Question

i need help with part two the open ended section

Patients with chronic hypoxia (low 0, levels) due to decreased lung function may adapt by increasing their circulating BPG levels, Predict which outcome will be TRUE for such a patient. "A) The R-state of hemoglobin will be favored. B) O, binding to hemoglobin will be hyperbolic. C) Pse for O2 will be increased D) Pse for O2 will be decreased. B) None of the answers is correct. 14. Which amino acid in hemoglobin is involved in coordinating iron in addition to the heme prosthetic group? A) Glu B) Cys C) Val D) Tyr His 15. Which molecule binds MOST strongly to the heme iron? A) CO2 BCO C) H2O D) N2 E) O2 PART TWO 1) The steady-state kinetics of an enzyme is studied in the absence and presence of inhibitor A. The initial rate is given as a function of substrate concentration in the following table. I [S] (mm) 1.25 1.67 2.50 4 5.00 10.00 Velocity in substrate only 3 Velocity in substrate + (MM/min) inhibitor A mM min) .$ 1.72 0.98 1.02 598 .60 2.04 1.17 . 5 2 .63 38 1.47 . .2 3.33 30 1 .96 51 - 4.17 24 2 .38 42 (a) Draw the Lineweaver-Burk plot on the graph paper provided using the kinetic data above. Label all the axes on your graph. (5 pts) k On back attachment to (b) Using the above Lineweaver-Burk plot, determine V. and Kin the absence and presence of inhibitor A. Also, determine the type of inhibition involved? (5 pts) Non compeli we inhibition Page 4

Answer 1

The plot of 1/v₀ versus 1/[S] both in the presence and in the absence of inhibitor can be drawn as follows.

Presence of inhibitor Absence of inhibitor 1v, (min/mm) 0.0 0.1 0.2 0.3 0.6 0.7 0.8 0.9 0.4 0.5 1[S] {mM")

Equations of the lines:

Absence of inhibitor: 1/v₀ = 0.48431*1/[S] + 0.19509

Presence of inhibitor: 1/v₀ = 0.85816*1/[S] + 0.33693

Now, compare the above two equations with the Line-weaver Burk equation 1/v₀ = (K_M/V_max)*1/[S] + (1/V_max)

Absence of inhibitor: K_M/V_max = 0.48431, 1/V_max = 0.19509, i.e. V_max = 5.126 mM/min and K_M = 2.482 mM

Presence of inhibitor: K_M/V_max = 0.85816, 1/V_max = 0.33693, i.e. V_max = 2.968 mM/min and K_M = 2.547 mM

Therefore, it is Non-competitive inhibition