DNA is an important biomolecule. In organisms, the four bases $A, T, C, G$ in DNA molecules have strict matching rules: the Watson-Crick base pairing rule. However, in rare organisms, or under in vitro conditions, bases may form combinations that do not conform to Watson-Crick pairing. For example, if a base is protonated, it can form unusual pairing patterns; in addition, DNA can also form triple helix structures, mainly divided into pyrimidine-purine-pyrimidine and purine-purine-pyrimidine pairing modes.

Replace one $\mathrm{H}$ atom on the $\mathrm{N}$ atom of acrylamide with the following two DNA single strands: (1)$5'-(AAA)_7-3'$; (2)$5'-(AAACCCC)_2-3'$; then copolymerize these two substituted acrylamides together with acrylamide to prepare polyacrylamide containing a certain number of DNA side chains. Studies have shown that the polymer undergoes different conformational changes in aqueous systems with different pH values and exists in the form of hydrogels or ordinary aqueous solutions. Please determine the form in which the polymer exists in each of the following scenarios.

(a) When the pH of the system is 1.1;

(b) When the pH of the system is 5.2;

(c) When the pH of the system is 7.2;

(d) Continue to add the following two DNA single strands to system (c) in sequence, $5'-(TTT)_7-3'$; $5'-(TTT)_{15}-3'$;

(e) When the pH of the system is 10.2;

(f) Continue to add the following two DNA single strands to system (e) in sequence, $5'-(TTT)_7-3'$; $5'-(TTT)_{15}-3'$;

When answering, please select six correct answers in sequence from the options below to form an answer sequence (such as AAABBB):

A. Hydrogel B. Aqueous solution
DNA是一种重要的生物分子。在生物体中，DNA分子中的四种碱基$A、T、C、G$之间有着严格的匹配规则：Watson-Crick 碱基互补配对规则。但是，在极少数的生物体中，或者在体外条件下，碱基有可能形成不符合Watson-Crick配对的组合模式。例如，若碱基被质子化，则可以形成不同寻常的配对形式；此外，DNA还可形成三螺旋结构，主要分为嘧啶-嘌呤-嘧啶和嘌呤-嘌呤-嘧啶两种配对模式。

将丙烯酰胺的$\mathrm{N}$原子上的1个$\mathrm{H}$原子用以下两种DNA单链取代，(1)$5'-(AAA)_7-3'$;(2)$5'-(AAACCCC)_2-3'$；然后将这两种取代丙烯酰胺和丙烯酰胺一起进行共聚，可以制得含有一定数量的DNA侧链的聚丙烯酰胺。研究表明，该聚合物在不同pH的水体系中会发生不同的变构反应，并以水凝胶或普通水溶液的形体存在。请分别判断下列几个情形中该聚合物的存在形式是什么。

（a）体系的pH=1.1时；

（b）体系的pH=5.2时；

（c）体系的pH=7.2时；

（d）继续向体系（c）依次加入以下两种DNA单链，$5'-(TTT)_7-3'$；$5'-(TTT)_{15}-3'$时；

（e）体系的pH=10.2时；

（f）继续向体系（e）依次加入以下两种DNA单链，$5'-(TTT)_7-3'$；$5'-(TTT)_{15}-3'$时；

作答时请从下面的选项中进行依次选择六个正确的答案组成答案序列（如AAABBB）：

A.水凝胶 B.水溶液
All other options are incorrect
AAAAAA
AABAAA
AABABA
ABAABB
BBABAA
AAAABB
ABAABA
AAAABA
BBBBBB
ABBABB
其他选项均不正确
AAAAAA
AABAAA
AABABA
ABAABB
BBABAA
AAAABB
ABAABA
AAAABA
BBBBBB
ABBABB
After the bases are protonated, they can form unusual pairing patterns. The bases given in the question are mainly $A$ and $C$, so consider the pairing situations after these two bases are protonated.

After A is protonated, it can produce $A\mathrm{H^+}-\mathrm{H^+}A$ pairing, and its structure is shown in the figure below:

<MultiModal>![The figure shows two [H][N+](C=NC1=C2N=CN1[R])=C2N([H])[H] forming hydrogen bonds with each other. The amine group at the 6-position of each purine ring forms a hydrogen bond with the nitrogen atom at the 7-position of the other purine ring.](/media/uploads/7b3dede3-3dc2-4547-b3a7-06cb59dda44e.png)</MultiModal>

<Checkpoint>[Protonated $A$ can form $A\mathrm{H^+}-\mathrm{H^+}A$ pairing](1)</Checkpoint>

Similarly, protonated $C$ can also form hydrogen bonds with unprotonated $C$, and its structure is as follows:

