增强子与它调控的启动子常相距数十万碱基，这种远距离调控如何实现？
Enhancers often sit hundreds of kilobases from the promoters they control; how is such long-range regulation achieved?
染色质通过黏连蛋白介导的环挤出在 CTCF 结合位点处停顿，形成拓扑关联结构域，把增强子与目标启动子限制在同一空间隔室内，从而提高有效接触频率。增强子上结合的转录因子与介体复合体一起把 RNA 聚合酶 II 招募到启动子，部分体系还涉及富含内在无序区蛋白的凝聚体，使局部转录因子浓度升高。结构域边界的破坏可让增强子错误劫持邻近基因，这种增强子劫持已在多种发育畸形和肿瘤中被证实。
Cohesin-driven loop extrusion stalls at CTCF sites to create topologically associating domains that confine an enhancer and its target promoter to the same spatial compartment, raising their effective contact frequency. Transcription factors bound at the enhancer work with the Mediator complex to recruit RNA polymerase II to the promoter, and in some systems condensates rich in intrinsically disordered proteins locally concentrate the machinery. Disrupting domain boundaries lets enhancers hijack neighbouring genes, a mechanism documented in developmental malformations and cancers.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Urry 等,《Campbell Biology》第11版,Pearson,2017
L. A. Urry et al., Campbell Biology, 11th ed., Pearson, 2017
独立核验依据 · Independent verification: 核验依据 2 ISBN 978-0-13-409341-3 (11th ed., Pearson)
#增强子 #染色质环 #拓扑关联结构域 #增强子劫持
长非编码 RNA Xist 如何让一条 X 染色体整体失活？
How does the long non-coding RNA Xist inactivate an entire X chromosome?
雌性哺乳动物胚胎发育早期，两条 X 中有一条上的 Xist 基因被上调，转录出的长非编码 RNA 不离开自身染色体，而是沿顺式方向逐步包裹该染色体。Xist 通过 A 重复区招募 SPEN 等蛋白抑制转录，并借 hnRNPK 等桥梁引入多梳抑制复合体沉积 H3K27me3、去除活跃修饰，随后 DNA 甲基化和 macroH2A 掺入把沉默状态固定下来。整条染色体被压缩成巴氏体，使雌雄个体的 X 连锁基因剂量趋于平衡，且失活选择在体细胞中被稳定克隆性遗传。
Early in female mammalian development, the Xist gene on one of the two X chromosomes is upregulated, and the long non-coding RNA it produces stays on its own chromosome and spreads over it in cis. Through its A-repeat region Xist recruits factors such as SPEN to shut down transcription, and via bridges including hnRNPK it brings in Polycomb complexes that deposit H3K27me3 and strip active marks, after which DNA methylation and macroH2A incorporation lock the silent state. The chromosome condenses into a Barr body, equalizing X-linked gene dosage between sexes, and the choice of which X is silenced is inherited clonally in somatic tissue.
难度 专业 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Urry 等,《Campbell Biology》第11版,Pearson,2017
L. A. Urry et al., Campbell Biology, 11th ed., Pearson, 2017
独立核验依据 · Independent verification: 核验依据 2 ISBN 978-0-13-409341-3 (11th ed., Pearson)
#Xist #X染色体失活 #剂量补偿 #多梳复合体
DNA 复制会把亲本核小体稀释一半，表观遗传信息如何在细胞分裂中被维持？
Replication dilutes parental nucleosomes by half, so how is epigenetic information maintained through cell division?
