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बनाया गया: 26 नवंबर 2025

का उपयोग करके उत्तर दिया गया GPT-5.1 Thinking द्वारा Chat01

प्रश्न

s41477-020-0688-1.pdf

阅读下这篇文章

विचार

Summarizing the article’s content

The user wants me to read and summarize the article they uploaded. Since it's in Chinese, I'll give a brief summary of the main findings and methods. I’ll make sure to explain in Chinese and mention any references properly. I won’t promise future work but will aim for a moderate explanation now.

Clarifying citation details

I need to include citations with line ranges (L10-L20), but the system doesn’t provide line numbers for the article. Since the tool spec mentions these line ranges, it seems I should still follow it. However, the system message says only the website citation is necessary, which is simpler. I'll stick to the website citation.

Summarizing the article briefly

The user only asked to read the article, but to be helpful, I’ll offer a concise summary of key points: the background, method, findings, and significance. The paper explores post-transcriptional splicing and intron retention in Arabidopsis using Nanopore and PacBio sequencing techniques. I'll keep it simple and provide bullet points in Chinese.

Summarizing the key findings

In Arabidopsis, over half of introns remain unspliced when Pol II has transcribed more than 1 kb past the 3’ splice site, displaying slower splicing compared to yeast. They track cotranscriptional splicing, revealing a correlation between upstream and downstream introns. Also, certain polyA+ chromatin-bound RNAs remain unspliced.

Summarizing key findings from the article

Pts introns show slower cotranscriptional splicing than non-pts. The study highlighted that intron retention in Arabidopsis is mostly related to pts introns, which are sensitive to splicing perturbation and environmental stress. They also identified regulatory roles of incompletely spliced, polyA+ RNAs in chromatin for rapid response.

Thought for 1m 28s

उत्तर

这篇文章我已经读完了,大致内容先帮你梳理一下👇 (s41477-020-0688-1.pdf)


1. 研究想回答什么问题?

  • 绝大多数真核基因的剪接被认为是在转录过程中“共转录剪接”完成的。
  • 但在多细胞生物(比如植物)里,剪接到底有多快、和 Pol II 延伸以及加尾(polyadenylation)是怎样配合的,其实还不清楚。
  • 同时,植物里普遍存在的 intron retention(IR,内含子保留),到底是“剪接失败的废料”,还是一种受调控的中间状态,也一直有争议。

这篇文章聚焦在 拟南芥,想系统回答:

  1. 共转录剪接在植物里的动力学是什么样(快还是慢)?
  2. 多内含子基因的剪接顺序和协同性如何?
  3. 某些内含子为什么在 poly(A)+ 转录本里仍然不剪接,却几乎不出现在细胞质里
  4. 这些“迟迟不剪”的内含子是否受到 剪接因子和环境信号 的调控?

2. 核心方法(Fig.1)

作者建立了一套 Nanopore 长读长的“全长 chromatin-bound RNA-seq”

  1. 分离拟南芥幼苗的 染色质结合 RNA(chromatin-bound nascent RNA),再去除 rRNA。
  2. 在 3′ 端连接接头,用模板切换反转录得到 全长 cDNA
  3. Nanopore MinION(两批生物学重复)和 PacBio 分别测序。
  4. 自己写了分析流程:
    • 区分 正在延伸的转录本(没有 polyA 尾)已经 polyadenylated 的转录本
    • 根据读长覆盖情况判断每个内含子是否剪接;
    • 开发了一个 PolyAcaller,从 Nanopore 原始电流信号估算 poly(A) 尾长。

得到约 350 万条高质量全长 reads,覆盖 2 万多个蛋白编码基因。(s41477-020-0688-1.pdf)


3. 主要发现

3.1 植物的共转录剪接并不快(Fig.2b)

  • 统计 Pol II 从某个内含子的 3′ splice site 往下走了多少碱基,该内含子的剪接比例是多少。
  • 结果:当 Pol II 已经走过 3′SS 约 1,100 nt 时,仍有超过一半内含子未被剪接。(s41477-020-0688-1.pdf)
  • 这比酵母里报道的“~45 nt 内完成剪接”要慢得多,更类似果蝇和人类。

结论: 在拟南芥中,共转录剪接存在,但整体偏慢。


3.2 多内含子基因中的剪接顺序和协同(Fig.2c–e)

  • 相邻两个内含子里,约 70% 的情况是上游内含子先被剪接,大体上遵循转录顺序。
  • 对含 5 个内含子的基因做“条形图热图”,发现:
    • 上游内含子的剪接比例普遍高于下游;
    • 某些内含子即使在下游都剪完、甚至转录本已经 polyadenylated 之后,仍然保持未剪接状态。(s41477-020-0688-1.pdf)
  • 邻近内含子的剪接状态高度相关,说明 早期剪接事件会促进同一转录本上其他内含子的剪接

3.3 出乎意料:大量 poly(A)+ 转录本仍滞留在染色质上、并且不完全剪接(Fig.1b, Fig.3)

  • 在染色质结合 RNA 中,大约:
    • 70% 是延伸中的 Pol II 转录本
    • 30% 已经加了 poly(A) 尾。其中又有约 1/3 在长读长上显示 “只剩一两个内含子没剪”。(s41477-020-0688-1.pdf)
  • 把同一批样品的 Illumina RNA-seq(PolyA+) 和长读长结果对比,内含子保留比例高度一致,说明这些不是测序伪影。(s41477-020-0688-1.pdf)
  • 进一步比较 染色质、核质、胞质 三个分馏:
    • 在染色质 PolyA+ 中内含子保留比例很高;
    • 到了核质和胞质,这些内含子几乎看不到(保留比例接近 0)。(s41477-020-0688-1.pdf)

解释:
这些带内含子的 poly(A)+ 转录本 被“扣留”在染色质上,只有在后续把特定内含子剪掉后,成熟 mRNA 才能释放到核质和胞质。

作者把这类 在染色质 poly(A)+ 中仍有明显保留,但在胞质里基本全剪掉的、构成型内含子,定义为:

pts introns(post-transcriptionally spliced introns,后转录剪接内含子)。(s41477-020-0688-1.pdf)


3.4 这些“迟剪内含子”不会走 NMD,而是真·中间状态(Fig.4b)

  • 从 NMD 途径的 upf1 upf3 双突变体的 RNA-seq 看:
    • 大多数这类未剪内含子的累积水平在 upf1 upf3 里并没有上升。
  • 说明这些带内含子的 RNA 不是被 NMD 清除的靶标,而更像是 还没进一步加工的“半成品”。(s41477-020-0688-1.pdf)

3.5 pts 内含子的剪接对剪接因子突变和环境胁迫特别敏感(Fig.4)

作者整理了 6,521 个公开 RNA-seq 文库(>1500 个“mutant vs WT” 或 “treatment vs control” 配对),系统寻找:

  • 哪些突变体或处理会出现大量 intron retention 上升?
  • 这些 IR 事件发生在 pts 内含子还是普通内含子上?

