Tandem mass spectrometry is a rapid method for peptide sequencing. This technique involves the formation of precursor ions and their fragmentation into smaller ions. In peptides, fragmentation often occurs on the peptide backbone. The naming of fragment ions depends on the site of fragmentation and which atom carries the positive charge. For example, some ions formed from alanyl-leucyl-glycine are shown below:

<MultiModal>![For the tripeptide alanyl-leucyl-glycine, cleavage of the amide bond between alanine and leucine yields an N-terminal fragment, the alanyl ion, denoted as $b_1$ ion. Cleavage of the amide bond between leucine and glycine yields an N-terminal fragment, the alanyl-leucyl ion, denoted as $b_2$ ion](/media/uploads/6ddec20b-b55c-46ce-93f0-a7fd36cafaf7.png)</MultiModal>

<MultiModal>![For the tripeptide alanyl-leucyl-glycine, cleavage of the amide bond between alanine and leucine yields a C-terminal fragment, the protonated leucyl-glycine ammonium ion, denoted as $y_2$ ion. Cleavage of the amide bond between leucine and glycine yields a C-terminal fragment, the protonated glycine ammonium ion, denoted as $y_1$ ion](/media/uploads/1a6bfa79-09d1-43ac-bae6-f57719b1c4e6.png)</MultiModal>

<MultiModal>![For the tripeptide alanyl-leucyl-glycine, cleavage of the carbon-carbon bond between the carbonyl carbon and the carbonyl $\alpha-$ carbon within alanine yields an N-terminal fragment, the protonated immonium ion, denoted as $a_1$ ion. Cleavage of the carbon-carbon bond between the carbonyl carbon and the carbonyl $\alpha-$ carbon within leucine-glycine yields an N-terminal fragment, the protonated alanyl immonium ion, denoted as $a_2$ ion](/media/uploads/38bd5889-416b-4ce3-b4fb-83d097f7f0f3.png)</MultiModal>

Scientists extracted a protein from a fossilized bone of a prehistoric organism. The mass spectrometry data of a 19-peptide fragment is shown in the following table:

| ion   | *m/z* | ion   | *m/z* | ion      | *m/z*  | ion      | *m/z*  |
| :---- | :---- | :---- | :---- | :------- | :----- | :------- | :----- |
| $y_1$ | 175.1 | $b_5$ | 715.3 | $y_8$    | 986.5  | $b_{12}$ | 1400.7 |
| $a_2$ | 249.1 | $y_6$ | 726.4 | $b_9$    | 1069.5 | $y_{14}$ | 1508.8 |
| $y_2$ | 272.2 | $a_6$ | 800.4 | $y_9$    | 1083.5 | $b_{14}$ | 1612.7 |
| $y_3$ | 401.2 | $y_7$ | 823.4 | $b_{10}$ | 1140.5 | $a_{15}$ | 1681.8 |
| $a_4$ | 501.2 | $b_6$ | 828.4 | $a_{11}$ | 1209.6 | $y_{15}$ | 1694.9 |
| $b_4$ | 529.2 | $b_7$ | 885.4 | $y_{11}$ | 1267.6 | $y_{16}$ | 1831.9 |
| $y_5$ | 611.4 | $a_8$ | 928.4 | $y_{12}$ | 1338.7 | $y_{17}$ | 1946.9 |
| $a_5$ | 687.3 | $b_8$ | 956.5 | $y_{13}$ | 1395.7 | $b_{17}$ | 1951.9 |

It is known that the first two amino acids of the peptide are Tyr-Leu, and the peptide sequence includes a hydroxyproline, represented as Hyp, with a mass of 131.1, and the structure is as follows:

<MultiModal>![$\text{C1[C@H](CN[C@@H]1C(=O)O)O}$](/media/uploads/90cf6b47-d140-42e5-937e-0c4c7f1989c9.png)</MultiModal>

Using the mass spectrometry ion mass table, determine the most accurate possible peptide sequence. In the peptide sequence, for amino acids whose exact type cannot be determined, consider all possibilities for multiple amino acids present at a position, based on calculations and reasoning.

