The incorrect proposition number among the following statements about genetic translation is:  

Proposition 1:  

Each amino acid is encoded by a codon (a segment consisting of three nucleotide residues in mRNA). Among all codons, the total number of codons that have or also possess special signal functions is consistent with the number of amino acid types that exhibit the highest codon degeneracy.  

Proposition 2:  

If the sequence 5'-AUG UCU UAA-3' is translated in an in vitro system lacking translation release factors, the ribosome will recognize the UAA codon and recruit a tRNA carrying leucine (L), thereby yielding the polypeptide M-S-L.  

Proposition 3:  

The molar molecular mass of an Escherichia coli protein is approximately $51~\mathrm{kDa}$. If the ribosome translates $20$ nucleotides per second, according to commonly used estimation methods in biochemistry, the time required for a cell to synthesize this protein is 1 minute and 15 seconds.  

Proposition 4:  

A synthetic polyribonucleotide consisting solely of A and C in a 1:5 ratio is used as messenger RNA. In the amino acid composition of the protein synthesized by this RNA, the first five amino acids, ordered from smallest to largest molecular weight, exhibit a composition ratio of 150:30:5:5:1.
以下关于基因翻译问题中，错误的命题序号是：

命题1：

每一个氨基酸都被一个密码子（mRNA 中一个三核苷酸残基组成的片段）所编码，在所有密码子中，具有或兼具特殊信号功能的密码子的总数、与拥有最高密码子简并度的氨基酸种类数一致。
 
命题2：

如果将序列 5'-AUG UCU UAA-3' 在一个缺乏翻译释放因子的体外体系中进行翻译，核糖体将会识别到UAA密码子并招募一个携带亮氨酸（L）的tRNA，从而可以得到多肽 M-S-L。

命题3：

大肠杆菌蛋白质的摩尔分子质量约为 $51~\mathrm{kDa}$。若核糖体每秒翻译 $20$ 个核苷酸，按照生物化学中常用的估算方法，一个细胞合成这个蛋白质需要时间为1分15秒。

命题4：

一段人工合成的只含 1:5 比例 A 和 C 的多核糖核苷酸作为信使 RNA，该RNA合成蛋白质的氨基酸组成中，各个氨基酸按照分子量的由小到大，前五个的组成比例为150:30:5:5:1。
1
2
3
4
1, 2
1, 3
1, 4
2, 3
2, 4
3, 4
1, 2, 3
2, 3, 4
1, 2, 4
1, 3, 4
1, 2, 3, 4
None of the above options is correct
1
2
3
4
1，2
1，3
1，4
2，3
2，4
3，4
1，2，3
2，3，4
1，2，4
1，3，4
1，2，3，4
以上选项均不正确
Based on the provided information, analyze these four propositions.

Analysis of Proposition 1: Among the 64 mRNA codons, there are 3 stop codons (UAA, UAG, UGA), 1 start codon (AUG, encoding methionine), and the remaining codons encode 20 amino acids. Therefore, the total number of codons with or additionally possessing special signal functions is 4.

<Checkpoint>[Proposition 1: The total number of codons with or additionally possessing special signal functions in mRNA codons is 4](0.5)</Checkpoint>

In mRNA codons, codon degeneracy refers to the number of codons encoding the same amino acid. The highest degeneracy is 6, observed for:

- Leucine (Leu/L): 6 (UUA, UUG, CUU, CUC, CUA, CUG)
- Arginine (Arg/R): 6 (CGU, CGC, CGA, CGG, AGA, AGG)
- Serine (Ser/S): 6 (UCU, UCC, UCA, UCG, AGU, AGC)

Thus, the number of amino acid types with the highest codon degeneracy is 3.

<Checkpoint>[Proposition 1: The number of amino acid types with the highest codon degeneracy in mRNA codons is 3](0.5)</Checkpoint>

Therefore, in Proposition 1, the total number of codons with or additionally possessing special signal functions (4) does not match the number of amino acid types with the highest codon degeneracy (3). Proposition 1 is incorrect.

<Checkpoint>[Proposition 1 is incorrect](1)</Checkpoint>

Analysis of Proposition 2: During translation, AUG encodes methionine (M), and UCU encodes serine (S), both of which are correct.

<Checkpoint>[Proposition 2: AUG encodes methionine (M), UCU encodes serine (S)](0.5)</Checkpoint>

However, when the ribosome encounters the stop codon UAA, under normal circumstances, this termination signal is recognized by release factor proteins, which cleave and release the synthesized polypeptide chain, thereby terminating translation. In a system lacking release factors, this termination step cannot be completed. More importantly, there is no leucine tRNA in the cell that can pair with the UAA stop codon (leucine codons are UUA/G or CUX). Therefore, the ribosome cannot recruit leucine tRNA to extend the peptide chain, nor can it release the existing methionine-serine (M-S) peptide chain. The final result is that the ribosome stalls at the UAA codon, and the entire translation process halts at this point, preventing the synthesis of the M-S-L polypeptide.

<Checkpoint>[The translation process stalls, and the M-S-L polypeptide is not synthesized. Proposition 2 is incorrect](1)</Checkpoint>

Analysis of Proposition 4: The protein has a mass of $51~\mathrm{kDa}$, and the average mass of a single amino acid is approximately $110~\mathrm{Da}$. Therefore, the number of amino acids is $\dfrac{51000}{110}=463.64\approx 464$, and the number of nucleotides required in the mRNA region encoding this protein is $464\times 3=1392$.

<Checkpoint>[Proposition 3: The number of nucleotides required in the mRNA region encoding this protein is $1392$](1)</Checkpoint>

Thus, synthesizing an E. coli protein with a molecular mass of approximately 51 kDa would take about $\dfrac{1392}{20}=69.6\approx70$ seconds. Proposition 3 is incorrect.

<Checkpoint>[Synthesizing this protein takes approximately 1 minute and 10 seconds, not 1 minute and 15 seconds. Proposition 3 is incorrect](1)</Checkpoint>

Analysis of Proposition 4: Since codons consist of three nucleotides, there are eight possible codons that can form on this messenger RNA. The amino acids encoded by these codons include lysine (encoded by AAA, molecular weight $146.19~\mathrm{g/mol}$), asparagine (AAC, $132.12~\mathrm{g/mol}$), threonine (ACA, ACC, $119.12~\mathrm{g/mol}$), glutamine (CAA, $146.14~\mathrm{g/mol}$), histidine (CAC, $155.16~\mathrm{g/mol}$), and proline (CCA, CCC, $115.13~\mathrm{g/mol}$). Their molecular weights, ordered from smallest to largest, are proline, threonine, asparagine, glutamine, lysine, histidine.

<Checkpoint>[Proposition 4: The molecular weights of the amino acids in the protein, ordered from smallest to largest, are proline, threonine, asparagine, glutamine, lysine, histidine](1)</Checkpoint>

For the frequency of occurrence of each amino acid:

- Proline: $P(\text{Pro}) = P(\text{CCA}) + P(\text{CCC}) = (\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}) + (\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}) = \frac{25+125}{216} = \frac{150}{216}$
- Threonine: $P(\text{Thr}) = P(\text{ACA}) + P(\text{ACC}) = (\frac{1}{6} \times \frac{5}{6} \times \frac{1}{6}) + (\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6}) = \frac{5+25}{216} = \frac{30}{216}$
- Histidine: $P(\text{His}) = P(\text{CAC}) = \frac{5}{6} \times \frac{1}{6} \times \frac{5}{6} = \frac{25}{216}$
- Asparagine and glutamine: $P(\text{Asn}) = P(\text{AAC}) = \frac{1}{6} \times \frac{1}{6} \times \frac{5}{6} = \frac{5}{216}$; $P(\text{Gln}) = P(\text{CAA}) = \frac{1}{6} \times \frac{1}{6} \times \frac{5}{6} = \frac{5}{216}$
- Lysine: $P(\text{Lys}) = P(\text{AAA}) = \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{216}$

Therefore, for the five amino acids with the smallest molecular weights, their frequency ratios are: 150 (proline) : 30 (threonine) : 5 (asparagine) : 5 (glutamine) : 1 (lysine). Thus, Proposition 4 is correct.

<Checkpoint>[The correct ratio is 150 (proline) : 30 (threonine) : 5 (asparagine) : 5 (glutamine) : 1 (lysine). Proposition 4 is correct](1)</Checkpoint>

Consequently, select option **K**.
根据所提供的信息，对这四个命题进行分析。

分析命题1：在64个mRNA密码子中，有3个密码子是终止密码子（UAA、UAG、UGA）、1 个密码子是起始密码子（AUG，编码甲硫氨酸）、其余编码20个氨基酸，故具有或兼具特殊信号功能的密码子的总数为4。

<Checkpoint>[命题1：mRNA密码子中具有或兼具特殊信号功能的密码子的总数为4](0.5)</Checkpoint>

mRNA密码子中，密码子简并度是指编码同一种氨基酸的密码子数量。其中简并度最高为6，分别为：

- 亮氨酸（Leu/L）：6（UUA、UUG、CUU、CUC、CUA、CUG）
- 精氨酸（Arg/R）：6（CGU、CGC、CGA、CGG、AGA、AGG）
- 丝氨酸（Ser/S）：6（UCU、UCC、UCA、UCG、AGU、AGC）

故拥有最高密码子简并度的氨基酸种类数为3。

<Checkpoint>[命题1：mRNA密码子中拥有最高密码子简并度的氨基酸种类数为3](0.5)</Checkpoint>

因此，命题1中具有或兼具特殊信号功能的密码子的总数（4）、与拥有最高密码子简并度的氨基酸种类数不一致（3），命题1错误。

<Checkpoint>[命题1错误](1)</Checkpoint>

分析命题2：在翻译过程中，AUG编码甲硫氨酸（M），UCU编码丝氨酸（S），这两步正确。

<Checkpoint>[命题2：AUG编码甲硫氨酸（M），UCU编码丝氨酸（S）](0.5)</Checkpoint>

然而，当核糖体遇到UAA这个终止密码子时，在正常情况下，该终止信号由释放因子蛋白质别，该蛋白质会切断并释放合成好的多肽链，从而结束翻译。而在一个缺少释放因子的体系中，这个终止步骤无法完成。更重要的是，细胞中并不存在能够与UAA终止密码子配对的亮氨酸tRNA（亮氨酸的密码子是UUA/G或CUX）。因此，核糖体既不能招募亮氨酸tRNA来延长肽链，也无法释放已有的甲硫氨酸-丝氨酸（M-S）肽链。最终的结果是，核糖体会在UAA密码子处停滞不前，整个翻译过程停滞于此，不会合成出M-S-L多肽。

<Checkpoint>[翻译过程停滞，不会合成出M-S-L多肽，命题2错误](1)</Checkpoint>

分析命题4：蛋白质质量为 $51~\mathrm{kDa}$，单个氨基酸的平均质量约为 $110~\mathrm{Da}$，因此，氨基酸数量为 $\dfrac{51000}{110}=463.64\approx 464$ 个，编码这个蛋白质的 mRNA 区域所需的核苷酸数量为 $464\times 3=1392$ 个。 

<Checkpoint>[命题3：编码这个蛋白质的 mRNA 区域所需的核苷酸数量为  $1392$ 个](1)</Checkpoint>

因此，合成一个分子质量约为 51 kDa 的大肠杆菌蛋白质大约需要 $\dfrac{1392}{20}=69.6\approx70$ 秒，命题3错误。

<Checkpoint>[合成该蛋白大约需要1分10秒而非1分15秒，命题3错误](1)</Checkpoint>

分析命题4：由于密码子由三个核苷酸组成，因此在这条信使RNA上可能形成的密码子共有八种，这些密码子所编码的氨基酸种类包括赖氨酸（由AAA编码，分子量$146.19~\mathrm{g/mol}$）、天冬酰胺（AAC，$132.12~\mathrm{g/mol}$）、苏氨酸（ACA、ACC，$119.12~\mathrm{g/mol}$）、谷氨酰胺（CAA，$146.14~\mathrm{g/mol}$）、组氨酸（CAC，$155.16~\mathrm{g/mol}$）和脯氨酸（CCA、CCC，$115.13~\mathrm{g/mol}$）。其分子量由小到大依次为脯氨酸、苏氨酸、天冬酰胺、谷氨酰胺、赖氨酸、组氨酸。

<Checkpoint>[命题4：组成蛋白质的氨基酸分子量由小到大依次为脯氨酸、苏氨酸、天冬酰胺、谷氨酰胺、赖氨酸、组氨酸](1)</Checkpoint>

对于每种氨基酸的出现频率：

- 脯氨酸：$P(\text{Pro}) = P(\text{CCA}) + P(\text{CCC}) = (\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}) + (\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}) = \frac{25+125}{216} = \frac{150}{216}$
- 苏氨酸：$P(\text{Thr}) = P(\text{ACA}) + P(\text{ACC}) = (\frac{1}{6} \times \frac{5}{6} \times \frac{1}{6}) + (\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6}) = \frac{5+25}{216} = \frac{30}{216}$
- 组氨酸：$P(\text{His}) = P(\text{CAC}) = \frac{5}{6} \times \frac{1}{6} \times \frac{5}{6} = \frac{25}{216}$
- 天冬酰胺和谷氨酰胺：$P(\text{Asn}) = P(\text{AAC}) = \frac{1}{6} \times \frac{1}{6} \times \frac{5}{6} = \frac{5}{216}$
；$P(\text{Gln}) = P(\text{CAA}) = \frac{1}{6} \times \frac{1}{6} \times \frac{5}{6} = \frac{5}{216}$
- 赖氨酸：$P(\text{Lys}) = P(\text{AAA}) = \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{216}$

因此，按照氨基酸分子量由小到大的前五个氨基酸，其依次的出现频率的比例为：150（脯氨酸）: 30（苏氨酸）: 5（天冬酰胺）: 5（谷氨酰胺）: 1（赖氨酸），故命题4正确。

<Checkpoint>[正确的比例关系为150（脯氨酸）: 30（苏氨酸）: 5（天冬酰胺）: 5（谷氨酰胺）: 1（赖氨酸），命题4正确](1)</Checkpoint>

从而选择选项 **K**。