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MCAT

Kurs: MCAT > Rozdział 2

Lekcja 1: Biological sciences practice passage questions

Glucose-6-phosphate dehydrogenase and the most important function of the pentose phosphate pathway

Zadanie

The pentose phosphate pathway plays an important role in the synthesis of lipids and nucleotides. It also helps cells produce antioxidants that detoxify hydrogen peroxide, a main source of oxidative stress in animal cells.
The rate-limiting enzyme of the pentose phosphate pathway (see figure 1) is glucose 6-phosphate dehydrogenase (G6PD). Scientists used cloning efficiency assays with cells that were modified to not express G6PD (“G6PD null cells”) to investigate how disrupting the pentose phosphate pathway impacts a cell’s ability to grow and reproduce. (A cloning efficiency assay quantifies the ability of a single cell to replicate and produce a colony of cells. For example, if 100 individual cells are plated and cultured, and 50 distinct colonies eventually appear, then cloning efficiency for the cell line in question is 50%.)
In the case of the experiment with G6PD null cells, mean cloning efficiency was 5.7% +/- 1.8%. Mean cloning efficiency for control cells was 51.2% +/- 9.4%. These results are depicted in figure 2. Another cloning efficiency assay was performed under culture conditions in which oxygen concentration was reduced by 40%. The results of this experiment are depicted in figure 2.
Figure 1: The pentose phosphate pathway converts glucose-6-phosphate to ribulose-5-phosphate and NADPH.
Figure 2: Mean cloning efficiency for G6PD null cells vs. control in 24% oxygen and 13% oxygen.
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.
Which of the following hypotheses best explains the low cloning efficiency of G6PD-null cells in standard cell culture conditions (figure 2, 24% oxygen)?
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