Question

notice that the branch on the left is reversible and nonoxidative. it involves exchanges of 2- and 3- carbon fragments among intermediates. a. considering the fact that no co2 is lost, how many moles of ribulose 5-p should be produced from 5 moles of glucose 6-p by these reactions? b. what types of enzymes catalyze the nonoxidative reactions (p. 549)?

Answer 1

In the given reactions, no CO2 is lost, and from 5 moles of glucose 6-P, we would produce 5/6 moles of ribulose 5-P. The nonoxidative reactions in the Calvin cycle involve the transfer of 2- and 3-carbon fragments among intermediates, which is facilitated by enzymes like transketolase and transaldolase.

Step-by-step explanation:

The nonoxidative reactions in the Calvin cycle involve exchanges of 2- and 3-carbon fragments among intermediates. In these reactions, no CO2 is lost. To calculate the moles of ribulose 5-P produced from 5 moles of glucose 6-P, we need to consider that each turn of the Calvin cycle fixes one molecule of CO2 and forms two molecules of 3-PGA. It takes three turns of the cycle to produce a single three-carbon glyceraldehyde-3-phosphate (GA3P) molecule, and six turns to produce a six-carbon glucose molecule. Therefore, from 5 moles of glucose 6-P, we would produce 5/2 moles of GA3P or 5/6 moles of ribulose 5-P.

The nonoxidative reactions of the Calvin cycle are catalyzed by several enzymes, including transketolase and transaldolase. Transketolase catalyzes the transfer of a two-carbon fragment while transaldolase catalyzes the transfer of a three-carbon fragment. These enzymes allow for the rearrangement of carbon fragments, facilitating the production of different intermediates in the cycle.