Question

Not sure how to trace the carbons for this problem.

A nment 3 150 pls) Due al the Bug Ur 8.) (5 pts) A liver extract is capable er extract is capable of carrying out all of the metabolic reactions involved in gluconeogenesis. Trace the following radio-labeled precursors through gluconcos location of all "C-labeled intermediates and in the product glucose. wing radio-labeled precursors through gluconeogenesis and indicate the ATP+H20 ADP+P+24+ H O - C- 0 0-14 7 fyruvat Carboxylale GTP LG GDP Po- O c= (- ilip H3C-C-0-0 Athao ADD+PL+ 2H+ అలు 14 He=c-c-2- 14 6 140 C-o- +- HD-C1-0 14 Pyro carboxylarc C-H H-C-OH + the-opo - . H-C-OH 4 HC-OH HSS-opos HC-OPO3²- Hac-OPO²- 보다 애 HE, -OPO32- C=0 H₂ C-OH c=0 H-C-OH HO-C-H H-C-OH HE-OPO H₂C-OPO} OH H or-p-o- CH2OH Ciha OH RH OH не он OH OH OM

Answer 1

Please find the figure. All the carbon atoms with ¹⁴C labeled are marked in red. Two steps are re-oriented for better understanding. First one is '[ ] bracket' and other one is in '{ } bracket'. Names of the compounds are written below the structures. Finally C-1 and C-6 carbons of glucose will be ¹⁴C labeled.

Po OFO O= اسے - غ =ع ,H H3c-eco (Pyrunater) HOME (bicarbonate) = 0 14 Coxaloacetate) (Phospho enol pyruvate) JH₂o - 8 h Post Topo de ATP cho H.C-OH Hq G-o-pos 2 HOHN HECH H, C- opon - H26-0-962 - It Go=0 F14 (Glyceraldehyde (1,3-Bisphospho (3-Phospho glyciate) Héro-por I 3-Phosphat) Hea 01 214 C2-Phospho. H₂C-OH glycerate) d=0 14 1 H₂ C-O-PO 2. Girdere H₂C-0-P02- PO prophets ucht - cao H-C-CH Hodoh H-d-ot 14& Hq -o-peg | Ct/ HEC-oh Oh (Fructose-1,6-Bisphosphate) Ho-oh City 0-10? 0-P02 CH₂OH P4 ÇH₂OH 14 CH2 Kombine 1 CH2 Jees 14 HO 1974 oh V oh! 10H ost Glucan) 04 Glucose-6 phosphate) Fruclose Co-phosphate)