There is now a sample of [1-$^{14}\mathrm{C}$] glucose, in which $\mathrm{C}-1$ is labeled with the $^{14}\mathrm{C}$ isotope. Bacterial culture medium is added to it, and after a period of time, the following metabolic intermediates $i=1\sim 8$ are separated (with atom numbering):

<MultiModal>![Numbering of intermediates and their corresponding SMILES: 1. O[C@@:1]1([C:2]P([O-])([O-])=O)[C@@:3]([H])(O)[C@@:4](O)([H])[C@:5]([C:6]OP([O-])([O-])=O)([H])O1
2. O=[C:1][C@H:2](O)[C:3]OP(O)(O)=O
3. [C:1]=[C:2](OP(O)(O)=O)[C:3](O)=O
4. [C:1][C:2](S[C:3][C:4]N[C:5]([C:6][C:7]N[C:8]([C:9]([C:10]([C:11])([C:12]OP(OP(O[C:13][C@@H:14]1[C@@H:15](OP([O-])([O-])=O)[C@@H:16](O)[C@H:17](N2[C:18]=N[C:19]3=[C:20]2N=[C:21]N=[C:22]3N)O1)([O-])=O)([O-])=O)[C:23])O)=O)=O)=O
5. O=[C:1](O)[C:2][C:3]([C:4](O)=O)(O)[C:5][C:6](O)=O
6. O=[C:1]([C:2](O)=O)[C:3][C:4][C:5](O)=O
7. O=[C:1](O)[C:2][C:3][C:4](O)=O
8. O=[C:1]([C:2](O)=O)[C:3][C:4](O)=O
](/media/uploads/210734d5-f0f9-4b80-b449-f04478d8a4a1.png)</MultiModal>

For each metabolic intermediate $i$, based on your understanding of the glucose metabolic pathway, find the atom numbers of the sites where the $^{14}\mathrm{C}$ label may appear, and define a set $P_i$, which represents the set composed of the atom numbers of all carbon atoms that may be labeled with $^{14}\mathrm{C}$ in the $i$-th metabolic intermediate.

* For example: If in product $i$, the label may appear on carbon atoms 2 and 3, then $P_i = \{2, 3\}$.

The calculation rule for $z_i$ is divided into two cases according to the size of the set $P_i$ (i.e., the number of possible labeling sites):

$$
z_i = \begin{cases}
\dfrac{p}{i^2 + 1} & \text{if there is only one possible labeling site (i.e. } |P_i| = 1 \text{, and } P_i = \{p\}\text{)} \\
\\
\dfrac{\prod_{p \in P_i} p}{i^2} & \text{if there are multiple possible labeling sites (i.e. } |P_i| > 1\text{)}
\end{cases}
$$

* In the second case, $\prod_{p \in P_i} p$ represents the product of all atom numbers in the set $P_i$.

Finally, calculate the sum $S$ of all $z_i$ values:

$$
S = \sum_{i=1}^{8} z_i
$$

Choose the correct option from the following options, keeping the final result to four decimal places, and choose the option whose deviation from your calculation result is within 1\%, otherwise choose option **A**: All other options are incorrect.
现有一份 [1-$^{14}\mathrm{C}$] 葡萄糖样品，其中 $\mathrm{C}-1$ 为 $^{14}\mathrm{C}$ 同位素标记。在其中加入细菌培养液，一段时间后分离得到如下代谢中间产物 $i=1\sim 8$（带原子编号）：

<MultiModal>![中间产物的编号及其对应的SMILES: 1. O[C@@:1]1([C:2]P([O-])([O-])=O)[C@@:3]([H])(O)[C@@:4](O)([H])[C@:5]([C:6]OP([O-])([O-])=O)([H])O1
2. O=[C:1][C@H:2](O)[C:3]OP(O)(O)=O
3. [C:1]=[C:2](OP(O)(O)=O)[C:3](O)=O
4. [C:1][C:2](S[C:3][C:4]N[C:5]([C:6][C:7]N[C:8]([C:9]([C:10]([C:11])([C:12]OP(OP(O[C:13][C@@H:14]1[C@@H:15](OP([O-])([O-])=O)[C@@H:16](O)[C@H:17](N2[C:18]=N[C:19]3=[C:20]2N=[C:21]N=[C:22]3N)O1)([O-])=O)([O-])=O)[C:23])O)=O)=O)=O
5. O=[C:1](O)[C:2][C:3]([C:4](O)=O)(O)[C:5][C:6](O)=O
6. O=[C:1]([C:2](O)=O)[C:3][C:4][C:5](O)=O
7. O=[C:1](O)[C:2][C:3][C:4](O)=O
8. O=[C:1]([C:2](O)=O)[C:3][C:4](O)=O
](/media/uploads/210734d5-f0f9-4b80-b449-f04478d8a4a1.png)</MultiModal>