突变体:

  • 找到 10 个突变体在 >500 个内含子里 IR 上升,这些基因都是已知剪接因子(如 PRMT5,SKIP 等)。(s41477-020-0688-1.pdf)
  • 虽然 pts 内含子只占所有内含子约 28%,但这些突变体中 ~80% 的 IR 事件发生在 pts 内含子上,远高于预期。(s41477-020-0688-1.pdf)
  • 在 prmt5 和 skip 中,pts 内含子 IR 明显增加,而非 pts 内含子影响较小。
  • 图 4c 给了一个 prmt5 的经典例子:AT2G17340 第一个内含子 IR 大幅增加。

环境处理:

  • 各种处理(冷、热、盐、感染等)下出现 IR 上升的内含子里,绝大多数也是 pts 内含子。(s41477-020-0688-1.pdf)
  • 冷/热处理尤其明显;例如 昼夜节律基因 TOC1 的某个内含子在低温下 IR 增高,并且这是一个 pts intron(Fig.4f)。

结论:

Arabidopsis 里大量观察到的 intron retention 事件,本质上大多发生在 pts 内含子上,是染色质滞留 + 后转录剪接所允许的一种可调控状态,而不是随机剪接失败。


3.6 与 poly(A) 位点选择(APA)的关系

  • 作者顺便利用长读长去看 剪接和 poly(A) 位点选择之间的关系
    • 在冷/热诱导的高 IR 基因中,发生 APA 的比例并不比其他基因高太多;
    • 对于有多个 poly(A) 位点的基因,主、次 PA 异构体上的内含子剪接比例整体高度相似,只在少数基因上有明显差别。(s41477-020-0688-1.pdf)
  • 提示:在大多数情况下,IR 的调控相对独立于 APA,但个别基因可能存在交叉影响。

4. 作者提出的模型(Fig.4g)

可以看成是一个“核内待命 mRNA 仓库”模型:

  1. 基因被转录,部分内含子共转录剪接,但某些 pts 内含子剪接速度刻意偏慢
  2. 转录终止并 polyadenylation 后,转录本已经是 poly(A)+,但因为还残留 pts 内含子:
    • 滞留在染色质/细胞核内
    • 不会被 NMD 清除。
  3. 环境信号或发育信号 来临(温度变化、应激、剪接因子活性改变……):
    • 刺激对特定 pts 内含子的剪接;
    • “一刀剪完”,瞬间把大批已有的核内 poly(A)+ 转录本转化为成熟 mRNA;
    • 再输出到胞质,快速翻译成蛋白,而不一定要重新启动大量转录。

这样,后转录剪接 + 核内滞留 就变成了一个可以快速释放响应的调控层,在植物和动物中可能都是保守机制。(s41477-020-0688-1.pdf)


如果你接下来想:

  • 看中文详细总结 / 画流程图;
  • 或者只针对某一部分(比如剪接动力学、数据分析方法、Figure 某一页)深入讲;

直接告诉我你更关心哪一块,我可以按那一块给你“逐段精读版”。

प्रश्न

Unexpectedly, we also detected a high percentage of polyad-
enylated transcripts on chromatin, and ~30% of these transcripts
remained incompletely spliced (Figs. 1b, 2a and Fig. 3a). Both the ratio
of incompletely splicing transcripts and the ratio of unspliced introns
are highly consistent between the two Nanopore libraries (r = 0.83
for genes and r = 0.85 for introns; Supplementary Figs. 6 and 7). The
splicing results are also highly consistent between the PacBio library
and the two Nanopore libraries (r = 0.87 and r = 0.92 for genes, and
r = 0.88 and r = 0.91 for introns; Supplementary Figs. 6 and 7). We
also constructed Illumina RNA-seq libraries of chromatin-bound
polyadenylated RNA with three biological replicates (Supplementary
Table 1). The ratios of unspliced introns are also highly consistent
between Illumina RNA-seq and Nanopore data (r = 0.91), and are
also consistent across three Illumina RNA-seq biological replicates
(r = 0.94, r = 0.95 and r = 0.96; Fig. 3b and Supplementary Fig. 7).
Furthermore, most of the incompletely spliced polyadenylated
transcripts contained only one or a few unspliced introns, whereas
the other introns had been spliced efficiently (Figs. 2d, 3a and 3c),
indicating the specificity of unspliced introns. To further analyse the
differences between introns with high and low unspliced ratios in
chromatin-bound polyadenylated RNA, we separated introns into
two groups using a cut-off value of 0.1. For introns in the group
with an unspliced ratio of chromatin-bound polyadenylated RNA
of higher than 0.1, the cotranscriptional splicing rates were much
slower (Fig. 3d), consistent with the result showing that these introns
remain unspliced after PA.
这一大段再说什么,我理解前一大段说这一段说明核内存在加了a尾的read仍然有一部分还有内含子,感觉后面说的是一样的事情呢

विचार

Clarifying new points in the paragraph

The user is asking about a paragraph that seems repetitive based on their understanding of an earlier section — they think the paragraph is already explaining polyA reads with introns in the nucleus. However, this part introduces new points: it discusses ruling out technical artifacts and quantifying reproducibility and specificity of the splicing data across different sequencing libraries. It highlights consistent results and the specificity of incompletely spliced transcripts.

Clarifying distinctions in the analysis of introns

The user is likely confused because they think the paragraph is repetitive, but there's an important distinction. Earlier, the presence of nuclear polyA reads with introns was observed, and here it's confirmed to be real and reproducible across platforms. The slow splicing is specifically tied to certain introns with high unspliced ratios, and the 0.1 threshold helps define the group of slow-spliced (pts) introns.

Clarifying new points in the paragraph

I can highlight three new insights here:

  1. Technical validation: Correlations were found across Nanopore replicates, as well as across Nanopore-PacBio and Nanopore-Illumina platforms.
  2. Pattern: Only specific introns retain unspliced regions, demonstrating that intron retention isn't random.
  3. Kinetics: Slow co-transcriptional splicing explains why high unspliced introns persist on polyA RNA.

I also need to explain the distinction between "incompletely spliced transcripts" per gene and "unspliced introns" per intron. The 0.1 unspliced ratio threshold ties to defining pts introns.

Breaking down the explanation and logic

I need to address what the correlation values mean: they represent comparisons using the Pearson correlation coefficient. The user's question about "是不是一样的事情呢" (Is it the same thing?) prompts me to emphasize that it's not a repetition, but a deeper exploration with method controls, replicates, and classifications. Initially, they show 30% of chromatin polyA transcripts are incompletely spliced, but then confirm that this finding is reproducible across platforms (Nanopore, PacBio, Illumina). They clarify "ratio of incompletely splicing transcripts" refers to the proportion of unspliced transcripts from a gene, while "ratio of unspliced introns" refers to unspliced portions at the intron level. The correlation values (r) demonstrate consistency across different replicates and technologies used.

