Biochemistry Li Enmin

有关我院医学本科生基础学习模块生物化学部分相关教师授课期间考察学生的决定以及实施参考意见 为了教师在授课期间，加强对教学过程的有效管理与灵活操作，充分调动学生学习生物化学课程的积极性与主动性，经征求科教处同意，决定在生物化学授课期间由授课教师对学生的学习情况实施考察，并对每一个听课学生记录一个成绩，并将该成绩最终记入期末总成绩中。

为了此项考查的客观公证性，使该举措真正发挥作用。特提出如下操作规则 无故不来上课者扣分。 上课时，未经请假提前早退者扣分。 上课时，影响教师授课氛围者扣分。 上课迟到者扣分。 对上课讨论时主动发言，积极提问或出色完成授课教师所布置的作业或其它相关学习任务，且无上述不良情况者,每位教师根据具体情况可给予这样的学生鼓励加分并记入总成绩，但最终总成绩以100分为限。

10. DNA replication* 11. RNA transcription* 12. Protein biosynthesis* 13. Regulation of gene expression** 14. Gene recombination and gene engineering

DNAs RNAs Proteins 10 DNA replication: entirety 11 RNA transcription: systematicness 12 Protein biosynthesis: individuality 10 12 11

Chapter ten DNA replication

Section 1* General features of DNA replication Section 2* Enzymes in DNA replication Section 3 Process of DNA replication Section 4 Reverse transcription and others Section 5* DNA damage and repair

Concept of DNA replication or DNA biosynthesis  The transmission of the genetic informations between DNA and DNA  The synthesis of a complementary DNA strand by forming phosphodiester linkages between nucleotides base-paired to template strand is catalyzed by large multienzyme complexes referred to as the DNA polymerases  Semiconservative replication

Section 1 General features of DNA replication 1) Experimental basis and the significance of semiconservative replication* 2) DNA replication is Bidirectional 3) Semidiscontinuous pattern of DNA replication

1) Experimental basis and the significance of semiconservative replication

A celebrated scientific experiment

total conserve commingle semi conserve parent DNA strand daughter DNA molecules parent DNA parent DNA strand Daughter or new DNA total conserve semi conserve commingle

The basic experiment condition A experimental table A centrifuge A kind of bacterium A few 14NH4Cl culture solutions A few 15NH4Cl culture solutions A few sucroses 7N14,15 / 1S22S22P3

The experiment procedure bacterium15 culture extraction bacterium DNA centrifugation DNA layer 14N 1 2 3 4

total conserve L L+H 15 H 14

14/15 15 L M M+L H semi conserve 14

The experiment procedure bacterium14 culture extraction bacterium DNA centrifugation DNA layer 15N 1 2 3 4

14/15 14 15 H L M M+H semi conserve

Genetic conservativeness Significance of semiconservative replication Genetic conservativeness ACGT TGCA

Section 2 Enzymes in DNA replication

Chemical reaction in DNA replication (dATP)m + (dCTP)n + (dGTP)y + (dGTP)z (dAMP + dCMP + dGMP + dGMP) m +n+ y+z + (m + n + y + z) PPi

single strand binding protein topoisomeraseⅡ helicase A C A G T G A T C A G A T G T C T C T 5’ O P P P H PP P HO O H 3’ A 3’ O OH G helicase DNA polymerase Ⅲ DNA polymerase Ⅰ RNase primase ligase single strand binding protein ……dGTP/dTTP/dCTP/dTTP

DNA polymerases prokaryote polymeras I polymeras II polymeras III* eukaryote polymeras * polymeras  polymeras  polymeras * polymeras  prokaryote polymeras I polymeras II polymeras III* veriest replicase substitute major action proof read, repair, filling in mitochondria DNA polymerases

DNA polymerase III of E. coli           ’   ’         DNA polymerase III of E. coli

DNA polymerase I of E. coli G I 1 109kD, 18  Helix 2 A-F: 323 aa, small fragment, 5’3’ exonuclease activity 3 G-R/c-terminal: 604 aa, Klenow fragment , 3’ 5’ exonuclease activity, 5’ 3’ polymerase activity, about 20 nucleotide. 50 aa H O E C D B R N P L M Q K J F

DNA polymerases function--1 5’3’ polymerase activity + N P OH3' 5'    3’ 5’ PPi DNA polymerases function--1 5’3’ polymerase activity

