Question

1. The genome of a newly discovered bacterial species (Bacillus sanfranciscus) is sequenced and found to have a circular genome of 8.7 x10⁶base pairs (bp).  Open reading frame (ORF) analysis indicated the presence of 7,250 ORFs that encode proteins with an average length of 360 aa.   

1. What is the information content of this genome (i.e.– how much information can be encoded in this length of DNA)?  Since the genetic code can be considered digital in nature, convert the information content of base pairs into bytes and compare this value with the information content of an iPhone operating system, which requires approximately 2 GB of information to perform its functions.  To compare the amount of information encoded in iOS8 and in the genome of B. sanfranciscus, the following assumptions about the digital content of DNA-encoded information might be helpful.  The double helix can potentially encode information in both strands but this is not usually the case; most stretches of DNA encode information in only one strand (although for any given gene it can be either of the two strands).  So, it is therefore reasonable to assume that each base pair of DNA encodes 2 bits of information (since there are 4 possible nucleotides).  Keeping in mind that 1 byte = 8 bits and 1 GB = 10⁹bytes, the calculation is pretty straightforward from there.  Express your answer as iPhone iOS units (i.e.- 1 iOS unit = 2GB).  Show your work.

2. Now calculate the percentage of the bacterial genome that encodes the cell’s complete proteome. Assume that: a) all of the predicted ORFs actually encode proteins, b) each gene is encoded by only one of the two strands of DNA, and c) there are no overlapping genes (i.e.- no region of DNA encodes more than a single gene).  Show your work

Answer 1

Total no. of base pairs= 8.7 x 10⁶

each base pair encodes 2 bits.

a) therefore, total digital information.= 8.7 x 10⁶ x 2 = 1.74 x 10⁷ bits = (1.74 x 10⁷)/8 bytes = (1.74 x 10⁷) x 10^-9/8 GB = 2.175 x 10^-3 GB = 2.17 MB (1GB = 10⁹ bytes approximately)

b) No. of ORFs = 7250

Average length of each ORFs = 360 aa

Total number of base pair associated with proteome = 7250 x 360 x 3 = 7.83 x 10⁶ base pairs. (since each codon have 3 bp)

Percentage of DNA associated with entire proteome = 100 x (7.83 x 10⁶)/(8.7 x 10⁶) = 90%.