对于每个代谢的中间产物 $i$，根据你对于葡萄糖代谢通路的理解，找出 $^{14}\mathrm{C}$ 标记可能出现的位点的原子编号，定义一个集合 $P_i$，它代表在第 $i$ 个代谢中间产物中，所有可能被 $^{14}\mathrm{C}$ 标记的碳原子的原子编号所构成的集合。

* 例如： 如果在产物 $i$ 中，标记可能出现在2号和3号碳原子上，则 $P_i = \{2, 3\}$。

$z_i$ 的计算规则根据集合 $P_i$ 的大小（即可能标记位的数量）分两种情况：

$$
z_i = \begin{cases}
\dfrac{p}{i^2 + 1} & \text{如果只有一个可能标记位 (即 } |P_i| = 1 \text{, 且 } P_i = \{p\}\text{)} \\
\\
\dfrac{\prod_{p \in P_i} p}{i^2} & \text{如果有多个可能标记位 (即 } |P_i| > 1\text{)}
\end{cases}
$$

* 在第二个情况中，$\prod_{p \in P_i} p$ 表示将集合 $P_i$ 中所有的原子编号相乘。


最终计算所有 $z_i$ 值的总和 $S$：

$$
S = \sum_{i=1}^{8} z_i
$$

从以下选项中选择正确的选项，最终结果保留小数点后四位，选择与你的计算结果偏差在 1\% 以内的选项，否则选择选项 **A**: 其他选项均不正确。
All other options are incorrect
1.836
2.100
2.210
2.276
2.311
2.353
2.418
2.436
2.670
3.604
其他选项均不正确
1.836
2.100
2.210
2.276
2.311
2.353
2.418
2.436
2.670
3.604
The core of this question is to track the fate of the $^{14}\mathrm{C}$ label from [1-$^{14}\mathrm{C}$]glucose in the major metabolic pathways (glycolysis in the cytoplasm, tricarboxylic acid cycle in the mitochondrial matrix). First, it is necessary to identify the names of the 8 intermediates in the picture and understand their reactions in the metabolic pathways:

1. Fructose-1,6-bisphosphate: This is a key intermediate in the glycolysis pathway. After consuming two molecules of ATP, glucose is converted into this highly activated 6-carbon sugar molecule. Its formation is an important and irreversible regulatory point in the glycolysis pathway.

<Checkpoint>[Intermediate 1 is Fructose-1,6-bisphosphate](0.5)</Checkpoint>

2. Glyceraldehyde-3-phosphate (G3P): Fructose-1,6-bisphosphate is cleaved by aldolase into two 3-carbon molecules, one of which is glyceraldehyde-3-phosphate. It is the direct substrate for subsequent reactions in glycolysis, marking the beginning of the 3-carbon stage.

<Checkpoint>[Intermediate 2 is Glyceraldehyde-3-phosphate (G3P)](0.5)</Checkpoint>

3. Phosphoenolpyruvate (PEP): A high-energy intermediate at the end of the glycolysis pathway. It generates one molecule of ATP by transferring a high-energy phosphate group to ADP (substrate-level phosphorylation), and itself is converted to pyruvate, which is the final product of glycolysis.

<Checkpoint>[Intermediate 3 is Phosphoenolpyruvate (PEP)](0.5)</Checkpoint>

4. Acetyl-CoA: Pyruvate produced by glycolysis enters the mitochondria and is catalyzed by the pyruvate dehydrogenase complex in the mitochondrial matrix, losing one carbon atom (released as $\rm CO_2$) and binding to coenzyme A to form 2-carbon acetyl-CoA. This is the key bridge molecule connecting glycolysis and the tricarboxylic acid cycle.

<Checkpoint>[Intermediate 4 is Acetyl-CoA](0.5)</Checkpoint>

5. Citrate: The first step of the tricarboxylic acid cycle. Acetyl-CoA (2 carbons) transfers its acetyl group to oxaloacetate (4 carbons), condensing to form a 6-carbon citrate molecule, thus starting the cycle.

<Checkpoint>[Intermediate 5 is Citrate](0.5)</Checkpoint>

6. α-Ketoglutarate: Citrate is converted into this 5-carbon intermediate through a series of reactions (including two decarboxylations, i.e., releasing two molecules of $\rm CO_2$). It is not only part of the cycle, but also an important precursor for the synthesis of amino acids such as glutamate.

