Please answer all parts. Thanks! B. (4 points) In order to elucidate the reactio

Question

Please answer all parts. Thanks!

B. (4 points) In order to elucidate the reactions involved in the TCA cycle, six different glucose samples were labeled singly with C at positions 1, 2, 3, 4, 5, and 6 respectively. Then each of these samples was incubated with muscle tissue suspensions and the rate of appearance odive CO, was measured (Lecture 8) with muscle tissue suspensions and the rate of appearance of radioactive CO2 was C1 C ()(1 ( HO-C-HC3 CH2OH C6 1. (1 point) Which carbon atoms of glucose end up in acetyl-CoA? 2. (1 point) Which carbon atoms of glucose are lost in the first pass of pyruvate into and through the TCA cycle?

Explanation / Answer

The oxidation of glucose through glycolysis pathway form pyruvate from C4, C5 and C6 of glucose. since during splitting of fructose-6-phosphate to DHAP and G3P the C1, C2 and C3 carbons of glucose split into DHAP whereas C4, C5, and C6 carbons of glucose form Glyceraldehyde-3-phosphate. With further steps last 3 carbons containing G3P produce pyruvic acid. The pyruvate undergo oxidative decarboxylation producing acetyl CoA. In the process the puruvate decarboxylate C4 carbon in form of CO2 thus producing acetyl CoA with C5 and C6 carbons in the glucose molecule.

2. During puruvate formation from glucose, C1, C2 and C3 carbons are lost in the form of DHAP leaving C4, C5 and C6 carbons of glucose in puruvate.

Later during acetyl CoA formation from pyruvate, C5 carbon is lost in the form of CO2.

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