PLEASE DO THIS EXERCISE REGARDING THE GRAPH Does the rate of glucose 6-phosphata

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PLEASE DO THIS EXERCISE REGARDING THE GRAPH

Does the rate of glucose 6-phosphatase activity change over time in isolated liver cells? Glucose 6-phosphatase, which is found in mammalian liver cells, is a key enzyme in control of blood glucose levels. The enzyme catalyzes the breakdown of glucose 6-phosphate into glucose and inorganic phosphate (Pi). These products are transported out of liver cells into the blood, increasing blood glucose levels. In this exercise, you will graph data from a time-course experiment that measured P concentration in the buffer outside isolated liver cells, thus indirectly measuring glucose 6-phosphatase activity inside the cells. Isolated rat liver cells were placed in a dish with buffer at physiological conditions (pH 7.4, 37°C). Glucose 6-phosphate (the substrate) was added to the dish, where it was taken up by the cells. Then a sample of buffer was removed every 5 minutes and the concentration of P, measured. The table shows the data collected at each time interval 400 300 (umol/ml) 200 100 0 Time (min) Concentration of (mol/mL) 10 90 180 270 330 355 355 355 [P 10 20 25 30 35 0 5 10 15 20 25 30 35 40 45 Data from s. R. Commerford et al., Diets enriched in sucrose or fat increase Time (min) and G-6-Pase but not basal glucose production in rats, American Physiology-Endocrinology and Metabolism 283:ES45-ESSS (2002). 8. A graph of the data is also shown. How would you describe the pattern you see in the graph? 9. Consider that the rate of enzyme activity is related to the slope of the line, Ay/Ax (the "rise" over the "run"), in 1mol/mL-min, with the steepest slope indicating the highest rate of enzyme activity. What part of the time course had the highest rate of enzvme activitv? 10. What is the approximate slope of the line between 5 and 20 minutes? When the substrate (glucose 6-phosphate) is first added to the liver cells, it is transported into the cells and processed by the enzyme (glucose 6-phosphatase). Then the product Pi is released from the cells into the buffer 11. Which of the following statements is the most reasonable biological explanation for the pattern vou see on the h?

Explanation / Answer

8. The amount of inorganic Phosphorus is released into buffer initially increases and after some time it attends a constant value. The constant value of Pi shows that no further glucose 6 phosphate is released into the blood stream as blood sugar (glucose) level is high.

9. Rate of enzyme activity = (conc. of Pi in umol/mL)/ Time period.

Hence rate of enzyme activity is maximum from 10-20.

10. The line of our interest in the graph is not a straight line. Hence we cannot find the slope directly. Therefore we will find the average slope of the lines.

Average slope of 3 lines, 5-10, 10-15, 15-20 (minutes). The slope of line between 5-10 is 16, 10-15 is 18 and 15-20 is 18. Hence the average slope is 17.33

Slope = (y1 - y2) / (x1 - x2). Note : This is an approximate value of slope.

11. Then the product Pi is released from the cells into the buffer. Reason - The graph shows concentration of Pi in buffer Vs time. Hence we are more concerned about conc. of Pi in buffer.

Time Period delta T delta conc. Pi Rate of enzyme activity 0-5 5 10 2 5-10 5 80 16 10-15 5 90 18 15-20 5 90 18 20-25 5 60 12 25-30 5 25 5 30-35 5 0 0 35-40 5 0 0

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