DNA replication and repair
Introduced| English | Meaning |
|---|---|
| replication fork | replication fork |
| leading strand/ˈliːdɪŋ strænd/ | The new DNA strand synthesised continuously in the same overall direction as a replication fork opens. |
| lagging strand/ˈlæɡɪŋ strænd/ | The new DNA strand synthesised discontinuously in short fragments as a replication fork opens. |
| Okazaki fragment | Okazaki fragment |
| proofreading | proofreading |
| mismatch repair | mismatch repair |
| thymine dimer | thymine dimer |
| nucleotide excision repair | nucleotide excision repair |
Six billion letters, almost no typos
- Every division, a cell copies 6.4 × 10⁹ letters of DNA, and only a handful of mistakes get through.
- The copying machine and the repair crews behind it are where inherited disease, mutation and most cancer begin.
The replication fork
- At the replication fork 复制叉, helicase unzips the double helix. Polymerase can only build 5′→3′, and the two template strands run in opposite directions.
- So the leading strand 前导链 is made in one piece, while the lagging strand 后随链 is made backwards in short Okazaki fragments 冈崎片段, each started by a primer and joined by ligase.
Why is there a lagging strand?
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Antiparallel templates and a 5′→3′-only enzyme mean one strand must be made in short pieces, backwards from the fork.
Why one start point is not enough
- Human S phase copies 6.4 × 10⁹ bp in about 8 h. Two forks from one origin at 50 bp/s would need $6.4\times10^{9}/100 = 6.4\times10^{7}$ s, about 2 years.
- About 30 000 origins fire in timed groups: $6.4\times10^{9}/(100\times3\times10^{4}) \approx 2100$ s, about 35 min of pure copying, spread across S phase.
A 1.2 × 10⁸ bp chromosome must be copied by forks moving at 50 bp/s each. With 2 forks per origin, how many origins are needed to finish in 1 hour of pure copying?
Difficulty 2/5 · 140 coins + a small bonus on first completion today
Each origin copies 100 bp/s, so 3.6 × 10⁵ bp per hour. 1.2 × 10⁸ / 3.6 × 10⁵ ≈ 333 origins.
The fidelity ladder
- Raw polymerase makes about one error per 10⁵ bp. Proofreading 校对功能 catches 99 in 100 of these: one per 10⁷.
- Mismatch repair 错配修复 catches 99 in 100 of what is left: one per 10⁹ bp. Three layers, one error in a billion.
Where does the final rate of one error per 10⁹ bp come from?
Difficulty 1/5 · 100 coins + a small bonus on first completion today
One per 10⁵ raw, one per 10⁷ after proofreading, one per 10⁹ after mismatch repair: three systems, one error in a billion.
Put the error rates in the order a new base meets them, starting with the raw polymerase.
Difficulty 3/5 · 200 coins + a small bonus on first completion today
Each layer removes about 99 in 100 of the errors the layer before let through.
Worked example: the mutation budget
- After all three layers: $6.4\times10^{9}\times10^{-9} \approx 6$ new mutations in each daughter cell. Genetics and oncology both build on this number.
- Without mismatch repair the rate is 100 times higher, about 640 per division. That is the mechanism of hereditary non-polyposis colon cancer (Lynch syndrome).
Losing mismatch repair raises the new mutations per division to about —
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Without mismatch repair the rate stops at one per 10⁷: 6.4 × 10⁹ / 10⁷ ≈ 640 mutations per division, the Lynch syndrome mechanism.
Worked example: sunlit skin without repair
- A (fictional) patient blisters after minutes in the sun and has skin cancers by twenty. Ultraviolet light joins neighbouring thymines into thymine dimers 胸腺嘧啶二聚体.
- Nucleotide excision repair 核苷酸切除修复 cuts them out; without it every sunbeam leaves mutations behind. Each repair pathway owns one kind of damage, and losing it names one syndrome.
Match each repair system to the damage it handles.
Difficulty 2/5 · 140 coins + a small bonus on first completion today
Each pathway owns one kind of damage, which is why losing one gives one syndrome (Lynch syndrome, xeroderma pigmentosum).
Three common slips
- "Polymerase builds 3′→5′." It moves 3′→5′ along the template while building the new strand 5′→3′.
- "One copy from one start." Without about 30 000 origins, S phase would take years.
- Quoting one in 10⁹ as the polymerase's own error rate. That is the rate after proofreading and mismatch repair.
After one round of replication, each daughter DNA molecule has one old strand and one new strand.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Replication is semiconservative; each old strand serves as the template for one new strand.
Recap
- Semiconservative copying, 5′→3′ only: one leading and one lagging strand.
- Fidelity: one per 10⁵, then 10⁷, then 10⁹; about 6 mutations survive each division.
- About 30 000 origins turn 2 years into 8 hours; each repair pathway owns one kind of damage.