6) (10 pts) S glycolysis and gluconeogenesis. (a) First, consider a scheme that maximizes
production of ribose-5-phosphate (R-5-P) via the PPP. Assume that flux through the PPP
converts all the G-6-P to R-5-P. Under these conditions, what is the maximum possible yield of
R-5-P and NADPH following the addition of 24 nmol of G-6-P to your extract? Show your
work. uppose you have a liver cell extract that contains all the enzymes for the pentose
phosphate pathway (PPP),
Solution
The needs for NADPH and R-5-P are balanced. The predominant reaction under these conditions
is the formation of two molecules of NADPH and one molecule of R-5-P from one molecule of
glucose-6-phosphare in the oxidative phase of the PPP.
One mole of a substance is equal to 6.023 x 1023 molecules. So 1 molecule will have 1.661 x 10-
24 moles of that substance.
24nmol is equals 24 x 10-9 moles.
The oxidation of 24nmol of G-6-P will yield 14.44 x 10-15 moles of R-5-P and 28.88 x 10-15
moles of NADPH
production of ribose-5-phosphate (R-5-P) via the PPP. Assume that flux through the PPP
converts all the G-6-P to R-5-P. Under these conditions, what is the maximum possible yield of
R-5-P and NADPH following the addition of 24 nmol of G-6-P to your extract? Show your
work. uppose you have a liver cell extract that contains all the enzymes for the pentose
phosphate pathway (PPP),
Solution
The needs for NADPH and R-5-P are balanced. The predominant reaction under these conditions
is the formation of two molecules of NADPH and one molecule of R-5-P from one molecule of
glucose-6-phosphare in the oxidative phase of the PPP.
One mole of a substance is equal to 6.023 x 1023 molecules. So 1 molecule will have 1.661 x 10-
24 moles of that substance.
24nmol is equals 24 x 10-9 moles.
The oxidation of 24nmol of G-6-P will yield 14.44 x 10-15 moles of R-5-P and 28.88 x 10-15
moles of NADPH
