Question

(c) Describe the hormone and signaling protein combination that produced the greatest and most prolonged response. For the phosphorylated STAT5 response that reached 100 percent at 10 minutes in figure 1A, calculate the rate of decrease (percent decrease per minute) in the detected level of phosphorylated STAT5 from 10 to 30 minutes.

Answer 1

Answer:

Insulin and protein kinase B interaction is the greatest and most prolonged response as is seen in the graph. Protein kinase B level reached to 100 after 30 minutes as insulin added while the level remained to near 80 even after 240 minutes.

The level of STAT 5 at the 10th minute = 100

The level of STAT 5 at 30th minute = 35

From 10 to 30 minutes, the level of STAT 5 = 100 – 35

From 10 to 30 minutes, the level of STAT 5 = 65

Duration = 30 – 10 minutes

Duration = 20 minutes

Rate of decrease = 65/20

Rate of decrease = 3.25 levels/min

Answer 2

Insulin and Protein kinase B interact, they so the greatest and prolonged response. The rate of decrease (percent decrease per minute) in the detected level of phosphorylated STAT5 from 10 to 30 minutes is 3.25 percent/min.

On the basis of given graph, the Figure B show the greatest and prolonged response.

• when insulin and Protein kinase B interact, they so the greatest and prolonged response.

• The STAT5 response is faster but not prolonged.

The decrease in the level of STAT5 from 10 to 30 minutes.

        STAT5 level at 10 minute = 100

        STAT5 level at 30 minute = 35

         Decrease between 10 to 30 minute = 65

         The rate of decrease =  $\frac{65}{20}$

Hence, Rate of decrease between 10 to 30 min = 3.25 percent/min

Therefore we can conclude that the rate of decrease in the level of STAT5 is 3.25 percent/minute.

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