Based on your inferred most likely peptide sequence, select the closest modern species from the partial sequences of homologous proteins obtained from the following modern species (in the following sequences, hydroxyproline and proline are both abbreviated as P):

| Species | Sequence                                              |
| :--- | :------------------------------------------------ |
| Carp | DLTVAQLESLKEVCEANLACEHMMDVSGIIAAYTAYYGPIPY        |
| Chicken   | HYAQDSGVAGAPPNPLEAQREVCELSPDCDELADQIGFQEAYRRFYGPV |
| Cow   | YLDHWLGAPAPYPDPLEPKREVCELNPDCDELADHIGFQEAYRRFYGPV |
| Horse   | YLDHWLGAPAPYPDPLEPRREVCELNPDCDELADHIGFQEAYRRFYGPV |
| Human | YLYQWLGAPVPYPDPLEPRREVCELNPDCDELADHIGFQEAYRRFYGPV |
| Rabbit | QLINGQGAPAPYPDPLEPKREVCELNPDCDELADQVGLQDAYQRFYGPV |
| Sheep   | YLDPGLGAPAPYPDPLEPRREVCELNPDCDELADHIGFQEAYRRFYGPV |
| Toad | SYGNNVGQGAAVGSPLESQREVCELNPDCDELADHIGFQEAYRRFYGPV |
串联质谱技术是多肽测序的快速方法，这项技术包括母离子的形成以及碎裂化成更小的离子。在多肽中，碎裂化经常在多肽骨架上发生。碎片离子的命名取决于碎裂化发生在哪个位点，还有哪个原子带正电荷。例如，丙氨酰-亮氨酰-甘氨酸形成的一些离子如下图所示：

<MultiModal>![对于三肽丙氨酰-亮氨酰-甘氨酸，从丙氨酰-亮氨酰之间的酰胺键断裂，多肽碎裂后的N端得到丙氨酰鎓离子记为 $b_1$ 离子，从亮氨酰-甘氨酸之间的酰胺键断裂，多肽碎裂后的N端得到丙氨酰-亮氨酰鎓离子记为 $b_2$ 离子](/media/uploads/6ddec20b-b55c-46ce-93f0-a7fd36cafaf7.png)</MultiModal>

<MultiModal>![对于三肽丙氨酰-亮氨酰-甘氨酸，从丙氨酰-亮氨酰之间的酰胺键断裂，多肽碎裂后的C端得到质子化的亮氨酰-甘氨酸铵离子记为 $y_2$ 离子，从亮氨酰-甘氨酸之间的酰胺键断裂，多肽碎裂后的C端得到质子化的甘氨酸铵离子记为 $y_1$ 离子](/media/uploads/1a6bfa79-09d1-43ac-bae6-f57719b1c4e6.png)</MultiModal>

<MultiModal>![对于三肽丙氨酰-亮氨酰-甘氨酸，从丙氨酰内部的羰基碳与羰基 $\alpha-$ 碳之间的碳碳键断裂，多肽碎裂后的N端得到质子化的亚铵离子记为 $a_1$ 离子，从亮氨酰-甘氨酸内部的羰基碳与羰基 $\alpha-$ 碳之间的碳碳键断裂，多肽碎裂后的N端得到质子化的丙氨酰亚铵离子记为 $a_2$ 离子](/media/uploads/38bd5889-416b-4ce3-b4fb-83d097f7f0f3.png)</MultiModal>



