全部书籍

专业

应用 关于 Coach 登录 开始阅读

书籍暂未推出简体中文版——当前显示英文版。

Biology · 术语表

什么是 Double-strand break repair？

Definition 3.7 University Biology — Year 3 · 第 3 章 — Genome Stability: DNA Damage, Repair and Recombination

A double-strand break severs the chromosome and has no intact strand to copy. Two pathways repair it. Non-homologous end joining (NHEJ): the Ku70–Ku80 ring binds each end, recruits the kinase DNA-PKcs, nucleases trim overhangs, and ligase IV joins the ends; it works at any stage of the cycle and in minutes, but loses or adds a few nucleotides at the junction and can join the wrong ends. Homologous recombination (HR): the MRN complex and nucleases resect the 5′' strands to leave long 3′' single-stranded tails, which are coated by RPA and then, with the help of BRCA2, by the recombinase Rad51; the Rad51 filament searches for a homologous duplex — the sister chromatid, in S and G2 — invades it, and the invading 3′' end primes synthesis on the intact copy. The joint molecules may be dissolved after a short synthesis (most mitotic repair, no exchange of flanking DNA) or mature into two four-way Holliday junctions, whose resolution by endonucleases yields either a crossover or a non-crossover product. HR is accurate because it copies a sister, but is available only when a sister exists.

The two fates of a double-strand break. Left: end joining, fast and always available, at the cost of a small change at the junction. Right: homologous recombination, which resects the ends, invades the sister chromatid with a Rad51 filament and copies what was lost.
The two fates of a double-strand break. Left: end joining, fast and always available, at the cost of a small change at the junction. Right: homologous recombination, which resects the ends, invades the sister chromatid with a Rad51 filament and copies what was lost.

示例

Example 3.9 (Why xeroderma is a disease of the fork)

A dose of sun leaves N=105N = 10^{5} pyrimidine dimers in a keratinocyte. With a nucleotide excision half-life of 2 h2\,\mathrm{h} (k=0.35 h−1k = 0.35\,\mathrm{h}^{-1}) and a fork arriving after 8 h8\,\mathrm{h}, Ne−kT=105×e−2.8≈6000N e^{-kT} = 10^{5}\times e^{-2.8} \approx 6000 dimers are still there to be copied; with the near-absent repair of an XP cell (k≈0.01 h−1k \approx 0.01\,\mathrm{h}^{-1}), 92 00092\,000 are. Each dimer met by the fork is bypassed by translesion synthesis (below), which is error-prone: the patient’s mutation load per division is fifteen times the normal one, and the skin cancers follow in childhood. The cure that works is to keep NN small — total avoidance of ultraviolet light.

Example 3.13 (Synthetic lethality: BRCA and PARP)

Women who inherit one defective copy of BRCA1 or BRCA2 have a lifetime risk of breast cancer of 50 to 80 %50\text{ to }80\,\%; the tumours arise in cells that have lost the second copy and can no longer do homologous recombination. Such cells repair their double-strand breaks by end joining alone and accumulate rearrangements. They also acquire a specific weakness. Single-strand breaks, some 10 00010\,000 a day, are repaired by a route that needs the enzyme PARP; when PARP is inhibited by a drug, single-strand breaks persist to S phase, where a fork converts each into a double-strand break with only one end — repairable only by recombination. A normal cell, with one good BRCA allele, copes; the tumour cell dies. Two defects, each harmless alone, are lethal together: the drug kills by synthetic lethality, and spares the patient’s other cells.

在原文中阅读 →

相关术语