Question

help!!!

Lab 3 Using the blood glucose calibration curve and your knowledge of what a diabetics glucose levels are, your job will now be to determine whether or not insulin is needed. You are going to imagine that you are a person with diabetes, but you've broken your meter. Luckily you had all of the reagents needed to test your blood glucose tests on your own. Please determine what your blood glucose level is at each time interval and determine whether or not you should take insulin. PRE Answer 1 of 1 Done 84 1.6 1- 1-2- 1.0 1 6-8 0.6 0.4 a-1+ ASD o 0 Lab 3 Fill in the glucose and insulin columns for each activity. Absorbance Glucose Insulin Time Activity (mg/dl) (nm) (Y/N) 12:00pm Eat lunch Send memo, also check meter 1:00pm 1.49 2:00pm Nap at desk 0.75 3:00pm Wake up to check meter Watch Youtube while pretending to work 0.65 4:00pm Drive home, check level in traffic 0.55 5:00pm Su can heip rigne 065 utube while pretending to work

Answer 1

To find the concentration from absorbance value, we have to find the equation of straight line.

C(mg/dL) Absorbance 50 0.4 75 0.5 100 0.8 0.9 125 150 1.1 175 1.3 200 1,5 Absorbance Versus C(mg/dL) 1.6 1.4 y 0.007x R2 0.991 1.2 1 0.8 Absorbance 0.6 Linear (Absorbance) 0.4 0.2 0 0 100 150 200 250 C(mg/dL)> Absorbance- 50 LD t

From the above graph, the equation of straight line is: Y = 0.007X + 0 ---- (1)

The formula for absorbance is A = EbC + A_o ------ (1)

where C = concentration of glucose in mg/dL

Ao = absorbance of the standard solution.

Comparing equation(1) and (2)

slope,m = Eb = 0.007

and A_o = Y-intercept = 0

1.00 PM:

Given A = 1.49

A = EbC + A_o

=> 1.49 = 0.007*C + 0

=> 0.007*C = 1.49

=> C = 1.49/0.007 = 212.9 mg/dL

Glucose concentration = 212.9 mg/dL

3.00 PM:

Given A = 0.75

A = EbC + A_o

=> 0.75 = 0.007*C + 0

=> 0.007*C = 0.75

=> C = 0.75/0.007 = 107.1 mg/dL

Glucose concentration = 107.1 mg/dL

4.00 PM:

Given A = 0.65

A = EbC + A_o

=> 0.65 = 0.007*C + 0

=> 0.007*C = 0.65

=> C = 0.65/0.007 = 92.9 mg/dL

Glucose concentration = 92.9 mg/dL

5.00 PM:

Given A = 0.55

A = EbC + A_o

=> 0.55 = 0.007*C + 0

=> 0.007*C = 0.55

=> C = 0.55/0.007 = 78.6 mg/dL

Glucose concentration = 78.6 mg/dL