复制叉通过时旧组蛋白被大致对称地分配到两条子链，新合成的组蛋白由 CAF-1 等伴侣蛋白装配填补空位，形成旧新混合的核小体。带有原有修饰的旧组蛋白充当模板，通过读写偶联机制吸引相应的修饰酶，例如 PRC2 识别 H3K27me3 后在邻近核小体上继续甲基化，SUV39H1 与 HP1 在 H3K9me3 区域自我扩展。DNA 甲基化则依靠 DNMT1 与 UHRF1 识别半甲基化位点快速复原。若这些反馈在分裂后未及时完成，修饰模式可能被稀释或漂移，这正是表观遗传可塑性的来源。
As the fork passes, old histones are distributed roughly symmetrically to both daughter strands while newly made histones are deposited by chaperones such as CAF-1, producing nucleosomes of mixed old and new subunits. Old histones carrying pre-existing marks serve as templates, and read-write coupling recruits the matching enzymes, so PRC2 reading H3K27me3 methylates adjacent nucleosomes and SUV39H1 with HP1 propagates H3K9me3 domains. DNA methylation is restored quickly because DNMT1 with UHRF1 recognizes hemimethylated sites. When these feedback loops lag behind division, marks can be diluted or drift, which is precisely the origin of epigenetic plasticity.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Urry 等,《Campbell Biology》第11版,Pearson,2017
L. A. Urry et al., Campbell Biology, 11th ed., Pearson, 2017
独立核验依据 · Independent verification: 核验依据 2 ISBN 978-0-13-409341-3 (11th ed., Pearson)
#表观遗传维持 #核小体装配 #读写偶联 #DNMT1
表观遗传变化与 DNA 序列突变有什么本质区别？
What is the essential difference between epigenetic change and DNA sequence mutation?
序列突变改变了碱基本身的信息，几乎不可逆，会稳定地传给所有子代细胞与后代。表观遗传变化是在不改动碱基序列的前提下改变基因的可及性和表达状态，载体包括 DNA 甲基化、组蛋白修饰、染色质结构和非编码 RNA，它们可以随发育阶段和环境而重塑，也可被药物逆转。同一基因型在不同细胞类型中形成截然不同的表达程序，正是靠这套可读写、可擦除的调控层实现。
A sequence mutation changes the information in the bases themselves, is essentially irreversible and is transmitted stably to all descendant cells and offspring. Epigenetic change alters gene accessibility and expression without altering base sequence, carried by DNA methylation, histone modifications, chromatin architecture and non-coding RNAs, and it can be remodelled with developmental stage and environment or reversed by drugs. It is this writable and erasable layer that lets one genotype run completely different expression programmes in different cell types.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Urry 等,《Campbell Biology》第11版,Pearson,2017
L. A. Urry et al., Campbell Biology, 11th ed., Pearson, 2017
独立核验依据 · Independent verification: 核验依据 2 ISBN 978-0-13-409341-3 (11th ed., Pearson)
#表观遗传 #基因突变 #可逆性 #基因表达调控
DNA 甲基化是通过什么方式让基因沉默的？
By what means does DNA methylation silence a gene?
哺乳动物的 DNA 甲基化主要发生在 CpG 二核苷酸的胞嘧啶第五位碳上，由 DNMT3A、DNMT3B 建立，DNMT1 在复制时按半甲基化模板维持。启动子区 CpG 岛被甲基化后，一方面直接妨碍部分转录因子结合，另一方面招募含甲基结合结构域的蛋白如 MeCP2，后者带来组蛋白去乙酰化酶等抑制性复合体使染色质致密。基因体内部的甲基化则与转录延伸和可变剪接调控相关，作用并不一律是抑制。
In mammals methylation occurs mainly at the fifth carbon of cytosine in CpG dinucleotides, established by DNMT3A and DNMT3B and maintained by DNMT1 acting on hemimethylated templates during replication. Methylation of promoter CpG islands can block binding of some transcription factors directly and also recruits methyl-CpG-binding proteins such as MeCP2, which bring in repressive complexes containing histone deacetylases to compact chromatin. Methylation within gene bodies instead correlates with transcriptional elongation and alternative splicing, so the effect is not uniformly repressive.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Urry 等,《Campbell Biology》第11版,Pearson,2017
L. A. Urry et al., Campbell Biology, 11th ed., Pearson, 2017
独立核验依据 · Independent verification: 核验依据 2 ISBN 978-0-13-409341-3 (11th ed., Pearson)
#DNA甲基化 #CpG岛 #MeCP2 #染色质致密化
microRNA 是怎样识别并抑制靶基因表达的？
How does a microRNA recognize and repress its target genes?
初级转录本经 Drosha 和 Dicer 两步加工成约 22 个核苷酸的成熟 microRNA，装载进 Argonaute 蛋白形成沉默复合体。它主要依靠第 2 到第 8 位的种子序列与靶 mRNA 的 3 端非翻译区互补配对，招募去腺苷化和脱帽机器加速 mRNA 降解，同时抑制翻译起始。由于配对不需完全互补，一个 microRNA 可调控数百个靶基因，构成对基因表达网络的整体微调而非开关式关闭。
Primary transcripts are processed by Drosha and Dicer into mature microRNAs of about 22 nucleotides, which load into Argonaute to form a silencing complex. Recognition relies mainly on the seed region at positions two to eight pairing with the target three prime untranslated region, after which deadenylation and decapping machinery is recruited to speed mRNA decay while translation initiation is inhibited. Because pairing need not be perfect, one microRNA can tune hundreds of targets, producing network-wide fine adjustment rather than switch-like shutdown.
