All GRE Subject Test: Biochemistry, Cell, and Molecular Biology Resources
Example Questions
Example Question #2 : Help With Inhibitors
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.
competitive . . . larger
non-competitive . . . larger
competitive . . . smaller
non-competitive . . . smaller
competitive . . . larger
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 #3 : Help With Inhibitors
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
I and II
II only
I, II, and III
I only
I and II
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 #21 : Enzymes
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?
That it binds the enzyme's active site
That is it an inorganic inhibitor
That it is a competitive inhibitor
That it is a noncompetitive inhibitor
That it is a kinase
That it is a noncompetitive inhibitor
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 #1 : Help With Other Regulatory Mechanisms
How do competitive inhibitors affect enzyme efficiency?
Lower the maximum rate of the enzymatic reaction
Raise the maximum rate of the enzymatic reaction
Lower the Michaelis constant
Raise the Michaelis constant
Raise the Michaelis constant
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 #2 : Help With Other Regulatory Mechanisms
How does a noncomeptitive inhibitor affect an enzyme?
Lowers the Michaelis constant of the enzyme
Raises the maximum rate of the enzymatic reaction
Raises the Michaelis constant of the enzyme
Lowers the maximum rate of the enzymatic reaction
Lowers the maximum rate of the enzymatic reaction
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 #64 : Biochemistry
How is pepsinogen activated in the stomach?
It is activated by the temperature change in the stomach lumen
It is phosphorylated by another enzyme
Cofactors bind to the enzyme, increasing its efficiency
A portion is cleaved, activating the enzyme
A portion is cleaved, activating the enzyme
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 #141 : Gre Subject Test: Biochemistry, Cell, And Molecular Biology
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?
Fuctose-6-phosphate
Fuctose-1,6-bisphosphate
Glucose-6-phosphate
Glucose
Fuctose-6-phosphate
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 #142 : Gre Subject Test: Biochemistry, Cell, And Molecular Biology
During the first step of glycolysis, glucose is phosphorylated by hexokinase. What is the purpose of this reaction?
The proton released via the phosphorylation reaction is necessary for the formation of NADH
The phophorylation of glucose creates a negative charge on the glucose molecule so that it cannot pass through the plasma membrane
The hydrolysis of ATP is necessary to start the glycolytic pathway
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
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 #1 : Cellular Respiration And Photosynthesis
What does it mean to say that glycolysis has an energy investment phase?
There is a net loss of ATP in glycolysis
ATP must be used in order to move the glucose into the cytosol
ATP must be used in order to create the NADH in glycolysis
ATP must be used in order to prepare the glucose molecule to be split
ATP must be used in order to prepare the glucose molecule to be split
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 #1 : Cellular Metabolism
What molecule is the critical product of fermentation that is reinvested in glycolysis?
Glucose
NADH
ADP
NAD+
NAD+
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.
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