* DNA polymerases function--2 exonuclease activity Polymerase I G 5’ 3’ Polymerase I Polymerase II Polymerase III Proof reading cut the fragment of primer and mutated fragments * DNA polymerases function--2 exonuclease activity C

high fidelity of DNA replication 5’ 3’ (1) The strict base complementary high fidelity of DNA replication

(2) the character of DNA polymerase selecting base The conformation of DNA polymerase III is changeable, as its affinity to nucleotide acids (2) the character of DNA polymerase selecting base A C G T 5’ 3’

(3) The proof reading function of DNA polymerase 5’ 3’ T C G proof reading

DNA helicase DNA topoisomerase DNA single strand binding protein

helicase Dna A, Dna B (rep), Dna C……Dna X Dna B Dna C Dna A Dna T

DNA topoisomerase normal helix helicase positive superhelix 1 2 3 4 5 6 7 8 9 10 11 helicase DNA polimerase positive superhelix negative superhelix 1 2 3 4 5 6 7 8 9 10 11 topoisomerase normal helix DNA topoisomerase

topoisomerase I: topoisomerase II: untwisting cut ligate breaks single strand DNA, not requires ATP topoisomerase II: breaks double strands DNA, requires ATP cut ligate

single stranded DNA-binding protein, SSB (1) SSB is consist of 177 amino acid residues in E.coli (2) tetrapolymer (3) a SSB= 32 nucleotides (4) Cooperative binding

Dna B SSB bank The role of SSB

the actions of primer and primase 3’ 5’ DNA polymerase III New DNA strand Parental DNA template primase SSB

Primase and Primosome different with that in the transcription.  The primase is a RNA polymerase, which is different with that in the transcription.  The primase synthesizes the primers, which is used in DNA synthesis  The primer is a fragment of RNA  The length of primer is 10-20bp approximately  The helicases and other replicated factors have binded with DNA, then the primase have binds with them and formed a primosome at last.

3’ 5’ ATP ADP+Pi ligase OH P

The compare of several enzymes, which in formation of phosphodiester bond enzymes provides 3’-OH provides 5’-P results 1 DNA primer or dNTP (dNMP) n+1 polymerase extending DNA strand 2 DNA no continued two single strands no continue ligase  continue 3 DNA to break and put in order two put in order to DNA topoisomerase DNA single strands superhelical structure 4 primase extending primer NTP primer

helicase primase single strand binding protein DNA polymerase Ⅲ RNase A C A G T G A T C A G A T G T C T C T 5’ O P P P H PP P HO O H 3’ A 3’ O OH G helicase DNA polymerase Ⅲ DNA polymerase Ⅰ RNase primase ligase single strand binding protein ……dGTP/dTTP/dCTP/dTTP

Preparation First, you will read the summary in every chapters. Second, To the best of your as much as, to understand the tables and the figures in every chapters. Last, Find out the problems in them.

The differences between the prokaryote and the eukaryote in DNA replication ?

Section 3 Process of DNA replication

M G1 G2 S cell cycle initiation extension termination 1 2 3 4

Initiation of DNA replication Prokaryotic cell ( for instance E.coli) 1 A fixed origin (ori) 2 Bidirectional replication Eukaryotic cell a lot of origin

theta Eukaryotic cell Prokaryotic cell origin D 3’ Direction O 2 1 21 2 1 O 2 1 21 3’ Direction D Prokaryotic cell Eukaryotic cell

leading strands Origin Helicase, SSB ... 1 2 lagging strands

The oriC of E coli.   palindrome structure DnaA DnaB DnaC DnaT 1 13 17 29 32 44 55 66 166 174 201 209 237 245 TGTGGATTA---TTATACACA------TTTGGATAA---TTATCCACA GATCTNTTTATTT---GATCTNTTNTATT---GATCTCTTATTAC The oriC of E coli. palindrome structure DnaA DnaB DnaC DnaT

The extension of DNA replication rate of extension 1 prokaryotic cell, 2500 bp/s。 2 eukaryotic cell a lot of origin

single strand binding protein polymerase III holoenzyme primer leading strand lagging strand polymerase III  subunit 5’ 3’ topoisomerase single strand binding protein

the termination of DNA replication the terminate point of E.coli and SV40 virus origin terminate point E.coli SV 40 82% 32% 0% 50%

the role of several enzyme in termination of DNA replication Ligase 3’ 5’ Pol I OH P RNase Parental DNA template strand RNA primer New DNA strand the role of several enzyme in termination of DNA replication degradation ligation filling

O origin 3’ D direction leading strand lagging strand

Okazaki fragments The 100--2000 bp Okazaki fragments was formed in DNA lagging strand in DNA replication .

