如何找到密码子的特定频率? [英] How to find specific frequency of a codon?

问题描述

我正在尝试在R中创建一个函数，该函数可以计算每个密码子的频率. 我们知道蛋氨酸是一种氨基酸，它只能由一组密码子ATG形成，因此它在每组序列中的百分比为1.而甘氨酸可以由GGT，GGC，GGA，GGG形成，因此，每个密码子将为0.25. 输入将是DNA序列，如-ATGGGTGGCGGAGGG，并且借助密码子表，它可以计算输入中每​​次出现的百分比.

I am trying to make a function in R which could calculate the frequency of each codon. We know that methionine is an amino acid which could be formed by only one set of codon ATG so its percentage in every set of sequence is 1. Where as Glycine could be formed by GGT, GGC, GGA, GGG hence the percentage of occurring of each codon will be 0.25. The input would be in a DNA sequence like-ATGGGTGGCGGAGGG and with the help of codon table it could calculate the percentage of each occurrence in an input.

请通过建议实现此功能的方法来帮助我.

please help me by suggesting ways to make this function.

例如， 如果我的论点是ATGTGTTGCTGG 那么，我的结果将是

for example, if my argument is ATGTGTTGCTGG then, my result would be

ATG=1
TGT=0.5
TGC=0.5
TGG=1

---

R的数据:

---

Data for R:

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")

推荐答案

首先，我得到了查询列表和序列.

First, I get my lookup list and sequence.

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")

MySeq <- "ATGTGTTGCTGG"

接下来，我加载stringi库并将序列分成三个字符的块.

Next, I load the stringi library and break the sequence into chunks of three characters.

# Load library
library(stringi)

# Break into 3 bases
seq_split <- stri_sub(MySeq, seq(1, stri_length(MySeq), by=3), length=3)

然后，我用table计算这三个基本块对应的字母.

Then, I count the letters that these three base chunks correspond to using table.

# Get associated letters
letter_count <- table(unlist(codon[seq_split]))

最后，我将序列与计数的倒数绑定在一起，并重命名我的数据框列.

Finally, I bind the sequences together with the reciprocal of the count and rename my data frame columns.

# Bind into a data frame
res <- data.frame(seq_split,
                  1/letter_count[match(unlist(codon[seq_split]), names(letter_count))])

# Rename columns
colnames(res) <- c("Sequence", "Letter", "Percentage")

#  Sequence Letter Percentage
#1      ATG      M        1.0
#2      TGT      C        0.5
#3      TGC      C        0.5
#4      TGG      W        1.0

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