科学家从一史前生物骨化石中，提取出了某蛋白质 ，其19肽的碎片质谱数据如下表所示：

| ion   | *m/z* | ion   | *m/z* | ion      | *m/z*  | ion      | *m/z*  |
| :---- | :---- | :---- | :---- | :------- | :----- | :------- | :----- |
| $y_1$ | 175.1 | $b_5$ | 715.3 | $y_8$    | 986.5  | $b_{12}$ | 1400.7 |
| $a_2$ | 249.1 | $y_6$ | 726.4 | $b_9$    | 1069.5 | $y_{14}$ | 1508.8 |
| $y_2$ | 272.2 | $a_6$ | 800.4 | $y_9$    | 1083.5 | $b_{14}$ | 1612.7 |
| $y_3$ | 401.2 | $y_7$ | 823.4 | $b_{10}$ | 1140.5 | $a_{15}$ | 1681.8 |
| $a_4$ | 501.2 | $b_6$ | 828.4 | $a_{11}$ | 1209.6 | $y_{15}$ | 1694.9 |
| $b_4$ | 529.2 | $b_7$ | 885.4 | $y_{11}$ | 1267.6 | $y_{16}$ | 1831.9 |
| $y_5$ | 611.4 | $a_8$ | 928.4 | $y_{12}$ | 1338.7 | $y_{17}$ | 1946.9 |
| $a_5$ | 687.3 | $b_8$ | 956.5 | $y_{13}$ | 1395.7 | $b_{17}$ | 1951.9 |

已知已知多肽的前两个氨基酸是Tyr-Leu，多肽序列中包括一种羟脯氨酸，表示为Hyp，质量数是131.1，结构如下：

<MultiModal>![$\text{C1[C@H](CN[C@@H]1C(=O)O)O}$](/media/uploads/90cf6b47-d140-42e5-937e-0c4c7f1989c9.png)</MultiModal>

利用质谱离子质量表，确定尽可能准确的多肽序列。在多肽序列中，对于无法确定确切种类的氨基酸，根据计算与推理，考虑当一个位置存在的多种氨基酸的所有可能。

根据你推断的最可能的多肽序列，从如下的一些现代物种中得到的同源蛋白质序列的部分序列中（以下序列中羟脯氨酸和脯氨酸都缩写为P），挑选出与之最为接近的一个现代物种。

| 物种 | 序列                                              |
| :--- | :------------------------------------------------ |
| 鲤鱼 | DLTVAQLESLKEVCEANLACEHMMDVSGIIAAYTAYYGPIPY        |
| 鸡   | HYAQDSGVAGAPPNPLEAQREVCELSPDCDELADQIGFQEAYRRFYGPV |
| 牛   | YLDHWLGAPAPYPDPLEPKREVCELNPDCDELADHIGFQEAYRRFYGPV |
| 马   | YLDHWLGAPAPYPDPLEPRREVCELNPDCDELADHIGFQEAYRRFYGPV |
| 人类 | YLYQWLGAPVPYPDPLEPRREVCELNPDCDELADHIGFQEAYRRFYGPV |
| 兔子 | QLINGQGAPAPYPDPLEPKREVCELNPDCDELADQVGLQDAYQRFYGPV |
| 羊   | YLDPGLGAPAPYPDPLEPRREVCELNPDCDELADHIGFQEAYRRFYGPV |
| 蟾蜍 | SYGNNVGQGAAVGSPLESQREVCELNPDCDELADHIGFQEAYRRFYGPV |
Carp
Chicken
Cow
Horse
Human
Rabbit
Sheep
Toad
鲤鱼
鸡
牛
马
人类
兔子
羊
蟾蜍
The mass of the ion $y_1$ can be used to determine the identity of the last amino acid in a polypeptide. The mass of the $y_1$ ion is one mass unit greater than the corresponding amino acid; therefore, the last amino acid must be arginine (Arg).

<Checkpoint>[Indicates that $y_1$ is arginine Arg](1)</Checkpoint>

The y-series ions are the most complete, and the sequence can be determined by comparing the masses of successive y ions:

