Isothermal Titration Calorimetry (ITC) is a classical thermodynamic method for determining biomolecular interactions, renowned as the "gold standard" for biomolecular interaction studies due to its high sensitivity and accuracy. Da Congming discovered in experiments that two chemically synthesized small molecules, $A$ and $B$, can inhibit the function of enzyme $E$. Anxious, he planned to directly use ITC to measure and compare the binding abilities of the two molecules to the enzyme, conducting the following experiments:

1.  Titration of $15$ ${\mu} M$ protein $E$ directly with $250$ ${\mu}M A$. The fitting results yielded a stoichiometry ratio of $n=1$, $ΔH =-7.6$ $kcal · mol ^{-1}$, but failed to fit $K_d$.
2.  Titration of $15$ ${\mu} M$ protein $E$ directly with $250$ ${\mu} B$. The fitting results yielded $n=1$, $ΔH = 7.3$ $kcal · mol ^{-1}$, $K_d = 0.41$ ${\mu}M$.
3.  Mixing a 10-fold excess of $B$ with $15$ ${\mu} M$ $E$, and then titrating with $250$ ${\mu} M$ $A$. Under these conditions, the measured values were $n=0.87$, $ΔH = -4.8$ $kcal · mol ^{-1}$, $K_d = 15.5$ $nM$.

Based on the above information, which of the following options is most likely the binding constant $K_a$ of $A$ and $E$?
等温量热滴定（ITC)是经典的利用热力学方法测定生物分子相互作用的方法之一，因其高灵敏度和准确性被誉为生物分子相互作用的“金标准”。大聪明在实验中发现两种化学合成的小分子$A$、$B$能够抑制酶$E$的功能，心急的他计划直接用ITC对两分子对酶的结合能力进行测定与比较，进行如下实验：

1. 用$250$ ${\mu}M$ $A$直接对$15$ ${\mu} M$的蛋白$E$进行滴定，拟合结果可得化学计量比$n=1$，$ΔH =-7.6$ $kcal · mol ^{-1}$，但未能拟合出$K_d$。
2. 用$250$ ${\mu}M$ $B$直接对$15$ ${\mu} M$的蛋白$E$进行滴定，拟合可得$n=1$，$ΔH = 7.3$ $kcal · mol ^{-1}$，$K_d = 0.41$ ${\mu}M$。
3. 将过量10倍的$B$与$15$ ${\mu} M$ $E$混和，然后用$250$ ${\mu} M$ $A$进行滴定。在此条件下测得$n=0.87$，$ΔH = -4.8$ $kcal · mol ^{-1}$，$K_d = 15.5$ $nM$。

根据以上信息，下列选项中最有可能是$A$与$E$的结合常数$K_a$的是？
$2.36\times10^{9}$
$2.42\times10^{9}$
$2.28\times10^{10}$
$2.37\times10^{10}$
$3.94\times10^{10}$
$3.94\times10^{11}$
Insufficient information to calculate.
$2.36\times10^{9}$
$2.42\times10^{9}$
$2.28\times10^{10}$
$2.37\times10^{10}$
$3.94\times10^{10}$
$3.94\times10^{11}$
条件不足，无法计算
This question examines the common competition analysis method in ITC analysis. When the binding between a protein and a small molecule is too strong, the ITC method may not be able to accurately fit $K_d$ due to sharp signal spikes. A feasible approach to solving this problem is to reduce the protein concentration to decrease the c-value. However, when the protein concentration is too low, the ITC heat signal will be masked by noise, making it impossible to obtain an accurate titration curve. Therefore, competing with a known inhibitor and then titrating the complex has become an effective analytical method.

Considering a side reaction caused by $B$, it is easy to write the side reaction coefficient caused by $B$ as ${\alpha}_{B} = \frac{[E]+[EB]}{[E]} = 1 +\frac{B}{K_d^B}$. The true $K_d$ of $A$ is $K_{app} \times {\alpha}_{B} = 4.2 \times 10^{-11} M$, so $K_a = 2.37 \times 10^{10}$. At this time, option C is satisfied.

