The five nucleotide bases are shown in the figure, where the atomic numbering of heavy atoms is indicated.

<MultiModal>![The SMARTS of the bases in the figure can be represented as `[N:1][C:2]([N:3][C:4]1=[O:5])=[N:6][C:7]2=[C:8]1[N:9]=[C:10][N:11]2[R].[C:12][C:13]3=[C:14][N:15]([R])[C:16]([N:17][C:18]3=[O:19])=[O:20].[N:21][C:22]4=[N:23][C:24]=[N:25][C:26]5=[C:27]4[N:28]=[C:29][N:30]5[R].[O:31]=[C:32]6[N:33]([R])[C:34]=[C:35][C:36]([N:37])=[N:38]6.[O:39]=[C:40]([N:41]7)[N:42]([R])[C:43]=[C:44][C:45]7=[O:46]`](/media/uploads/27031065-54ea-4403-a1c2-78707e2417d9.png)</MultiModal>

Identify and match the above bases. In all base-pairing relationships in DNA/RNA, the base-pairing relationships with fewer hydrogen bonds are denoted as \( D_1/R_1 \), while those with more hydrogen bonds are denoted as \( D_2/R_2 \). For each hydrogen bond in the paired bases \( D_1/R_1/D_2/R_2 \), the atomic number of the hydrogen bond donor is denoted as \( h_n \) (\( n=1\sim 2/3 \)), and the atomic number of the hydrogen bond acceptor is denoted as \( a_n \) (\( n=1\sim 2/3 \)), where \( n \) is sorted in ascending order of \( a \times h \). For the hydrogen bond pairing in the bases \( D_{1}/R_{1} \), calculate \( x_{D_{1}/R_{1}} = (a_2 - a_1) \times (h_2 - h_1) \). For the hydrogen bond pairing in the bases \( D_{2}/R_{2} \), calculate \( y_{D_{2}/R_{2}} = (a_3 - a_1) \times (h_3 - h_2) - a_2 \times h_2 \).

Finally, compute the value of \( z = \dfrac{x_{R_1} - y_{D_2}}{x_{D_1} - y_{R_2}} \).
现有以下五个核苷酸的碱基如图所示，其中重原子的原子编号在图中表示。

<MultiModal>![图中的碱基的SMARTS可以表示为 `[N:1][C:2]([N:3][C:4]1=[O:5])=[N:6][C:7]2=[C:8]1[N:9]=[C:10][N:11]2[R].[C:12][C:13]3=[C:14][N:15]([R])[C:16]([N:17][C:18]3=[O:19])=[O:20].[N:21][C:22]4=[N:23][C:24]=[N:25][C:26]5=[C:27]4[N:28]=[C:29][N:30]5[R].[O:31]=[C:32]6[N:33]([R])[C:34]=[C:35][C:36]([N:37])=[N:38]6.[O:39]=[C:40]([N:41]7)[N:42]([R])[C:43]=[C:44][C:45]7=[O:46]`](/media/uploads/27031065-54ea-4403-a1c2-78707e2417d9.png)</MultiModal>

将以上碱基进行识别与匹配，记在 DNA/RNA 中所有的碱基配对关系中，包含氢键个数较少的碱基配对关系分别记为 $D_1/R_1$，包含氢键个数较多的碱基配对关系分别记为 $D_2/R_2$；对于每一对配对的碱基 $D_1/R_1/D_2/R_2$ 中的氢键，氢键给体的原子编号记为 $h_n$（$n=1\sim 2/3$），氢键受体的原子编号记为 $a_n$（$n=1\sim 2/3$），其中 $n$ 的排序按照 $a\times h$ 由小到大排序；对于配对的碱基 $D_{1}/R_{1}$ 中的氢键配对关系，计算 $x_{D_{1}/R_{1}} =(a_2-a_1)\times(h_2-h_1)$；对于配对的碱基 $D_{2}/R_{2}$ 中的氢键配对关系，计算 $y_{D_{2}/R_{2}} =(a_3-a_1)\times(h_3-h_2)-a_2\times h_2$。

最终，计算 $z=\dfrac{x_{R_1}-y_{D_2}}{x_{D_1}-y_{R_2}}$ 的值。
0.55
0.11
1.40
-0.14
-0.55
0.14
-1.23
-0.20
0.55
0.11
1.40
-0.14
-0.55
0.14
-1.23
-0.20
The five bases in the figure are, from left to right and top to bottom: guanine (G, atoms 1-11), thymine (T, atoms 12-20), adenine (A, atoms 21-30), cytosine (C, atoms 31-38), and uracil (U, atoms 39-46).