<Checkpoint>[Intermediate 6 is α-Ketoglutarate](0.5)</Checkpoint>

7. Succinate: α-Ketoglutarate is decarboxylated again to produce 4-carbon succinate. This step is accompanied by the production of one molecule of GTP (or ATP), which is another substrate-level phosphorylation in the tricarboxylic acid cycle.

<Checkpoint>[Intermediate 7 is Succinate](0.5)</Checkpoint>

8. Oxaloacetate (OAA): Succinate is regenerated into 4-carbon oxaloacetate through several oxidation reactions. Oxaloacetate, as the end point of the cycle, is also the starting point of the next cycle. It will combine with new acetyl-CoA again to start a new round of the cycle.

<Checkpoint>[Intermediate 8 is Oxaloacetate (OAA)](0.5)</Checkpoint>

These intermediates constitute a continuous chain of biochemical reactions: Glucose → [Glycolysis] → Fructose-1,6-bisphosphate → Glyceraldehyde-3-phosphate → Phosphoenolpyruvate → Pyruvate → [Link Reaction] → Acetyl-CoA + Oxaloacetate → [Tricarboxylic Acid Cycle] → Citrate → α-Ketoglutarate → Succinate → ... → (Regenerated) Oxaloacetate.

Based on the atomic labeling in the picture, combined with the metabolic pathway of glucose, the carbon atom number that may contain isotopic labeling in each intermediate can be inferred:

1. Fructose-1,6-bisphosphate: The label is located on the carbon at the 1 position of fructose in 1,6-bisphosphate, which is carbon atom number 2 in the figure.

$$
z_1 = \dfrac{p}{i^2 + 1} = \dfrac{2}{1^2 + 1} = \dfrac{2}{2} = 1
$$

<Checkpoint>[The label of Fructose-1,6-bisphosphate is located on carbon atom number 2, $z_1=1$](1)</Checkpoint>

2. Glyceraldehyde-3-phosphate: The label is located on carbon atom number 3, which is linked to phosphate.

$$
z_2 = \dfrac{p}{i^2 + 1} = \dfrac{3}{2^2 + 1} = \dfrac{3}{5} = 0.6
$$

<Checkpoint>[The label of Glyceraldehyde-3-phosphate is located on carbon atom number 3, $z_2=0.6$](1)</Checkpoint>

3. Phosphoenolpyruvate (PEP): The label is located at the $\beta$ position of the unsaturated ketone, which is carbon atom number 1 in the figure.

$$
z_3 = \dfrac{p}{i^2 + 1} = \dfrac{1}{3^2 + 1} = \dfrac{1}{10} = 0.1
$$

<Checkpoint>[The label of Phosphoenolpyruvate (PEP) is located on carbon atom number 1, $z_3=0.1$](1)</Checkpoint>

4. Acetyl-CoA: The label is located on the methyl group of the acetyl group, which is carbon atom number 1 in the figure.

$$
z_4 = \dfrac{p}{i^2 + 1} = \dfrac{1}{4^2 + 1} = \dfrac{1}{17}
$$

<Checkpoint>[The label of Acetyl-CoA is located on carbon atom number 1, $z_4=\dfrac{1}{17}$](1)</Checkpoint>

5. Citrate: The label is located on the carbon $\alpha$ to the carboxyl group with two methylene groups, which are carbon atoms number 2 and 5 in the figure.

$$
z_5 = \dfrac{\prod_{p \in P_5} p}{i^2} = \dfrac{2 \times 5}{5^2} = \dfrac{10}{25} = 0.4
$$

<Checkpoint>[The label of Citrate is located on carbon atoms number 2 and 5, $z_5=0.4$](1)</Checkpoint>

6. α-Ketoglutarate: The label is located on the carbon $\alpha$ to the carboxyl group with a methylene group, which is carbon atom number 4 in the figure.

$$
z_6 = \dfrac{p}{i^2 + 1} = \dfrac{4}{6^2 + 1} = \dfrac{4}{37}
$$

<Checkpoint>[The label of α-Ketoglutarate is located on carbon atom number 4, $z_6=\dfrac{4}{37}$](1)</Checkpoint>

7. Succinate: The label is located on the carbon $\alpha$ to the carboxyl group with two methylene groups, which are carbon atoms number 2 and 3 in the figure.

$$
z_7 = \dfrac{\prod_{p \in P_7} p}{i^2} = \dfrac{2 \times 3}{7^2} = \dfrac{6}{49}
$$

<Checkpoint>[The label of Succinate is located on carbon atoms number 2 and 3, $z_7=\dfrac{6}{49}$](1)</Checkpoint>

