We denote the position of the amino acid residue at the $C$-terminus of a peptide chain cleaved by a protease as 0, and sequentially forward as -1, -2, -3, and backward as +1, +2, +3, and so on. For example, the substrate recognized by trypsin is a residue at position 0 that is lysine or arginine, and the residue at position +1 is not proline.

Subtilisin is a serine/threonine protease. Subtilisin has the ability to recognize longer substrate series and has certain selectivity for different residues at positions -3 to 0. We denote the substrate pocket recognizing the 0 residue as S1, the pocket for the -1 residue as S2, -2 as S3, and -3 as S4.

In wild-type subtilisin, the S4 pocket residue is Tyr104, the S2 pocket is Asn62, and the S1 pocket is Gly166. Scientists tested the selectivity of wild-type subtilisin for different substrates using a C-terminally substituted tetrapeptide substrate XXXX-p-nitroanilide, and found that the fastest reacting sequence was Ala-Ala-Pro-Phe-p-nitroanilide.

Now we consider a batch of subtilisin mutants:

A. G166D, N62D

B. G166L, Y104D

C. N62V, Y104K

And some substrates:

1. AAPF-p-nitroanilide
2. AAPA-p-nitroanilide
3. AAPQ-p-nitroanilide
4. AAPY-p-nitroanilide
5. DAKF-p-nitroanilide
6. RAKR-p-nitroanilide
7. RGKE-p-nitroanilide
8. DAPR-p-nitroanilide
9. DAVF-p-nitroanilide
10. FGKR-p-nitroanilide
11. LGFR-p-nitroanilide
12. AAKR-p-nitroanilide
13. RGAA-p-nitroanilide
14. DAHF-p-nitroanilide
15. RGAG-p-nitroanilide

Then, matching these mutants with the substrates with which they react fastest, which of the following options has the numbers from front to back representing the substrate numbers corresponding to subtilisin A, B, and C, respectively, with completely correct results?
我们记肽链中$C$端被蛋白酶切割的氨基酸残基的位置为0，向前依次为-1，-2，-3，向后为+1，+2，+3，依此类推。例如胰蛋白酶识别的底物是0号残基为赖氨酸或者精氨酸且+1号残基不为脯氨酸。

枯草杆菌蛋白酶是一种丝氨酸/苏氨酸蛋白酶，枯草杆菌蛋白酶具有识别更长底物系列的能力，对-3到0号位置的不同残基都具有一定的选择性。我们记其识别0号残基的底物口袋为S1，-1号残基的口袋为S2，-2号为S3，-3号为S4.

在野生型枯草杆菌蛋白酶中，S4口袋残基为Tyr104，S2口袋为Asn62，S1口袋为Gly166。科学家用C端取代的四肽底物XXXX-对硝基苯胺测试了野生型枯草杆菌蛋白酶对不同底物的选择性，发现反应最快的序列是Ala-Ala-Pro-Phe-对硝基苯胺。

下面我们考虑一批突变体的枯草杆菌蛋白酶：

A. G166D, N62D

B. G166L, Y104D

C. N62V, Y104K

和一些底物：

1. AAPF-对硝基苯胺
2. AAPA-对硝基苯胺
3. AAPQ-对硝基苯胺
4. AAPY-对硝基苯胺
5. DAKF-对硝基苯胺
6. RAKR-对硝基苯胺
7. RGKE-对硝基苯胺
8. DAPR-对硝基苯胺
9. DAVF-对硝基苯胺
10. FGKR-对硝基苯胺
11. LGFR-对硝基苯胺
12. AAKR-对硝基苯胺
13. RGAA-对硝基苯胺
14. DAHF-对硝基苯胺
15. RGAG-对硝基苯胺

则将这些突变体与其在这些底物中反应速率最快的相匹配，下列选项的数字从前往后分别代表突变体的枯草杆菌蛋白酶A,B,C对应的底物序号，结果完全正确的是？
6, 13, 5
6, 13, 9
6, 13, 14
6, 15, 5
6, 15, 14
10, 13, 5
10, 13, 9
10, 13, 14
10, 15, 5
10, 15, 14
12, 13, 14
12, 13, 5
12, 15, 14
12, 15, 9
All of the above options are incorrect.
6，13，5
6，13，9
6，13，14
6，15，5
6，15，14
10，13，5
10，13，9
10，13，14
10，15，5
10，15，14
12，13，14
12，13，5
12，15，14
12，15，9
以上选项均不正确
The problem gives 14 different substrates. First, we roughly classify these substrates according to the residue type at different positions, as shown in the table below. Note that the order of pocket numbers and amino acid residue numbers is reversed. S4 corresponds to the first amino acid in the substrate.

|Position|Hydrophobic Amino Acids|Basic Amino Acids|Neutral Amino Acids|Acidic Amino Acids| Aromatic Amino Acids |
| --- |--- |--- |--- |---|---| 
|S4 | 1, 2, 3, 4, 11, 12| 6, 7, 13, 15 | /| 5, 8, 9, 14 |10|
|S2| 1, 2, 3, 4, 8, 9, 13, 15 | 5, 6, 7, 10, 12, 14| / | / |11 |
|S1 |2, 13, 15|6, 8, 9, 10, 11, 12|3 |7 | 1, 4, 5, 14|

Then, we examine the selectivity of each mutant for the substrates. The S1 and S2 pockets of mutant A are mutated to negatively charged amino acids, while the others remain unchanged. Therefore, 0 and -1 in the substrate should select basic amino acid residues.

<Checkpoint>[The S1 and S2 pockets of mutant A are mutated to negatively charged amino acids](1)</Checkpoint>

<Checkpoint>[Mutant A prefers positively charged amino acids at positions 0 and -1, and the substrates that meet the conditions are 6, 10, and 12](1)</Checkpoint>

Referring to the table, substrates 6, 10, and 12 meet the conditions. The other two positions of the substrate should be the same as the optimal substrate of WT. Therefore, mutant A should catalyze substrate number 12 the fastest.

<Checkpoint>[The other two positions of the substrate should be the same as the optimal substrate of the wild-type enzyme](1)</Checkpoint>

<Checkpoint>[Mutant A catalyzes substrate number 12 the fastest](1)</Checkpoint>

The S4 pocket of mutant B is mutated to a negatively charged residue, and the S1 pocket is mutated to a hydrophobic Leu with a certain volume of side chain. Therefore, the preferred substrate is -3 positively charged, and 0 is a hydrophobic amino acid.

<Checkpoint>[The S4 pocket of mutant B is mutated to a negatively charged residue, and the S1 pocket is mutated to a hydrophobic residue](1)</Checkpoint>

Substrates 13 and 15 meet the conditions.

<Checkpoint>[Mutant B prefers substrates with hydrophobic amino acids at position 0 and positively charged amino acids at position -3, and substrates 13 and 15 meet the conditions](1)</Checkpoint>

The only difference between 13 and 15 is the residue at position 0, one is Ala and the other is Gly. Considering the wild-type sequence, S1 is Gly, and the corresponding substrate is Phe; and the side chain size of Leu is significantly smaller than Phe. At this time, the side chain size of residue 0 should be slightly larger than Gly. In addition, the methyl group of Ala is more conducive to the formation of hydrophobic interactions between side chains than the hydrogen atom of Gly. Therefore, substrate 13 is the most suitable.

<Checkpoint>[Ala at position 0 of substrate 13 is more favorable for interacting with the S1 pocket of mutant B than Gly at position 0 of substrate 15](1)</Checkpoint>

<Checkpoint>[Mutant B catalyzes substrate number 13 the fastest](1)</Checkpoint>

Finally, we examine mutant C. The N62V mutation in S2 is a mutation that is difficult to directly assess interactions. We first examine Y104K, which is a mutation of a basic residue.

<Checkpoint>[The S4 pocket of mutant C is mutated to a basic amino acid](1)</Checkpoint>

Therefore, it is more suitable for the substrate -3 position to be an acidic amino acid. The substrates that meet this condition are 5, 8, 9, 14.

<Checkpoint>[Mutant C prefers substrates with acidic amino acids at position -3, and the substrates that meet the conditions are 5, 8, 9, 14](1)</Checkpoint>

Since S1 is not mutated, residue 0 is not affected and should be F. Based on this, substrate 8 can be excluded.

<Checkpoint>[S1 is not mutated, and mutant C prefers substrates with F at position 0](1)</Checkpoint>

The S2 positions of substrates 5, 9, and 14 are K, V, and H, respectively; the WT match is N-P; since the side chain of V is smaller than N in length, from a spatial perspective, something slightly larger than P may be a better substrate. However, if the side chain volume is too large, it may collide with V and affect binding. Based on this, exclude number 5, because the side chain volume of K is too large.

<Checkpoint>[The side chain volume of substrate 5 is too large, and there may be repulsion with the pocket](1)</Checkpoint>

Next, we examine the side chains of V and H. From the perspective of interaction, the interaction between the positively charged H with the hydrophobic pocket is less favorable than the hydrophobic side chain V; from a spatial perspective, H>V>P, H is slightly larger than V. By comparison, substrate 9 is the most suitable for mutant V.

<Checkpoint>[From the perspective of the hydrophobicity matching of the side chain with the S2 pocket of mutant C and the size of the side chain, V at position -1 is more suitable than H](1)</Checkpoint>

<Checkpoint>[The most suitable substrate for mutant C is substrate number 9](1)</Checkpoint>

Therefore, the optimal substrates for the three mutants are numbers 12, 13, and 9, respectively. Option O is selected for this question.

<Checkpoint>[Option O is correct](1)</Checkpoint>
题目给出了14种不同的底物，首先我们将这些底物根据不同位置的残基类型进行大概的分类，如下表。这里要注意一下口袋的编号和氨基酸残基编号的变化顺序是相反的，S4对应底物里第一个氨基酸。

|位置|疏水氨基酸|碱性氨基酸|中性氨基酸|酸性氨基酸| 芳香氨基酸 |
| --- |--- |--- |--- |---|---| 
|S4 |  1, 2, 3, 4, 11, 12| 6, 7, 13, 15 | /| 5,8, 9, 14 |10|
|S2| 1, 2, 3, 4, 8, 9, 13, 15 | 5, 6, 7, 10, 12, 14| /  | / |11 |
|S1 |2,13, 15|6, 8, 9, 10, 11, 12|3 |7 | 1, 4, 5, 14|

然后考察每一个突变体对底物的选择性。突变体A的S1和S2口袋都被突变为带负电的氨基酸，其他不变，故0和-1在底物中应当选择碱性氨基酸残基。

<Checkpoint>[突变体A的S1和S2口袋都被突变为带负电的氨基酸](1)</Checkpoint>

<Checkpoint>[突变体A喜好0位和-1位是带正电的氨基酸，符合条件的是6、10、12三个底物](1)</Checkpoint>

参考表格符合条件的是6、10、12三个底物。底物的其他两个位置应该与WT的最适底物相同，故突变体A催化最快的应该是12号底物。

<Checkpoint>[底物的其他两个位置应该与酶野生型的最适底物相同](1)</Checkpoint>

<Checkpoint>[突变体A催化效率最快的是12号底物](1)</Checkpoint>

突变体B的S4口袋突变为带负电的残基，S1口袋突变为疏水且侧链有一定体积的Leu。因此首选-3带正电，0为疏水氨基酸的底物。

<Checkpoint>[突变体B的S4口袋突变为带负电的残基，S1口袋突变为疏水残基](1)</Checkpoint>

符合条件的是13、15两个底物。

<Checkpoint>[突变体B喜欢0位疏水，-3位带正电的底物，符合条件的是13号和15号底物](1)</Checkpoint>

13和15只有0位的残基不同，一个是Ala，另一个是Gly。考虑野生型序列，S1为Gly，对应的底物为Phe；而Leu的侧链尺寸明显小于Phe，此时0号残基的侧链尺寸应当稍大于Gly，此外Ala的甲基也比Gly的氢原子更有利于侧链之间形成疏水相互作用，因此13号底物最合适。

<Checkpoint>[13号底物0位的Ala比15号底物0位的Gly更有利于与突变体B的S1口袋作用](1)</Checkpoint>

<Checkpoint>[突变体B催化最快的是13号底物](1)</Checkpoint>

最后考察突变体C，S2的N62V的突变是一个难以直接考量相互作用的突变，我们先考察Y104K，这是一个碱性残基的突变.

<Checkpoint>[突变体C的S4口袋被突变为碱性氨基酸](1)</Checkpoint>

因此底物-3位是酸性氨基酸比较合适，符合这一条件的底物有5, 8, 9, 14.

<Checkpoint>[突变体C喜好-3位是酸性氨基酸的底物，符合条件的底物是5，8，9，14](1)</Checkpoint>

由于S1没有突变，因此0号残基不受影响，应该是F，据此可以排除底物8.

<Checkpoint>[S1没有突变，突变体C喜好0位是F的底物](1)</Checkpoint>

5、9、14号底物的S2位分别是K，V，H；WT的匹配是N-P；由于V的侧链从长度上比N更小，因此从空间角度来说，比P稍大一些的可能是更好的底物。而侧链体积太大则可能与V发生碰撞，影响结合。据此排除5号,因为K的侧链体积太大。

<Checkpoint>[5号底物侧链体积太大，与口袋可能存在排斥](1)</Checkpoint>

下面我们考察一下V和H的侧链，从相互作用角度而言，侧链基带正电的H与疏水口袋的相互作用相比疏水侧链V来说不利一些；从空间角度而言，H>V>P，H稍稍比V更大一些。比较可知突变体V最合适的是9号底物。

<Checkpoint>[从侧链与突变体C S2口袋的亲疏水性匹配和侧链尺寸的角度，-1位的V比H更合适](1)</Checkpoint>

<Checkpoint>[突变体C最适的是9号底物](1)</Checkpoint>

故三种突变体的最适底物分别是12、13、9号，本题选择选项O。

<Checkpoint>[O 选项正确](1)</Checkpoint>