THANK YOU SO MUCH, LAST MINUTE QUESTIONS I\'M NOT TOO SURE ABOUT, EXAM TOMORROW!

Question

THANK YOU SO MUCH, LAST MINUTE QUESTIONS I'M NOT TOO SURE ABOUT, EXAM TOMORROW! THNANK YOU SO MUCH

1. I manage to recreate the glycolytic pathway in vitro using only purified components. I start the pathway by adding ATP that is radiolabeled using radioactive Phosphorus to the in vitro reaction. MOST LIKELY, predict all the molecules that will get radiolabeled when the pathway reaches equilibrium (assuming I am not replenishing the NAD+ present in the reaction). Explain your answer.

2. The enzyme Phosphoglucomutase (PGM) converts Glu-6-PO4 to Glu-1-PO4 and is the first step for the formation of glycogen. It’s Km is 60?M. The Km of Phosphoglucoisomerase (PGI), on the other hand, is 7 ?M. I make a mutation in PGM that changes its Km to 6?M. What would be the physiological consequences of this mutation?

3. In order to make DNA for replication, a cell must use up some TCA intermediates, which would decrease the TCA. But making DNA also requires energy. So how does the cell overcome this paradox of needing energy, but taking away intermediates from the cycle that provides energy? Explain your answer. ?

Explanation / Answer

1. During glycolysis, only 2 molecules are phosphorylated using ATP, which are glucose and fructose 6-phosphate, but the 2 radioactive phosphorus atoms added will remain in the molecules even if the molecules are transformed to something else, which means the following molecules will be radiolabeled:

glucose-6-phosphate, fructose-6-phosphate, fructose-1,6-biphosphate, dihydroxyacetone phosphate, glyceraldehyde-3-phosphate, 1,3-bisphosphoglycerate, 3-phosphoglycerate, 2-phosphoglycerate, and phosphoenolpyruvate.

2. Phosphoglucomutase (PGM) -> 1st step of glycogen synthesis (glycogenolysis) -> Km = 60?M. Mutated PGM -> Km = 6?M. Phosphoglucoisomerase (PGI) -> 1st step of glycolysis-> Km = 7 ?M. Km = Michaelis Menten constant, Km is inversely proportional to substrate affinity, an enzyme with high low Km has a high affinity for its substrate, therefore the mutation in PGM decreased the Km from 60?M to 6?M, increasing 10 times its affinity for glucose-6-phosphate. As the Km of the mutated PGM is lower than the Km of PGI, glycogenolysis will be favored over glycolysis, which is contrary to how it is.

3.  When intermediates of the tricarboxylic acid cycle are used for biosynthetic processes and therefore they don't enter the electron transport chain to produce a lot of ATP, the cell enters an anaerobic glycolysis to still be able to produce energy. 9The cell can still get sufficient energy (ATP) from the transformation of glucose to pyruvate (glycolysis), which yields 2 ATP molecules while still being able to use TCA intermediated molecules for DNA synthesis.

4. To see a growth over time in the pyruvate:glucose ratio, the amount of pyruvate must be higher than the amount of glucose. From the glycolysis of just one glucose molecule 2 pyruvates and 2 ATP molecules are produced which would be a 2:0 ratio (0 because the glucose was used), but when too much glucose is transformed into pyruvate the high amounts of ATP start to inhibit glycolysis by inhibiting the enzyme phosphofructokinase-1, therefore we either need an enzyme that uses the ATP and avoids its excessive accumulation for glycolysis to keep happening or an enzyme that reduces the glucose left. So the added protein is either phosphoglucomutase to reduce glucose molecules or pyruvate carboxylase that uses ATP to produce oxalacetate. It could also be insulin, but this assay is in vitro, an insulin increases glycolysis by allowing the entry of glucose to the cell.

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