# A question about ligand binding

I have done part a and part b but I keep getting a negative concentration for part c.

• For part a I got $$\pu{2 \mu M}$$
• For part b I got $$\pu{5.051 \mu M}$$

For part c, I get $$2=3.2 + \text{[bound ligand]}$$, but obviously that is wrong as you cannot have negative concentration for the bound ligand.

• I roughly calculate it as $\pu{0.793 \mu M}$. Is that closed to your answer (if given)? Oct 15 '20 at 22:11
• Hey there thanks for your reply. Did you use the same values I calculated in part a and part b for your answer of 0.793 cause I keep getting a negative answer when I try. Oct 15 '20 at 22:44
• No, I used the graph. The answers A and B are correct, but you must use the graph to answer Q3. Oct 15 '20 at 23:18
• You know the total amount of ligand, 5 micromole. You also know that 3.2 micromole are unbound. Isn't the remaining 1.8 micromole bound? And it is in 10 mL, so the concentration is 0.18 mM? Which is weird because the protein concentration is much lower. Oct 16 '20 at 2:26

You have calculated correctly for parts $$\bf{a}$$ and $$\bf{b}$$. However, to answer part $$\bf{c}$$, you need to use the given plot, the interception and slope of which can be calculated as follows:

The slope is $$\tan \alpha = - \frac{1}{K_d}$$:

$$\tan \alpha = \frac{1-0.2}{0.1-0.8} = -1.14 \ \Rightarrow \ \therefore \ \frac{1}{K_d} = 1.14$$

According to the given Scatchard equation (where each receptor has a single ligand binding site), interception of the plot is also equal to $$\frac{1}{K_d} = 1.14$$. Therefore, the Scatchard equation can be rewritten as:

$$\frac{r}{[A]} = 1.14(1-r) \tag1$$

where $$[A]$$ is actually free ligand concentration (Wikipedia). Therefore, when $$[A]= \pu{3.2 \mu M}$$, find the relevant $$r$$ using the equation $$(1)$$, which is the average amount of $$A$$ per $$\pu{mol}$$ of $$P$$ $$(r = \pu{0.785 \mu M})$$.

This means: Total number of binding site per receptor is $$1$$ and only $$0.789$$ is bound by $$A$$. Protein concentration in the dialysis tube $$= \pu{2.0 \mu M}$$

Therefore, $$A$$ bound concentration of $$P$$ $$= \pu{2.0 \mu M} \times 0.785 = \pu{1.57 \mu M}$$