8. Oxaloacetate: The label is located on the two middle carbons of the carbon chain, which are carbon atoms number 1 and 3 in the figure.

$$
z_8 = \dfrac{\prod_{p \in P_8} p}{i^2} = \dfrac{1 \times 3}{8^2} = \dfrac{3}{64}
$$

<Checkpoint>[The label of Oxaloacetate is located on carbon atoms number 1 and 3, $z_8= \dfrac{3}{64}$](1)</Checkpoint>

Finally, we add up all the calculated z_i values to get the sum S.

$$
S = \sum_{i=1}^{8} z_i = z_1 + z_2 + z_3 + z_4 + z_5 + z_6 + z_7 + z_8
$$
$$
S = 1 + 0.6 + 0.1 + \frac{1}{17} + 0.4 + \frac{4}{37} + \frac{6}{49} + \frac{3}{64}
$$
$$
S \approx 1 + 0.6 + 0.1 + 0.0588 + 0.4 + 0.1081 + 0.1224 + 0.0469
$$
$$S \approx 2.4362
$$
Therefore, the sum S of z_i of all intermediates is approximately 2.4362.

<Checkpoint>[The sum of z_i of all intermediates $S \approx 2.4362$](1)</Checkpoint>
本题的核心是追踪来自 [1-$^{14}\mathrm{C}$]葡萄糖的 $^{14}\mathrm{C}$ 标记在主要代谢通路（细胞质中进行的糖酵解、线粒体基质中的三羧酸循环）中的去向。首先，需要识别出图片中8个中间产物的名称，并理解其在代谢通路中的反应：

1. 果糖-1,6-二磷酸：这是糖酵解途径中的一个关键中间产物。在消耗了两分子ATP后，葡萄糖被转化为这个高度活化的6碳糖分子。它的生成是糖酵解途径中一个重要的、不可逆的调控点。

<Checkpoint>[中间产物1是果糖-1,6-二磷酸](0.5)</Checkpoint>

2. 3-磷酸甘油醛（G3P）：果糖-1,6-二磷酸被醛缩酶裂解成两个3碳分子，其中一个就是3-磷酸甘油醛。它是糖酵解后续反应的直接底物，标志着3碳阶段的开始。

<Checkpoint>[中间产物2是3-磷酸甘油醛（G3P）](0.5)</Checkpoint>

3. 磷酸烯醇式丙酮酸（PEP）：糖酵解途径末端的一个高能中间产物。它通过将高能磷酸基团转移给ADP来生成一分子ATP（底物水平磷酸化），自身则转化为丙酮酸，这是糖酵解的最终产物。

<Checkpoint>[中间产物3是磷酸烯醇式丙酮酸（PEP）](0.5)</Checkpoint>

4. 乙酰辅酶A：糖酵解产生的丙酮酸进入线粒体，在线粒体基质中被丙酮酸脱氢酶复合体催化，脱去一个碳原子（以 $\rm CO_2$ 形式释放），并与辅酶A结合，形成2碳的乙酰辅酶A。这是连接糖酵解和三羧酸循环的关键桥梁分子。

<Checkpoint>[中间产物4是乙酰辅酶A](0.5)</Checkpoint>

5. 柠檬酸：三羧酸循环的第一步。乙酰辅酶A（2碳）将其乙酰基转移给草酰乙酸（4碳），缩合形成一个6碳的柠檬酸分子，循环由此开始。

<Checkpoint>[中间产物5是柠檬酸](0.5)</Checkpoint>

6. α-酮戊二酸：柠檬酸经过一系列反应（包括两次脱羧，即释放两分子 $\rm CO_2$），转变成这个5碳的中间产物。它不仅是循环的一部分，也是合成谷氨酸等氨基酸的重要前体。

<Checkpoint>[中间产物6是α-酮戊二酸](0.5)</Checkpoint>

7. 琥珀酸：α-酮戊二酸再次脱羧，生成4碳的琥珀酸。这一步伴随着一分子GTP（或ATP）的生成，是三羧酸循环中又一次底物水平磷酸化。

<Checkpoint>[中间产物7是琥珀酸](0.5)</Checkpoint>

8. 草酰乙酸（OAA）：琥珀酸经过几步氧化反应，最终再生为4碳的草酰乙酸。草酰乙酸作为循环的终点，同时也是下一次循环的起点，它会再次与新的乙酰辅酶A结合，启动新一轮循环。

<Checkpoint>[中间产物8是草酰乙酸（OAA）](0.5)</Checkpoint>

这些中间产物构成了一个连续的生化反应链： 葡萄糖 → [糖酵解] → 果糖-1,6-二磷酸 → 3-磷酸甘油醛 → 磷酸烯醇式丙酮酸 → 丙酮酸 → [连接反应] → 乙酰辅酶A + 草酰乙酸 → [三羧酸循环] → 柠檬酸 → α-酮戊二酸 → 琥珀酸 → ... → (再生)草酰乙酸。

