Thank You 07. A scientist makes a mutant of a cell line and monitors the product
ID: 302680 • Letter: T
Question
Thank You
07. A scientist makes a mutant of a cell line and monitors the products of glycolysis in both the wildtype and the mutant versions under anerobic conditions. The scientist notiles the concentration of mutant cells do not survive while the wildtype do. To investigate, the scientist determines of the concentration of ATP vs time 1.2 0.8 shown in the graph for each cell line. What is the MOST LIKELY explanation for the results. Which enzyme is affected by the mutation? (15 points) OT 0.4 0.2 TimeExplanation / Answer
Anaerobic glycolysis is the process of breakdown of glucose in the absence of oxygen. It is a less efficient process which involves net gain of only two ATP molecules per glucose molecule. In this process ATPis gained when-
*1,3-bisphosphoglyceric acid is convered into 3-phosphoglyceric acid by enzyme phosphoglyceric trans phosphorylase.
* 2- phosphoenolpyruvic acid is converted into pyruvic acid by enzyme pyruvate kinase.
According to given graph ATP production is more in case of wild cells than mutant cells which tells that enzymes which are involved in ATP production are defective in mutant cells.
As we know that glycolysis process runs in series of steps therefore defect in any of these enzymes will not only reduces ATP production but also will not let the glycolysis to proceed further.
The enzymes defective in mutants are metioned above in * points by bold letter.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.