The pentose phosphate pathway and G6PD deficiency
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The pentose phosphate pathway and G6PD deficiency
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[post_date] => 2024-12-23 10:40:00
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[post_content] => Practice Passage (Question 1-6)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
The pentose phosphate pathway (PPP) can be divided into two phases, as shown in Figure 1. The oxidative phase is crucial for producing NADPH, which essential for both biosynthetic processes and antioxidant defenses. This phase regenerates glutathione into its active forms (GSH), enabling GSH to neutralize reactive species like hydrogen peroxide (H₂O₂) and combat oxidative stress within cells. The nonoxidative phase comprises a series of reversible reactions catalyzed by enzymes such as transketolase and transaldolase; these reactions facilitate the rearrangement of carbon atoms among various sugar phosphates, generating critical metabolic intermediates like glyceraldehyde-3-phosphate and fructose-6-phosphate. The rate limiting enzyme of the oxidative phase is glucose-6-phosphate dehydrogenase (G6PD).
Figure 1 The pentose phosphate pathway
To investigate how disrupting the PPP impacts a cell’s ability to grow and reproduce, researchers used cloning efficiency assays with cells that were modified to not express G6PD (“G6PD null cells”). A cloning efficiency assay quantifies the ability of a single cell to replicate and produce a colony of cells; cloning efficiency (%) equals the percentage of plated cells that produce colonies. Figure 2 shows the mean cloning efficiency for control and G6PD null cells under different oxygen conditions.
Figure 2 Mean cloning efficiency for control and G6PD null cells
Researchers further assessed cloning efficiency after exposing cells to increasing levels of diamide, a sulfhydryl containing compound that depletes GSH in the cell. Figure 3 shows the mean cloning efficiency for control and G6PD null groups exposed to varying concentrations of diamide.
Figure 3 Mean cloning efficiency for control and G6PD null cells
Pandolfi, P. P., Sonati, F., Rivi, R., Mason, P., Grosveld, F., & Luzzatto, L. (1995). Targeted disruption of the housekeeping gene encoding glucose 6-phosphate dehydrogenase (G6PD): G6PD is dispensable for pentose synthesis but essential for defense against oxidative stress. EMBO J, 14(21), 5209-5215.
[post_title] => The pentose phosphate pathway and G6PD deficiency
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[question] => Which of the following hypotheses best explains results shown in Figure 2?
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[answer] => 1
[description] => Reason for Correct Answer:
Low cloning efficiency in 24% oxygen suggests that G6PD null cells are at a survival disadvantage compared to control cells, with have functional G6PD.
If disruption of nucleotide or lipid synthesis was causing the survival disadvantage for G6PD null cells, one would not expect a decreased oxygen concentration to improve cloning efficiency.
Cloning efficiency for G6PD null cells is comparable to cloning efficiency for control cells when cultured in 13% oxygen.
This suggests that absence of the rate-limiting enzyme of the pentose phosphate pathway sensitizes cells to harmful oxidative stress such that they cannot survive well in higher oxygen conditions
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[answers] => Array
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[0] => Array
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[each_answer] => A. Absence of the rate-limiting enzyme of the pentose phosphate pathway sensitizes cells to harmful oxidative stress.
)
[1] => Array
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[each_answer] => B. Absence of the rate-limiting enzyme of the pentose phosphate pathway prevents production of electron-acceptor molecules.
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[2] => Array
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[each_answer] => C. Absence of the rate-limiting enzyme of the pentose phosphate pathway prevents synthesis of nucleotides needed for cell division.
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[each_answer] => D. Absence of the rate-limiting enzyme of the pentose phosphate pathway prevents synthesis of lipids needed for cell division.
)
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[1] => Array
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[quiz_unique_key] => 1403770772
[question] => Diamide was likely used in the experiment because it is a:
[value] => Array
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[answer] => 2
[description] => Reason for Correct Answer:
Increasing concentrations of diamide made it harder for G6PD cells to survive, and the passage states that diamide depletes GSH in the cell.
Note that for many biological molecules, more -H bonds means that it’s in a reduced state and more -OH groups means it’s in an oxidized state.
GSH is the reduced state of glutathione; as shown in Figure 1, it’s able to act as a reducing agent to remove harmful oxidants like H₂O₂.
Diamide therefore likely depletes GSH because it can oxidize GSH, acting as an oxididizing agent. Diamide induces the oxidation of GSH by promoting the formation of disulfide bonds (-S-S-) between two GSH molecules, resulting in the generation of oxidized glutathione (GSSG). Cells that are G6PD deficient cannot regenerate the reduced glutathione.
