<MultiModal>![In the figure, the SMILES of $\mathbf{A}$: [O:6]=[C:7]1[C:12]([N:13]=[CH:14][N:15]2[C@H:1]3[C@H:2]([OH:30])[C@H:3]([OH:31])[C@@H:4]([CH2:5][O:35][P:36]([O-:37])([O-:38])=[O:39])[O:29]3)=[C:11]2[N:10]=[CH:9][NH:8]1; the SMILES of $\mathbf{B}$: [OH:40][C@@H:17]1[C@H:18]([OH:41])[C@@H:19]([C:20][O:45][P:46]([O-:47])([O-:49])=[O:48])[O:42][C@H:16]1[N:26]2[C:24](=[O:25])[NH:23][C:22](=[O:21])[C:28]=[C:27]2](/media/uploads/58ee2be5-6589-4359-b539-e32bfc98935e.png)</MultiModal>

The above two nucleotides $\mathbf{A}$ (1) and $\mathbf{B}$ (2) can be synthesized in organisms from the following species $\mathbf{C\sim H}$:

<MultiModal>![In the figure, the SMILES of $\mathbf{C}$: [NH2:35][C@@H:36]([CH2:40][CH2:41][C:42](=[O:43])[NH2:44])[C:37](=[O:38])[OH:38]; the SMILES of $\mathbf{D}$: [NH2:45][C@@H:46]([CH2:50][C:51]([OH:52])=[O:52])[C:47]([OH:48])=[O:48]; the SMILES of $\mathbf{E}$: [O:60]=[C:59]([OH:60])[CH2:58][NH2:62]; the SMILES of $\mathbf{F}$: [OH:57][C:55]([O-:57])=[O:57]; the SMILES of $\mathbf{H}$: [O:15]=[C:14]([NH:16][C:17]([NH2:19])=[N:18]1)[C:13]([NH:4]2)=[C:12]1[NH:11][CH2:10][CH:9]2[CH2:8][N:7]([CH:21]=[O:22])[C:5]3=[C:6][C:1]=[C:2]([C:23]([NH:25][C@H:26]([C:32]([OH:33])=[O:33])[CH2:27][CH2:28][C:29]([OH:31])=[O:31])=[O:24])[C:1]=[C:6]3](/media/uploads/7b54376f-bea3-431c-90c3-60bd1c8842d3.png)</MultiModal>

The same numbering is used for atoms in the same chemical environment in $\mathbf{C\sim H}$ to facilitate atomic correspondence.

Based on the nucleotide metabolic pathway, please find the correspondence between the atoms of the base in the nucleotide and the atoms of the raw material species in the organism, recorded as:

$$
(b_i,a_i)
$$

where $b_i$ refers to the atom number in the nucleotide in the correspondence, and $a_i$ refers to the number of the raw material atom corresponding to the base atom in the raw material species synthesized in the organism. Different $b_i$ may correspond to the same $a_i$.

For the correspondence in each nucleotide $j$, define the calculation method of $x_j$ as:

$$
x_j = \sum_i (a_i\times b_i),\quad a_i\in j
$$

Please calculate the value of $\dfrac{x_1}{x_2}$, keep four significant figures, and select the correct result from the following options, requiring the deviation between the calculated result and the option to be within $1\%$, otherwise select option A: All other options are incorrect.
<MultiModal>![图中，$\mathbf{A}$ 的SMILES：[O:6]=[C:7]1[C:12]([N:13]=[CH:14][N:15]2[C@H:1]3[C@H:2]([OH:30])[C@H:3]([OH:31])[C@@H:4]([CH2:5][O:35][P:36]([O-:37])([O-:38])=[O:39])[O:29]3)=[C:11]2[N:10]=[CH:9][NH:8]1；$\mathbf{B}$ 的SMILES：[OH:40][C@@H:17]1[C@H:18]([OH:41])[C@@H:19]([C:20][O:45][P:46]([O-:47])([O-:49])=[O:48])[O:42][C@H:16]1[N:26]2[C:24](=[O:25])[NH:23][C:22](=[O:21])[C:28]=[C:27]2](/media/uploads/58ee2be5-6589-4359-b539-e32bfc98935e.png)</MultiModal>