Thought for 58s

उत्तर

你理解得对:这段开头那句,确实是在重复前面那个结论——
“染色质上有不少已经 polyA 的转录本,其中大概 30% 还是没剪完内含子”。

但这整段的 重点不是再说一遍现象本身,而是做两件事:


1. 先证明:这个现象是“真”的,不是测序噪音

后面一串 r = 0.xx,都是在做“交叉验证”:

  1. Nanopore 两个生物学重复之间

    • 看每个基因:不完全剪接的转录本比例在两次实验之间相关性 r = 0.83
    • 看每个内含子:未剪接比例在两次实验之间 r = 0.85
  2. PacBio 和 Nanopore 之间

    • 基因层面 r = 0.87、0.92
    • 内含子层面 r = 0.88、0.91
  3. Illumina 的短读长 vs Nanopore 长读长

    • 染色质结合 polyA RNA 里,未剪接内含子比例:Illumina 和 Nanopore 之间 r = 0.91
    • 三个 Illumina 生物学重复之间也非常一致:r = 0.94, 0.95, 0.96 (s41477-020-0688-1.pdf)

意思就是:

“不管用哪个平台(Nanopore / PacBio / Illumina),用哪次重复,我看到的‘polyA+ 但有内含子保留’这个图像都一样。
所以这绝不是某个平台特有的错误或某一次实验的偶然现象。”

这一步是在 打消读者的技术质疑
“Nanopore 错配/错判内含子吧?” —— 他们用不同平台 + 多重复来证明 不是技术 artifact。


2. 再说明:哪些内含子被保留,是“有选择性”的 + 和剪接动力学有关

后半段又做了两件事情:

2.1 不是“整条 pre-mRNA 都没剪完”,而是“只剩少数特定内含子”

“most of the incompletely spliced polyadenylated transcripts contained only one or a few unspliced introns, whereas the other introns had been spliced efficiently … indicating the specificity of unspliced introns.”

翻成白话就是:

一条 polyA+ 转录本如果没剪完,通常只剩一两个内含子没剪
其余内含子都剪得很好,
所以被保留的那些内含子是“被特别点名留下来的”,不是随机漏剪。

这在图 2d, 3a, 3c 里都有可视化。

2.2 把“容易被保留的内含子”单独拎出来:这些内含子本来就剪得慢

他们设了一个阈值 0.1 来分组:

  • 如果某个内含子在“染色质结合 polyA RNA”里,未剪接比例 > 0.1(>10% 的 polyA+ 转录本还带着这段内含子),
    → 把它归为“高 unspliced ratio 组”
  • 其余的归为“低组”

然后去看这两组内含子在 共转录剪接时的速度曲线(Fig.3d):

结果:高 unspliced ratio 组的共转录剪接明显更慢
也就是:在 Pol II 走过 3′SS 很远之后,这些内含子仍然不太容易被剪掉。

这句话的逻辑是:

“之所以它们在 polyA 之后还留着内含子,是因为它们从一开始共转录剪接时就剪得比较慢。”

这一步是在把 “后转录还没剪完” 和 “共转录剪接动力学偏慢” 联系起来
给后面“pts intron = 本来就剪得慢的一类内含子”这个概念打基础。


小结一下这段话在干嘛

  • 前一段:告诉你——“我们发现:染色质上有不少 polyA+ 但没剪完的转录本。”
  • 这一大段
    1. 证明这个现象在不同平台、不同实验、不同测序技术之间都高度一致 → 现象可信
    2. 指出:
      • 未剪接只集中在少数特定内含子上 → 有选择性
      • 这些内含子在共转录阶段就剪得慢 → 是一类“天生慢剪”的内含子,不是随机漏剪。

所以它看上去好像在“重复”,其实是在 从“有这个现象”走到“这个现象是可靠的,而且指向一类特殊慢剪内含子”
后面把这类内含子命名为 pts introns,再去分析它们在突变体/环境胁迫中的调控。

प्रश्न

We next examined whether these incompletely spliced transcripts
can be released from the chromatin. We constructed next-generation
sequencing (NGS) RNA-seq libraries using polyadenylated RNAs
from different cell fractions, and found that the introns with an
unspliced ratio of higher than 0.1 in chromatin-bound polyadenyl-
ated RNA were barely detectable in the nucleoplasm and cytoplasm
(Fig. 3e), suggesting that the intron-containing transcripts were
still bound to chromatin after PA. Moreover, the RNA accumula-
tion level (fragments per kilobase of transcript per million mapped
reads (FPKM) value) of genes with a higher incompletely spliced
ratio in chromatin-bound polyadenylated RNA was notable higherWe next examined whether these incompletely spliced transcripts
can be released from the chromatin. We constructed next-generation
sequencing (NGS) RNA-seq libraries using polyadenylated RNAs
from different cell fractions, and found that the introns with an
unspliced ratio of higher than 0.1 in chromatin-bound polyadenyl-
ated RNA were barely detectable in the nucleoplasm and cytoplasm
(Fig. 3e), suggesting that the intron-containing transcripts were
still bound to chromatin after PA. Moreover, the RNA accumula-
tion level (fragments per kilobase of transcript per million mapped
reads (FPKM) value) of genes with a higher incompletely spliced
ratio in chromatin-bound polyadenylated RNA was notable higher

我还是有一点疑问,未剪切比不是read吗,怎么能是内含子呢? 第二点在核质和胞质看不见,核质不是我们传统意义上的核内吗

विचार

Clarifying the unspliced ratio concept

The user seems to be misunderstanding the unspliced ratio of introns, thinking of it as a read-specific issue, whereas it's about the proportion of introns in a transcript that remain unspliced. I'll make sure to explain the difference between "per read" and "per intron" and how this applies to both short- and long-read sequencing.

Clarifying intron retention and compartmentation

The unspliced ratio of introns reflects the retention of introns in a chromatin polyA fraction. It's important to clarify that 'unspliced ratio' is per intron, and 'reads' are used to estimate this. Regarding compartmentation, nucleoplasm is nuclear but not chromatin-associated, differing from chromatin-bound material in fractionation processes.

Clarifying intron retention and addressing citations

The intron-containing polyA transcripts' enrichment in the chromatin fraction suggests they're tied to chromatin and not exported to nucleoplasm or cytoplasm. Additionally, the user set an unspliced ratio cutoff for chromatin-bound polyA, with barely detectable intron presence elsewhere. I'll incorporate the correct citations and clarify these points.

Approximating the citation and explaining ratio confusion

We’ll use the citation as "(website)" since line numbers weren’t visible. For the user’s confusion, I'll explain that the unspliced ratio is calculated per intron, with the ratio coming from reads that either show retention or splicing. It’s about the intron being the unit of measurement.

Explaining metrics and compartmentation

Intron retention ratios and splicing ratios need to be differentiated by context. For example, gene-level ratios refer to chromatin-bound polyA reads with unspliced introns. I'll clarify that PIR is about aggregate intron retention, while unspliced ratio applies per read. I'll also explain the nucleoplasm vs chromatin compartmentation and provide method details for clarity.

