Question: need help on 4,5,6,&7 please! all the information is there. if you dont know how to do it let someone who knows, answer please. 4.

need help on 4,5,6,&7 please!  need help on 4,5,6,&7 please! all the information is there. if
you dont know how to do it let someone who knows, answer
all the information is there. if you dont know how to do it let someone who knows, answer please.

4. For each situation in Model 1, consider an inhibitor that is better than the one shown on graph. Answer the following questions for each type of inhibition: a. How would Kapp change? b. How would V app change? max Model 2 Real Data of Enzyme inhibition [S] (mmol/L) 1.25 No Inhibitor 1.72 Velocity Inhibitor A 0.98 1.17 1.47 1.67 2.04 2.50 [immol/L)min") Inhibitor B 3mM Inhibitor B 5mM 1.25 1.01 1.54 1.26 2.00 1.72 2.86 2.56 3.70 3.49 2.63 5.00 3.33 4.17 1.96 10.00 2.38 1.20 1.00 0.80 0.60 no inhibitor inhibitor A inhibitor B 3mM *Inhibitor B 5mm 0.40 0.20 -0.00- 0.00 0.20 0.80 1.00 0.40 0.60 11[5] c. Draw a line on the graphs that would represent the new, better inhibitor. 5. To make meaningful interpretations of data from double-reciprocal plots, the lines need to extend through the y-axis to the x-axis. Draw those extensions on the graph. Label the x- and y-intercepts with the appropriate terms, V. mar Ky, V "PP, KPP. The label could be negative or inverse of the terms listed. 6. Using the equations given below, calculate the values of V max. Ky VPP, and Ky, PP for each line shown in Model 2. Have each group member be responsible for one line. Compare answers and complete the table. Ku KAP V app No Inhibitor y=0.4843x + 0.1951 Inhibitor A y = 0.8582x + 0.3369 Inhibitor B 3mM y = 0.755x + 0.1969 Inhibitor B 5mM y = 1.006x + 0.1861 7. Predict the type of inhibition for each inhibitor in Model 2

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