A certain single-stranded $\rm{DNA}$ to be sequenced is truncated to obtain a segment, and then the $5^{'}$ end of the segment is labeled with $^{32}\mathrm{P}$, and then placed in $4$ different reaction tubes for the following two-step treatment.

In the first step, the $4$ reaction tubes are treated as follows, respectively: Tube (1) is treated with hot, nearly neutral dimethyl sulfate aqueous solution; Tube (2) is treated with formic acid aqueous solution; Tube (3) is treated with alkaline hydrazine aqueous solution; Tube (4) is treated with alkaline hydrazine aqueous solution containing $\rm{1.5~mol/L}$ $\mathrm{NaCl}$.

In the second step, the above $4$ reaction tubes are treated with $9 0 ^ { \circ } \mathrm { C }$ piperidine solution; by controlling the amount of reagents and reaction conditions, on average, only about $1$ site on each $\rm{DNA}$ single strand is cleaved.

The products obtained from the four reaction tubes are separated by gel electrophoresis and autoradiography under the same conditions, and the corresponding bands are displayed on the $\rm{X}$-ray film, as shown in the figure below.

<MultiModal>![The image shows the result of a gel electrophoresis, containing four lanes, numbered (1), (2), (3), and (4) from left to right. There is a vertical black arrow on the right side of the image, with the text "Direction of Electrophoretic Migration" next to the arrow, pointing downwards. There are different numbers and positions of horizontal bands on the gel. The electrophoresis results can be represented in a table as follows:
|(1)|(2)|(3)|(4)|
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According to the electrophoresis results, which of the following polypeptide fragments could not be synthesized from this $\rm{DNA}$ segment as the coding strand:
将某种待测序的单链$\rm{DNA}$截取其中一段，然后将该片段的$5^{'}$端用$^{32}\mathrm{P}$标记，然后置于$4$个不同的反应管中，进行如下两步处理。

第一步，分别对$4$个反应管进行以下处理： (1)号管用热的近中性硫酸二甲酯水溶液处理；(2)号管用甲酸水溶液处理； (3)号管用碱性肼水溶液处理；(4)号管用含$\rm{1.5~mol/L}$的$\mathrm{NaCl}$的碱性肼水溶液处理。

第二步，将以上$4$个反应管用 $90 ^ { \degree } \mathrm { C }$ 六氢吡啶溶液处理；通过控制试剂用量和反应条件，使每条$\rm{DNA}$单链平均只有$1$个左右的位点发生断裂。

将四个反应管所得产物在相同条件下进行凝胶电泳分离和放射自显影，在$\rm{X}$线片底板上显示相应谱带，如下图所示。

<MultiModal>![该图像显示了一个凝胶电泳的结果，包含四条泳道，从左到右依次编号为（1）、（2）、（3）、（4）。图像的右侧有一条垂直的黑色箭头，箭头旁边标注有“电泳移动方向”的文字，箭头指向下方。凝胶上有不同数量和位置的水平条带。电泳结果用表格可以表示为：
|(1)|(2)|(3)|(4)|
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根据电泳结果，以下哪个多肽片段的氨基酸残基序列不可能由这段$\rm{DNA}$作为编码链合成得到：
All other options are incorrect
S-C-Q-A
R-V-S-A
F-V-S-A
L-V-S-A
I-V-S-A
V-V-S-A
S-R-L-C
V-S-A-L
其他选项均不正确
S-C-Q-A
R-V-S-A
F-V-S-A
L-V-S-A
I-V-S-A
V-V-S-A
S-R-L-C
V-S-A-L
Since DNA with a larger volume experiences greater resistance during electrophoresis, its migration distance is shorter. Also, because the 5' end of single-stranded DNA is labeled with $^{32}\mathrm{P}$, fragments with cleavage sites closer to the 5' end migrate further after being cleaved.

<Checkpoint>[Fragments with cleavage sites closer to the 5' end migrate further](1)</Checkpoint>

The nitrogen atom on the guanine pentagonal ring has stronger nucleophilicity, so dimethyl sulfate can selectively cleave guanine groups. Aqueous formic acid solution can cleave all purine groups. Treatment with alkaline hydrazine aqueous solution can cleave all pyrimidine groups, while alkaline hydrazine aqueous solution containing 1.5 mol/L $\mathrm{NaCl}$ can selectively cleave cytosine groups.

