Question

我有一个DNA序列作为我的论点。

sequence<-c("ATGAATTTTGATTTA")

我想找到ATG重复和其他64个密码子的次数，编码特定氨基酸的64个密码子是

codon <- list(ATA = "I", ATC = "I", ATT = "I", ATG = "M", ACA = "T", 
              ACC = "T", ACG = "T", ACT = "T", AAC = "N", AAT = "N", AAA = "K", 
              AAG = "K", AGC = "S", AGT = "S", AGA = "R", AGG = "R", CTA = "L", 
              CTC = "L", CTG = "L", CTT = "L", CCA = "P", CCC = "P", CCG = "P", 
              CCT = "P", CAC = "H", CAT = "H", CAA = "Q", CAG = "Q", CGA = "R", 
              CGC = "R", CGG = "R", CGT = "R", GTA = "V", GTC = "V", GTG = "V", 
              GTT = "V", GCA = "A", GCC = "A", GCG = "A", GCT = "A", GAC = "D", 
              GAT = "D", GAA = "E", GAG = "E", GGA = "G", GGC = "G", GGG = "G", 
              GGT = "G", TCA = "S", TCC = "S", TCG = "S", TCT = "S", TTC = "F", 
              TTT = "F", TTA = "L", TTG = "L", TAC = "Y", TAT = "Y", TAA = "stop", 
              TAG = "stop", TGC = "C", TGT = "C", TGA = "stop", TGG = "W")

然后，我想计算形成特定氨基酸的密码子的百分比，并希望以下列方式得出。

codon   count   amino_acids  percentage
CTC     19666       L           0.18
CTT     27340       L           0.13
CTA     31534       L           0.20
CTG     76644       L           0.49

请帮我解决这个问题。

Answer 1

只要您的密码子没有任何移位或间隙，

sequence<-c("ATGAATTTTGATTTAATG")

#split into 3-character codons
splitseq<-substring(sequence, seq(1, nchar(sequence)-1, 3), seq(3, nchar(sequence), 3))

[1] "ATG" "AAT" "TTT" "GAT" "TTA" "ATG"

#table them to get the frequency
x<-as.data.frame(table(splitseq))

#match up codon translation
x$codon<-codon[match(x$splitseq, names(codon))]

#get percentage
x$percentage<-x$Freq / sum(x$Freq)

x
splitseq Freq codon percentage
1      AAT    1     N  0.1666667
2      ATG    2     M  0.3333333
3      GAT    1     D  0.1666667
4      TTA    1     L  0.1666667
5      TTT    1     F  0.1666667

如何计算字符串中出现字符的次数？

1 个答案: