The following DNA fragment $\mathbf A$ is available, combined with a $\rm ^{32}P$ labeled primer:

<MultiModal>![The figure shows the DNA fragment bound to the primer. In the fragment, there is a sequence at the 5' end that is not bound to the primer, denoted as $\rm X\cdots\cdots X$](/media/uploads/1b992784-6d18-4e87-b3e3-fff892cf8aa3.png)</MultiModal>

Where $\mathbf X$ represents one of $\rm A/T/C/G$.

DNA polymerase, $\rm ^{32}P$ labeled primer, $\rm Mg^{2+}$, and four dNTPs were added to four test tubes, and a small amount of the following substrates were added separately:

1. Test tube 1: ddTTP
2. Test tube 2: ddGTP
3. Test tube 3: ddATP
4. Test tube 4: ddCTP

After the reaction is completed, the product fragments are separated by PAGE gel and autoradiographed to obtain the following band diagram:

<MultiModal>![The band diagram has four lanes, labeled 1/2/3/4 respectively. In the figure, the lane numbers in which each band appears from top to bottom are 1413312144112421331](/media/uploads/ea9a4d5c-c855-474a-9a73-1acbb2cb658f.png)</MultiModal>

Based on this band diagram, please find out which polypeptide sequence cannot be synthesized from the following options for the part of DNA fragment $\mathbf A$ that is not bound to the primer.
现有如下DNA片段 $\mathbf A$，结合了 $\rm ^{32}P$ 标记的引物：

<MultiModal>![图中展示了结合引物的DNA片段，片段中，5'端有一段未结合引物的序列，记为 $\rm X\cdots\cdots X$](/media/uploads/1b992784-6d18-4e87-b3e3-fff892cf8aa3.png)</MultiModal>

其中 $\mathbf X$ 表示 $\rm A/T/C/G$ 中的一种。

 在四个试管中加入DNA聚合酶、$\rm ^{32}P$ 标记的引物、$\rm Mg^{2+}$、四种dNTP，并分别加入少量的如下底物：

1. 试管1：ddTTP	
2. 试管2：ddGTP	
3. 试管3：ddATP	
4. 试管4：ddCTP

反应完成后用PAGE胶分离产物片段并进行放射性自显影，得到如下条带图：

<MultiModal>![条带图有四条泳道，分别标有1/2/3/4，图中从上到下每个条带出现的泳道标号分别为1413312144112421331](/media/uploads/ea9a4d5c-c855-474a-9a73-1acbb2cb658f.png)</MultiModal>

请你根据这个条带图，从以下选项中找出DNA片段 $\mathbf A$ 中未结合引物的部分不可能合成那种多肽序列。
All other options are incorrect
Gln-Ile-Thr-Gly-Thr-His
Met-Arg-Ser-Cys-Gly-Ser
Arg-Leu-Gln-Glu-Arg-Ile
Asp-Tyr-Arg-Asn-Ala-Leu
Asn-Ala-Phe-Leu-Ser-Ile
Met-Arg-Ser-Cys-Gln-Ser
Unable to form a polypeptide chain.
其他选项均不正确
Gln-Ile-Thr-Gly-Thr-His
Met-Arg-Ser-Cys-Gly-Ser
Arg-Leu-Gln-Glu-Arg-Ile
Asp-Tyr-Arg-Asn-Ala-Leu
Asn-Ala-Phe-Leu-Ser-Ile
Met-Arg-Ser-Cys-Gln-Ser
无法形成多肽链
The method designed for this question is a typical application of Sanger dideoxy chain termination sequencing using PAGE.

<Checkpoint>[The question involves Sanger dideoxy chain termination sequencing using PAGE](0.5)</Checkpoint>

This method has the following characteristics:
- For single-stranded DNA $\mathbf A$, its 3' end is bound to a
 P-labeled primer (i.e., the primer serves as the 5'→3' starting point for the newly synthesized strand)
- Four reaction systems, with ddTTP, ddGTP, ddATP, and ddCTP added separately, generate DNA fragments terminated by each dideoxy
- PAGE electrophoresis runs from top to bottom, short to long (the top end is short chains, close to the upper edge)
- Each band represents the primer plus extension until encountering the corresponding termination (e.g., ddTTP, terminates at the next A)
- The sequence read should be in the 5' to 3' direction of the newly synthesized strand

Read the PAGE information in the second image, and analyze each row (from bottom to top) according to the lane bands:

| Position        | Tube 1 (T) | Tube 2 (G) | Tube 3 (A) | Tube 4 (C) |
| :---------- | :-------- | :-------- | :-------- | :-------- |
| 1 (Bottom) | +         |           |           |           |
| 2           |           |           | +         |           |
| 3           |           |           | +         |           |
| 4           | +         |           |           |           |
| 5           |           | +         |           |           |
| 6           |           |           |           | +         |
| 7           |           | +         |           |           |
| 8           | +         |           |           |           |
| 9           | +         |           |           |           |
| 10          |           |           |           | +         |
| 11          |           |           |           | +         |
| 12          | +         |           |           |           |
| 13          |           | +         |           |           |
| 14          | +         |           |           |           |
| 15          |           |           | +         |           |
| 16          |           |           | +         |           |
| 17          | +         |           |           |           |
| 18          |           |           |           | +         |
| 19          | +         |           |           |           |

<Checkpoint>[The determined sequence of the newly synthesized strand is: 5'-TAATGCGTTCCTGTAATCTG-3'](1)</Checkpoint>

The question requires giving the sequence of the un-primer-bound portion of DNA fragment $\mathbf A$, which is the 5' end sequence of the template, and the newly synthesized strand is complementary to the template. Therefore, the sequence of the $\rm X\cdots\cdots X$ strand (template) is the complement of the new strand: 5'-CAGATTACAGGAACGCATTA-3'

<Checkpoint>[The $\rm X\cdots\cdots X$ strand is 5'-CAGATTACAGGAACGCATTA-3'](1)</Checkpoint>

(1) The $\rm X\cdots\cdots X$ strand is the coding strand, and the corresponding mRNA is 5'-CAGAUUACAGGAACGCAUUA-3', which has three possible reading frame forms:

<Checkpoint>[In $\mathbf A$, the $\rm X\cdots\cdots X$ strand may be the coding or template strand during translation](0.5)</Checkpoint>, consider the possibility of each case:

1. Starting from the 1st position: CAG (Gln) - AUU (Ile) - ACA (Thr) - GGA (Gly) - ACG (Thr) - CAU (His), producing the polypeptide: Gln-Ile-Thr-Gly-Thr-His, i.e., option B
2. Starting from the 2nd position: AGA (Arg) - UUA (Leu) - CAG (Gln) - GAA (Glu) - CGC (Arg) - AUU (Ile) producing the polypeptide: Arg-Leu-Gln-Glu-Arg-Ile, i.e., option D
3. Starting from the 3rd position: GAU (Asp) - UAC (Tyr) - AGG (Arg) - AAC (Asn) - GCA (Ala) - UUA (Leu) producing the polypeptide: Asp-Tyr-Arg-Asn-Ala-Leu, i.e., option E

<Checkpoint>[When the $\rm X\cdots\cdots X$ strand in $\mathbf A$ is the coding strand, the corresponding polypeptides are Gln-Ile-Thr-Gly-Thr-His, Arg-Leu-Gln-Glu-Arg-Ile, Asp-Tyr-Arg-Asn-Ala-Leu](1)</Checkpoint>

(2) The $\rm X\cdots\cdots X$ strand is the template strand, and the corresponding mRNA is 5'- UAAUGCGUUCCUGUCAAUCUG-3', which has three possible reading frame forms:

1. Starting from the 1st position: UAA (Stop codon) Result: Translation will terminate immediately, and a polypeptide chain cannot be formed.
2. Starting from the 2nd position: AAU (Asn) - GCG (Ala) - UUC (Phe) - CUG (Leu) - UCA (Ser) - AUC (Ile) producing the polypeptide: Asn-Ala-Phe-Leu-Ser-Ile, i.e., option F
3. Starting from the 3rd position: AUG (Met, Start codon) - CGU (Arg) - UCC (Ser) - UGU (Cys) - CAA (Gln) - UCU (Ser) producing the polypeptide: Met-Arg-Ser-Cys-Gln-Ser, i.e., option G

<Checkpoint>[When the $\rm X\cdots\cdots X$ strand in $\mathbf A$ is the template strand, the corresponding polypeptides are "a polypeptide chain cannot be formed", Asn-Ala-Phe-Leu-Ser-Ile, Met-Arg-Ser-Cys-Gln-Ser](1)</Checkpoint>

Therefore, only option C: Met-Arg-Ser-Cys-Gly-Ser is an impossible polypeptide to form.
这道题目设计的方法，是典型使用PAGE的Sanger双脱氧链终止法测序。

<Checkpoint>[题目涉及使用PAGE的Sanger双脱氧链终止法测序](0.5)</Checkpoint>

该方法有以下特点：
- 对于单链DNA $\mathbf A$，其3'末端结合了
 P标记引物（即引物作为新合成链的5'→3'起始点）
- 四个反应体系，分别加入ddTTP、ddGTP、ddATP、ddCTP，生成以各dideoxy为终止的DNA片段
- PAGE电泳中，DNA片段由上（负极）向下（正极）迁移，片段越短，迁移距离越远，因此条带越靠下。读序时应从凝胶底部向上读取，这对应新合成链的5'→3'方向

读取第二幅图的PAGE信息，按照泳道条带，在每一行（从下到上）分析：

| 位置        | 1号管 (T) | 2号管 (G) | 3号管 (A) | 4号管 (C) |
| :---------- | :-------- | :-------- | :-------- | :-------- |
| 1（最下方） | +         |           |           |           |
| 2           |           |           | +         |           |
| 3           |           |           | +         |           |
| 4           | +         |           |           |           |
| 5           |           | +         |           |           |
| 6           |           |           |           | +         |
| 7           |           | +         |           |           |
| 8           | +         |           |           |           |
| 9           | +         |           |           |           |
| 10          |           |           |           | +         |
| 11          |           |           |           | +         |
| 12          | +         |           |           |           |
| 13          |           | +         |           |           |
| 14          | +         |           |           |           |
| 15          |           |           | +         |           |
| 16          |           |           | +         |           |
| 17          | +         |           |           |           |
| 18          |           |           |           | +         |
| 19          | +         |           |           |           |

<Checkpoint>[确定得到新合成链的序列是：5'-TAATGCGTTCCTGTAATCTG-3'](1)</Checkpoint>

题目要求给出DNA片段$\mathbf A$中未结合引物的部分序列，也就是模板的5'端序列，而新合成链与模板互为互补。得到所以 $\rm X\cdots\cdots X$ 链（模板）序列为新链补链：5'-CAGATTACAGGAACGCATTA-3'

<Checkpoint>[$\rm X\cdots\cdots X$ 链为 5'-CAGATTACAGGAACGCATTA-3'](1)</Checkpoint>

（一）$\rm X\cdots\cdots X$ 链为编码链，对应的mRNA为5'-CAGAUUACAGGAACGCAUUA-3'，有三种可能的阅读框形式：

<Checkpoint>[$\mathbf A$ 中 $\rm X\cdots\cdots X$ 链在翻译时可能是编码链或模板链](0.5)</Checkpoint>，考虑其每一种情况的可能性：

1. 从第1位开始：CAG (Gln) - AUU (Ile) - ACA (Thr) - GGA (Gly) - ACG (Thr) - CAU (His)，产生多肽：Gln-Ile-Thr-Gly-Thr-His，即选项 B
2. 从第2位开始：AGA (Arg) - UUA (Leu) - CAG (Gln) - GAA (Glu) - CGC (Arg) - AUU (Ile) 产生多肽: Arg-Leu-Gln-Glu-Arg-Ile，即选项 D
3. 从第3位开始：GAU (Asp) - UAC (Tyr) - AGG (Arg) - AAC (Asn) - GCA (Ala) - UUA (Leu) 产生多肽: Asp-Tyr-Arg-Asn-Ala-Leu，即选项 E

<Checkpoint>[$\mathbf A$ 中 $\rm X\cdots\cdots X$ 链为编码链时，对应的多肽为Gln-Ile-Thr-Gly-Thr-His、Arg-Leu-Gln-Glu-Arg-Ile、Asp-Tyr-Arg-Asn-Ala-Leu](1)</Checkpoint>

（二）$\rm X\cdots\cdots X$ 链为模板链，对应的mRNA为5'- UAAUGCGUUCCUGUCAAUCUG-3'，有三种可能的阅读框形式：

1. 从第1位开始：UAA (终止密码子) 结果: 翻译会立即终止，无法形成多肽链。
2. 从第2位开始: AAU (Asn) - GCG (Ala) - UUC，即选项 H
(Phe) - CUG (Leu) - UCA (Ser) - AUC (Ile) 产生多肽: Asn-Ala-Phe-Leu-Ser-Ile，即选项 F
3. 从第3位开始：AUG (Met, 起始密码子) - CGU (Arg) - UCC (Ser) - UGU (Cys) - CAA (Gln) - UCU (Ser) 产生多肽: Met-Arg-Ser-Cys-Gln-Ser，即选项 G

<Checkpoint>[$\mathbf A$ 中 $\rm X\cdots\cdots X$ 链为模板链时，对应的多肽为“无法形成多肽链”、Asn-Ala-Phe-Leu-Ser-Ile、Met-Arg-Ser-Cys-Gln-Ser](1)</Checkpoint>

因此，只有选项 C: Met-Arg-Ser-Cys-Gly-Ser 为不可能形成的多肽。