I have three questions. I don't even know where to begin.
1.
a) If an in vitro fatty acid synthesizing system is supplied with [2-14C]-acetyl CoA as the only source of radioactive label along with all other necessary cofactors/reactants for net fatty acid synthesis, in what position(s) will the label appear in newly synthesized palmitic acid? (Note that the in vitro system lacks acetyl CoA carboxylase)
b) If instead 2,2-dideuteromalonyl CoA was supplied as the only source of label, how many deuterium atoms would be incorporated into palmitate and where would they appear?
c) If [3-14C]-malonyl CoA were used in the above reaction, which atom(s) in palmitate would be labeled?
d) Describe the pathway by which some of the carbons in an odd-chain fatty acid can undergo net conversion to carbohydrate.
2. A mammal on a particular diet expends 10,000 kJ of energy per day while excreting 40g of urea. Assuming that protein is 16% N by weight and that its catabolism yields 18kJ of energy per gram, what percentage of the mammal's energy needs are met by the degradation of protein?
3. [3-14C]-Pyruvate was administered to isolated liver cells in the presence of sufficient malonate to completely block succinate dehydrogenase. After a 30-minute incubation period, isocitrate was isolated from the reaction mixture and was found to contain 14C at both C2 and C4: Explain the presence of isotopic label at the indicated positions.
Any help would be appreciated! Thanks!
Sure, I'll try my best to help you with each of your questions.
1.
a) To determine the position(s) where the label will appear in newly synthesized palmitic acid, you need to understand the process of fatty acid synthesis. In the absence of acetyl CoA carboxylase, this process is catalyzed by fatty acid synthase (FAS). FAS has several enzymatic activities, including acetyl transferase, malonyl transferase, and reduction enzymes.
In the initial step of fatty acid synthesis, acetyl CoA and malonyl CoA condense to form acetoacetyl ACP (acyl carrier protein). Then, a series of enzyme-catalyzed reactions occur, resulting in the elongation of the fatty acid chain.
Since [2-14C]-acetyl CoA is supplied, the labeled carbon will be incorporated at the second position of the growing fatty acid chain.
b) If 2,2-dideuteromalonyl CoA is supplied, it means that the malonyl CoA used in the synthesis contains two deuterium atoms. During the condensation reaction, one of the deuterium atoms will be incorporated at the β-carbon of the malonyl CoA. As a result, one deuterium atom will appear at the β-carbon of palmitate.
c) If [3-14C]-malonyl CoA is used, the labeled carbon will be incorporated at the third position of the growing fatty acid chain.
d) The net conversion of some carbons in an odd-chain fatty acid to carbohydrate occurs through a process called gluconeogenesis. In this pathway, the odd-numbered fatty acid is first converted to propionyl-CoA. Propionyl-CoA is then carboxylated to form methylmalonyl-CoA, followed by rearrangement to form succinyl-CoA. Succinyl-CoA enters the citric acid cycle and undergoes further processing to ultimately produce oxaloacetate, a key intermediate in gluconeogenesis.
2. To determine the percentage of energy needs met by protein degradation, we need to calculate the energy content of urea and protein.
Given: 40g of urea is excreted, and the protein is 16% nitrogen by weight, yielding 18kJ of energy per gram.
The first step is to calculate the grams of nitrogen excreted by urea: Since urea contains 46.7% nitrogen, we can calculate:
40g urea * (0.467 * (1/14)) = 0.995g nitrogen.
To calculate the protein that yields this amount of nitrogen:
0.995g nitrogen * (1 / 0.16) = 6.21875g protein.
Finally, to calculate the energy provided by protein degradation:
6.21875g protein * 18kJ/g = 112.139 kJ.
Therefore, the percentage of the mammal's energy needs met by protein degradation would be:
(112.139 kJ / 10,000 kJ) * 100% = 1.12%.
3. The presence of isotopic label at both C2 and C4 of isocitrate indicates the presence of label from [3-14C]-pyruvate in these positions. The synthesis of isocitrate involves several steps, including the conversion of pyruvate to oxaloacetate (via pyruvate carboxylase), then the conversion of oxaloacetate to citrate (via citrate synthase). Citrate undergoes isomerization to isocitrate through the enzymatic activity of aconitase.
In this case, the presence of label at C2 and C4 suggests that the labeled pyruvate was incorporated during the formation of oxaloacetate. The labeling pattern suggests that both carbon atoms of pyruvate were retained during the carboxylation reaction, leading to the labeled oxaloacetate. This labeled oxaloacetate then proceeds through the citrate synthase and aconitase reactions to produce labeled isocitrate with label incorporation at both C2 and C4.
I hope these explanations help you understand how to approach each question! Let me know if you have any further doubts.