Michaelis constant.

Example Question #3 : Enzyme Regulation

How is pepsinogen activated in the stomach?

Possible Answers:

It is activated by the temperature change in the stomach lumen

A portion is cleaved, activating the enzyme

It is phosphorylated by another enzyme

Cofactors bind to the enzyme, increasing its efficiency

Correct answer:

A portion is cleaved, activating the enzyme

Explanation:

Once in the stomach lumen, pepsinogen finds itself in a very acidic environment. The acidic environment cleaves an amino acid sequence from pepsinogen, turning it into the active enzyme pepsin. This type of activation causes pepsin to only activate in the stomach lumen where it is needed.

Example Question #61 : Biochemistry

The carbohydrate mannose is not present in the standard glycolytic pathway. It can, however, enter glycolysis by first being converted into another sugar. Which of the following choices represents the point at which mannose first enters the glycolytic pathway?

Possible Answers:

Fuctose-6-phosphate

Glucose

Fuctose-1,6-bisphosphate

Glucose-6-phosphate

Correct answer:

Fuctose-6-phosphate

Explanation:

Mannose enters glycolysis by first being phosphorylated by hexokinase. The newly formed mannose-6-phosphate is then isomerized into fructose-6-phosphate by the enzyme phosphomannose isomerase. The sugar is now in a form that can follow the normal glycolytic pathway.

Example Question #62 : Biochemistry

During the first step of glycolysis, glucose is phosphorylated by hexokinase. What is the purpose of this reaction?

Possible Answers:

The hydrolysis of ATP is necessary to start the glycolytic pathway

The proton released via the phosphorylation reaction is necessary for the formation of NADH

The phophorylation of glucose changes the structure of glucose so that it can be isomerized in the next step

The phophorylation of glucose creates a negative charge on the glucose molecule so that it cannot pass through the plasma membrane

Correct answer:

The phophorylation of glucose creates a negative charge on the glucose molecule so that it cannot pass through the plasma membrane

Explanation:

If glucose was not phosphorylated, it would be free to diffuse through the plasma membrane and leave the cell. This situation would not be good for the cell because the reaction cannot continue outside of the cytosol. The negative charge created by the phosphorylation prevents the glucose molecule from crossing the plasma membrane due to the similar charge at the plasma membrane. 

Example Question #2 : Cellular Respiration And Photosynthesis

What does it mean to say that glycolysis has an energy investment phase?

Possible Answers:

ATP must be used in order to create the NADH in glycolysis

There is a net loss of ATP in glycolysis

ATP must be used in order to prepare the glucose molecule to be split

ATP must be used in order to move the glucose into the cytosol

Correct answer:

ATP must be used in order to prepare the glucose molecule to be split

Explanation:

Glycolysis can be divided into two parts: the energy investment phase and the energy payoff phase. The energy investment phase comes first when glucose is phosphorylated twice, requiring the use of two molecules of ATP. After the glucose is split, four molecules of ATP will be made in the final steps. This results in a net gain of two ATP in glycolysis, but ATP must be spent prior to being made.

Example Question #3 : Cellular Respiration And Photosynthesis

What molecule is the critical product of fermentation that is reinvested in glycolysis?

Possible Answers:

ADP

Glucose

NADH

NAD+

Correct answer:

NAD+

Explanation:

During glycolysis, a total of two molecules of NAD+ are reduced in order to form two NADH molecules. These NAD+ molecules need to be regenerated in order for more glycolytic reactions to take place; otherwise, the process would come to a halt. Fermentation takes care of this problem in anaerobic environments by oxidizing excess NADH (since it is no longer utilized in the electron transport chain) into NAD+, which is then returned to the cytosol where it can be used again in glycolysis.

All GRE Subject Test: Biochemistry, Cell, and Molecular Biology Resources

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