| Ion     | m/z    | Mass difference between $b_n$ and $b_{n-1}$ | Corresponding amino acid | Amino acid mass |
| :------- | :----- | :------------------------------ | :----------- | :--------- |
| $y_1$    | 175.1  |                                 |              |            |
| $y_2$    | 272.2  | 97.1                            | 18           | 115.1      |
| $y_3$    | 401.2  | 129.0                           | 17           | 147.0      |
| $y_4$    |        |                                 |              |            |
| $y_5$    | 611.4  |                                 |              |            |
| $y_6$    | 726.4  | 115.0                           | 14           | 133.0      |
| $y_7$    | 823.4  | 97.1                            | 13           | 115.1      |
| $y_8$    | 986.5  | 163.1                           | 12           | 181.1      |
| $y_9$    | 1083.5 | 97.1                            | 11           | 115.1      |
| $y_{10}$ |        |                                 |              |            |
| $y_{11}$ | 1267.6 |                                 |              |            |
| $y_{12}$ | 1338.7 | 71.0                            | 8            | 89.0       |
| $y_{13}$ | 1395.7 | 57.0                            | 7            | 75.0       |
| $y_{14}$ | 1508.8 | 113.1                           | 6            | 131.1      |
| $y_{15}$ | 1694.9 | 186.1                           | 5            | 204.1      |
| $y_{16}$ | 1831.9 | 137.1                           | 4            | 155.1      |
| $y_{17}$ | 1946.9 | 115.0                           | 3            | 133.0      |

<Checkpoint>[Analyze the sequence of y-series ions](1)</Checkpoint>

Based on the y-series, the sequence is:
Tyr-Leu-Asp-His-Trp-Leu/Ile/Hyp-Gly-Ala-xxx-xxx-Pro-Tyr-Pro-Asp-xxx-xxx-Glu-Pro-Arg

<Checkpoint>[The polypeptide sequence obtained from the y-series ions is Tyr-Leu-Asp-His-Trp-Leu/Ile/Hyp-Gly-Ala-[undetermined]-[undetermined]-Pro-Tyr-Pro-Asp-[undetermined]-[undetermined]-Glu-Pro-Arg](5)</Checkpoint>

The identity of the 15th amino acid in the sequence can be determined by the mass difference between $b_{14}$ and $a_{15}$:

$$M_r(\text{amino acid 15}) = \text{mass}(a_{15}) - \text{mass}(a_{14}) + M_r(\text{C}) + 2 M_r(\text{O}) + 2 M_r(\text{H}) = 115.0$$
Therefore, amino acid 15 must be proline (Pro).

<Checkpoint>[Determine that amino acid number 15 is proline Pro](1)</Checkpoint>

The mass difference between ions $y_3$ and $y_5$ gives the fragment mass corresponding to amino acids 15 and 16.
$$M_r(\text{15-16 dipeptide}) = \text{mass}(y_5) - \text{mass}(y_3) + M_r(H_2O)$$
$$M_r(\text{amino acid 16}) = M_r(\text{15-16 dipeptide}) - M_r(\text{amino acid 15}) + M_r(H_2O) = 131.1$$
Therefore, amino acid 16 must be isoleucine (Ile), leucine (Leu), or hydroxyproline (Hyp).

<Checkpoint>[Determine that amino acid number 16 is isoleucine Ile, leucine Leu, or hydroxyproline Hyp](1)</Checkpoint>

The mass of amino acid 10 can be determined by the mass difference between $b_9$ and $b_{10}$:
$$M_r(\text{amino acid 10}) = \text{mass}(b_{10}) - \text{mass}(b_9) + M_r(H_2O)$$
Amino acid 10 is alanine (Ala).

<Checkpoint>[Determine that amino acid number 10 is alanine Ala](1)</Checkpoint>

The mass difference between ions $y_{11}$ and $y_9$ gives the fragment mass corresponding to amino acids 9 and 10.
$$M_r(\text{9-10 dipeptide}) = \text{mass}(y_{11}) - \text{mass}(y_9) + M_r(H_2O)$$
$$M_r(\text{amino acid 9}) = M_r(\text{9-10 dipeptide}) - M_r(\text{amino acid 10}) + M_r(H_2O)$$
The mass of amino acid 9 is 131.1, so it must be isoleucine (Ile), leucine (Leu), or hydroxyproline (Hyp).