Rolling circle replication 1 3’ OH 5’ 3 5 8 10 2 3’5’ 9 4 3’ 6 7

telomere and telomerase of eukaryotes structure of telomere TTAGGGTTAGGG……TTAGGGTTAGGG sequence of telomere

The end of linear DNA in replication 5’

structure of telomerase 3’ 5’ CCCUAA GGGATTGGGATT RNA 5’ 3’ RTase TP1

mechanism of telomerase 3’ 5’ CCCUAA GGGATTGGGATT mechanism of telomerase TTGGGATT 1 2 3 4

3’ 5’ CCCUAA TTGGGATT 5 n 1 2 3

RNA-DNA double strands Section 4 Reverse transcription and others single RNA strand RTase RNA-DNA double strands RNase H single DNA strand RTase DNA double strands

RNA 3’ 5’ RTase RNase H integration

? synthesize cDNA from mRNA S1 nuclease mRNA DNA polymerase I reverse AAA…AAA 5’ 3’ mRNA TTT...TTT primer: oligo dT reverse transcriptase basic hydrolysis ? DNA polymerase I S1 nuclease

significance of reverse transcription ?

Section 5 DNA damage and repair

DNA damage mutations They are heritable permanent changes in the base sequence of DNA. DNA damage

the effect of mutation genotype, and no or not detectable 3 Lethal 4 Only lost some function 1 Useful 2 There may be some effect to the genotype, and no or not detectable effect to the phenotype

cause of mutation Induced mutation 1 spontaneous mutation mechanism? frequency 1/109 2 physical 3 chemical 4 biological Induced mutation

mutational types 1 mismatch / point mutation 2 deletion 3 insertion 4 rearrangement / inversion

point mutation 1) transition 2) transversion

transition A given pyrimidine is changed to the other pyrimidine, or a given purine is changed to the other purine.

transition C T T C G A A G C G T 1 A 2 3 4

Sequence of NGAL gene in SHEEC cell 223 AG, 75 IV ATG CCC CTA GGT CTC CTG TGG CTG GGC CTA GCC CTG TTG GGG GCT CTG CAT GCC CAG GCC CAG GAC TCC ACC TCA GAC CTG ATC CCA GCC CCA CCT CTG AGC AAG GTC CCT CTG CAG CAG AAC TTC CAG GAC AAC CAA TTC CAG GGG AAG TGG TAT GTG GTA GGC CTG GCA GGG AAT GCA ATT CTC AGA GAA GAC AAA GAC CCG CAA AAG ATG TAT GCC ACC G*TC TAT GAG CTG AAA GAA GAC AAG AGC TAC AAT GTC ACC TCC GTC CTG TTT AGG AAA AAG AAG TGT GAC TAC TGG ATC AGG ACT TTT GTT CCA GGT TGC CAG CCC GGC GAG TTC ACG CTG GGC AAC ATT AAG AGT TAC CCT GGA TTA ACG AGT TAC CTC GTC CGA GTG GTG AGC ACC AAC TAC AAC CAG CAT GCT ATG GTG TTC TTC AAG AAA GTT TCT CAA AAC AGG GAG TAC TTC AAG ATC ACC CTC TAC GGG AGA ACC AAG GAG CTG ACT TCG GAA CTA AAG GAG AAC TTC ATC CGC TTC TCC AAA TCT CTG GGC CTC CCT GAA AAC CAC ATC GTC TTC CCT GTC CCA ATC GAC CAG TGT ATC GAC GGC TGA Sequence of NGAL gene in SHEEC cell 223 AG, 75 IV

transvertion Transversions are changes from a pyrimidine to either of the two purines, and or the changes of a purine into either of the two pyrimidines.