<Checkpoint>[Aqueous formic acid solution cleaves all purine groups, and alkaline hydrazine aqueous solution cleaves all pyrimidine groups](2)</Checkpoint>

<Checkpoint>[Dimethyl sulfate selectively cleaves guanine groups, and alkaline hydrazine aqueous solution containing 1.5 mol/L $\mathrm{NaCl}$ selectively cleaves cytosine groups](2)</Checkpoint>

Therefore, (1) causes cleavage at guanine, (2) causes cleavage at all purines, (3) causes cleavage at all pyrimidines, and (4) causes cleavage at cytosine.

<Checkpoint>[(1) and (2) lanes both have signals for G, (2) lane has a unique signal for A, (3) lane has a unique signal for T, and (3) and (4) lanes both have signals for C](2)</Checkpoint>
The electrophoresis results are:
|(1)|(2)|(3)|(4)|
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Therefore, the sequence from bottom to top is the base sequence from 5' to 3', and it is easy to obtain the sequence as

5'-TCGTGTCAGCGCT-3'

<Checkpoint>[DNA sequence is 5'-TCGTGTCAGCGCT-3'](2)</Checkpoint>

The following analyzes possible composition methods:

5'-TCG-TGT-CAG-CGC-T-3' corresponds to S-C-Q-R

5'-T-CGT-GTC-AGC-GCT-3' corresponds to R-V-S-A

5'-TC-GTG-TCA-GCG-CT-3' has V-S-A in the middle, -TC- at the 5' end can be F, L, I, V, and -CT- at the 3' end can only be L

Therefore, the possible situations are (F, L, I, V)-V-S-A and V-S-A-L

<Checkpoint>[Possible situations are S-C-Q-R, R-V-S-A, (F, L, I, V)-V-S-A and V-S-A-L](2)</Checkpoint>

Therefore, only option H is impossible in the options.
由于体积较大的DNA在电泳时受到的阻力较大，因此移动距离较短。又因为单链DNA的5'端用$^{32}\mathrm{P}$标记，所以在切断后切断位点越靠近5'端的片段移动距离更远。

<Checkpoint>[切断位点越靠近5'端的片段移动距离更远](1)</Checkpoint>

鸟嘌呤五元环上的氮原子亲核性更强，因此使用硫酸二甲酯能选择性的使鸟嘌呤基团断裂。甲酸水溶液则可以使所有嘌呤基团断裂。碱性肼水溶液处理可以使所有嘧啶基团断裂，含1.5 mol/L的$\mathrm{NaCl}$的碱性肼水溶液则可以选择性的使胞嘧啶基团断裂。

<Checkpoint>[甲酸水溶液使所有嘌呤基团断裂，碱性肼水溶液使所有嘧啶基团断裂](2)</Checkpoint>

<Checkpoint>[硫酸二甲酯选择性使鸟嘌呤基团断裂，含1.5 mol/L的$\mathrm{NaCl}$的碱性肼水溶液选择性的使胞嘧啶基团断裂](2)</Checkpoint>

因此（1）处理后在鸟嘌呤处断裂，（2）处理后在所有嘌呤处断裂，（3）处理后在所有嘧啶处断裂，（4）处理后在胞嘧啶处断裂。

<Checkpoint>[（1）（2）号泳道都有的信号为G，（2）号泳道特有的信号为A，（3）号泳道特有信号为T，（3）（4）号泳道都有的信号为C](2)</Checkpoint>
电泳结果为：
|(1)|(2)|(3)|(4)|
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因此从下往上的顺序即为5'到3'的碱基序列，很容易得到序列为

5'-TCGTGTCAGCGCT-3'

<Checkpoint>[DNA序列为5'-TCGTGTCAGCGCT-3'](2)</Checkpoint>

下面分析可能的组成方式：

5'-TCG-TGT-CAG-CGC-T-3'对应S-C-Q-R

5'-T-CGT-GTC-AGC-GCT-3'对应R-V-S-A

5'-TC-GTG-TCA-GCG-CT-3'中间三个为V-S-A，5'端的-TC-可能的情况为F，L，I，V，3'端的-CT-只可能为L

从而可能的情况为(F，L，I，V)-V-S-A和V-S-A-L

<Checkpoint>[可能的情况有S-C-Q-R，R-V-S-A，(F，L，I，V)-V-S-A和V-S-A-L](2)</Checkpoint>

因此选项中只有选项H不可能