Question

3. You are given 90 mg of the following radioactively labeled carbohydrate, which contains a total of 3 uCi of 14C (14C = *). Draw the specific degradation pathway required to metabolize this molecule into a 3-Carbon subunit (pyruvate). Noting in this pathway we also incubated with a triose phosphate isomerase inhibitor (assume 100% inhibition). Specifically label the radioactive position(s) in the final product(s) and account for all radioactivity observed as this starting compound progresses through the pathway. Finally, calculate the number of moles of all the intermediates and final products produced. (all the structures, full name of each intermediate and names of enzymes are required) (10 marks) CH2OH - OH

Answer 1

The given molecule is glucose with carbon no. 1 labelled. Glucose is degraded into pyruvate via glycolysis or Embden-Meyerhof Pathway through a series of enzyme catalysed reactions.

Full explanation for the problem is given below alongwith all the intermediates and final product with their structure:

• Solutions C Given molecule is glucose Pe rtandingan par labelled te OH celysis Glucose is broken down into pyruvate through glycolysis (Embden - Meyerhof pathways in a series of enzyme - catalysed reactions. All the steps of this process are carried out in the cytoplasm of a cell. There are total 10 steps in glucose degradation pathway: 6 CH2OH ist OH PILA 10 HOY labelled's w OH Steb-I. Phosphorylation (Hexokinase) TATP ADP+Pi 아 1 Gbc - 6-phosphate CH Labelled 'c Hoy on -Step-2 - Isomerization Phosphogle.co Conc, OH Fructose-6-phosphate) CH2OH Step-3. - Phosphorylation ATP phos pligfreucto L kinase ADP+ PE Fructose –1,6 - bisphosphate? Dove gode - slek_4. - cleavage. Aedalade

Glyceraldehyde-3-phosphate? (DHAD) Dihydroxyacetone phos plate Home -cu, OP Labelled c. 애 poh¢-CH-CHO 31e6-5Isomerization enzyme & triose phosphate isomerase But, it is given in question that triose phosphate isomerase inhibitor (with 100.4. inhibition) is present, so , Step-5 will be indubited, ie., alihydłonyacetone phosphate is net converted into Glyceraldehyde - 3-phosphates 163-P) Usually, in glycolysis 2 molecules of G-3-p enter in step-6 and leads to formation of a pyruvate melecules. But in this case only 1 G-3-p will enter in Step-6 and further due to inhibition of triose phosphate isomerase. 4CHO séH-OH (Glyceraldehyden 3-phosphate 66170 Step-6. Oxidative phosphorylation dehydrogendlie 0 (1,3 – bisphosphoglycerate @ow, e-CH-0-00). 0 1

Stebe?, 6pP+ ། Phosphoglycesale substrate level ATRE TL kinase phosphorylation . 13- phosphoglycerate Step-8 plos bloglycerate I samavizaten mutase pino DO Hz C-cn-6-0-. 2- phosphoglycerate Hoke- CH - 200 Steb-9. Enol formation - | Pelesettyust g Are Thyruat kerane ADPUR Steb-10. Substrate level te kenase ТАТА a phosphorylation . 1111 given weight - Number of moles of glucose = molecular weight E = go mg I 1803 - = 9980 90 x103 10.0005 moles forms Here, 1 mole glucose to 1 mile of pyruvate 0,0005 and glucose Tow 0.0005 mole pyruvete a to No. of nicles of all the intermediates = 0.0005 moles. * No carbon is labelled in final product (pyruvate), was the labelled carbon is lost in DHAP (steps).