<Checkpoint>[Determine that amino acid number 9 is isoleucine Ile, leucine Leu, or hydroxyproline Hyp](1)</Checkpoint>

Therefore, the sequence of the polypeptide is:
Tyr-Leu-Asp-His-Trp-Leu/Ile/Hyp-Gly-Ala-Leu/Ile/Hyp-Ala-Pro-Tyr-Pro-Asp-Pro-Leu/Ile/Hyp-Glu-Pro-Arg

In order to verify the relationship between the prehistoric protein A and modern species, the inferred 19-peptide sequence was aligned with the homologous protein sequences of various modern organisms given. The purpose of the alignment is to find the sequence with the highest degree of matching.

First, represent the ancient protein sequence finally inferred from the question. This sequence contains three uncertain amino acid sites (positions 6, 9, 16), which may be leucine (L), isoleucine (I), or hydroxyproline (Hyp). According to the information in the question, hydroxyproline (Hyp) and proline (P) are both represented by **P** in modern species sequences. Therefore, in the alignment, these three ambiguous sites **`X`** can match **L**, **I**, or **P**.

Ancient 19-peptide sequence pattern:
    `Y-L-D-H-W-`**`X`**`-G-A-`**`X`**`-A-P-Y-P-D-P-`**`X`**`-E-P-R`

The above ancient sequence pattern is aligned one by one with the first 19 amino acids of the homologous protein sequence of each modern species, and the number of matched amino acids is calculated. Mismatched amino acids are indicated in bold.

| Species | Alignment sequence (first 19 amino acids) | Number of mismatches | Similarity |
| :--- | :--- | :--- | :--- |
| **Ancient sequence pattern** | `Y L D H W X G A X A P Y P D P X E P R` | - | - |
| **Horse** | `Y L D H W L G A P A P Y P D P L E P R` | **0** | **19/19** |
| Cow | `Y L D H W L G A P A P Y P D P L E P` **`K`** | 1 | 18/19 |
| Sheep | `Y L D` **`P`** `G` **`L`** `G A P A P Y P D P L E P R` | 2 | 17/19 |
| Human | `Y L` **`Y`** **`Q`** `W L G A P` **`V`** `P Y P D P L E P R` | 3 | 16/19 |
| Rabbit | **`Q`** `L` **`I`** **`N`** `G` **`Q`** `G A P A P Y P D P L E P` **`K`** | 6 | 13/19 |
| Chicken | **`H`** **`Y`** **`A`** **`Q`** **`D`** **`S`** `G` **`V`** **`A`** **`G`** **`A`** **`P`** `P` **`N`** `P L E` **`A`** **`Q`** | 14 | 5/19 |
| Toad | **`S`** **`Y`** **`G`** **`N`** **`N`** **`V`** `G` **`Q`** **`G`** **`A`** **`A`** **`V`** `G` **`S`** `P L E` **`S`** **`Q`** | 15 | 4/19 |
| Carp | **`D`** `L` **`T`** **`V`** **`A`** **`Q`** **`L`** **`E`** **`S`** **`L`** **`K`** **`E`** **`V`** **`C`** **`E`** `A` **`N`** `L` **`A`** | 16 | 3/19 |

<Checkpoint>[Perform sequence alignment and obtain the correct results, the order of polypeptide sequence similarity is horse>cow>sheep>human>rabbit>chicken>toad>carp](3)</Checkpoint>

From the alignment results in the table above, it can be seen that the sequence is most similar to that of the horse.

<Checkpoint>[Conclude that the most similar sequence is that of the horse](1)</Checkpoint>
离子 $y_1$ 的质量可用于确定多肽中最后一个氨基酸的身份。$y_1$ 离子的质量比相应的氨基酸大一个质量单位；因此最后一个氨基酸必定是精氨酸(Arg)。

<Checkpoint>[指出 $y_1$ 是精氨酸Arg ](1)</Checkpoint>

y系列离子是最完整的，通过比较连续y离子的质量可以确定序列：

| 离子     | m/z    | $b_n$ 和 $b_{n-1}$ 之间的质量差 | 对应的氨基酸 | 氨基酸质量 |
| :------- | :----- | :------------------------------ | :----------- | :--------- |
| $y_1$    | 175.1  |                                 |              |            |
| $y_2$    | 272.2  | 97.1                            | 18           | 115.1      |
| $y_3$    | 401.2  | 129.0                           | 17           | 147.0      |
| $y_4$    |        |                                 |              |            |
| $y_5$    | 611.4  |                                 |              |            |
| $y_6$    | 726.4  | 115.0                           | 14           | 133.0      |
| $y_7$    | 823.4  | 97.1                            | 13           | 115.1      |
| $y_8$    | 986.5  | 163.1                           | 12           | 181.1      |
| $y_9$    | 1083.5 | 97.1                            | 11           | 115.1      |
| $y_{10}$ |        |                                 |              |            |
| $y_{11}$ | 1267.6 |                                 |              |            |
| $y_{12}$ | 1338.7 | 71.0                            | 8            | 89.0       |
| $y_{13}$ | 1395.7 | 57.0                            | 7            | 75.0       |
| $y_{14}$ | 1508.8 | 113.1                           | 6            | 131.1      |
| $y_{15}$ | 1694.9 | 186.1                           | 5            | 204.1      |
| $y_{16}$ | 1831.9 | 137.1                           | 4            | 155.1      |
| $y_{17}$ | 1946.9 | 115.0                           | 3            | 133.0      |

<Checkpoint>[分析y系列离子的序列](1)</Checkpoint>

根据y系列，序列为：
Tyr-Leu-Asp-His-Trp-Leu/Ile/Hyp-Gly-Ala-xxx-xxx-Pro-Tyr-Pro-Asp-xxx-xxx-Glu-Pro-Arg

<Checkpoint>[根据y系列离子得到多肽序列为Tyr-Leu-Asp-His-Trp-Leu/Ile/Hyp-Gly-Ala-[未确定]-[未确定]-Pro-Tyr-Pro-Asp-[未确定]-[未确定]-Glu-Pro-Arg](5)</Checkpoint>

序列中第15个氨基酸的身份可以通过 $b_{14}$ 和 $a_{15}$ 之间的质量差来确定：

$$M_r(\text{氨基酸 15}) = \text{质量}(a_{15}) - \text{质量}(a_{14}) + M_r(\text{C}) + 2 M_r(\text{O}) + 2 M_r(\text{H}) = 115.0$$
因此，氨基酸15必定是脯氨酸(Pro)。

<Checkpoint>[确定第15号氨基酸为脯氨酸Pro](1)</Checkpoint>

离子 $y_3$ 和 $y_5$ 之间的质量差给出了对应于氨基酸15和16的片段质量。
$$M_r(\text{15-16 二肽}) = \text{质量}(y_5) - \text{质量}(y_3) + M_r(H_2O)$$
$$M_r(\text{氨基酸 16}) = M_r(\text{15-16 二肽}) - M_r(\text{氨基酸 15}) + M_r(H_2O) = 131.1$$
因此，氨基酸16必定是异亮氨酸(Ile)、亮氨酸(Leu)或羟脯氨酸(Hyp)。

<Checkpoint>[确定第16号氨基酸为异亮氨酸Ile、亮氨酸Leu或羟脯氨酸Hyp](1)</Checkpoint>

氨基酸10的质量可以通过 $b_9$ 和 $b_{10}$ 之间的质量差来确定：
$$M_r(\text{氨基酸 10}) = \text{质量}(b_{10}) - \text{质量}(b_9) + M_r(H_2O)$$
氨基酸10是丙氨酸(Ala)。

<Checkpoint>[确定第10号氨基酸为丙氨酸Ala](1)</Checkpoint>

离子 $y_{11}$ 和 $y_9$ 之间的质量差给出了对应于氨基酸9和10的片段质量。
$$M_r(\text{9-10 二肽}) = \text{质量}(y_{11}) - \text{质量}(y_9) + M_r(H_2O)$$
$$M_r(\text{氨基酸 9}) = M_r(\text{9-10 二肽}) - M_r(\text{氨基酸 10}) + M_r(H_2O)$$
氨基酸9的质量为131.1，因此必定是异亮氨酸(Ile)、亮氨酸(Leu)或羟脯氨酸(Hyp)。

<Checkpoint>[确定第9号氨基酸为异亮氨酸Ile、亮氨酸Leu或羟脯氨酸Hyp](1)</Checkpoint>

因此，该多肽的序列是：
Tyr-Leu-Asp-His-Trp-Leu/Ile/Hyp-Gly-Ala-Leu/Ile/Hyp-Ala-Pro-Tyr-Pro-Asp-Pro-Leu/Ile/Hyp-Glu-Pro-Arg

为了验证史前生物蛋白质A与现代物种的亲缘关系，将推断出的19肽序列与给出的多种现代生物的同源蛋白质序列进行比对。比对的目的是找出匹配度最高的序列。

首先，表示出题目最终推断出的古代蛋白质序列。该序列包含三个不确定的氨基酸位点（位置6, 9, 16），可能是亮氨酸(L)、异亮氨酸(I)或羟脯氨酸(Hyp)。根据题目信息，在现代物种序列中，羟脯氨酸(Hyp)和脯氨酸(P)均用 **P** 表示。因此，在比对中，这三个模糊位点 **`X`** 可与 **L**、**I** 或 **P** 匹配。

古代19肽序列模式：
    `Y-L-D-H-W-`**`X`**`-G-A-`**`X`**`-A-P-Y-P-D-P-`**`X`**`-E-P-R`

将上述古代序列模式与每个现代物种同源蛋白序列的前19个氨基酸进行逐一比对，并计算匹配的氨基酸数量。不匹配的氨基酸用粗体标出。

| 物种 | 比对序列 (前19个氨基酸) | 错配数量 | 相似度 |
| :--- | :--- | :--- | :--- |
| **古代序列模式** | `Y L D H W X G A X A P Y P D P X E P R` | - | - |
| **马 (Horse)** | `Y L D H W L G A P A P Y P D P L E P R` | **0** | **19/19** |
| 牛 (Cow) | `Y L D H W L G A P A P Y P D P L E P` **`K`** | 1 | 18/19 |
| 羊 (Sheep) | `Y L D` **`P`** `G` **`L`** `G A P A P Y P D P L E P R` | 2 | 17/19 |
| 人类 (Human) | `Y L` **`Y`** **`Q`** `W L G A P` **`V`** `P Y P D P L E P R` | 3 | 16/19 |
| 兔子 (Rabbit) | **`Q`** `L` **`I`** **`N`** `G` **`Q`** `G A P A P Y P D P L E P` **`K`** | 6 | 13/19 |
| 鸡 (Chicken) | **`H`** **`Y`** **`A`** **`Q`** **`D`** **`S`** `G` **`V`** **`A`** **`G`** **`A`** **`P`** `P` **`N`** `P L E` **`A`** **`Q`** | 14 | 5/19 |
| 蟾蜍 (Toad) | **`S`** **`Y`** **`G`** **`N`** **`N`** **`V`** `G` **`Q`** **`G`** **`A`** **`A`** **`V`** `G` **`S`** `P L E` **`S`** **`Q`** | 15 | 4/19 |
| 鲤鱼 (Carp) | **`D`** `L` **`T`** **`V`** **`A`** **`Q`** **`L`** **`E`** **`S`** **`L`** **`K`** **`E`** **`V`** **`C`** **`E`** `A` **`N`** `L` **`A`** | 16 | 3/19 |

<Checkpoint>[进行序列比对并得出正确结果，多肽序列相似性顺序为马>牛>羊>人类>兔子>鸡>蟾蜍>鲤鱼](3)</Checkpoint>

从上表的比对结果可以看出，该序列与马的序列最为相似。

<Checkpoint>[得出最相似序列为马的结论](1)</Checkpoint>