A peptide compound with antitumor activity was isolated from a marine alga. The compound was subjected to sequence analysis using chemical methods, and the experiments and results are as follows:

1. After treating the compound with 2-mercaptoethanol, sequencing was performed using Edman degradation. The results obtained were:  
   a) AVCRTGCKNFL  
   b) LDCKGGAYKCFRHTKCS  

2. After treating the compound with 2-mercaptoethanol, the thiol groups were blocked with iodoacetamide, followed by trypsin digestion. MS/MS analysis was performed to determine the molecular weight of each amino acid in the peptide fragments (all listed from N- to C-terminus):  
   a) 71.04, 99.07, 160.03, 156.01  
   b) 160.03, 87.03  
   c) 101.04, 57.02, 160.03, 128.09  
   d) 114.04, 147.07, 113.08  
   e) 137.06, 101.05, 128.09  
   f) 160.04, 147.07, 156.10  
   g) 57.02, 57.02, 71.03, 163.06, 128.09  
   h) 113.08, 115.03, 160.03, 128.09  
   i) 114.04, 128.06  
   j) 159.04, 114.04, 115.03, 99.07, 160.03, 156.10  

3. After treating the compound with a demethylase and trypsin digestion, the resulting fragments were (Cys2 indicates cystine):  
   a) Ala, Val, Cys2, Arg, Ser  
   b) His, Thr, Lys  
   c) Thr, Gly, Cys2, Lys, Phe, Arg  
   d) Asn, Phe, Leu  
   e) Gly, Gly, Ala, Tyr, Lys  
   f) Asn, Gln  
   g) Leu, Asp, Cys2, Lys, Arg, Val, Asp, Asn, Met  

For different peptide chains within the compound, they are named in order of length from longest to shortest as **Chain A**, **Chain B**, **Chain C**, **Chain D**, and so on. Residues are denoted in the format **PeptideChain-ResidueNumber**, e.g., A3 represents the third residue of **Chain A**, and B9 represents the ninth residue of **Chain B**. Assuming only standard residues and their modifications are considered, and no unnatural amino acids are present, which of the following statements about this peptide compound is correct?
从海洋中的一种藻类分离得到某种具有抗肿瘤活性的多肽化合物，现采用化学方法对该化合物进行序列分析，进行的实验和结果如下：