<Checkpoint>[If A and B conform to the competition model, then $K_a^A = 2.37 \times 10^{10}$](1)</Checkpoint>

However, it should be noted that the question does not specify whether there is a competitive relationship between small molecules $A$ and $B$. We can focus on the $\Delta H$ values in the ITC experiment. The binding of $A$ has $ΔH =-7.6 kcal · mol ^{-1}$, and the binding of $B$ has $ΔH = 7.3 kcal · mol ^{-1}$. If the two have a competitive relationship, then the reaction formula is $EB \longrightarrow B+E$, $E+ A \longrightarrow EA$, and the theoretical $ΔH =-7.3 + (-7.6) = -14.9 kcal · mol ^{-1}$. The measured $ΔH = -4.8 kcal · mol ^{-1}$ indicates that the binding of $B$ has a weak effect on the thermal effect of $A$, which is obviously inconsistent with the competition model. A more reasonable explanation is that the two have a synergistic effect. The binding of molecule $B$ causes a certain large conformational change in the protein, which affects the binding of $A$ at another site, resulting in a change in the thermal effect and a weakening of the binding, which can be measured by ITC. Therefore, it is obviously incorrect to directly apply the ITC formula for competitive binding. The impatient and clever student did not verify the competitive relationship between the two compounds before performing the ITC experiment. This is unscientific, so the $K_d$ of $A$ cannot be calculated in this question.

<Checkpoint>[The thermal effect data in ITC indicate that the two molecules should not conform to the competition model, so it cannot be calculated](1)</Checkpoint>
本题考察的是ITC分析中常见的竞争分析法。当蛋白与小分子的结合过强时，ITC方法会由于信号突跃尖锐而无法准确拟合$K_d$。一种可行的解决问题的思路是降低蛋白浓度以减小c值，但是当蛋白浓度过低时，ITC的热量信号会被噪音掩盖，无法获得准确的滴定曲线。因此用一种已知的抑制剂进行竞争，然后对该复合物进行滴定成为了一种行之有效的分析方法。

考虑一个由$B$引起的副反应，很容易写出$B$引起的副反应系数${\alpha}_{B} = \frac{[E]+[EB]}{[E]} = 1 +\frac{B}{K_d^B}$，$A$的真实$K_d$为$K_{app} \times {\alpha}_{B} = 4.2 \times 10^{-11}$ $M$，故$K_a = 2.37 \times 10^{10}$。 此时C选项满足。

假如A与B符合竞争模型，此时$K_a^A = 2.37 \times 10^{10}$

然而注意到，题目中并未说明小分子$A$与$B$是否存在竞争关系.我们可以关注到ITC实验中的$\Delta H$数值。$A$的结合$ΔH =-7.6$ $kcal · mol ^{-1}$，$B$的结合$ΔH = 7.3$ $kcal · mol ^{-1}$；若两者为竞争关系，则反应式为$EB \longrightarrow  B+E$，$E+ A \longrightarrow EA$，理论$ΔH =-7.3 + (-7.6) = -14.9$ $kcal · mol ^{-1}$，实测$ΔH = -4.8$ $kcal · mol ^{-1}$，说明$B$的结合对$A$的热效应有微弱影响，显然与竞争模型不符合。

<Checkpoint>[假如A与B符合竞争模型，此时理论$ΔH = -14.9$ $kcal · mol ^{-1}$](1)</Checkpoint>

<Checkpoint>[ITC中的热效应数据表明两分子不应符合竞争模型，故不能计算](1)</Checkpoint>

更合理的解释是两者存在协同效应。分子$B$的结合导致蛋白出现某种较大的构象变化，影响了$A$在另一位点的结合，导致热效应变化，同时结合变弱，可以被ITC测出。因此直接套用竞争结合的ITC公式显然是不正确的。心急的大聪明同学没有验证两个化合物的竞争关系，就进行了ITC实验。这是不科学的，因此本题无法计算$A$的$K_d$。