For the following two polypeptide fragments ($\rm{P1/P2}$) in a protein:

<MultiModal>![Fragment 1 (P1) SMARTS: [*]-[N:2]-[C@@:1]([C:28]([C:29])[C:30])-[C:3](=[O:4])-[N:5]-[C@@:6]([C@@:7]([C:8])[C:9]-[C:10])-[C:11](=[O:12])-[N:13]-[C@@:14]([C@:15]([O:16])[C:17])-[C:18](=[O:19])-[N:20]-[C@@:21]([C:22]([C:23])[C:24])-[C:25](=[O:26])-[N:27]-[*]](/media/uploads/3f271512-c312-4c47-a0aa-66ea72f04e6f.png)</MultiModal>

<MultiModal>![Fragment 2 (P2) SMARTS: [*]-[N:3]-[C@@:2]([C:1]-[c:33]1[c:34][c:35][c:36]([O:39])[c:37][c:38]1)-[C:4](=[O:5])-[N:6]-[C@@:7]([C:10]([C:8])[C:9])-[C:11](=[O:12])-[N:13]-[C@@:14]([C:15]-[C:16]([C:17])[C:18])-[C:19](=[O:20])-[N:21]-[C@@:22]([C:23]-[c:24]1[c:28][c:29][c:30][c:31][c:32]1)-[C:25](=[O:26])-[N:27]-[*]](/media/uploads/edbc6759-2279-4209-b423-81937ecbd09e.png)</MultiModal>

The two polypeptide fragments in the protein can undergo parallel $\beta$-sheet and antiparallel $\beta$-sheet formation. It is known that in each possible folding pairing mode, only the inter-amide hydrogen bond pairing links between the fragments are considered, and the hydrogen bond links with amino acid residues outside the fragments are not considered. At the same time, the number of hydrogen bonds formed between the two polypeptide fragments is the same in each folding pairing mode.

In each of the two polypeptide fragments paired by $\beta$-sheet, define:

$$S_1 = \sum_ia_{1,i}-\sum_jh_{1,j}$$

$$S_2 = \sum_ia_{2,i}-\sum_jh_{2,j}$$

where the subscript $_{1,2}$ is used to distinguish the two polypeptide fragments, $a_{1,i}$ represents the atomic number of the hydrogen bond acceptor in polypeptide fragment $1$, $h_{1,i}$ represents the atomic number of the hydrogen bond donor in polypeptide fragment $1$, $a_{2,i}$ represents the atomic number of the hydrogen bond acceptor in polypeptide fragment $2$, and $h_{2,i}$ represents the atomic number of the hydrogen bond donor in polypeptide fragment $2$.

Further define the parameter $z_n$, the value of which is calculated as follows:

$$
z_n = \begin{cases}
\dfrac{S_1 \cdot S_2}{\left(\min\{S_1, S_2\}\right)^2}\quad\quad \text{if parallel }\beta\text{-sheet}\\
\dfrac{S_1 \cdot S_2}{\left(\max\{S_1, S_2\}\right)^2}\quad\quad \text{if antiparallel }\beta\text{-sheet}\\
\end{cases}
$$

The subscript $_{n}$ represents different $\beta$-sheet pairing combinations forming different hydrogen bonds, and its value is $1,2,\cdots$.

Please observe the two polypeptide fragments and calculate the value of $\max{(z_n)},\,n=1,2,\cdots$.
对于蛋白质中的以下两个多肽片段（$\rm{P1/P2}$）：

片段1（P1）: <MultiModal>![SMILES:[*]-[N:2]-[C@@:1]([C:28]([C:29])[C:30])-[C:3](=[O:4])-[N:5]-[C@@:6]([C@@:7]([C:8])[C:9]-[C:10])-[C:11](=[O:12])-[N:13]-[C@@:14]([C@:15]([O:16])[C:17])-[C:18](=[O:19])-[N:20]-[C@@:21]([C:22]([C:23])[C:24])-[C:25](=[O:26])-[N:27]-[*]](/media/uploads/3f271512-c312-4c47-a0aa-66ea72f04e6f.png)</MultiModal>

片段2（P2）: <MultiModal>![SMILES:[*]-[N:3]-[C@@:2]([C:1]-[c:33]1[c:34][c:35][c:36]([O:39])[c:37][c:38]1)-[C:4](=[O:5])-[N:6]-[C@@:7]([C:10]([C:8])[C:9])-[C:11](=[O:12])-[N:13]-[C@@:14]([C:15]-[C:16]([C:17])[C:18])-[C:19](=[O:20])-[N:21]-[C@@:22]([C:23]-[c:24]1[c:28][c:29][c:30][c:31][c:32]1)-[C:25](=[O:26])-[N:27]-[*]](/media/uploads/edbc6759-2279-4209-b423-81937ecbd09e.png)</MultiModal>

两个多肽片段在蛋白质中，可以发生平行 $\beta$-折叠与反平行 $\beta$-折叠。已知每种可能的折叠配对方式中，只考虑片段之间的酰胺间氢键配对链接，而不考虑与片段外的氨基酸残基的氢键链接，同时，两个多肽片段之间形成的氢键数量在每种折叠的配对方式中均相同。

在每个通过$\beta$-折叠配对的两个多肽片段中，定义：

$$S_1 = \sum_ia_{1,i}-\sum_jh_{1,j}$$

$$S_2 = \sum_ia_{2,i}-\sum_jh_{2,j}$$

其中，下标 $_{1,2}$ 用以表示区分两个多肽片段，$a_{1,i}$ 表示多肽片段 $1$ 中的氢键受体的原子序号，$h_{1,i}$ 表示多肽片段 $1$ 中的氢键给体的原子序号，$a_{2,i}$ 表示多肽片段 $2$ 中的氢键受体的原子序号，$h_{2,i}$ 表示多肽片段 $2$ 中的氢键给体的原子序号。

进一步定义参数 $z_n$，其值计算方式如下：

$$
z_n = \begin{cases}
\dfrac{S_1 \cdot S_2}{\left(\min\{S_1, S_2\}\right)^2}\quad\quad \text{if 平行 }\beta\text{-折叠}\\
\dfrac{S_1 \cdot S_2}{\left(\max\{S_1, S_2\}\right)^2}\quad\quad \text{if 反平行 }\beta\text{-折叠}\\
\end{cases}
$$

下标 $_{n}$ 表示形成不同氢键的$\beta$-折叠配对组合，取值为 $1,2,\cdots$。

请你观察两个多肽片段，计算 $\max{(z_n)},\,n=1,2,\cdots$ 的值。
All other options are incorrect
0.750
1.333
0.875
3.00
0.890
1.500
1.866
1.145
1.000
其他选项均不正确
0.750
1.333
0.875
3.00
0.890
1.500
1.866
1.145
1.000
First consider that for these two polypeptide fragments, there are four possible $\beta$-sheet pairing modes:

1. Parallel $\beta$-sheet:
    - Hydrogen bonds ($\rm{P2}$ atom -> $\rm{P1}$ atom): $27->26$, $20<-20$, $13->12$, $5<-5$

<Checkpoint>[Folding mode 1: Parallel sheet, hydrogen bonds ($\rm{P2}$ atom -> $\rm{P1}$ atom): $27->26$, $20<-20$, $13->12$, $5<-5$](0.5)</Checkpoint>

2. Parallel $\beta$-sheet:
    - Hydrogen bonds ($\rm{P2}$ atom -> $\rm{P1}$ atom): $26<-27$, $21->19$, $12<-13$, $6->4$

<Checkpoint>[Folding mode 2: Parallel sheet, hydrogen bonds ($\rm{P2}$ atom -> $\rm{P1}$ atom): $26<-27$, $21->19$, $12<-13$, $6->4$](0.5)</Checkpoint>

3. Antiparallel $\beta$-sheet:
    - Hydrogen bonds ($\rm{P2}$ atom -> $\rm{P1}$ atom): $26<-5$, $21->12$, $12<-20$, $6->26$

<Checkpoint>[Folding mode 3: Antiparallel sheet, hydrogen bonds ($\rm{P2}$ atom -> $\rm{P1}$ atom): $26<-5$, $21->12$, $12<-20$, $6->26$](0.5)</Checkpoint>

4. Antiparallel $\beta$-sheet:
    - Hydrogen bonds ($\rm{P2}$ atom -> $\rm{P1}$ atom): 
$20<-2$, $13->4$, $5<-13$, $3->19$

<Checkpoint>[Folding mode 4: Antiparallel sheet, hydrogen bonds ($\rm{P2}$ atom -> $\rm{P1}$ atom): $20<-2$, $13->4$, $5<-13$, $3->19$](0.5)</Checkpoint>

Further calculate the $z_n$ value for each pairing combination.

1. Pairing combination one ($z_1$, parallel $\beta$-sheet)

- Hydrogen bonds (P2 atom -> P1 atom): $27->26$, $20<-20$, $13->12$, $5<-5$
- Fragment 1 (P1) atoms:
  - Donors ($h_1$): 20, 5
  - Acceptors ($a_1$): 26, 12
- Fragment 2 (P2) atoms:
  - Donors ($h_2$): 27, 13
  - Acceptors ($a_2$): 20, 5
- Calculate $S_1$ and $S_2$:
  - $S_1 = (26 + 12) - (20 + 5) = 38 - 25 = 13$
  - $S_2 = (20 + 5) - (27 + 13) = 25 - 40 = -15$
- Calculate $z_1$:
  $$
  z_1 = \dfrac{S_1 \cdot S_2}{\left(\min\{S_1, S_2\}\right)^2} = \dfrac{13 \times (-15)}{(\min\{13, -15\})^2} = \dfrac{-195}{(-15)^2} = \dfrac{-195}{225} \approx -0.8667
  $$

<Checkpoint>[$z=-0.8667$ for one pairing mode](1)</Checkpoint>

2. Pairing combination two ($z_2$, parallel $\beta$-sheet)

- Hydrogen bonds (P2 atom -> P1 atom): $26<-27$, $21->19$, $12<-13$, $6->4$
- Fragment 1 (P1) atoms:
  - Donors ($h_1$): 27, 13
  - Acceptors ($a_1$): 19, 4
- Fragment 2 (P2) atoms:
  - Donors ($h_2$): 21, 6
  - Acceptors ($a_2$): 26, 12
- Calculate $S_1$ and $S_2$:
  - $S_1 = (19 + 4) - (27 + 13) = 23 - 40 = -17$
  - $S_2 = (26 + 12) - (21 + 6) = 38 - 27 = 11$
- Calculate $z_2$:
  $$
  z_2 = \dfrac{S_1 \cdot S_2}{\left(\min\{S_1, S_2\}\right)^2} = \dfrac{-17 \times 11}{(\min\{-17, 11\})^2} = \dfrac{-187}{(-17)^2} = \dfrac{-187}{289} \approx -0.6471
  $$

<Checkpoint>[$z=-0.6471$ for one pairing mode](1)</Checkpoint>

3. Pairing combination three ($z_3$, antiparallel $\beta$-sheet)

- Hydrogen bonds (P2 atom -> P1 atom): $26<-5$, $21->12$, $12<-20$, $6->26$
- Fragment 1 (P1) atoms:
  - Donors ($h_1$): 5, 20
  - Acceptors ($a_1$): 12, 26
- Fragment 2 (P2) atoms:
  - Donors ($h_2$): 21, 6
  - Acceptors ($a_2$): 26, 12
- Calculate $S_1$ and $S_2$:
  - $S_1 = (12 + 26) - (5 + 20) = 38 - 25 = 13$
  - $S_2 = (26 + 12) - (21 + 6) = 38 - 27 = 11$
- Calculate $z_3$:
  $$
  z_3 = \dfrac{S_1 \cdot S_2}{\left(\max\{S_1, S_2\}\right)^2} = \dfrac{13 \times 11}{(\max\{13, 11\})^2} = \dfrac{143}{13^2} = \dfrac{143}{169} \approx 0.8462
  $$