Thought for 1m 18s

उत्तर

好问题,这两点一开始确实很容易绕晕 🤝 我按两个小问题分别说。


一、为什么说“未剪切比例高的内含子”,未剪切比不是 read 吗?

这里有两个层次的量,文章里都有提到:

  1. ratio of incompletely spliced transcripts

    • 这是“转录本层面”的量:
    • 比如某个基因在 chromatin polyA+ 里有 100 条 reads,其中 40 条还有至少一个内含子没剪完 →
      • 这个基因的 “incompletely spliced transcript ratio” = 40/100 = 0.4
  2. ratio of unspliced introns(未剪切内含子比例)

    • 这是“内含子这个 feature 层面”的量:
    • 对于基因里某一个 intron(例如 intron 2),你看所有覆盖它的 polyA+ reads:
      • 有 30 条还保留这个 intron(unspliced)
      • 有 70 条这个 intron 已经被剪掉(spliced)
    • 那么这个内含子的 unspliced ratio = 30 / (30+70) = 0.3

也就是说:

“introns with an unspliced ratio of higher than 0.1”
指的是:在 chromatin-bound polyA RNA 中,有 ≥10% 的 reads 仍然保留这个 intron 的那些 intron。(s41477-020-0688-1.pdf)

read 只是用来“数”,真正被打分的是 每一个内含子位点
所以“未剪切比例高的内含子”这句话是没问题的。

在 Illumina 数据里,他们用 PIR(percent intron retention)做这个 “unspliced ratio”,方法部分也写得很清楚:拿 E1–I、I–E2 的 junction reads 代表保留内含子,E1–E2 的 junction 代表已剪接,然后算一个百分比。(s41477-020-0688-1.pdf)


二、“核质 nucleoplasm” 不也是核内吗?为啥说“从染色质 release 出来”就应该在 nucleoplasm 里?

对,这里是分级亚细胞分馏的问题:作者把“核内”再细分成了两块。

他们的操作大致是这样(Methods 页有详细 protocol):(s41477-020-0688-1.pdf)

  1. 先得到三个大 fraction:
    • Cytoplasmic fraction(胞质):第一次离心上清
    • Nuclei pellet(细胞核沉淀)
  2. 再对核做第 2 步:
    • 高盐 / 尿素 / 去污剂洗掉核里“可溶”的成分 → 这是 nucleoplasmic fraction(核质)
    • 剩下紧紧和 DNA / 核骨架结合的部分 → chromatin fraction(染色质部分)

所以:

  • chromatin fraction:还是“核内”,但指的是紧贴染色质 / 转录位点那一撮东西(包括 nascent RNA、被 tether 的 polyA+ RNA)
  • nucleoplasm:同样在核内,但已经从染色质上“脱落”到可溶核质里的 RNA / 蛋白

因此他们的逻辑是:

如果这些带内含子的 polyA+ 转录本已经从染色质上释放下来,
那至少应该在 nucleoplasm(核质) 里能看到它们的踪迹,
最后才有机会出口到 cytoplasm(胞质)

但他们看到的是(Fig.3e):

  • 在 chromatin polyA+ 里,那些 unspliced ratio ≥ 0.1 的内含子 很多;
  • 到了 nucleoplasm 和 cytoplasm,这些内含子对应的 IR 几乎为 0,基本检测不到。(s41477-020-0688-1.pdf)

加上 Fig.3f 里这些基因在 chromatin 里的表达量高于 nucleoplasm / cytoplasm,更说明:

这些“不完全剪接的 polyA+ 转录本”
停留在“染色质这一层的核内”,而不是核质 / 胞质

所以你可以这样理解他们的几个 compartment:

  • 核内 = nucleoplasm + chromatin
  • 他们是在这俩之间再做了一个“是否还 tether 在染色质上”的区分:
    • 还 tether 在染色质上 → chromatin fraction
    • 已经从染色质上释放到核质 → nucleoplasm
    • 再往外运输出去 → cytoplasm

小复盘

  • “未剪切比”这个 ratio 是 按内含子算 的:看某个内含子在多少条 read 里是 retained,从而给这个内含子一个 0–1 的“保留概率”。
  • nucleoplasm 的确也是“核内”,但文章在做的是 核内再分层
    • 染色质结合 vs “自由核质” vs 胞质。
  • 结论:那些在 chromatin polyA+ 里有高 IR 的内含子,在 nucleoplasm/ cytoplasm 的 polyA+ 里几乎看不见 → 说明这批带内含子的 polyA+ RNA 一直没能从染色质层跑出来

如果你愿意,我可以给你画一个小示意流程图(文字版),把 “chromatin → nucleoplasm → cytoplasm” 三层和剪接状态对应上,一眼就能看到这些 pts intron 卡在哪一层。

प्रश्न

这一段就是 这些不剪的intro在核外几乎看不到,意思就是到核外会被剪,是吗

विचार

Clarifying the splicing location

The user’s question is about whether the unspliced introns are spliced once they move out of the nucleus, based on the description in the text. In essence, it seems that these introns remain in the nucleus, possibly attached to chromatin, until they've been spliced. Once fully spliced, they’re released into the nucleoplasm/cytoplasm.

Refining the interpretation of splicing and export

There’s a possibility that polyadenylation hasn’t finished yet. However, the main idea is that incomplete splicing causes transcripts to remain bound to chromatin. Once fully spliced, these transcripts are found in the nucleoplasm and cytoplasm. This suggests splicing must happen before leaving the chromatin or cytoplasm.

Clarifying the nuances of splicing and export

I need to explain that while we infer splicing happens before export, we can’t state it as a certainty since we can't observe the same molecule before and after. I’ll briefly address the user’s confusion on “核外” (outside the nucleus, meaning the cytoplasm) and clarify the concept of nucleoplasm being still within the nucleus.

Clarifying "核外" and the splicing process

It sounds like the user might be using "核外" to refer to areas outside the chromatin region, possibly referring to the nucleoplasm or cytoplasm. I’ll explain that within the nucleus, intron-retaining transcripts are bound to chromatin and only move to the nucleoplasm once spliced, suggesting splicing happens within the chromatin.

Clarifying chromatin retention and splicing process

To explain clearly: within the nucleus, intron-retaining transcripts are bound to chromatin, while spliced transcripts move to nucleoplasm and cytoplasm for export. The key point is that chromatin retention of incompletely spliced transcripts is a critical step in mRNA maturation. I should also include the proper citation from the text provided.

Clarifying citation format

It seems the user wants a citation with line numbers, but I don't have them right now in this format. The instructions are clear to use line number ranges, but the prompt says to include "(website)" which I would normally question as conflicting. Given that they're asking to include it, I’ll lean toward satisfying the prompt directly.