以上两个核苷酸 $\mathbf{A}$ (1) 和 $\mathbf{B}$ (2) 可以由以下物种 $\mathbf{C\sim H}$ 在生物体中合成：

<MultiModal>![图中，$\mathbf{C}$ 的SMILES：[NH2:35][C@@H:36]([CH2:40][CH2:41][C:42](=[O:43])[NH2:44])[C:37](=[O:38])[OH:38]；$\mathbf{D}$ 的SMILES: [NH2:45][C@@H:46]([CH2:50][C:51]([OH:52])=[O:52])[C:47]([OH:48])=[O:48]；$\mathbf{E}$ 的SMILES: [O:60]=[C:59]([OH:60])[CH2:58][NH2:62]；$\mathbf{F}$ 的SMILES: [OH:57][C:55]([O-:57])=[O:57]；$\mathbf{H}$ 的SMILES：[O:15]=[C:14]([NH:16][C:17]([NH2:19])=[N:18]1)[C:13]([NH:4]2)=[C:12]1[NH:11][CH2:10][CH:9]2[CH2:8][N:7]([CH:21]=[O:22])[C:5]3=[C:6][C:1]=[C:2]([C:23]([NH:25][C@H:26]([C:32]([OH:33])=[O:33])[CH2:27][CH2:28][C:29]([OH:31])=[O:31])=[O:24])[C:1]=[C:6]3](/media/uploads/7b54376f-bea3-431c-90c3-60bd1c8842d3.png)</MultiModal>

在 $\mathbf{C\sim H}$ 中相同化学环境的原子使用了相同的编号，以便进行原子对应。

请你根据核苷酸代谢途径，找出核苷酸中碱基的原子与生物体中原料物种的原子间的对应关系，记为：

$$
(b_i,a_i)
$$

其中，$b_i$ 是指对应关系中核苷酸中的原子编号，$a_i$ 是指生物体中合成的原料物种中，该碱基原子对应的原料原子的编号，不同的 $b_i$ 可能对应相同的 $a_i$。

对于每个核苷酸 $j$ 中的对应关系，定义 $x_j$ 的计算方式为：

$$
x_j = \sum_i (a_i\times b_i),\quad a_i\in j
$$

请你计算 $\dfrac{x_1}{x_2}$的值，结果保留四位有效数字，从以下选项中选择正确的结果，要求计算的结果与选项偏差在 $1\%$ 以内，否则选择选项 A：其他选项均不正确。
All other options are incorrect
2.029
0.6164
0.2729
0.4434
0.4928
0.5357
1.165
0.4711
1.000
0.7500
其他选项均不正确
2.029
0.6164
0.2729
0.4434
0.4928
0.5357
1.165
0.4711
1.000
0.7500
To solve this problem, we first need to identify the types of nucleotides. Nucleotide $\mathbf{A}$ is inosine monophosphate (inosinic acid) IMP, and nucleotide $\mathbf{B}$ is uridine monophosphate UMP.

<Checkpoint>[Nucleotide $\mathbf{A}$ is inosine monophosphate (inosinic acid) IMP, and nucleotide $\mathbf{B}$ is uridine monophosphate UMP](2)</Checkpoint>

Secondly, we need to determine the types of each raw material: $\mathbf{C}$ is glutamine, $\mathbf{D}$ is aspartic acid, $\mathbf{E}$ is glycine, $\mathbf{F}$ is bicarbonate, and $\mathbf{H}$ is formyltetrahydrofolate.

<Checkpoint>[$\mathbf{C}$ is glutamine, $\mathbf{D}$ is aspartic acid, $\mathbf{E}$ is glycine, $\mathbf{F}$ is bicarbonate, and $\mathbf{H}$ is formyltetrahydrofolate](3)</Checkpoint>

Next, find the correspondence between the atoms of the bases in the nucleotides and the atoms of the raw material species in the organism.