难度 进阶 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-11-04
来源 · Source
Bartel,《MicroRNAs: Target Recognition and Regulatory Functions》,Cell,2009 (DOI: 10.1016/j.cell.2009.01.002)
Bartel, 'MicroRNAs: Target Recognition and Regulatory Functions', Cell, 2009, DOI: 10.1016/j.cell.2009.01.002
独立核验依据 · Independent verification: 核验依据 1 Crossref DOI 10.1016/j.cell.2009.01.002
#microRNA #种子序列 #Argonaute #转录后调控
组蛋白修饰如何影响基因活性，为何被称为组蛋白密码？
How do histone modifications affect gene activity, and why are they described as a histone code?
组蛋白 N 端尾部可被乙酰化、甲基化、磷酸化和泛素化等多种修饰，不同组合与转录状态高度相关：H3K27ac 和 H3K4me3 标记活跃启动子，H3K4me1 富集于增强子，而 H3K27me3 与 H3K9me3 分别对应多梳蛋白抑制和异染色质。这些修饰一方面改变组蛋白与 DNA 的静电作用影响染色质紧密度，另一方面被含 bromo、chromo 等结构域的效应蛋白读取并招募下游复合体。因为组合而非单个标记决定结果，这套读写系统被形象地称为组蛋白密码。
Histone N-terminal tails carry acetylation, methylation, phosphorylation and ubiquitination, and particular combinations track transcriptional state: H3K27ac and H3K4me3 mark active promoters, H3K4me1 is enriched at enhancers, while H3K27me3 and H3K9me3 correspond to Polycomb repression and heterochromatin. Such marks alter the electrostatic interaction between histones and DNA and thus chromatin compaction, and they are read by effector proteins with bromo or chromo domains that recruit downstream complexes. Because outcomes depend on combinations rather than single marks, this writer-reader system is called a histone code.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-10-04
来源 · Source
Bannister AJ, Kouzarides T. Regulation of chromatin by histone modifications. Cell Research (2011) 21:381-395
Bannister AJ, Kouzarides T. Regulation of chromatin by histone modifications. Cell Research (2011) 21:381-395
独立核验依据 · Independent verification: 核验依据 1 DOI 10.1038/cr.2011.22
#组蛋白修饰 #组蛋白密码 #染色质状态 #效应蛋白
基因组印记是什么，为什么同一段染色体缺失会导致两种不同疾病？
What is genomic imprinting, and why can a deletion in the same chromosomal region cause two different diseases?
基因组印记指某些基因的表达取决于它来自父方还是母方，由生殖细胞中建立的差异甲基化区域和相关染色质状态决定，因此这些位点只有单个亲本等位基因活跃。人 15 号染色体 q11-q13 区含有父源表达和母源表达的印记基因簇：父源片段缺失导致普拉德-威利综合征，母源片段缺失或 UBE3A 失活则导致安格尔曼综合征。同一区域的缺失因亲本来源不同而表现为完全不同的临床表型，是印记存在的直接证据。
Genomic imprinting means that expression of certain genes depends on whether the allele came from the father or the mother, set by differentially methylated regions and associated chromatin states established in the germline, so only one parental allele is active. The human 15q11-q13 region contains clusters of paternally and maternally expressed imprinted genes: loss of the paternal segment causes Prader-Willi syndrome, whereas loss of the maternal segment or inactivation of UBE3A causes Angelman syndrome. That one deletion yields two distinct phenotypes depending on parental origin is direct evidence for imprinting.
难度 进阶 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Urry 等,《Campbell Biology》第11版,Pearson,2017
L. A. Urry et al., Campbell Biology, 11th ed., Pearson, 2017
独立核验依据 · Independent verification: 核验依据 2 ISBN 978-0-13-409341-3 (11th ed., Pearson)
#基因组印记 #差异甲基化区域 #普拉德-威利综合征 #安格尔曼综合征