C G C A T G T A transversion C G 1 A 2 T 3 4 A C A T G C G T 5 6 7 8

HbA  HbS  gene peptide Glu Val CTC CAC GAG GTG 17TA, 6GluVal

point mutation transversions transitions C A C G T A T G A C A T G C C T T C A G G A

deleted mutation

inserted mutation

inversed mutation 5’ 3’ OH P

rearrangement  1 2   Hb  genes family in 11 chromosome Hb anti-Lepore Hb Lepore two gene types of Med anaemia

repair of DNA damaged 1 photoreactivation repair 2 excision repair 3 recombination repair 4 SOS repair

Photoreactivation repair CH3 UV photolyase 300-600nm

excision repair 5’ affect factor pol 1 3’ ligase specific endo- nucleases 3’ 5’ ligase pol 1 Uvr A Uvr B Uvr C

recombination repair pol I ligase RecA

SOS repair regulation.  The DNA is damaged gravely and extensively.  The replication of DNA had been stoped.  The SOS repair is a emergency step.  The SOS repair is a complicated network of regulation.  The rate of DNA mutation is still very high.  The SOS repair is called error prone repair.

1、DNA复制时，下列哪一种酶是不需要的？ 练 习 题 一 1、DNA复制时，下列哪一种酶是不需要的？ A DNA指导下的DNA聚合酶 B DNA连接酶 C 拓扑异构酶 D 解链酶 E 限制性内切酶

2、合成DNA的原料是 A dNMA B dNDP C dNTP D NTP E NMP

3、真核细胞进行DNA复制的部位是 原核细胞进行DNA复制的部位是其拟核区 B 细胞浆 C 细胞核 D 核蛋白体 E 微粒体 原核细胞进行DNA复制的部位是其拟核区

4、DNA复制中的引物是 A 由DNA为模板合成的DNA片段 B 由DNA为模板合成的RNA片段 C 由RNA为模板合成的DNA片段 D 由RNA为模板合成的RNA片段

5、DNA复制时，模板序列 5’-TAGA-3’ 将合成哪种互补序列？ A 5’-TCTA-3’ B 5’-ATCT-3’ C 5’-UCUA-3’ D 5’-GCGA-3’ E 3’-TCTA-5’

6、关于DNA复制中DNA聚合酶的错误说法是 A 底物是dNTP B 必须有DNA模板 C 合成方向是5’ 3’ D 需要Mg2+参与 E 需要ATP参与

7、下列关于大肠杆菌DNA聚合酶 的叙述哪一项是正确的 A 具有3’ 5’外切酶活性 B 不需要引物 C 需要四种不同的三磷酸核苷 D dUTP是它的一种作用底物 E 可以将两个DNA片段连接起来

8、DNA连接酶： A 使DNA形成超螺旋 B 使DNA链 上切口的两个末端连接起来 C 合成引物 D 将双螺旋解链 E 去除引物，填补空缺

9、DNA复制需要 (1) helicase，(2) primase， (3) DNA polymerase，(4) DNA topoisomerase， (5) DNA ligase。其作用顺序是： A (1) (5) (2) (4) (3) B (1) (2) (3) (4) (5) C (1) (2) (4) (3) (5) D (2) (1) (3) (4) (5) E (3) (1) (4) (5) (2) F (4) (1) (2) (3) (5)

10、Okazaki 片段是 A DNA模板上的DNA片段 B 引物酶催化合成的RNA片段 C 随从链上合成的DNA片段 D 领头链上合成的DNA片段

11、DNA复制时，子链的合成方向是 A 一条子链 5’ 3’，另一条子链 3’ 5’ B 两条子链均为 5’ 3’ C 两条子链均为 3’ 5’

12、镰刀状红细胞性贫血，其  链有关的突变是 A insertion B deletion C inversion D point mutation transition ? transversion ?

13、DNA复制的特点是 A 半保留复制 B 需合成RNA引物 C 形成复制叉 D 有半不连续性 E 形成Okazaki 片段 F 多点复制 G 双向复制

14、DNA复制时，连续合成的链为 ， 领头链 而不连续合成的链为 。 随从链 15、DNA合成的原料是 。 dNTP 16、DNA复制时，子链DNA合成的方向为 。 5’ 3’ 17、DNA复制时，亲代模板链与子代合成链之间碱 基配对的原则是：A与 配对，C与 配对。 T G

18、DNA复制时，改变模板DNA超螺旋结构的酶是 拓扑异构酶 解链酶 19、DNA的半不连续复制是指 的不连续合 成和 的连续合成。 随从链 领头链 20、DNA的半保留复制是指复制所生成的子代DNA 分子中，其中有一条链是 ，而 另一条是 。 来自亲代DNA 新合成的