Thought for 31s

उत्तर

对,你现在的理解可以这样微调一下,会更准确一点👇


1. 这一段到底在说什么?

introns with an unspliced ratio of higher than 0.1 in chromatin-bound polyadenylated RNA were barely detectable in the nucleoplasm and cytoplasm … suggesting that the intron-containing transcripts were still bound to chromatin after PA. (s41477-020-0688-1.pdf)

翻成白话:

  • 染色质结合的 polyA+ RNA(chromatin fraction) 里,那些 保留比例 ≥0.1 的内含子 很多;
  • 但在 核质(nucleoplasm)胞质(cytoplasm) 的 polyA+ RNA 里,这些内含子几乎检测不到;
  • 同时,这些基因在染色质那一层的表达量高,在核质/胞质反而低 → 这些“没剪完的 polyA+ 转录本”积在染色质上。(s41477-020-0688-1.pdf)

2. “是不是说明到了核外就会被剪掉?”

可以这么理解,但要注意两个层次:

  1. 实验现象(他们真的看到的)

    • “带内含子的 polyA+ 转录本” 只在 chromatin fraction 里出现
    • 一旦到了 nucleoplasm 和 cytoplasm,对应位置基本都是 已剪接状态,几乎看不到 IR。
  2. 作者的推论 / 模型

    • 说明这些带内含子的 polyA+ RNA 在还没剪完之前,就被卡在染色质上,不会被释放到核质/胞质
    • 要想“离开染色质 → 进入核质 → 出核到胞质”,中间那几个 pts introns 必须先被剪掉(发生在转录后期、而且大概率仍然在核内)。
    • 加上 upf1 upf3 NMD 双突变也没看到这些未剪内含子在胞质里堆积,所以它们不是跑到 cytoplasm 被 NMD 清掉,而是压根没出去。(s41477-020-0688-1.pdf)

所以你问的那句:

“意思就是到核外会被剪,是吗?”

更精确一点说:

  • 不是说“出了核才剪”,而是:
    • 染色质这一层的核内,它们是 polyA+ 但还有 pts intron → 被扣留在 chromatin;
    • 剪完 pts intron 之后,这些转录本才会从 chromatin 释放到 核质(仍然在核内),再进一步输出到胞质;
    • 等你在 核质/胞质 抓 polyA+ RNA 时,看见的基本都是 已经全剪好的版本

也就是:
👉 “带内含子的 polyA+ 转录本不会离开染色质;能到核质/胞质的,基本都已经把这些特定内含子剪掉了。”

如果你愿意,我可以用“时间线”方式帮你把一个 pts intron 所经历的几个阶段画成 1→2→3→4 的流程(比如:共转录剪一部分 → 加 polyA 但还留一个内含子 → 染色质滞留 → 剪完 → 释放/出核)。

प्रश्न

更精确一点说是必须剪完才能出核?

विचार

Clarifying concept of RNA splicing and nuclear export

The user wants clarification about whether RNA must be fully spliced before exiting the nucleus. According to the paper, transcripts with retained introns are chromatin-bound and not released. These intron-containing polyA+ RNAs are spliced before being exported into the cytoplasm.

Considering alternative explanations for RNA splicing and export

They suggest an alternative view: splicing might continue after polyadenylation, and transcripts may be fully spliced before export. While the paper infers that most intron-containing transcripts are retained in the nucleus until spliced and then exported, they acknowledge a possibility of rare exceptions. This aligns with earlier findings in Arabidopsis.

Responding to user's question on splicing and export of pts introns

For the user's question, yes, in the model, transcripts need to have the specific pts introns spliced away to leave the nucleus. There's a small possibility that a fraction of these transcripts with introns might still be exported, but it's rare. In this study, most of the intron-containing transcripts stay in the nucleus until fully spliced.

Clarifying RNA export and splicing requirements

To clarify, for transcripts with pts introns, evidence shows they stay in chromatin until these introns are spliced out, which aligns with the idea that splicing must occur before nuclear export. However, the term “必须” (must) is too strong – there’s a correlation, but not direct causal proof. In this context, it’s more accurate to say: "Mostly, polyA+ RNAs with pts introns are not released into the nucleoplasm/cytoplasm before splicing." Support for this includes data from chromatin expression, NMD

Thought for 45s

उत्तर

可以这么理解,但要把“必须”说得稍微柔一点 😊

在这篇文章的语境里

对这些 pts intron 所在的转录本,作者的模型基本就是:

只要还带着 pts intron,polyA+ 转录本就被扣在染色质上;
能到核质 / 胞质的,几乎全都是把 pts intron 剪掉后的版本。
(s41477-020-0688-1.pdf)

证据主要有三块:

  1. 分级分馏:

    • chromatin polyA+ 里,某些内含子 IR≥0.1 很常见;
    • 同一批细胞的 nucleoplasmcytoplasm polyA+ 里,这些内含子的 IR 几乎为 0 → “带内含子的 polyA+ 转录本只在染色质这一层出现”。(s41477-020-0688-1.pdf)
  2. 表达量分布:

    • 这些“incompletely spliced ratio 高的基因”,在 chromatin 里的 FPKM 明显高于 nucleoplasm / cytoplasm,说明转录本堆在染色质,不是在外面被降解掉。(s41477-020-0688-1.pdf)
  3. NMD 双突变 upf1 upf3:

    • 如果有不少“带 intron 的 mRNA 出核再被 NMD 清除”,那在 upf1 upf3 里应该堆积起来;
    • 实际上大部分这些未剪内含子在 upf1 upf3 里没有明显上升 → 说明它们压根就没大量出核。(s41477-020-0688-1.pdf)

所以,在这套数据的统计意义上,可以很接地气地说一句:

👉 对这些 pts intron 来讲,“基本上是剪完了才能走出染色质 → 进核质 → 再出核”。


但“必须”这个词为什么要稍微保留一点?

  • 他们并没有 单分子实时追踪“某条 RNA 从染色质剪完 → 出核”的全过程,而是做了分馏 + 测序,看到的是“哪里多、哪里几乎没有”这种 分布差异
  • 文中也自己提了一句 alternative explanation,说也可能和 PA 完成过程的慢快纠缠在一起,技术上现在还没法直接测“PA 完成的那一刻”。(s41477-020-0688-1.pdf)

所以更严谨一点的说法是:

它们的结果“强烈暗示”——
带 pts intron 的 polyA+ 转录本在剪掉这些 intron 之前,一般不会被释放到核质/胞质;
实际上我们几乎只在核外看到全剪接版本。

如果你以后在写综述/汇报,可以这样概括一句:

“对于 pts introns,核输出与否基本上取决于这些内含子是否被剪完,是一个关键的后转录‘质检/闸门’步骤。”

这样既保留了作者想表达的生物学模型,又不过度绝对化“必须剪完”这个词。

प्रश्न

Furthermore, if these
intron-containing transcripts were to be released from chromatin,
they would probably be targeted by the nonsense-mediated decay
(NMD) pathway, but we found that the accumulation level of most
of these unspliced introns remained the same in the up frameshift
1 (upf1) upf3 double mutant, which disrupts the NMD pathway
(Fig. 4b), suggesting that most of these incompletely spliced tran-
scripts are not degraded by NMD, consistent with previous reports
that most transcripts with intron retention (IR) are resistant to
NMD in plants23. Thus, our results suggest that chromatin retention
of incompletely spliced transcripts represents an important step in
mRNA maturation and has a critical role in post-transcriptional
gene regulation. Misannotation of introns are frequent and would
appear as IR events when analysing the RNA-seq data. To obtain
introns that are being actively spliced, we followed a recent study
that characterized splicing kinetics in Drosophila and humans, and
we used only constitutively spliced introns for our splicing analy-
sis (see Methods)22
. To simplify the analysis, we hereafter define pts
introns as constitutively spliced introns with an unspliced ratio of
at least 0.1 in the chromatin-bound polyadenylated RNA fraction.
Furthermore, if these
intron-containing transcripts were to be released from chromatin,
they would probably be targeted by the nonsense-mediated decay
(NMD) pathway, but we found that the accumulation level of most
of these unspliced introns remained the same in the up frameshift
1 (upf1) upf3 double mutant, which disrupts the NMD pathway
(Fig. 4b), suggesting that most of these incompletely spliced tran-
scripts are not degraded by NMD, consistent with previous reports
that most transcripts with intron retention (IR) are resistant to
NMD in plants23. Thus, our results suggest that chromatin retention
of incompletely spliced transcripts represents an important step in
mRNA maturation and has a critical role in post-transcriptional
gene regulation. Misannotation of introns are frequent and would
appear as IR events when analysing the RNA-seq data. To obtain
introns that are being actively spliced, we followed a recent study
that characterized splicing kinetics in Drosophila and humans, and
we used only constitutively spliced introns for our splicing analy-
sis (see Methods)22
. To simplify the analysis, we hereafter define pts
introns as constitutively spliced introns with an unspliced ratio of
at least 0.1 in the chromatin-bound polyadenylated RNA fraction.

विचार

Interpreting the paragraph for explanation

It seems like the user wants me to explain the pasted paragraph, particularly about NMD, chromatin retention, misannotation, constitutively spliced introns, and PTS introns. Given the context of their previous questions, I need to provide a bilingual explanation, primarily in Chinese, while making it clear and concise.

Explaining the paragraph's key points

This paragraph covers NMD, intron retention (IR), and mRNA maturation. It explains how NMD typically degrades transcripts with premature stop codons (PTCs), but this doesn’t happen in certain mutants, suggesting chromatin retention plays a crucial role. It also mentions how misannotations can cause confusion with IR and highlights the use of constitutively spliced introns in analysis. Lastly, it defines PTS introns as those with a certain ratio in chromatin-bound polyA.

Clarifying NMD and nuclear export process

The passage describes how certain introns are retained in the chromatin fraction and spliced post-transcriptionally, not degraded by NMD, and act as a regulatory pool. If the user inquires about NMD and nuclear export, I'll clarify how they relate to this process. I also need to ensure citations are included for the relevant sections.

Clarifying citation process and user inquiry

The snippet I have reflects a collapsed version of the file, with no visible line numbers in the UI, so I'll estimate the line range, like L140-L180, although this isn't verified. I also need to confirm the user's previous interpretation: unspliced transcripts aren't degraded by NMD, meaning nuclear/chromatin retention plays a regulatory role, not an error.

Thought for 47s

उत्तर

我给你把这段拆成三块说,一句一句对上来。


1. 先用 NMD 来“排除”一种解释

if these intron-containing transcripts were to be released from chromatin, they would probably be targeted by the nonsense-mediated decay (NMD) pathway …

意思是:
按照常识,如果这些还带内含子的 mRNA 真出了核(离开染色质到细胞质),大概率会因为带有提前终止密码子被 NMD 降解。

所以他们做了一个“反证”实验:

… but we found that the accumulation level of most of these unspliced introns remained the same in the upf1 upf3 double mutant, which disrupts the NMD pathway …

  • upf1 upf3 双突变 = NMD 基本废掉 的背景。
  • 如果这些“带内含子的 mRNA”经常跑到细胞质被 NMD 吃掉,那一旦 NMD 坏掉,它们就应该大量堆积
  • 实际观察:这些未剪内含子的水平 在 upf1 upf3 里并没有明显升高。(s41477-020-0688-1.pdf)

… suggesting that most of these incompletely spliced transcripts are not degraded by NMD …

所以结论是:

👉 这些 incompletely spliced transcripts 不是“出核→被 NMD 清除”的那一类
而是压根没有大规模走 NMD 这条路

这和前面你问的“是不是必须剪完才能出核?”是呼应的:
数据支持的模型是——没剪完的那批基本就卡在染色质,不是出去被 NMD 杀。


2. 于是:染色质上的“扣留”本身就是一个成熟步骤 / 调控层

Thus, our results suggest that chromatin retention of incompletely spliced transcripts represents an important step in mRNA maturation and has a critical role in post-transcriptional gene regulation.

翻成白话:

  • 这些没剪完的 polyA+ 转录本 被留在染色质上
  • 这个“留着不放行”的过程 是 mRNA 成熟过程中的一个重要步骤
  • 同时也是一个很重要的 转录后调控层 ——什么时候把它们剪完、放出去,是个可以调控的开关。(s41477-020-0688-1.pdf)

跟你之前的总结可以合并成一句话:

对 pts intron 来说,剪完 / 不剪完 = 出不出核,这个“卡在染色质上等剪完”的状态被用来做基因表达调控。


3. 处理“假 IR”:只看真正会被剪掉的内含子

接下来这几句是在说 怎么避免把注释错误当成 IR

Misannotation of introns are frequent and would appear as IR events when analysing the RNA-seq data.

有些“被标成 intron 的片段”其实生理上就是 exon 或 UTR,永远不会被剪掉;
如果不小心把它们也算进来,就会 “看上去都是 IR”,但那其实只是注释错了。

所以他们做了一个筛选:

we used only constitutively spliced introns for our splicing analysis …

  • constitutively spliced introns = 在胞质 mRNA 里 几乎总是被剪掉 的那类内含子
  • 方法上,就是选那些在 cytoplasm 里 PSI > 0.8 的 intron(s41477-020-0688-1.pdf)

这样就确保:

我们讨论的 IR 事件,都是发生在“正常情况下必剪”的内含子上,
不是因为基因注释错导致的“永远 IR”。


4. 正式给 pts intron 下定义

To simplify the analysis, we hereafter define pts introns as constitutively spliced introns with an unspliced ratio of at least 0.1 in the chromatin-bound polyadenylated RNA fraction.

翻译成公式就是:

pts intron 满足两个条件:

  1. constitutively spliced intron:在胞质成熟 mRNA 里几乎总被剪掉(说明它真的是一个要剪掉的 intron);
  2. chromatin-bound polyA+ RNA 里,unspliced ratio ≥ 0.1:
    • 也就是:在紧贴染色质的 polyA+ 转录本中,至少有 10% 的 reads 保留了这个 intron。(s41477-020-0688-1.pdf)

结合前面的结果:

  • 在染色质 polyA+ fraction 里:这些 intron 经常是保留的;
  • 在核质 / 胞质 polyA+ 里:它们几乎总是被剪掉;
  • 在 NMD 失活背景(upf1 upf3)里:它们也不会在胞质大量堆积。

所以 pts intron 就是你现在脑子里的那个形象:

“在染色质上拖着不剪,作为一个可调控的中间状态;剪完之后才放行出核”的那一类内含子。


如果你愿意,下一步我们可以把“普通 intron vs pts intron”的区别整理成一张小对照表(剪接速度、所在 compartment、对 mutant / stress 的敏感性),方便你以后做汇报 / 画模型图。

प्रश्न

所以整个figure3这一部分就是在说这些加了a尾巴在核内的read含保留一些特殊的intron,且这些intron如果出核,就不会存在在read里,然后根据这个假设连续的nmd途径,如果他是被nmd途径干掉了,那我沉默nmd途径他应该大量上升,但是实验发现并没有类似的现象出现

विचार

Clarifying figure 3 and NMD hypothesis

Yes, you're on the right track. Figure 3 shows polyA+ nuclear reads with retained introns. If these introns exit the nucleus, they'd either be spliced or degraded by NMD. The absence of NMD-mediated degradation is seen in upf1/upf3 mutants. Be sure to highlight nuclear compartments and NMD implications.

Handling citation requirements and system message guidance

Okay, it seems the system message wants me to cite just "(website)" without needing line ranges—maybe that’s shorthand for the whole file. To be safe, I’ll follow that directly. For clarity, I’ll include the relevant content for Fig 3 & NMD, noting the connection with the incomplete splicing and NMD.

Clarifying figure 3 and NMD results

Figure 3 suggests that polyA+ chromatin-bound reads retain specific introns. These introns aren't found in the nucleoplasm or cytoplasm but are instead stuck in the chromatin. Genes with more retained introns show higher expression in chromatin. NMD mutants show these transcripts aren't exported or degraded, just retained. A precise correction: if exported, spliced isoforms appear without introns, with rare exceptions.

Refining the explanation of nucleocytoplasmic dynamics

Your summary is nearly there! Here's a more precise version:

  1. Figure 3 panels a-d show introns with an unspliced ratio ≥0.1 as potential PTC candidates and slower co-transcriptional splicing.
  2. Based on NMD inhibition, the retention of the intron-containing transcripts supports the idea that they likely stay within the chromatin until spliced or otherwise processed.

Thought for 1m 16s

उत्तर

对,你这段话的理解大方向是对的,我帮你把逻辑再“校正 + 补几句细节”,变成一个更严谨版👇 (s41477-020-0688-1.pdf)


1. Figure 3 这一块在讲什么?

可以拆成三步:

  1. 现象:染色质上的 polyA+ 转录本里,有一批内含子经常被保留

    • chromatin-bound polyA+ RNA 里,某些内含子有明显的 unspliced ratio(≥0.1),而同一个基因的其他内含子剪得很好 → 这些就是候选的 pts introns(Fig.3a,c,d)。
    • 说明:不是整个 pre-mRNA 都没剪完,而是“少数特定内含子”慢剪、经常被保留。
  2. 分馏对比:这些保留的内含子只出现在染色质那一层

    • 做了 chromatin / nucleoplasm / cytoplasm 三个分馏的 polyA+ RNA-seq:
      • 这些在 chromatin polyA+ 里 unspliced ratio ≥0.1 的内含子,到了 nucleoplasm 和 cytoplasm 里几乎检测不到 IR(Fig.3e)。
    • 同时,这类基因在 chromatin 里的 FPKM 明显高于 nucleoplasm / cytoplasm(Fig.3f)。
    • 综合起来的意思是:

      “带这些 intron 的 polyA+ 转录本主要堆在染色质上,几乎不会以‘带 intron 形态’出现在核质和胞质里。”

  3. 推断:那它们如果真的跑出去了,会怎样?——引出 NMD 这条线


2. 你的总结哪里需要微调?

你说的版本大概是:

“这些加了 A 尾的核内 read 含保留一些特殊 intron;
如果这些 intron 出核,就不会存在在 read 里;
可能是被 NMD 干掉;
如果真是 NMD 干掉,那敲掉 NMD 它应该上升,但实验没看到。”

我帮你稍微修一下重点和用词:

2.1 “核内 read”→ 更准确是“染色质上的 polyA+ read”

  • 这里他们分的是:chromatin(染色质) / nucleoplasm(核质) / cytoplasm(胞质) 三个层次,而不是简单“核内 vs 核外”。
  • 他们强调的是:

    chromatin fraction 中的 polyA+ RNA 里,有很多带 pts intron 的;
    但在 nucleoplasm 和 cytoplasm 的 polyA+ 里几乎看不到这些 intron。

所以更精确的说法是:

这些带 pts intron 的 polyA+ 转录本主要被扣在染色质上,没有作为“带 intron 的形式”进入核质和胞质。

2.2 “出核就不会存在在 read 里” → 更严谨一点

从数据看:

  • 核质和胞质的 polyA+ 库里,同一段 intron 的 IR 基本为 0:
    👉 能被检测到的几乎全是 已经把这段 intron 剪掉的 fully spliced mRNA

所以比较好的说法是:

只要我们在核质/胞质里抓到的 polyA+ mRNA,基本都不再带这些 pts intron。

至于“是不是每一条出核之前都剪完”,严格来说他们是从群体分布推断,而不是做单分子追踪,所以我们说“基本上需要剪完才能出去”会更恰当。


3. NMD 那一段在逻辑上怎么接上?

这里确实是你说的那条推理链,但可以拆得更清楚一点:

  1. 很多带 intron 的转录本会引入 提前终止密码子(PTC)按经典 NMD 模型应该是 NMD 靶标
  2. 如果这些带 intron 的 polyA+ 转录本 真的大量跑到胞质,那在 NMD 功能正常时应该被清掉;
  3. 一旦 NMD 被破坏(upf1 upf3 双突变)
    • 理论预期:这些转录本应该 大量累积
    • 实际观察:大部分未剪内含子的水平在 upf1 upf3 里没明显升高(Fig.4b)。

→ 所以:

这些 incompletely spliced transcripts 并不是主要在胞质被 NMD 清除,而是根本就没怎么出核(更准确:没离开 chromatin 层)。

然后顺势得出那句结论:

chromatin retention of incompletely spliced transcripts represents an important step in mRNA maturation and has a critical role in post-transcriptional gene regulation.

也就是:

把带 pts intron 的 polyA+ 转录本锁在染色质上、等它们后转录剪完再放行,这是 mRNA 成熟过程中的一个关键步骤,也是一层重要的转录后调控。


小结成一句话版本给你背:

Fig.3 先证明:有一批 polyA+ 转录本在染色质上带着特定 intron(pts intron)堆积,但这些 intron 在核质和胞质里几乎看不到;
再结合 upf1 upf3 不会让这些 intron 大量上升,说明它们不是出核后被 NMD 清除,而是作为“染色质滞留的、等待后转录剪接的中间状态”存在。

प्रश्न

我希望我们以后这个有一批 polyA+ 转录本就说成加了a尾的rnaread 我明白它的具体意思,但是你那种说法泰国拗口,我每次看到都要反应一下,你可以在我给你的说法后面加个括号备注一下

विचार

Thought for 19s

उत्तर

好哒,收到!以后我就按你说的来 👇

  • 我原来写的 “polyA+ transcripts”,以后统一说成:
    “加了 A 尾的 RNA read(polyA+ transcripts)”
  • 每次第一次出现的时候,我在你这套说法后面给个小括号备注英文,后面就可以直接用“加了 A 尾的 RNA read”了。

比如这篇文章里,我们就可以总结成:

在染色质上,有一批 加了 A 尾的 RNA read(polyA+ transcripts) 还保留着特定的内含子(pts introns),这些带内含子的加 A 尾 read 在核质和胞质里几乎看不到,说明它们还扣在染色质上,没被释放出去。(s41477-020-0688-1.pdf)

你后面要是对别的术语也有“顺嘴版本”,随时跟我说,我都按你的习惯来写~

प्रश्न

To determine whether the splicing of pts introns is affected in
certain mutants or regulated by environment signals, we selected
6,521 high-quality, publicly available RNA-seq libraries that account
for 1,512 matched groups with at least two biological replicates of
each sample, including 713 groups of mutant versus wild type (WT)
and 799 groups of treatment versus mock/CK, to search for dif-
ferentially retained introns at genome-wide scale (Supplementary
Tables 2 and 3). From the mutant versus WT analysis, we identi-
fied ten mutant samples with an increased retention ratio (that
is, increased unspliced ratio) in more than 500 introns compared
with their matched WT controls (see Methods). The corresponding
genes for these mutants were all previously reported to be involved
in splicing, such as PROTEIN ARGININE METHYLTRANSFERASE
5 (PRMT5) and SKI-INTERACTING PROTEIN (SKIP)24–27

विचार

Explaining experimental design for intron retention

Okay, first, they selected 6,521 RNA-seq libraries, which include rRNA-depleted and polyA+ short read libraries. However, I don't need to emphasize the polyA+ aspect. The goal is to analyze splicing sensitivity to mutants and the environment. They created a dataset with mutant vs WT and treatment vs mock samples, then searched for differentially retained introns across the genome.

Summarizing experiment and bridging details

The paragraph discusses points about how 28% of all introns are pts, but around 80% of enhanced intron retention in mutants are pts. It emphasizes the strong suppression of pts intron splicing in prmt5 and skip mutants. The user may want a summary of numbers like 6,521 libraries and 1,512 groups, as well as a connection to prior topics like pts introns and NMD.

Understanding retention ratio terminology and citation requirements

Increased retention ratio just means the higher unspliced ratio, so they're describing the same thing. I should remember to include a citation. Since the snippet doesn't have line numbers, I can use the citation

Thought for 54s

उत्तर

这段主要是在说 “我们怎么证明 pts 内含子的剪接确实会被剪接因子和环境信号调控”,以及他们具体是怎么挖这些调控事件的。(s41477-020-0688-1.pdf)

我给你拆成几步看:


1. 先从全网把能用的 RNA-seq 都捞一遍

we selected 6,521 high-quality, publicly available RNA-seq libraries … 1,512 matched groups … 713 groups of mutant versus wild type (WT) and 799 groups of treatment versus mock/CK …

意思就是:

  • 在 NCBI 上找了 6,521 个质量好的公共 RNA-seq 文库
  • 把它们整理成 1,512 个成对/成组的对比(每组至少有 2 个生物学重复):
    • 713 组:突变体 vs 野生型
    • 799 组:各种处理(冷、热、盐、病原等)vs 对应的对照 mock/CK
  • 目的是在 全基因组尺度 去找:哪些内含子在这些条件下出现了 “保留比例改变”(IR 改变)

可以理解成:

“我们把公开数据库里所有像样的 Arabidopsis RNA-seq 都翻出来,当一个超大规模的 stress/mutant screen,用来查 pts 内含子剪接有没有被影响。”


2. 在这些“突变体 vs WT”里,筛出“剪接整体出大问题”的那几类突变体

From the mutant versus WT analysis, we identified ten mutant samples with an increased retention ratio (that is, increased unspliced ratio) in more than 500 introns compared with their matched WT controls …

这里做的是:

  • 对每一对 “mutant vs WT”:
    • 算每个内含子的 retention ratio = unspliced ratio(未剪接比例)
    • 看:相比 WT,有多少内含子在 mutant 里 IR 上升
  • 然后他们挑出那种:
    • 有 >500 个内含子 IR 上升的突变体
      → 说明这个突变体对剪接的影响是 全局性的,而不是零星几个基因。

最后筛出来 10 个这样的突变体样本


3. 发现:这 10 个突变体都是“剪接机器本身”出了问题

The corresponding genes for these mutants were all previously reported to be involved in splicing, such as PRMT5 and SKIP …

也就是说:

  • 这 10 个“让几百个内含子同时 IR 上升”的突变体,对应的基因 全部都是已知的剪接相关因子
    • 比如:
      • PRMT5:蛋白精氨酸甲基转移酶,帮助招募 NineTeen Complex 到剪接体;
      • SKIP:剪接体组分,之前就被证明跟 Arabidopsis 时钟和可变剪接有关。
  • 这说明他们的筛选是靠谱的:

    大规模 IR 上升 → 基本就指向剪接因子突变,而不是别的乱七八糟的基因。

后面紧接着的一段(你上一条没贴完的部分)就进一步说:

  • 虽然 pts 内含子只占所有内含子的 ~28%
  • 但在这 10 个突变体里,~80% 的 IR 事件都发生在 pts 内含子上
  • 而且 prmt5、skip 这些突变体对 pts 内含子的影响远大于对非 pts 内含子。(s41477-020-0688-1.pdf)

你可以把这段记成一句话:

他们用 6,521 个公共 RNA-seq 做了一个超大规模筛选,
找到 10 个“让很多内含子都更容易被保留”的剪接因子突变体(比如 PRMT5、SKIP),
后面再证明:这些突变主要是 让 pts 内含子更难被剪完
说明 pts 内含子的剪接确实是一个可以被剪接因子和环境信号专门调控的层级

如果你愿意,下一步我可以帮你把 Fig.4a–e 每一个 panel 和这一段对应起来,看具体是怎么算“>500 introns IR 上升”、以及 pts intron 在图上是怎么体现出来的。

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