DNA is composed of long strands of four basic molecules: Adenine, Cytosine, Guanine and Thymine. These four bases coil together to form a double-helix structure, and we have 23 such structures, called chromosomes, in every cell of our body. The bases have the interesting property of pairing up across the double-helix, such that Adenine always matches Thymine, and Cytosine always matches Guanine. For our purposes as computer scientists, we can abbreviate these bases as A, C, G, and T.
DNA contains segments called genes detailing how our cells should work. A gene becomes a protein through a process known as the Central dogma of molecular biology: DNA is transcribed into RNA (Ribonucleic Acid) and then translated into proteins. Proteins are the vital machinery that makes our cells work; they form the cell walls and transport molecules around the cell. It is this process of DNA to RNA to Proteins that we will simulate in this project by creating appropriate classes and objects.
Because of how genes are encoded, most genes have a balanced number of Gs and Cs in relation to the number of As and Ts. To determine the G-C content of a DNA strand, we can sum the number of Gs and Cs we find, and divide this by the total number of bases in the string. Most genes have between 25% and 75% Gs and Cs.
Our next step is to transcribe the DNA into RNA. RNA is a single-stranded molecule as opposed to the double-stranded DNA. This means it cannot replicate itself as does DNA, but RNA still plays many useful roles in cell function. Primarily, RNA gets translated into proteins. Having these steps be separate eases the burden on DNA to process all the work. In the transcription from DNA to RNA, every Thymine molecule is replace with Uracil (denoted with U).
When you transcribing DNA into a strand of RNA, there are two important locations to find: where to start reading and where to stop reading. This process is rather complex, and so we will simplify here to only transcribe the exact location of a gene (in reality there are buffers transcribed before and after the entire gene).
Just like words are composed of individual letters, genes are divided up into segments called codons. A codon consists of three sequential bases, so with 4 choices for each base (from A, C, G, and T), this gives us 4 X 4 X 4 = 64 different combinations. So there are actually three different ways to change a strand into amino acids. We could start with the first base at index 0, such that our codons are substrings from 0:3, 3:6, 6:9, etc. Staring with index 1 makes our substrings 1:4, 4:7, 7:10, etc, and starting with index 2 gives us 2:5, 5:8, 8:11, etc.
These initial indices (0, 1, 2) define the reading frame for this gene. For example, in English we start every sentence with a capital letter, and stop each sentence with a period. DNA uses an interesting way of signaling the start of a gene by designating one particular codon, ATG, as the start codon. For our purposes, wherever you find ATG, this marks the start of a gene; this also defines our reading frame, which can be found by modding the initial index of the start codon by 3.
For robustness, there are three codons which tell us the end of a gene: TAG, TGA and TAA. As soon as any one of these three stop codons is found in the same reading frame after the start codon, we have found the end of the gene and our RNA molecule.
With our strand now in RNA form, we are ready to make the final step to proteins. This works by reading the Genetic Code of RNA. Each of the 64 possibile codons translates into one of 20 amino acids (redundancy is built into this system to decrease the effects of random mutations), and it is the conjunction of these amino acids that ultimately fold up and create proteins. This can be represented in a Codon Table, pairing codons such as UCU, UCC, UCA and UCG with Serine. See the paragraph before the Testing section for more details on how to read this Codon Table. (Note, the amino acids are commonly abbreviated to a single letter, like DNA molecules, so in the case of Serine, this would be S.)
In this project, you will simulate this process as described above using object-oriented programming in Python to create three classes, DNA, RNA and Protein, along with a driver program for the cell.
pgsm.py(which stands for Protein, Gene and Small Molecule) will contain your class definitions for DNA, RNA and Protein. Each of these classes is described below.
The DNA class will represent a chromosome within a cell. It will bring in a string of DNA and be able to return a list of all genes contained within this DNA string or its reverse complement.
The constructor brings in a string called strand, which will be saved as a data member component of this DNA object.
This method will be called by str() to produce a string representing the DNA object. If the strand is longer than 33 bases, return the first 15 bases, followed by "..." and then the last 15 bases, otherwise, return the whole DNA strand.
This method will return the length of the strand for this DNA object.
Calculate the G-C content of the DNA object and return as a floating point number. This
will represent the percent of Gs and Cs in the strand as compared to the complete length of
the strand. You should implement the search for Gs and Cs yourself such that you run through the
strand only once, as opposed to using the
find method of string objects.
For half credit, you can use the
DNA is a double-stranded molecule. The strand passed in through the constructor is only one half of the molecule. This method will replace that strand with it's reverse complement. You can find the opposite strand of DNA by replacing all Gs with Cs and vice versa, and all As with Ts and vice versa, and then reversing the string. Using a recursive definition to reverse the string will be infeasible when handling 1 million base-pair DNA molecules; instead, you should use a for-loop for this reversal.
This method returns a list of RNA objects representing transcribed genes based on current the inversion of the DNA strand. To find the RNA objects, walk through the string searching for the start codon "ATG" and record these indicies in a list. Then for each start codon, search the rest of the string to find a stop codon "TAG", "TGA" or "TAA" in the same reading frame. Your RNA strands will be the base pairs between each start and stop codon; use these to create RNA objects to add to your list. You should substitute a U for each T found in this substring before creating the RNA object. It is possible that genese will overlap, containing some of the same DNA but in different reading frames. If you reach the end of a DNA strand before you find a Stop codon in the same reading frame, do not create an RNA molecule.
__init__(self, r_strand, start, stop)
The constructor for an RNA object will bring in a string called r_strand, along with the start and stop index of transcription from the original DNA molecule. All of these will be saved as a data member components of this RNA object.
This method will be called by str() to produce a string representing the RNA object. First, the string will contain the start and stop indicies of the RNA, separated by a dash. If the strand is longer than 33 bases, return the first 15 bases, followed by "..." and then the last 15 bases, otherwise, return the whole RNA strand.
This method will return a Protein object based on the RNA strand and the dictionary
codon_table. For each three letter codon, translate it to the appropriate
amino acid based on the
codon_table. Create a Protein object with this
amino acid string and the start and stop index, and return it.
__init__(self, aa, start, stop)
The constructor for a Protein object will bring in a string called aa, along with the start and stop index of transcription from the original DNA molecule. All of these will be saved as a data member components of this Protein object.
This method will be called by str() to produce a string representing the Protein object. First, the string will contain the start and stop indicies of the Protein, separated by a dash. If the strand is longer than 33 amino acids, return the first 15 amino acids, followed by "..." and then the last 15 amino acids, otherwise, return the whole Protein strand.
cell.pywill contain the main function for the program. You will implement the following pseudocode:
The DNA file will be in a format such that the data might be split over multiple lines.
The RNA Codon Table file is organized to have the letter of an amino acid first, followed by
the codons which encode for this amino acid separated by commas.
The Codon table should be loaded such that the keys are 3 base pairs of RNA (such as GCU, AUU) and
the values are the single letters representing an amino acid. Note that not all methods
requested above, such as
invert, are used in
cell.py. These methods
will be tested separately during grading.
DNA Molecule is TTAATAGCGTGGAAT...GTCCTAAAGATAACA Length is 99 base pairs GC Content is 0.42424242424242425 Found 2 RNA molecules The Proteins are: 1: 16-48 ILIKECHEESE 2: 58-87 EATVEGGIES
Next, download the file:
According to current analysis, E. coli UTI89 should contain approximately 5000 genes total, many less than what would be predicted by our program. The discrepancy is due to the simplifications we made in our model of DNA; in reality, there are many other factors beyond the start codon which determine if a segment of DNA encodes a gene, such as Transcription Factors and Promoters. You must hand in: