我正在尝试将一些 png 图形添加到文档中。图形部分本身可以编译，文档的其余部分也可以。结合起来，它总是在第一次出错，但有时只需重新粘贴文本并重新编译即可。我在 Windows 8 上使用 texworks 和 miktex。
我收到的错误消息是“扫描使用 \end 时文件结束”。
这是完整的文档。
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\usepackage{amsmath}
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\author{Ian Fiddes}
\title{BME 205 Assignment \#5: Null Models}
\date{\today}
\begin{document}
\maketitle
\section*{Introduction}
In this assignment, we utilize four null models to determine the likelihood of a open-reading frame (ORF) to be found on the opposite strand of a gene. While the programs are generalized to work for any gene and genome,
we will focus on the \textit{S. solfataricus} genome, and in particular the THSA\_SULSH gene. The impetus of this assignment is the discovery of a putative 388 codon long ORF on the opposite strand of the THSA\_SULSH gene.
Is this a real gene? What is the likelihood of a ORF of that length being found randomly in a genome? What is the best way to model this? We will move from simpler to more complicated models to try and answer this question.
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The first model relies purely on GC content of the species' genome to determine the likelihood of an ORF of a given size. The second model relies on the species'
codon bias, as found from codon bias databases. Codon bias is the amount different codons are used in annotated ORFs for the species. Not all codons are used equally between
different species. The third model determines the probability of a ORF being found on the opposite strand of a DNA sequence generated with a weighted random using a provided
codon bias table. The fourth model determines the probability of a ORF being found on the opposite strand of a provided protein sequence by generating a representative DNA sequence based on provided
codon bias.
\section*{Model 1 - No Selection or Selection for G+C Content}
The probability of a ORF to be found in a random sequence based only on GC content can be calculated by the following equation.
\begin{equation}
P(\mbox{ORF}) = 2N*P(\mbox{ATG})*(1-P(\mbox{TAA})-P(\mbox{TAG})-P(\mbox{TGA}))^{n-1}
\end{equation}
Where N is the number of nucleotides in the genome and n is the size of the ORF in amino acids. The individual probabilities are calculated simply using the GC content. Note that N is multiplied by 2 to account for both strands of the genome.
\begin{equation}
\begin{aligned}
P(A,T) &= \frac{(1-\mbox{\%GC})}{2} \\
P(G,C) &= \frac{\mbox{\%GC}}{2}
\end{aligned}
\end{equation}
For the putative ORF found on the opposite strand of the THSA\_SULSH gene in \textit{S. solfataricus}, which is 388 codons long, $P(\mbox{ORF})$ is $7.05190*10^{-08}$.
\section*{Model 2 - Non-specific Selection}
The probability of an ORF to be found in a random sequence based on kmer counts can also be calculated directly. After counting the 3-mers present in a genome the following equation,
which is identical to equation 1. The only difference is how the individual probabilities are calculated.
\begin{equation}
\begin{aligned}
P(\mbox{ORF}) &= 2N*P(\mbox{ATG})*(1-P(\mbox{TAA})-P(\mbox{TAG})-P(\mbox{TGA}))^{n-1}
\\
P(\mbox{XXX}) &= \frac { \mbox{Count}(\mbox{XXX})}{2N}
\end{aligned}
\end{equation}
XXX is any codon.
For the putative ORF found on the opposite strand of the THSA\_SULSH gene in \textit{S. solfataricus}, which is 388 codons long, $P(\mbox{ORF})$ is $4.87122*10^{-08}$.
\section*{Model 3 - Opposite strand of a protein coding gene}
For this model, simulations must be used and the result cannot be calculated directly. To this end, I wrote a Python program to generate weighted random protein-coding ORFs and then look for the longest ORF on the opposite strand.
For this, an ORF was defined as a sequence that began with 'ATG' and contained no stop codons ('TAA','TAG','TGA') in the frame of the sequence. The random codon generation is weighted by using a codon bias table for the species.
\begin{figure}
\centering
\includegraphics[scale=0.5]{model3_lognormal_evalue.png}
\caption{Model 3 fitted to a log normal distribution. 10,000 sequences were tested.}
\end{figure}
\end{document}