Description
3. 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 equilibrium expression you propose.
cell receptor + free insulin cell receptor-insulin complex
or in abbreviated terms
R + I R-I
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 34oC, 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
Experiment 1
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
Experiment 1
Experimental Control
[free modified insulin] = 35.0 uM
[free normal insulin] = 50.0 uM
A. Write the equilibrium expression for this reaction using [I] to represent the insulin concentration, [R] the cell receptor concentration, and [I-R] to represent the cell receptor-insulin complex concentration. (10 pts) Show all work.
B. Determine an equilibrium constant for both the modified insulin and the normal insulin reaction at 34oC based on the data above. (10 pts)
[free modified insulin]
[receptor]
[Insulin-receptor]
I
C
E
[free normal insulin]
[receptor]
[Insulin-receptor]
I
C
E
C. Which type of insulin binds more strongly to the cell receptor? Explain your answer. (10 pts)
D. Suppose the concentration of the receptor in the stock solution dropped to 50.0 uM. If you ran the experiment again with the same initial concentration of free modified insulin as above, what would you expect the free modified insulin's equilibrium concentration to be? (10 pts)
[free modified insulin]
[receptor]
[Insulin-receptor]
I
C
E
4. The equilibrium results on your modified insulin look promising, but it seems like it took forever for the reaction to reach equilibrium. Your supervisor now wants you to study the reaction rate for the binding of your modified insulin to the cell receptor. Show all work.
- Using the chemical equation from Question 3 write the mathematical relationship between the rate of disappearance of free insulin and the rate of appearance of the insulin-receptor complex. (10 pts)
- As mentioned above, you devised a method for measuring the concentration of free insulin in solution. You are able to get an estimate of the initial rate of reaction by adding a known concentration of insulin, [I]o,to a known concentration of receptor, [Ro] and then taking a measurement of the free insulin, [It], 60 seconds after mixing. From the data in Table 3 calculate the initial rate for each experiment in uM/s. Show all work. (10 pts)
Table 3: Initial rate data for insulin/receptor binding. [It] measured after 60 seconds.
[Io]
[Ro]
[It]
Initial Rate (uM/s)
Experiment 1
100 uM
70.0 uM
95.0 uM
Experiment 2
100 uM
35.0 uM
95.0 uM
Experiment 3
50.0 uM
70.0 uM
47.5 uM
- How does the initial rate of reaction change with changing insulin concentration? How does it change with changing receptor concentration? (5 pts)
- Write the rate law for this reaction with its correct reaction orders and calculate the rate constant. Show your calculation. (10 pts)
- If modified insulin has a higher equilibrium constant than normal insulin does that mean the modified insulin will have a higher reaction rate than normal insulin? Explain your answer. (5 pts
Explanation & Answer
Hey buddy, uploading the answers again. :)
2. It is your first week in the lab and you are given the task of modifying insulin to make it more hydrophilic
by modifying one or more of its amino acids. You notice there are several serine amino acids in the
structure of insulin. Show all work.
A. Show the chemical reaction for the oxidation of serine to form the amino acid aspartic acid. Look up
the reagent that could be used for this reaction (citation). (5 pts)
Aspartic acid has a carboxylic acid side chain and a carbon longer than serine. For the oxidation of serine,
we can think of a series of organic reactions where the hydroxyl group of serine is initially converted into
a reactive species. This reactive species can then undergo a substitution reaction to generate a carboxylic
acid which is a carbon longer from the original substrate.
The following figures below were obtained from the 6th edition of Organic Chemistry by Paula Y. Bruice
pp 416, 417, and 746, respectively.
To convert the alcohol group of serine, tosylation reaction can be first performed using common tosylating
reagents such as p-toluenesulfonyl chloride, to generate a sulfonate ester similar to the figure shown
below: (Bruice, p.416)
After tosylation, nucleophilic substitution can then be performed using NaCN to generate the nitrile
derivative of serine. The reaction should be similar to this one (Bruice, p. 417).
Finally, once the nitrile derivative is obtained, acid-catalyzed nitrile hydrolysis can be performed to
generate the Aspartic acid. The reaction should be similar to the reaction shown below (Bruice, p 746).
Reagents:
(1) p-toluene sulfonyl chloride
(2) NaCN
(3) H2O, HCl, heat
B. You need to make up 50 mL a 0.25 M solution of the oxidizing reagent used in part A to perform the
reaction. Show the calculations and describe the procedure for making up this solution. (5 pts)
Since the reaction above is not a straightforward oxidation process as you might have expected, I will
just use NaCN for the calculation, the species that will undergo oxidation to generate the carboxylic
acid group.
𝑚𝑜𝑙 𝑁𝑎𝐶𝑁 = 0.25 𝑀 ∗ 0.050 𝐿 = 0.0125 𝑚𝑜𝑙 𝑁𝑎𝐶𝑁
𝑚𝑎𝑠𝑠 𝑁𝑎𝐶𝑁 = 0.0125 𝑚𝑜𝑙 𝑁𝑎𝐶𝑁 ∗
49.01 𝑔 𝑁𝑎𝐶𝑁
= 0.613 𝑔 𝑁𝑎𝐶𝑁
1 𝑚𝑜𝑙 𝑁𝑎𝐶𝑁
To prepare 50 mL of 0.25 M NaCN solution, carefully dissolve 0.613 g of NaCN in 50 mL water.
C. Why would this reaction make insulin more hydrophilic? (5 pts)
The alcohol group in serine is converted to a carboxylic acid functionality. This functionality is highly
polar and it can form at least two hydrogen bonds with water as opposed to serine, in which its OH
group can only form one. More H-bonds, higher interaction with water, increase in hydrophilicity.
D. Would you have to be concerned with forming and geometric isomers in converting serine to aspartic
acid? Explain your answer. (5 pts)
No. Based on the reaction shown above, there is only one way to attache the carboxylic acid
functionality of aspartic acid to the CH2 group of serine. Thus, the cha...
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