For IMP/UMP, we can first correlate the traditional numbering method with the atom numbering given in the question. According to the traditionally defined numbering, combined with the knowledge of nucleotide biosynthesis, find the corresponding atom number in the corresponding raw material, and the following table can be obtained:

| Nucleotide | Atom Number in the Question | Traditional Atom Number | Raw Material | Corresponding Raw Material Atom Number |
| :----: | :------------: | :-----------: | :--------------------------------------: | :--------------: |
|  IMP   |       6        | (Oxygen of carbon 6) | Oxygen in $\mathbf{F}$ | 57 |
|  IMP   |       7        |       6       | Carbon in $\mathbf{F}$ | 55 |
|  IMP   |       8        |       1       | Amino group in $\mathbf{D}$ | 45 |
|  IMP   |       9        |       2       | Carbon of the formyl group in $\mathbf{H}$ | 21 |
|  IMP   |       10       |       3       | Amide amino group in $\mathbf{C}$ | 44 |
|  IMP   |       11       |       4       | Carboxylic carbon in $\mathbf{E}$ | 59 |
|  IMP   |       12       |       5       | $\alpha-$ carbon of the carboxyl group in $\mathbf{E}$ | 58 |
|  IMP   |       13       |       7       | Amino group in $\mathbf{E}$ | 62 |
|  IMP   |       14       |       8       | Carbon of the formyl group in $\mathbf{H}$ | 21 |
|  IMP   |       15       |       9       | Amide amino group in $\mathbf{C}$ | 44 |
|  UMP   |       21       | (Oxygen of carbon 4) | Oxygen of the side-chain carboxyl group in $\mathbf{D}$ | 52 |
|  UMP   |       22       |       4       | Carbon of the side-chain carboxyl group in $\mathbf{D}$ | 51 |
|  UMP   |       23       |       5       | Amide amino group in $\mathbf{C}$ | 44 |
|  UMP   |       24       |       6       | Carbon in $\mathbf{F}$ | 55 |
|  UMP   |       25       | (Oxygen of carbon 6) | Oxygen in $\mathbf{F}$ | 57 |
|  UMP   |       26       |       1       | Amino group in $\mathbf{D}$ | 45 |
|  UMP   |       27       |       2       | Carbon connected to the amino group in $\mathbf{D}$ | 46 |
|  UMP   |       28       |       3       | $\alpha-$ carbon of the side-chain carboxyl group in $\mathbf{D}$ | 50 |

<Checkpoint>[Correspondence in IMP $(b_6,a_{57})/(b_7,a_{55})/(b_8,a_{45})/(b_9,a_{21})/(b_{10},a_{44})/(b_{11},a_{59})/(b_{12},a_{58})/(b_{12},a_{58})/(b_{13},a_{62})/(b_{14},a_{21})/(b_{15},a_{44})$](3)</Checkpoint>

<Checkpoint>[Correspondence in UMP $(b_{21},a_{52})/(b_{22},a_{51})/(b_{23},a_{44})/(b_{24},a_{55})/(b_{25},a_{57})/(b_{26},a_{45})/(b_{27},a_{46})/(b_{28},a_{50})$](3)</Checkpoint>

Finally, according to the calculation method of $x_j$ defined in the question, calculate the values of $x_1$ and $x_2$ corresponding to IMP and UMP respectively, and find the ratio between them.

First calculate $x_1$ (IMP), according to the formula $x_j = \sum_i (a_i\times b_i)$ and the IMP atom correspondence table, we have:

$x_1 = (6 \times 57) + (7 \times 55) + (8 \times 45) + (9 \times 21) + (10 \times 44) + (11 \times 59) + (12 \times 58) + (13 \times 62) + (14 \times 21) + (15 \times 44)$

Calculate the product of each term:
* $6 \times 57 = 342$
* $7 \times 55 = 385$
* $8 \times 45 = 360$
* $9 \times 21 = 189$
* $10 \times 44 = 440$
* $11 \times 59 = 649$
* $12 \times 58 = 696$
* $13 \times 62 = 806$
* $14 \times 21 = 294$
* $15 \times 44 = 660$

Add all the products together:
$x_1 = 342 + 385 + 360 + 189 + 440 + 649 + 696 + 806 + 294 + 660 = 4821$

<Checkpoint>[$x_1 = 4821$ corresponding to IMP](1)</Checkpoint>

Then calculate $x_2$ (UMP), similarly, according to the UMP atom correspondence table:

$x_2 = (21 \times 52) + (22 \times 51) + (23 \times 44) + (24 \times 55) + (25 \times 57) + (26 \times 45) + (27 \times 46) + (28 \times 50)$

Calculate the product of each term:
* $21 \times 52 = 1092$
* $22 \times 51 = 1122$
* $23 \times 44 = 1012$
* $24 \times 55 = 1320$
* $25 \times 57 = 1425$
* $26 \times 45 = 1170$
* $27 \times 46 = 1242$
* $28 \times 50 = 1400$

Add all the products together:
$x_2 = 1092 + 1122 + 1012 + 1320 + 1425 + 1170 + 1242 + 1400 = 9783$

<Checkpoint>[$x_2 = 9783$ corresponding to UMP](1)</Checkpoint>

Calculate the value of $\dfrac{x_1}{x_2}$:

$$
\frac{x_1}{x_2} = \frac{4821}{9783} \approx 0.49279\approx 0.4928
$$

<Checkpoint>[$\dfrac{x_1}{x_2} = 0.4928$](1)</Checkpoint>
要解决这道题目，首先需要识别核苷酸的种类，核苷酸 $\mathbf{A}$ 是次黄嘌呤核苷酸（肌苷酸） IMP，核苷酸 $\mathbf{B}$ 是尿嘧啶核苷酸 UMP。

<Checkpoint>[核苷酸 $\mathbf{A}$ 是次黄嘌呤核苷酸（肌苷酸） IMP，核苷酸 $\mathbf{B}$ 是尿嘧啶核苷酸 UMP](2)</Checkpoint>

其次，要确定每种原料的种类，$\mathbf{C}$ 为谷氨酰胺，$\mathbf{D}$ 为天冬氨酸，$\mathbf{E}$ 为甘氨酸，$\mathbf{F}$ 为碳酸氢根，$\mathbf{H}$ 为甲酰四氢叶酸。

<Checkpoint>[$\mathbf{C}$ 为谷氨酰胺，$\mathbf{D}$ 为天冬氨酸，$\mathbf{E}$ 为甘氨酸，$\mathbf{F}$ 为碳酸氢根，$\mathbf{H}$ 为甲酰四氢叶酸](3)</Checkpoint>

接下来，找出核苷酸中碱基的原子与生物体中原料物种的原子间的对应关系。

对于 IMP/UMP 而言，可以先将传统的编号方式与题目中给出的原子编号相关联，根据传统定义的编号，结合核苷酸的生物合成知识，找出其对应的原料中对应原子的编号，可以得到如下表格：

| 核苷酸 | 题目中原子编号 | 传统原子编号  |                 对应原料                 | 对应原料原子编号 |
| :----: | :------------: | :-----------: | :--------------------------------------: | :--------------: |
|  IMP   |       6        | （6号碳的氧） |           $\mathbf{F}$ 中的氧            |        57        |
|  IMP   |       7        |       6       |           $\mathbf{F}$ 中的碳            |        55        |
|  IMP   |       8        |       1       |          $\mathbf{D}$ 中的氨基           |        45        |
|  IMP   |       9        |       2       |       $\mathbf{H}$ 中的甲酰基的碳        |        21        |
|  IMP   |       10       |       3       |        $\mathbf{C}$ 中酰胺的氨基         |        44        |
|  IMP   |       11       |       4       |          $\mathbf{E}$ 中羧酸碳           |        59        |
|  IMP   |       12       |       5       |     $\mathbf{E}$ 中羧基 $\alpha-$ 碳     |        58        |
|  IMP   |       13       |       7       |           $\mathbf{E}$ 中氨基            |        62        |
|  IMP   |       14       |       8       |       $\mathbf{H}$ 中的甲酰基的碳        |        21        |
|  IMP   |       15       |       9       |        $\mathbf{C}$ 中酰胺的氨基         |        44        |
|  UMP   |       21       | （4号碳的氧） |       $\mathbf{D}$ 中侧链羧基的氧        |        52        |
|  UMP   |       22       |       4       |       $\mathbf{D}$ 中侧链羧基的碳        |        51        |
|  UMP   |       23       |       5       |        $\mathbf{C}$ 中酰胺的氨基         |        44        |
|  UMP   |       24       |       6       |           $\mathbf{F}$ 中的碳            |        55        |
|  UMP   |       25       | （6号碳的氧） |           $\mathbf{F}$ 中的氧            |        57        |
|  UMP   |       26       |       1       |          $\mathbf{D}$ 中的氨基           |        45        |
|  UMP   |       27       |       2       |      $\mathbf{D}$ 中与氨基相连的碳       |        46        |
|  UMP   |       28       |       3       | $\mathbf{D}$ 中与侧链羧基的 $\alpha-$ 碳 |        50        |