1. 将该化合物用2-巯基乙醇处理后，使用Edman降解对该多肽进行测序，测得的结果为：

a) AVCRTGCKNFL
 
b) LDCKGGAYKCFRHTKCS

2. 将该化合物用2-巯基乙醇处理后，用碘乙酰胺封闭巯基，再用胰蛋白酶消化，并用MS/MS分析肽段中每个氨基酸的分子量，得到的结果如下(均为从N端到C端排序）：

a) 71.04, 99.07, 160.03, 156.01 

b) 160.03,87.03 

c) 101.04, 57.02, 160.03, 128.09 

d) 114.04, 147.07, 113.08 

e) 137.06, 101.05, 128.09 

f) 160.04, 147.07, 156.10 

g) 57.02, 57.02, 71.03, 163.06, 128.09 

h) 113.08, 115.03, 160.03, 128.09 

i) 114.04, 128.06 

j) 159.04, 114.04, 115.03, 99.07, 160.03, 156.10 

3. 将该化合物用去甲基化酶处理后，用胰蛋白酶处理后得到的片段为（Cys2表示胱氨酸）：

a) Ala, Val, Cys2, Arg, Ser 

b) His, Thr, Lys 

c) Thr, Gly, Cys2, Lys, Phe, Arg 

d)Asn, Phe, Leu 

e) Gly, Gly, Ala, Tyr, Lys 

f) Asn, Gln 

g) Leu, Asp, Cys2, Lys, Arg, Val, Asp, Asn, Met

规定对于一个化合物中不同的多肽链，按照长度顺序从长到短依次命名为**A**链，**B**链，**C**链，**D**链，以此类推。残基用多肽-序数的格式表示，例如A3表示**A**链的第三个残基，B9表示**B**链的第9个残基。假设不考虑非天然氨基酸，仅考虑标准残基及其修饰，则以下关于该多肽化合物的说法中，正确的是？
The compound has three polypeptide chains, with disulfide bond positions at A3-C5, B3-A16, and B7-A10, and a molar ratio of 1:1 in reaction with cyanogen bromide. The compound contains a total of 15 methyl groups.
The compound consists of three polypeptide chains, with disulfide bond positions at A3-C5, B3-A16, and B7-A10. The molar ratio of reaction with cyanogen bromide is 1:1. The compound contains a total of 13 methyl groups.
The compound has three polypeptide chains, with disulfide bond positions at A15-C5, B3-A2, and B7-A8. The molar ratio of its reaction with cyanogen bromide is 1:1. The compound contains a total of 15 methyl groups.
The compound has three polypeptide chains, with disulfide bond positions at A15-C5, B3-A2, and B7-A8, and a molar ratio of 1:1 in the reaction with cyanogen bromide. The compound contains a total of 13 methyl groups.
The compound has three polypeptide chains, with disulfide bond positions at A3-C5, B3-A16, and B7-A10, and does not react with cyanogen bromide. The compound contains a total of 15 methyl groups.
The compound has three polypeptide chains, with disulfide bond positions at A3-C5, B3-A16, and B7-A10, and does not react with cyanogen bromide. The compound contains a total of 13 methyl groups.
The compound has three polypeptide chains, with disulfide bond positions at A15-C5, B3-A2, and B7-A8, and does not react with cyanogen bromide. The compound contains a total of 15 methyl groups.
The compound consists of 3 polypeptide chains, with disulfide bond positions at A15-C5, B3-A2, and B7-A8, and does not react with cyanogen bromide. The compound contains a total of 13 methyl groups.
The compound has two polypeptide chains, with disulfide bond positions at B3-A16 and B7-A10, and does not react with cyanogen bromide. The compound contains a total of 10 methyl groups.
The compound consists of two polypeptide chains, with disulfide bond positions at B3-A2 and B7-A8, and does not react with cyanogen bromide. There are a total of 10 methyl groups in the compound.
None of the above options is correct
该化合物有3条多肽链，二硫键的位置为A3-C5，B3-A16，B7-A10，与溴化氰反应的摩尔比为1:1。化合物中共计有15个甲基。
该化合物有3条多肽链，二硫键的位置为A3-C5，B3-A16，B7-A10，与溴化氰反应的摩尔比为1:1。化合物中共计有13个甲基。
该化合物有3条多肽链，二硫键的位置为A15-C5，B3-A2，B7-A8，与溴化氰反应的摩尔比为1:1。化合物中共计有15个甲基。
该化合物有3条多肽链，二硫键的位置为A15-C5，B3-A2，B7-A8，与溴化氰反应的摩尔比为1:1。化合物中共计有13个甲基。
该化合物有3条多肽链，二硫键的位置为A3-C5，B3-A16，B7-A10，与溴化氰不反应。化合物中共计有15个甲基。
该化合物有3条多肽链，二硫键的位置为A3-C5，B3-A16，B7-A10，与溴化氰不反应。化合物中共计有13个甲基。
该化合物有3条多肽链，二硫键的位置为A15-C5，B3-A2，B7-A8，与溴化氰不反应。化合物中共计有15个甲基。
该化合物有3条多肽链，二硫键的位置为A15-C5，B3-A2，B7-A8，与溴化氰不反应。化合物中共计有13个甲基。
该化合物有2条多肽链，二硫键的位置为B3-A16，B7-A10，与溴化氰不反应。化合物中共计有10个甲基。
该化合物有2条多肽链，二硫键的位置为B3-A2，B7-A8，与溴化氰不反应。化合物中共计有10个甲基。
以上选项均不正确
First, let's examine Experiment 1. Edman degradation yielded the sequences of two peptide chains: AVCRTGCKNFL and LDCKGGAYKCFRHTKCS. Since BME was used to disrupt disulfide bonds in both Experiments 1 and 2, we can temporarily disregard the positions of the disulfide bonds.

We first write down the products of the two peptide chains in Experiment 1 after trypsin digestion and match the sequences with the molecular weights from Experiment 2.

| Digest Product | Amino Acid Molecular Weights | Mass Spectrometry Results |
|-------|-------|-------|
| AVCR | 71.03, 99.07, 160.03, 156.10 | 2a |
| TGCK | 101.05, 57.02, 160.03, 128.09 | 2c |
| NFL | 114.04, 147.07, 113.08 | 2d |
| LDCK | 113.08, 115.03, 160.03, 128.09 | 2h |
| GGAYK | 57.02, 57.02, 71.04, 163.06, 128.09 | 2g |
| CFR | 160.03, 147.07, 156.10 | 2f |
| HTK | 137.06, 101.05, 128.09 | 2e |
| CS | 160.03, 87.03 | 2b |

2i and 2j could not be matched to any products of the two peptide chains sequenced in Experiment 1, indicating the presence of a third peptide chain composed of 2i and 2j in the compound.

<Checkpoint>[There is a third peptide chain composed of 2i and 2j, meaning the compound includes three peptide chains](1)</Checkpoint>

The reason this peptide chain was not sequenced may be due to cyclization, N-terminal post-translational modifications, or other factors. Since Edman degradation relies on reacting with the free N-terminal amino group of the protein, it cannot sequence the peptide when the amino group is blocked.

Next, we deduce the sequences of these two fragments. 2i is clearly NQ, while 2j can be partially inferred as NDVCR. However, the first residue with a molecular weight of 159.04 does not correspond to any standard residue. Note that the question states non-natural amino acids should not be considered, so this residue must be a modified form of a standard residue.

<Checkpoint>[The residue with a molecular weight of 159.04 is a modified form of a standard residue](1)</Checkpoint>

Additionally, as mentioned earlier, Edman degradation could not sequence this peptide chain because its N-terminal amino group was blocked. Therefore, 159.04 corresponds to an amino acid with a blocked N-terminal amino group. Cyclic peptides can be ruled out here because i and j have only one cleavage site (the C-terminal R of j). If it were a cyclic peptide, digestion would yield a single linear peptide, not two. Furthermore, the mass spectrometry results also indicate that this peptide could not be linked to other peptide chains via isopeptide bonds, as the digestion products would otherwise appear together with those sequenced in Experiment 1.

Examining common N-terminal modifications, such as acetylation or methylation, we note that 159.04 = 131.04 + 28, corresponding to methionine + two methyl groups. Thus, the N-terminal methionine of this peptide is blocked by dimethylation, and the third peptide chain can be written as $(CH_2)_2MNDVCRQN$. The connectivity here is uniquely determined by the digestion results of 2i and 2j.

<Checkpoint>[The N-terminal methionine of this peptide is blocked by dimethylation](1)</Checkpoint>

<Checkpoint>[The sequence of the third peptide chain is MNDVCRQN, with the amino group of M dimethylated](1)</Checkpoint>

Now, let’s examine Experiment 3. Note that Experiment 3 begins with the use of a demethylase, which aligns with our reasoning in Experiment 2 and follows the progressive logic of Experiments 1–3.

We match the disulfide bond positions identified in Experiment 3 with the peptide chains A, B, and C. Here, LDCKGGAYKCFRHTKCS is chain A, AVCRTGCKNFL is chain B, and MNDVCRQN is chain C. For example, 3a indicates a disulfide bond between Cys3 of chain B and Cys16 of chain A (i.e., AVC-CR), while 3c indicates a disulfide bond between Cys7 of B and Cys10 of A. Additionally, the results of 3f and 3g are consistent with the sequences we deduced in Experiment 2, and one disulfide bond is located between Cys3 of A and Cys5 of C.

<Checkpoint>[The disulfide bond positions are: A3-C5, B3-A16, B7-A10](1)</Checkpoint>

Cyanogen bromide reacts specifically with the side chain and C-terminus of methionine to form homoserine lactone. Since the compound contains only one methionine in chain C, the reaction stoichiometry is 1:1.

<Checkpoint>[The compound contains 1 methionine, and the reaction stoichiometry with cyanogen bromide is 1:1](1)</Checkpoint>

Finally, let’s count the number of methyl groups. The three peptide chains are: AVCRTGCKNFL, LDCKGGAYKCFRHTKCS, and MNDVCRQN. The residues with methyl groups in their side chains are: A, T, M (1 each), V, L (2 each). Thus, the side chain methyl count is: $1+2+1+2+2+1+1+1+2=13$, plus the two methyl groups at the N-terminus of chain C, totaling 15.

<Checkpoint>[There are 15 methyl groups in total: 13 on side chains and 2 on the N-terminus of the main chain](1)</Checkpoint>

Option A is correct.
首先考察实验1，Edman降解给出了两条肽链的序列为AVCRTGCKNFL, LDCKGGAYKCFRHTKCS。由于1、2中均用BME破坏了二硫键，因此可暂不考虑二硫键的位置；

我们首先写出1中两条肽链被胰酶酶切后的产物，并将序列与实验2中的式量对应。

|酶切产物| 氨基酸分子量 | 质谱结果 |
|-------|-------|-------|
| AVCR | 71.03, 99.07, 160.03, 156.10 | 2a |
| TGCK | 101.05, 57.02, 160.03, 128.09 | 2c |
|NFL| 114.04, 147.07, 113.08 | 2d |
| LDCK | 113.08, 115.03, 160.03, 128.09 | 2h |
| GGAYK | 57.02, 57.02, 71.04, 163.06, 128.09 | 2g |
| CFR | 160.03, 147.07, 156.10 | 2f |
| HTK | 137.06, 101.05, 128.09| 2e |
| CS | 160.03, 87.03 | 2b |

2i和2j无法匹配1中测序得到两条多肽链的任何产物，说明化合物中还存在2i、2j组成的第三条多肽。

<Checkpoint>[存在2i、2j组成的第三条多肽，化合物中包括3条肽链](1)</Checkpoint>

这条多肽没有被测序到的原因可能是存在环肽、N端翻译后修饰等原因；因Edman测序的原理是与蛋白质的N端游离氨基反应，当氨基被封闭时，Edman降解自然无法测得序列。

我们下面推断这两条序列，2i很明显为NQ，2j可以推断部分序列为NDVCR，然而第一个分子量为159. 04的残基无法对应到任何标准残基上。注意到题干说明暂不考虑非天然氨基酸，因此该残基应当为某一标准残基的修饰结果。

<Checkpoint>[分子量为159. 04的残基为某一标准残基的修饰结果](1)</Checkpoint>

此外，我们注意到前述，Edman测序无法测得这条肽链，原因是该肽链的N端氨基被封闭，因此159. 04为某一氨基酸的N端氨基被封闭的结果。这里可以排除环肽，因为i、j仅有一处酶切位点（j的C端R），如果是环肽，酶切得到的应当是一条线性肽，而非两条。此外，质谱结果也说明了该多肽不可能是以异肽键的形式连接在其他多肽链上，否则酶切后的质谱结果里会与1中测序得到的产物一起出现。

考察常见的N端修饰，如乙酰化、甲基化等，可以注意到159.04 = 131.04+28，即甲硫氨酸+2个甲基，故该多肽的N端甲硫氨酸被两个甲基化封闭，第三条肽链可以写成$(CH_2)_2MNDVCRQN$。这里的连接方式是由2i、2j的酶切结果唯一确定的。

<Checkpoint>[该多肽的N端甲硫氨酸被两个甲基化封闭](1)</Checkpoint>

<Checkpoint>[第三条肽链的序列为MNDVCRQN，M的氨基被二甲基化](1)</Checkpoint>

下面考察实验3。注意到实验3一开始就使用了去甲基化酶，与我们在实验2中的推理结果吻合，也符合实验1-3递进式的推理顺序。

我们分别将实验3中测得的二硫键位置与多肽A、B、C匹配。其中LDCKGGAYKCFRHTKCS是A链，AVCRTGCKNFL是B链，MNDVCRQN是C链。例如3a说明二硫键的位置在B的3号Cys与A的16号Cys中，即AVC-CR；3c说明二硫键在B7与A10之间.此外3f、3g的结果也与我们在2中分析得到的序列吻合，并且有一条二硫键在A3-C5。

<Checkpoint>[二硫键的位置分别为：A3-C5，B3-A16，B7-A10](1)</Checkpoint>


溴化氰专一地与蛋白甲硫氨酸的侧链和C端反应生成高丝氨酸内酯，化合物中仅C链有一个甲硫氨酸，故反应摩尔比为1:1.

<Checkpoint>[化合物含1个甲硫氨酸，与溴化氰反应摩尔比为1:1](1)</Checkpoint>

最后来计数甲基的数量。三条多肽链分别为：AVCRTGCKNFL, LDCKGGAYKCFRHTKCS, MNDVCRQN。其中侧链含有甲基的残基为：A、T、M（1个），V、L（2个）。故侧链甲基数：$1+2+1+2+2+1+1+1+2=13$，加上C链N端的甲基，一共有15个。

<Checkpoint>[一共有15个甲基，其中侧链13个，主链N端2个](1)</Checkpoint>

A选项正确。