<Checkpoint>[Atoms 1-11 are guanine (G), atoms 12-20 are thymine (T), atoms 21-30 are adenine (A), atoms 31-38 are cytosine (C), and atoms 39-46 are uracil (U)](2)</Checkpoint>

In DNA, the base pair with two hydrogen bonds, $D_1$, is adenine (A, atoms 21-30) and thymine (T, atoms 12-20).

<Checkpoint>[$D_1$ is the pairing of adenine and thymine](0.5)</Checkpoint>

The hydrogen bonds between them are $h=21, a=19$ and $h=17, a=23$. For sorting, we calculate the product $a \times h$: $23 \times 17 = 391$ and $19 \times 21 = 399$. Since 391 is smaller, the first pair is $a_1=23, h_1=17$, and the second pair is $a_2=19, h_2=21$.

<Checkpoint>[$D_1$ has $a_1=23, h_1=17$ and $a_2=19, h_2=21$](0.5)</Checkpoint>

Substituting into the formula $x =(a_2-a_1)\times(h_2-h_1)$, we obtain $x_{D_1} = (19 - 23) \times (21 - 17)=-16$.

<Checkpoint>[$x_{D_1}=-16$](1)</Checkpoint>

In RNA, the base pair with two hydrogen bonds, $R_1$, is adenine (A, atoms 21-30) and uracil (U, atoms 39-46).

<Checkpoint>[$R_1$ is the pairing of adenine and uracil](0.5)</Checkpoint>

Its hydrogen bonds are $h=21, a=46$ and $h=41, a=23$. The corresponding products are $23 \times 41 = 943$ and $46 \times 21 = 966$. After sorting, the first pair is $a_1=23, h_1=41$, and the second pair is $a_2=46, h_2=21$.

<Checkpoint>[$R_1$ has $a_1=23, h_1=41$ and $a_2=46, h_2=21$](0.5)</Checkpoint>

Calculation yields $x_{R_1} = (46 - 23) \times (21 - 41)=-460$.

<Checkpoint>[$x_{R_1} =-460$](1)</Checkpoint>

The base pair with three hydrogen bonds, $D_2$ and $R_2$, is guanine (G, atoms 1-11) and cytosine (C, atoms 31-38).

<Checkpoint>[$D_2/R_2$ is the pairing of guanine and cytosine](1)</Checkpoint>

This pairing has the same structure in both DNA and RNA, so the values of $y_{D_2}$ and $y_{R_2}$ are identical. The three given hydrogen bonds are $a=5, h=37$; $a=38, h=3$; and $a=31, h=1$. Their $a \times h$ products are $5 \times 37 = 185$, $38 \times 3 = 114$, and $31 \times 1 = 31$. Sorting from smallest to largest, we get: the first pair $a_1=31, h_1=1$; the second pair $a_2=38, h_2=3$; and the third pair $a_3=5, h_3=37$.

<Checkpoint>[$D_2/R_2$ has $a_1=31, h_1=1$, $a_2=38, h_2=3$, and $a_3=5, h_3=37$](1)</Checkpoint>

Using the formula $y =(a_3-a_1)\times(h_3-h_2)-a_2\times h_2$, the calculation is $y_{D_2} = (5 - 31) \times (37 - 3) - (38 \times 3)=-884 - 114=-998$. Thus, the value of $y_{R_2}$ is also **-998**.

<Checkpoint>[$y_{D_2}=y_{R_2}=-998$](1)</Checkpoint>

Finally, we substitute all calculated values into the final formula $z=\dfrac{x_{R_1}-y_{D_2}}{x_{D_1}-y_{R_2}}$. The calculation proceeds as $z = \dfrac{-460 - (-998)}{-16 - (-998)} = \dfrac{-460 + 998}{-16 + 998} = \dfrac{538}{982}\approx0.55$.