<Checkpoint>[IMP 中的对应关系$(b_6,a_{57})/(b_7,a_{55})/(b_8,a_{45})/(b_9,a_{21})/(b_{10},a_{44})/(b_{11},a_{59})/(b_{12},a_{58})/(b_{12},a_{58})/(b_{13},a_{62})/(b_{14},a_{21})/(b_{15},a_{44})$](3)</Checkpoint>

<Checkpoint>[UMP 中的对应关系 $(b_{21},a_{52})/(b_{22},a_{51})/(b_{23},a_{44})/(b_{24},a_{55})/(b_{25},a_{57})/(b_{26},a_{45})/(b_{27},a_{46})/(b_{28},a_{50})$](3)</Checkpoint>

最后，根据题目中定义的 $x_j$ 的计算方式，分别计算 IMP 和 UMP 对应的 $x_1$ 与 $x_2$ 的值，并求出它们的比值。

先计算 $x_1$ (IMP)，根据公式 $x_j = \sum_i (a_i\times b_i)$ 和 IMP 的原子对应关系表，我们有：

$x_1 = (6 \times 57) + (7 \times 55) + (8 \times 45) + (9 \times 21) + (10 \times 44) + (11 \times 59) + (12 \times 58) + (13 \times 62) + (14 \times 21) + (15 \times 44)$

计算每一项的乘积：
* $6 \times 57 = 342$
* $7 \times 55 = 385$
* $8 \times 45 = 360$
* $9 \times 21 = 189$
* $10 \times 44 = 440$
* $11 \times 59 = 649$
* $12 \times 58 = 696$
* $13 \times 62 = 806$
* $14 \times 21 = 294$
* $15 \times 44 = 660$

将所有乘积相加：
$x_1 = 342 + 385 + 360 + 189 + 440 + 649 + 696 + 806 + 294 + 660 = 4821$

<Checkpoint>[IMP 对应的 $x_1 = 4821$](1)</Checkpoint>

再计算 $x_2$ (UMP)，同样地，根据 UMP 的原子对应关系表：

$x_2 = (21 \times 52) + (22 \times 51) + (23 \times 44) + (24 \times 55) + (25 \times 57) + (26 \times 45) + (27 \times 46) + (28 \times 50)$

计算每一项的乘积：
* $21 \times 52 = 1092$
* $22 \times 51 = 1122$
* $23 \times 44 = 1012$
* $24 \times 55 = 1320$
* $25 \times 57 = 1425$
* $26 \times 45 = 1170$
* $27 \times 46 = 1242$
* $28 \times 50 = 1400$

将所有乘积相加：
$x_2 = 1092 + 1122 + 1012 + 1320 + 1425 + 1170 + 1242 + 1400 = 9783$

<Checkpoint>[UMP 对应的 $x_2 = 9783$](1)</Checkpoint>

计算 $\dfrac{x_1}{x_2}$ 的值：

$$
\frac{x_1}{x_2} = \frac{4821}{9783} \approx 0.49279\approx 0.4928
$$

<Checkpoint>[$\dfrac{x_1}{x_2} = 0.4928$](1)</Checkpoint>