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$$\ce{Biotin + Avidin → Complex}$$

Biotin is a vitamin derivative that binds extremely tightly to avidin (a protein found in egg white) with a $K_\mathrm{d}$ of $\pu{1e-15 M}$.

If you make a $\pu{2 mL}$ solution of $\pu{10 μL}$ biotin and $\pu{2 μL}$ avidin, what fraction of biotin molecules will remain free at equilibrium?

Answer Key

The TA found the

$$K_\mathrm{d,new} = \frac{[\text{Biotin}][\text{Avidin}]}{[\text{Complex}]}$$

using the given initial reactant concentration values and making an ICE table to find the equilibrium concentrations and found $K_\mathrm{d,new} = \pu{1e-9}.$

$$K_\mathrm{d} = \frac{(10 - δ)(2 - δ)}{δ} = 10^{-9}$$

Where does the TA get that $K_\mathrm{d} = \pu{1e-9 μL}$? Don't you need to either know the delta term or have the value of $ΔG$ of reaction? Neither of which was given in the problem statement. Also, why did TA find another $K_\mathrm{d,new}$? Doesn't the provided $K_\mathrm{d}$ already describe the dissociation, which gives the concentration of reactants on top?

I understand that $K_\mathrm{d}$ is the equilibrium constant which describes dissociation.

My solution

My solution was to find

$$f_\mathrm{empty} = \frac{[\ce{RL}]}{[\ce{R}] + [\ce{RL}]}$$

I rewrote the $f_\mathrm{empty}$ relation to be in terms of the known values, the provided $K_\mathrm{d}$ and the concentration values.

I found $f_\mathrm{empty}$ to be $0.0009999$ which is also essentially zero, meaning that $f_\mathrm{bound}$ is essentially $1$, which is also what the TA found.

The TA's approach wasn't intuitive for me, so if someone could help clarify it?

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  • $\begingroup$ Why not to ask TA directly? TAs are here to help you understand stuff, not to invent riddles:) $\endgroup$ – andselisk May 7 at 4:07
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    $\begingroup$ They didn't respond to my questions so that's why I'm asking the internet. They've been spending the last week grading the last hw, and haven't responded to questions regarding that either. $\endgroup$ – ThermoRestart May 7 at 4:08
  • $\begingroup$ If you invert the equation then $\displaystyle K=\frac{\delta}{(10-\delta)(2-\delta)}$. If you solve with different $K$ then even if $K$ is small, such as $K=100 ,\, \delta=1.998$. So with a huge $K,\, \delta=2$ and the amount of reactant present is infinitesimal. Thus it makes no difference $10^9, \to 10^{15}$ $\endgroup$ – porphyrin May 7 at 8:38
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    $\begingroup$ It's a question of units: $K_d=10^{15}$ is defined with a M reference state. Presumably it has to do with you defining concentrations in "μL" which leads to a factor of $10^6$ somewhere, but I am not familiar with such use of volume units in place of concentrations, so I am a little confused how you can use the equation as written without further assumptions. Sometimes "units" is used as the unit for enzymes, but in this case I am guessing "L" means liters. $\endgroup$ – Buck Thorn May 7 at 10:03

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