<Checkpoint>[$z=0.8462$ for one pairing mode](1)</Checkpoint>

4. Pairing combination four ($z_4$, antiparallel $\beta$-sheet)

- Hydrogen bonds (P2 atom -> P1 atom): $20<-2$, $13->4$, $5<-13$, $3->19$
- Fragment 1 (P1) atoms:
  - Donors ($h_1$): 2, 13
  - Acceptors ($a_1$): 4, 19
- Fragment 2 (P2) atoms:
  - Donors ($h_2$): 13, 3
  - Acceptors ($a_2$): 20, 5
- Calculate $S_1$ and $S_2$:
  - $S_1 = (4 + 19) - (2 + 13) = 23 - 15 = 8$
  - $S_2 = (20 + 5) - (13 + 3) = 25 - 16 = 9$
- Calculate $z_4$:
  $$
  z_4 = \dfrac{S_1 \cdot S_2}{\left(\max\{S_1, S_2\}\right)^2} = \dfrac{8 \times 9}{(\max\{8, 9\})^2} = \dfrac{72}{9^2} = \dfrac{72}{81} \approx 0.8889
  $$

<Checkpoint>[$z=0.8889$ for one pairing mode](1)</Checkpoint>

Finally, all $z_n$ values are as follows:

- $z_1 \approx -0.8667$
- $z_2 \approx -0.6471$
- $z_3 \approx 0.8462$
- $z_4 \approx 0.8889$

Comparing these four values, the maximum value can be obtained.

$$\max{(z_n)} = z_4 \approx \bf{0.890}$$

Therefore, choose option **F**.

<Checkpoint>[Maximum value $\max{(z_n)}=0.890$, choose option **F**](1)</Checkpoint>
首先考虑对于这两个多肽片段，共有四种可能的 $\beta$-折叠配对方式：

1. 平行 $\beta$-折叠：
    - 氢键 ($\rm{P2}$原子 -> $\rm{P1}$原子): $27->26$, $20<-20$, $13->12$, $5<-5$

<Checkpoint>[折叠方式1：平行折叠、氢键 ($\rm{P2}$原子 -> $\rm{P1}$原子): $27->26$, $20<-20$, $13->12$, $5<-5$](0.5)</Checkpoint>

2. 平行 $\beta$-折叠：
    - 氢键 ($\rm{P2}$原子 -> $\rm{P1}$原子): $26<-27$, $21->19$, $12<-13$, $6->4$

<Checkpoint>[折叠方式2：平行折叠、氢键 ($\rm{P2}$原子 -> $\rm{P1}$原子): $26<-27$, $21->19$, $12<-13$, $6->4$](0.5)</Checkpoint>

3. 反平行 $\beta$-折叠：
    - 氢键 ($\rm{P2}$原子 -> $\rm{P1}$原子): $26<-5$, $21->12$, $12<-20$, $6->26$

<Checkpoint>[折叠方式3：反平行折叠、氢键 ($\rm{P2}$原子 -> $\rm{P1}$原子): $26<-5$, $21->12$, $12<-20$, $6->26$](0.5)</Checkpoint>

4. 反平行 $\beta$-折叠：
    - 氢键 ($\rm{P2}$原子 -> $\rm{P1}$原子): 
$20<-2$, $13->4$, $5<-13$, $3->19$

<Checkpoint>[折叠方式4：反平行折叠、氢键 ($\rm{P2}$原子 -> $\rm{P1}$原子): $20<-2$, $13->4$, $5<-13$, $3->19$](0.5)</Checkpoint>

进一步计算每种配对组合的 $z_n$ 值。

1. 配对组合一 ($z_1$, 平行 $\beta$-折叠)

- 氢键 (P2原子 -> P1原子): $27->26$, $20<-20$, $13->12$, $5<-5$
- 片段1 (P1) 原子:
  - 给体 (Donors, $h_1$): 20, 5
  - 受体 (Acceptors, $a_1$): 26, 12
- 片段2 (P2) 原子:
  - 给体 (Donors, $h_2$): 27, 13
  - 受体 (Acceptors, $a_2$): 20, 5
- 计算 $S_1$ 和 $S_2$:
  - $S_1 = (26 + 12) - (20 + 5) = 38 - 25 = 13$
  - $S_2 = (20 + 5) - (27 + 13) = 25 - 40 = -15$
- 计算 $z_1$:
  $$
  z_1 = \dfrac{S_1 \cdot S_2}{\left(\min\{S_1, S_2\}\right)^2} = \dfrac{13 \times (-15)}{(\min\{13, -15\})^2} = \dfrac{-195}{(-15)^2} = \dfrac{-195}{225} \approx -0.8667
  $$