<Checkpoint>[$z =0.55$](1)</Checkpoint>

Therefore, the correct choice is option **A**.
图中五个碱基从左到右、从上到下依次为鸟嘌呤（G, 原子1-11）、胸腺嘧啶（T, 原子12-20）、腺嘌呤（A, 原子21-30）、胞嘧啶（C, 原子31-38）、尿嘧啶（U, 原子39-46）。

<Checkpoint>[原子1-11为鸟嘌呤（G）、原子12-20为胸腺嘧啶（T）、原子21-30为腺嘌呤（A）、原子31-38为胞嘧啶（C）、原子39-46为尿嘧啶（U）](2)</Checkpoint>

在DNA中，含有两个氢键的配对关系 $D_1$ 是腺嘌呤（A, 原子21-30）和胸腺嘧啶（T, 原子12-20）。

<Checkpoint>[$D_1$ 为腺嘌呤和胸腺嘧啶配对](0.5)</Checkpoint>

它们之间的氢键为 $h=21, a=19$ 和 $h=17, a=23$。为了排序，我们计算 $a \times h$ 的乘积：$23 \times 17 = 391$ 以及 $19 \times 21 = 399$。由于391较小，第一对为 $a_1=23, h_1=17$，第二对为 $a_2=19, h_2=21$。

<Checkpoint>[$D_1$ 中 $a_1=23, h_1=17$、$a_2=19, h_2=21$](0.5)</Checkpoint>

代入公式 $x =(a_2-a_1)\times(h_2-h_1)$，得到 $x_{D_1} = (19 - 23) \times (21 - 17)=-16$。

<Checkpoint>[$x_{D_1}=-16$](1)</Checkpoint>

在RNA中，含有两个氢键的配对关系 $R_1$ 是腺嘌呤（A, 原子21-30）和尿嘧啶（U, 原子39-46）。

<Checkpoint>[$R_1$ 为腺嘌呤和尿嘧啶配对](0.5)</Checkpoint>

其氢键为 $h=21, a=46$ 和 $h=41, a=23$。对应的乘积为 $23 \times 41 = 943$ 和 $46 \times 21 = 966$。排序后，第一对为 $a_1=23, h_1=41$，第二对为 $a_2=46, h_2=21$。

<Checkpoint>[$R_1$ 中 $a_1=23, h_1=41$、$a_2=46, h_2=21$](0.5)</Checkpoint>

计算可得 $x_{R_1} = (46 - 23) \times (21 - 41)=-460$。

<Checkpoint>[$x_{R_1} =-460$](1)</Checkpoint>

含有三个氢键的配对关系 $D_2$ 和 $R_2$ 则是鸟嘌呤（G, 原子1-11）和胞嘧啶（C, 原子31-38）。

<Checkpoint>[$D_2/R_2$ 为鸟嘌呤和胞嘧啶配对](1)</Checkpoint>

此配对在DNA和RNA中结构相同，因此 $y_{D_2}$ 和 $y_{R_2}$ 的值也相同。给定的三个氢键为 $a=5, h=37$；$a=38, h=3$ 以及 $a=31, h=1$。它们的 $a \times h$ 乘积分别为 $5 \times 37 = 185$，$38 \times 3 = 114$ 和 $31 \times 1 = 31$。按从小到大排序后得到：第一对 $a_1=31, h_1=1$；第二对 $a_2=38, h_2=3$；第三对 $a_3=5, h_3=37$。

<Checkpoint>[$D_2/R_2$ 中 $a_1=31, h_1=1$、$a_2=38, h_2=3$、$a_3=5, h_3=37$](1)</Checkpoint>

根据公式 $y =(a_3-a_1)\times(h_3-h_2)-a_2\times h_2$ 进行计算，$y_{D_2} = (5 - 31) \times (37 - 3) - (38 \times 3)=-884 - 114=-998$ 因此，$y_{R_2}$ 的值也是 **-998**。

<Checkpoint>[$y_{D_2}=y_{R_2}=-998$](1)</Checkpoint>

最后，我们将所有计算出的值代入最终公式 $z=\dfrac{x_{R_1}-y_{D_2}}{x_{D_1}-y_{R_2}}$。计算过程为 $z = \dfrac{-460 - (-998)}{-16 - (-998)} = \dfrac{-460 + 998}{-16 + 998} = \dfrac{538}{982}\approx0.55$。

<Checkpoint>[$z =0.55$](1)</Checkpoint>

因此，选择选项 **A**。