Question 3 (6 points) A) T2R proteins are bitter taste receptors that signal thr
ID: 85718 • Letter: Q
Question
Question 3 (6 points) A) T2R proteins are bitter taste receptors that signal through Go Gustducin, and there are at least 25 known T2Rs. You have discovered a new T2R and find that it binds to the toxic plant compound quinine with a Kd of 45 nM. a) You create a cell line expressing this receptor and expose it to a solution of 90 nM quinine. What percentage of receptor proteins are bound to quinine at equilibrium? If there are 1000 of this T2R per cell, how many per cell are bound? (6 points) In contrast to your new T2R, the Kd off the known TAS2R quinine receptor is 300 nM. You discover that your new T2R gene is expressed predominantly in an indigenous population in Australia where the local diet includes a unique species of fruit. This fruit is highly nutritious, although it has a non-toxic level of quinine of approximately 5 nM, but there is a closely related species of fruit that looks similar but contains a number of toxic compounds, including a quinine content of approximately 30 nM. Provide a quantitative explanation for why this population might express your new receptor. (6 points)Explanation / Answer
This question has multiple questions. As per Chegg’s policy, I should answer the first question. Here I am answering the first two questions
a)
Kd is 45 nM.
Concentration of ligand is 90 nM.
So, ligand concentration is two times of Kd.
Kd = [R][L]/[RL]
Kd = [R]2Kd/[RL]
1/2 = [R]/[RL]
[RL] = 2[R]
The fraction of receptor molecules in complex will be:
[RL]/[R]total
= [RL]/([RL]+[R])
= 2[R]/3[R]
= 2/3
= 0.666
Therefore, 66.6% of receptor molecules will be in complex.
If there are 1000 T2R molecules on cell, 666 will be bound to ligand.
b)
Kd of TAS2R is 300 nM.
Kd = [R][L]/[RL]
So, higher the Kd means, higher the concentration of free ligand and receptor. So, it requires more concentration of ligand to saturate the receptor.
Since this population feeds on fruit that has 30 nM quinine concentration, they should have a low affinity receptor like this. Otherwise, they will suffer from quinine toxicity.
Kd of TAS2R is 300 nM.
Quinine concentration in fruit is 30 nM. It is 10 times lesser than Kd.
Kd = [R][L]/[RL]
Kd = [R]0.1Kd/[RL]
1/0.1 = [R]/[RL]
10 = [R]/[RL]
[RL] = [R]/10
The fraction of receptor molecules in complex will be:
[RL]/[R]total
= [RL]/([RL]+[R])
= ([R]/10)/(([R]/10)+[R])
= ([R]/10)/(11[R]/10)
= [R]/11[R]
= 1/11
= 0.0909
Therefore, 9.09% of TAS2R receptor molecules will be in complex at 30nM concentration.
Let us calculate the occupied T2R receptors at 5 nM. It is non-toxic level.
Kd is 45 nM.
Concentration of ligand is 5 nM. This is 9 times less than Kd.
Kd = [R][L]/[RL]
Kd = [R]1/9Kd/[RL]
1/1/9 = [R]/[RL]
9 = [R]/[RL]
[RL] = [R]/9
The fraction of receptor molecules in complex will be:
[RL]/[R]total
= [RL]/([RL]+[R])
= ([R]/9)/(([R]/9)+[R])
= ([R]/9)/(10[R]/9)
= [R]/10[R]
= 1/10
= 0.1
Therefore, 10% of T2R receptor molecules will be in complex at 30nM concentration.
We can understand that with new receptor, eating more toxic fruit keeps 9% receptors bound whereas 10% receptors being in bound form is considered non-toxic. So, this population has TAS2R receptor which has low affinity (higher Kd) for ligand. So, they are protected from quinine toxicity.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.