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[answers] => Array
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[each_answer] => A. G6PD inhibitor
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[each_answer] => B. oxidizing agent.
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[each_answer] => C. reducing agent.
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[each_answer] => D. reactive oxygen species.
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[2] => Array
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[quiz_unique_key] => 1403770772
[question] => What characteristic of the transketolase and transaldolase-catalyzed nonoxidative reactions would enable them to produce ribose 5-phosphate in cells that are G6PD deficient?
[value] => Array
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[answer] => 3
[description] => Reason for Correct Answer:
Normally, ribulose-5-phosphate is produced by the oxidative phase and the nonoxidative phase of the PPP uses it to create glycolytic intermediates.
If the nonoxidative reactions catalyzed by transketolase and transaldolase can produce ribulose-5-phosphate from glycolytic intermediates, this means that the
Reactions with delta G close to 0 are reversible.
Therefore, the nonoxidative reactions having a delta G near 0 will allow them to operate in the reverse direction under conditions in which ribulose-5-phosphate is not being generated from the oxidative reactions.
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[answers] => Array
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[0] => Array
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[each_answer] => A. The enzymes have High Kₘ values.
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[1] => Array
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[each_answer] => B. The enzymes have ow k꜀ₐₜ values.
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[each_answer] => C. The reactions operate at a delta G close to 0.
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[each_answer] => D. The reactions operate at a delta G much less than 0.
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[quiz_unique_key] => 1403770772
[question] => NADPH is also produced by enzymatic reactions in mitochondria; Based on this fact, which of the following cell types would be most susceptible to oxidative stress, assuming G6PD deficiency?
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[answer] => 1
[description] => Reason for Correct Answer:
Mitochondria are membrane-bound organelles where much of cellular respiration occurs.
A cell which contained neither mitochondria nor G6PD would have very limited NADPH-producing abilities, and thus would be extremely susceptible to oxidative stress.
Red blood cells are unique amongst animal cells because they do not contain the normal complement of organelles: vertebrate red blood cells contain none, while other animal cells contain only a nucleus.
Red blood cells with G6PD deficiency would therefore be most susceptible to oxidative stress.
)
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[each_answer] => A. Red blood cells
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[each_answer] => B. Hepatocytes (liver cells)
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[each_answer] => C. Muscle cells
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[each_answer] => D. Neurons
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[quiz_unique_key] => 1403770772
[question] => Given the metabolic role of the pentose phosphate pathway, which of the following is least likely to be an inhibitor of G6PD?
[value] => Array
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[answer] => 1
[description] => Reason for Correct Answer:
The pentose phosphate pathway produces molecules that are important for the biosynthetic activities of cells.
The products of metabolic pathways commonly inhibit the further activity of the pathway in question. For instance, NADPH and ribulose-5-phosphate would inhibit G6PD, which in turn would downregulate the pentose phosphate pathway.
Insulin, released in response to high glucose levels, downregulates catabolic pathways (pathways that release energy) and upregulates anabolic pathways (pathways that store energy and synthesize molecules).
Therefore, insulin is most likely to promote the PPP and is least likely to be an inhibitor of G6PD.
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[each_answer] => A. Insulin
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[each_answer] => B. Epinephrine
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[each_answer] => C. NADPH
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[each_answer] => D. Ribulose-5-phosphate
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[quiz_unique_key] => 2103402213
[question] => Given ribulose-5-phosphate’s role in nucleotide synthesis, which of the following molecules would likely expected be depleted in a cell with low levels of ribulose-5-phosphate production?
I. ATP
II. NADH
III. DNA
[value] => Array
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[answer] => 4
[description] => Reason for Correct Answer:
Nucleotides have a nitrogenous base, a five carbon sugar, and one or more phosphate groups.
Deoxyribonucleic acid consists of repeating sequences of nucleotides.
Adenosine triphosphate consists of adenosine (an adenine nucleotide attached via a glycosidic bond to the five carbon sugar ribose) bonded to three phosphate molecules.
NADH is the reduced form of nicotinamide adenine dinucleotide.
A cell with low levels of ribulose-5-phosphate production would therefore also likely have low levels of ATP, NADH, and DNA.
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[answers] => Array
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[each_answer] => A. I only
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[each_answer] => B. III only
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[each_answer] => C. I and III only
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[each_answer] => D. I, II, and III
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[553958|1] => A
[553958|2] => B
[553958|3] => C
[553958|4] => A
[553958|5] => A
[553958|6] => D
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Figure 1
Figure 2
Figure 3