根据图片中的原子标记，结合葡萄糖的代谢通路，可以推断每个中间产物中可能存在同位素标记的碳原子序号：

1. 果糖-1,6-二磷酸：标记位于1,6-二磷酸中果糖的1号位的碳上，即图中的2号碳原子。

$$
z_1 = \dfrac{p}{i^2 + 1} = \dfrac{2}{1^2 + 1} = \dfrac{2}{2} = 1
$$

<Checkpoint>[果糖-1,6-二磷酸的标记位于2号碳原子上，$z_1=1$](1)</Checkpoint>

2. 3-磷酸甘油醛：标记位于与磷酸相连的3号碳原子上。

$$
z_2 = \dfrac{p}{i^2 + 1} = \dfrac{3}{2^2 + 1} = \dfrac{3}{5} = 0.6
$$

<Checkpoint>[3-磷酸甘油醛的标记位于3号碳原子上，$z_2=0.6$](1)</Checkpoint>

3. 磷酸烯醇式丙酮酸 (PEP)：标记位于不饱和酮的 $\beta$ 位，即图中1号碳原子。

$$
z_3 = \dfrac{p}{i^2 + 1} = \dfrac{1}{3^2 + 1} = \dfrac{1}{10} = 0.1
$$

<Checkpoint>[磷酸烯醇式丙酮酸 (PEP)的标记位于1号碳原子上，$z_3=0.1$](1)</Checkpoint>

4. 乙酰辅酶A：标记位于乙酰基的甲基上，即图中的1号碳原子。

$$
z_4 = \dfrac{p}{i^2 + 1} = \dfrac{1}{4^2 + 1} = \dfrac{1}{17}
$$

<Checkpoint>[乙酰辅酶A的标记位于1号碳原子上，$z_4=\dfrac{1}{17}$](1)</Checkpoint>

5. 柠檬酸：标记位于羧酸 $\alpha$ 位为两个亚甲基的碳，即图中2号和5号碳原子。

$$
z_5 = \dfrac{\prod_{p \in P_5} p}{i^2} = \dfrac{2 \times 5}{5^2} = \dfrac{10}{25} = 0.4
$$

<Checkpoint>[柠檬酸的标记位于2号和5号碳原子上，$z_5=0.4$](1)</Checkpoint>

6. α-酮戊二酸：标记位于羧酸 $\alpha$ 位为亚甲基的碳，即图中的4号碳原子。

$$
z_6 = \dfrac{p}{i^2 + 1} = \dfrac{4}{6^2 + 1} = \dfrac{4}{37}
$$

<Checkpoint>[α-酮戊二酸的标记位于4号碳原子上，$z_6=\dfrac{4}{37}$](1)</Checkpoint>

7. 琥珀酸：标记位于羧酸 $\alpha$ 位两个亚甲基的碳，即图中的2号和3号碳原子。

$$
z_7 = \dfrac{\prod_{p \in P_7} p}{i^2} = \dfrac{2 \times 3}{7^2} = \dfrac{6}{49}
$$

<Checkpoint>[琥珀酸的标记位于2号和3号碳原子上，$z_7=\dfrac{6}{49}$](1)</Checkpoint>

8. 草酰乙酸：标记位于碳链的中间两个碳上，即图中的1号和3号碳原子。

$$
z_8 = \dfrac{\prod_{p \in P_8} p}{i^2} = \dfrac{1 \times 3}{8^2} = \dfrac{3}{64}
$$

<Checkpoint>[草酰乙酸的标记位于1号和3号碳原子上，$z_8= \dfrac{3}{64}$](1)</Checkpoint>

最终，我们将所有计算出的 z_i 值相加，得到总和 S。

$$
S = \sum_{i=1}^{8} z_i = z_1 + z_2 + z_3 + z_4 + z_5 + z_6 + z_7 + z_8
$$
$$
S = 1 + 0.6 + 0.1 + \frac{1}{17} + 0.4 + \frac{4}{37} + \frac{6}{49} + \frac{3}{64}
$$
$$
S \approx 1 + 0.6 + 0.1 + 0.0588 + 0.4 + 0.1081 + 0.1224 + 0.0469
$$
$$S \approx 2.4362
$$
因此，所有中间产物的 z_i 之和 S 约为 2.4362。

<Checkpoint>[所有中间产物的 z_i 之和 $S \approx 2.4362$](1)</Checkpoint>