W 9 - U = CHE112 Mid Term Exam (Last saved by user) - Microsoft Word non-commerc
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W 9 - U = CHE112 Mid Term Exam (Last saved by user) - Microsoft Word non-commercial use File Home Insert Page Layout References Mailings Review View Add-ins O X Insulin resistance occurs when cells no longer bind insulin in a normal manner, thus inhibiting the cells from transporting glucose inside the cell. Without intercellular glucose the cells cannot produce energy to support life. Let's suppose part of your project requires you to study the equilibrium reaction between the new variation of insulin you synthesized and cell receptors on the walls of cells. This is the equ expression you propose. cell receptor + free insulin f cell receptor-insulin complex (Equation 1) or in abbreviated terms R+it R- You devise a method of measuring the free insulin in solution. You run two experiments under the same conditions, except one experiment uses your modified insulin and the other experiment uses normal insulin. Table 1 below provides the initial conditions of your experiment and Table 2 provides the data you collected after equilibrium was established. Initial Conditions for both experiments-all environmental conditions are kept constant (e.g. temperature at 34°C, pressure at 1 atm, pH at 7.4). A single stock nutrient and cell receptor solution was prepared with a concentration of 70 uM. 50 mL the solution was used in all experiments. The insulins were added as powders, so the volumes essentially remained unchanged. Table 1: Initial Experimental Conditions Experiment1 Experiment Control [free modified insulin] = 100.0 uM [free normal insulin] = 100.0 UM (cell receptor) = 70.0 uM (cell receptor) = 70.0 uM Table 2: Unbound Free Modified Insulin and Unbound Free Normal Insulin Levels at Equilibrium Experimenti Experimental Control [free modified insulin] = 35.0 um [free normal insulin] = 50.0 uM Page: 1 of 5 Words: 26/1,128 0 E R 2 = 100% 0-0 12:25 PM O Type here to search O O O O 3 O Ö N D A ~ 40 d* 3/8/2018Explanation / Answer
The equilibrium conditions specify that the rate of formation of the insulin-receptor complex is equal to the rate of dissociation of insulin receptor complex.
The rate of formation of Insulin-receptor complex =k [I] [R], where k is rate constant for the forward reaction.
The rate of dissociation of Insulin-receptor complex = k' [I-R], where k' is rate constant for the reverse reaction.
Hence, k [ I ] [ R ] = k' [ I-R ]
or k / k' = [ I ] [ R ] / [ I-R ]
or K = k/k' = [ I ] [ R ] / [ I-R ]
or Keq = [ I ] [ R ] / [ I-R ]
Answer B.
The equilibrium constant for modified insulin can be calculated as,
Kqe mod = [ 35 ] [ 70 ] / [ 65 ] = 37.69
Kqe nor = [ 50 ] [ 70 ] / [ 50 ] = 70.00
Answer C.
The modified type of insulin binds more tightly because it has a low equilibrium constant which means the time reached to bind insulin to the receptor is less as compared to dissociation of insulin-receptor binding.
Answer D.
The new equilibrium concentration of free modified insulin would be,
37.69 = [ I ] [ 50 ] / [ 50 ]
[ I ] = 37.69 mol/L
Question 4
Answer A.
The rate of disappearance of insulin can be written as;
- d [ I ] / dt = k [ Io ]
The rate of appearance of insulin-receptor complex can be written as;
d [ I-R ] / dt = k [ Io ] [ R ]
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