<MultiModal>![The figure shows three hydrogen bonds formed between O=C1N([R])C=CC(N([H])[H])=[N+]1[H] and O=C2N([R])C=CC(N([H])[H])=N2. The carbonyl group at the 2-position of the pyrimidine ring forms a hydrogen bond with the amine group at the 4-position of the other pyrimidine ring, and the nitrogen atoms at the 3-position of the two pyrimidine rings form a hydrogen bond through a proton.](/media/uploads/2b3b9506-fc2d-4e19-a2b4-72c2d4854c9e.png)</MultiModal>

<Checkpoint>[Protonated $C$ can form $\mathrm{H^+}C:C$ pairing with unprotonated $C$](1)</Checkpoint>

In addition, the question also mentions a triple helix structure. In this question, a pyrimidine-purine-pyrimidine pairing method can be formed, such as $T\cdot A-T$ pairing, and its structure is shown in the figure below:

<MultiModal>![The figure shows the three-base pairing structure formed between two CC(C(N1[H])=O)=CN(C1=O)[R] and one [H]N(C1=NC=NC2=C1N=CN2[R])[H]. The nitrogen atom at the 1-position and the amine group at the 6-position of the purine ring form two hydrogen bonds with the amine group at the 3-position and the carbonyl group at the 4-position of one pyrimidine ring, respectively. The nitrogen atom at the 7-position and the amine group at the 6-position of the same purine ring form two hydrogen bonds with the amine group at the 3-position and the carbonyl group at the 4-position of the other pyrimidine ring, respectively.](/media/uploads/fe5c10ee-ef40-40ff-8033-20ed8f1b29de.png)</MultiModal>

<Checkpoint>[$T$ can form $T\cdot A-T$ pairing with $A$](1)</Checkpoint>

With these three pairing methods, we can start analyzing the options.

(a) When the pH of the system is 1.1, the bases $A$ in the acrylamide side chain $5'-(AAA)_7-3'$ are largely protonated to $A\mathrm{H^+}$, and cross-linking between chains is achieved through $A\mathrm{H^+}-\mathrm{H^+}A$ pairing, resulting in its existence in the form of a hydrogel.

<Checkpoint>[When pH=1.1, cross-linking between chains is achieved through $A\mathrm{H^+}-\mathrm{H^+}A$ pairing, existing in the form of a hydrogel](1)</Checkpoint>

(b) When the pH of the system is 5.2, the bases $C$ in the acrylamide side chain $5'-(AAACCCC)_2-3'$ are partially protonated to $C\mathrm{H^+}$, and cross-linking between chains is achieved through $\mathrm{H^+}C:C$ pairing, resulting in its existence in the form of a hydrogel.

<Checkpoint>[When pH=5.2, cross-linking between chains is achieved through $\mathrm{H^+}C:C$ pairing, existing in the form of a hydrogel](1)</Checkpoint>

(c) When the pH of the system is 7.2, $A$ and $C$ are hardly protonated, and cross-linking between chains is difficult, resulting in its existence in the form of an aqueous solution.

<Checkpoint>[When the pH of the system is 7.2, $A$ and $C$ are hardly protonated, and cross-linking between chains is difficult](1)</Checkpoint>

(d) After adding $5'-(TTT)_7-3'$; $5'-(TTT)_{15}-3'$, a triple helix structure of $T\cdot A-T$ pairing is formed in the system. Since the length of $5'-(AAA)_7-3'$ and $5'-(TTT)_7-3'$ are equal, and the length of $5'-(TTT)_{15}-3'$ is slightly larger than twice the length of $5'-(AAA)_7-3'$, each $5'-(TTT)_{15}-3'$ can pair with two $5'-(AAA)_7-3'$ of the side chain (as well as two $5'-(TTT)_{15}-3'$) to achieve cross-linking between chains, resulting in its existence in the form of a hydrogel.

<Checkpoint>[Each $5'-(TTT)_{15}-3'$ can pair with two $5'-(AAA)_7-3'$ of the side chain to achieve cross-linking between chains, forming a triple helix structure of $T\cdot A-T$ pairing](2)</Checkpoint>

(e) (f) When the pH of the system is 10.2, the $\mathrm{N-H}$ of the base $T$ is deprotonated, making it difficult to form a $T\cdot A-T$ pairing structure, resulting in its existence in the form of an aqueous solution.