<Checkpoint>[一种配对方式的$z=-0.8667$](1)</Checkpoint>

2. 配对组合二 ($z_2$, 平行 $\beta$-折叠)

- 氢键 (P2原子 -> P1原子): $26<-27$, $21->19$, $12<-13$, $6->4$
- 片段1 (P1) 原子:
  - 给体 (Donors, $h_1$): 27, 13
  - 受体 (Acceptors, $a_1$): 19, 4
- 片段2 (P2) 原子:
  - 给体 (Donors, $h_2$): 21, 6
  - 受体 (Acceptors, $a_2$): 26, 12
- 计算 $S_1$ 和 $S_2$:
  - $S_1 = (19 + 4) - (27 + 13) = 23 - 40 = -17$
  - $S_2 = (26 + 12) - (21 + 6) = 38 - 27 = 11$
- 计算 $z_2$:
  $$
  z_2 = \dfrac{S_1 \cdot S_2}{\left(\min\{S_1, S_2\}\right)^2} = \dfrac{-17 \times 11}{(\min\{-17, 11\})^2} = \dfrac{-187}{(-17)^2} = \dfrac{-187}{289} \approx -0.6471
  $$

<Checkpoint>[一种配对方式的$z=-0.6471$](1)</Checkpoint>

3. 配对组合三 ($z_3$, 反平行 $\beta$-折叠)

- 氢键 (P2原子 -> P1原子): $26<-5$, $21->12$, $12<-20$, $6->26$
- 片段1 (P1) 原子:
  - 给体 (Donors, $h_1$): 5, 20
  - 受体 (Acceptors, $a_1$): 12, 26
- 片段2 (P2) 原子:
  - 给体 (Donors, $h_2$): 21, 6
  - 受体 (Acceptors, $a_2$): 26, 12
- 计算 $S_1$ 和 $S_2$:
  - $S_1 = (12 + 26) - (5 + 20) = 38 - 25 = 13$
  - $S_2 = (26 + 12) - (21 + 6) = 38 - 27 = 11$
- 计算 $z_3$:
  $$
  z_3 = \dfrac{S_1 \cdot S_2}{\left(\max\{S_1, S_2\}\right)^2} = \dfrac{13 \times 11}{(\max\{13, 11\})^2} = \dfrac{143}{13^2} = \dfrac{143}{169} \approx 0.8462
  $$

<Checkpoint>[一种配对方式的$z=0.8462$](1)</Checkpoint>

4. 配对组合四 ($z_4$, 反平行 $\beta$-折叠)

- 氢键 (P2原子 -> P1原子): $20<-2$, $13->4$, $5<-13$, $3->19$
- 片段1 (P1) 原子:
  - 给体 (Donors, $h_1$): 2, 13
  - 受体 (Acceptors, $a_1$): 4, 19
- 片段2 (P2) 原子:
  - 给体 (Donors, $h_2$): 13, 3
  - 受体 (Acceptors, $a_2$): 20, 5
- 计算 $S_1$ 和 $S_2$:
  - $S_1 = (4 + 19) - (2 + 13) = 23 - 15 = 8$
  - $S_2 = (20 + 5) - (13 + 3) = 25 - 16 = 9$
- 计算 $z_4$:
  $$
  z_4 = \dfrac{S_1 \cdot S_2}{\left(\max\{S_1, S_2\}\right)^2} = \dfrac{8 \times 9}{(\max\{8, 9\})^2} = \dfrac{72}{9^2} = \dfrac{72}{81} \approx 0.8889
  $$

<Checkpoint>[一种配对方式的$z=0.8889$](1)</Checkpoint>

最终所有的 $z_n$ 值如下：

- $z_1 \approx -0.8667$
- $z_2 \approx -0.6471$
- $z_3 \approx 0.8462$
- $z_4 \approx 0.8889$

比较这四个值，可以得出最大值。

$$\max{(z_n)} = z_4 \approx \bf{0.890}$$

因此，选择选项 **F**。

<Checkpoint>[最大值$\max{(z_n)}=0.890$，选择选项 **F**](1)</Checkpoint>