The genetic code Second letter of codon C G 1965年由 UUU Phe UCU Ser UAU Tyr UGU Cys Crick绘制的 UUC Phe UCC Ser UAC TyrUGC Cys U 密码表,UGA UUA Leu UCA Ser UAA Stop UGA Stop UAA和UAG因 UUG Leu UCG Ser UAG Stop UGG Trp 为没有对应的 CUU Leu CCu Pro CAU His CGU Arg CUC Leu CCc Pro CAC His CGC Arg 氨基酸而导致 C CUA CCA Pro CAA GInCGA Arg 肽链合成终止, First CUG CCG Pro CAG GIn CGG Arg letter of 因此定义为终 codon AUU Acu Thr AAu Asn AGu Ser AUC ACC Thr AAC Asn AGC Ser 止密码子(stop5end)A codon ),而从 AUA AUG Met ACA Thr AAA Lys AGA Arg ACG Thr AAG LysAGG Arg 蛋白质的N未 GUU Val GCU Ala GAU Asp GGU Gly 端测定的数据 GUC Val GCc Ala GAC Asp GGC Gly 表明Me对应 GUa Val Gca Ala GAa Glu GGA Gly GUG Val GCG Ala GAG Glu GGG Gly 的是起始密码 子 Reading frame 1 5'--G U AA G UA A GU A AG U AA G UA A---3 Reading frame 2 GUA AG U AA G UA A GU A AG U AA Reading frame 3 G UA A GUA AG U AJA G UA A GU AA
1965年由 Crick绘制的 密码表，UGA, UAA和UAG因 为没有对应的 氨基酸而导致 肽链合成终止， 因此定义为终 止密码子(stop codon), 而从 蛋白质的N末 端测定的数据 表明Met对应 的是起始密码 子。 The genetic code
The present genetic code is universal in all organisms on the earth 唯独在线粒体、支原体等中有少量变化 table I Known Variant Codon Assignments in Mitochondria Codons AGA UGA AUA AGG CUN CGG Normal code assignment Stop lle Arg Leu Arg Animals vertebrates Met Stop Drosophila Tr Met Yeasts Saccharomyces cerevisiae酿酒酵母Tr Met Torulopsis glabrata光滑球母 Tr Met Thr Schizosaccharomyces pom rp Filamentous fungi丝状真菌彭贝裂殖酵母rp Trypanosomes锥虫 T Higher plants Trp Chlamydomonas reinhardtii莱茵衣藻 A question mark Indicates that the codon has not been observed in the indicated mitochondrial genome; N, any nucleotide: + the codon has the same meaning as in the normal code
The present genetic code is universal in all organisms on the earth 唯独在线粒体、支原体等中有少量变化 酿酒酵母 光滑球酵母 丝状真菌彭贝裂殖酵母 锥虫 莱茵衣藻
密码子的简并 ble 27-4 ( degeneracy)是预 Degeneracy of the Genetic Code Amino acid Number of codons 料之中的事。事实上 Ala Arg 所对应的密码子的多 As sn Asp 少与该氨基酸在生物 Cys GIn 中的利用度成比例。 Gly Per cent of residues in protein His Met Phe Pro Pro Arg 4622222423621246412 Asn Ser Phe Thr TrD Cys TI 3
密码子的简并 (degeneracy)是预 料之中的事。事实上 所对应的密码子的多 少与该氨基酸在生物 中的利用度成比例
并不是一个tRNA 对应一个密码子, tRNA 般只有三十几种 tRNA来对应61个 密码子,因此需有 个机制让tRNA偷 懒”。这就是 Crick Anticodon 的 Wobble假说 mrna 5 A U C 3 123 Codon 321 321 Anticodon (3’)G-C-I G-C-I G-C-I (5) 二二 二 Codon(5’)C-G-A C-G-U C-G-C(3 123 123 123 (b)
并不是一个tRNA 对应一个密码子， 一般只有三十几种 tRNA来对应61个 密码子，因此需有 个机制让tRNA“偷 懒”。这就是Crick 的Wobble假说
table27-5 How the wobble base of the anticodon determines the number of Codons a tRNa Can recognize* I. One codon recognized Anticodon (3)X-Y-c(5′) (3)X-Y-A(5) Codon (5)Y-X-G(3) (5)Y-X-U(3”) 2. Two codons recognized Anticodon (3)X-Y-U(5) (3)X-Y-G(5") Codon (5)Y-X-4(3) (5)Y-X-6(3) 3. Three codons recognized Anticodon (3)X-Y-1(5) Cod (5)Y-X-u(3) X and Y denote complementary bases capable of strong Watson-Crick base pairing with each other. The bases in the wobble positions-the 3'position of codons and 5 position of anticodons-are shaded in red