<Checkpoint>[When pH=10.2, the $\mathrm{N-H}$ of the base $T$ is deprotonated, making it difficult to form a $T\cdot A-T$ pairing structure](1)</Checkpoint>

Therefore, the answer is AABABB, and option A is correct.
碱基被质子化后可以形成不同寻常的配对形式，题中所给的碱基主要为$A$和$C$，因此考虑这两种碱基质子化后的配对情况。

A被质子化后可以产生$A\mathrm{H^+}-\mathrm{H^+}A$配对，其结构如下图所示：

<MultiModal>![图中展示了两个[H][N+](C=NC1=C2N=CN1[R])=C2N([H])[H]互相形成氢键，每个嘌呤环6号位上的胺基与另一个嘌呤环上7号位的氮原子形成氢键。](/media/uploads/7b3dede3-3dc2-4547-b3a7-06cb59dda44e.png)</MultiModal>

<Checkpoint>[质子化的$A$之间可以形成$A\mathrm{H^+}-\mathrm{H^+}A$配对](1)</Checkpoint>

类似的，质子化的$C$与未质子化的$C$之间也可以形成氢键，其结构如下：

<MultiModal>![图中展示了O=C1N([R])C=CC(N([H])[H])=[N+]1[H]与O=C2N([R])C=CC(N([H])[H])=N2之间形成三根氢键，嘧啶环2号位的羰基与另一个嘧啶环4号位的胺基形成氢键，两个嘧啶环3号位的氮原子通过一个质子形成一根氢键。](/media/uploads/2b3b9506-fc2d-4e19-a2b4-72c2d4854c9e.png)</MultiModal>

<Checkpoint>[质子化的$C$与未质子化的$C$之间可以形成$\mathrm{H^+}C:C$配对](1)</Checkpoint>

除此之外，题目中还提到了三螺旋结构，在此题中可以形成嘧啶-嘌呤-嘧啶型配对方式，如$T\cdot A-T$配对，其结构如下图所示：

<MultiModal>![图中展示了两个CC(C(N1[H])=O)=CN(C1=O)[R]和一个[H]N(C1=NC=NC2=C1N=CN2[R])[H]之间形成的三碱基配对结构，嘌呤环1号位的氮原子和6号位的胺基分别与一个嘧啶环3号位的胺基和4号位的羰基形成两根氢键，同一个嘌呤环7号位的氮原子和6号位的胺基分别与另一个嘧啶环3号位的胺基和4号位的羰基形成两根氢键。](/media/uploads/fe5c10ee-ef40-40ff-8033-20ed8f1b29de.png)</MultiModal>

<Checkpoint>[$T$与$A$之间可以形成$T\cdot A-T$配对](1)</Checkpoint>

有了这三种配对方式后就可以开始分析选项了。

（a）体系的pH=1.1时，丙烯酰胺侧链的$5'-(AAA)_7-3'$中的碱基$A$大量被质子化为$A\mathrm{H^+}$，并通过$A\mathrm{H^+}-\mathrm{H^+}A$配对实现链间交联，导致其以水凝胶的形体存在。

<Checkpoint>[pH=1.1时通过$A\mathrm{H^+}-\mathrm{H^+}A$配对实现链间交联，以水凝胶的形体存在](1)</Checkpoint>

（b）体系的pH=5.2时，丙烯酰胺侧链的$5'-(AAACCCC)_2-3'$中的碱基$C$部分被质子化为$C\mathrm{H^+}$,并通过$\mathrm{H^+}C:C$配对实现链间交联，导致其以水凝胶的形体存在。

<Checkpoint>[pH=5.2时通过$\mathrm{H^+}C:C$配对实现链间交联，以水凝胶的形体存在](1)</Checkpoint>

（c）体系的pH=7.2时，$A,C$几乎不发生质子化，链间难以交联，导致其以水溶液的形体存在。

<Checkpoint>[体系的pH=7.2时$A,C$几乎不发生质子化，链间难以交联](1)</Checkpoint>

（d）加入$5'-(TTT)_7-3'$；$5'-(TTT)_{15}-3'$后，体系内形成$T\cdot A-T$配对的三螺旋结构，又由于$5'-(AAA)_7-3'$与$5'-(TTT)_7-3'$的长度相等，$5'-(TTT)_{15}-3'$的长度稍大于$5'-(AAA)_7-3'$长度的2倍，因此每条$5'-(TTT)_{15}-3'$可以与两条侧链的$5'-(AAA)_7-3'$（以及两条$5'-(TTT)_{15}-3'$）进行配对实现链间交联，导致其以水凝胶的形体存在。

<Checkpoint>[每条$5'-(TTT)_{15}-3'$可以与两条侧链的$5'-(AAA)_7-3'$进行配对实现链间交联，形成$T\cdot A-T$配对的三螺旋结构](2)</Checkpoint>

（e）（f）体系的pH=10.2时，碱基$T$的$\mathrm{N-H}$被去质子化，难以形成$T\cdot A-T$配对结构，导致其以水溶液的形体存在。

<Checkpoint>[pH=10.2时碱基$T$的$\mathrm{N-H}$被去质子化难以形成$T\cdot A-T$配对结构](1)</Checkpoint>

因此答案为AABABB，选项A正确。