Question

The genome of a newly discovered bacterial species (Bacillus sanfranciscus) is sequenced and found to have a circular genome of 8.7 x 10⁶ base pairs (bp). Open reading frame (ORF) analysis indicated the presence of 7,250 ORFs that encode proteins with an average length of 360 aa.  

a. 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 = 10^9 bytes , the calculation is pretty straightforward from here. Express your answer as iPhone iOS units (i.e - 1 iOS unit = 2GB).

b. 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).

Answer 1

Date. Siz Anscere to Part (a): s genome = 8.7x10 bl. Acc. to question, I bp = 2 bits Also I byte = 8 bits or 1 bit = ( 1 ) bytes. Also, 1GB = 109 bytes p 2 GB = 1:05 unit. Step 1 ( Conversion of base paires into Bile of informa bp= 2 bits 8.7x100 bp (8.78 108 x 2 ) bito Step-2 ( conversion of bito to Bytes) - I bit = ( bylo So, ( 8.7 x 168 x 2) 66= (3.7x10# x2 ) bytes. Stefs (conversion of Boyles o GB) 1 GB = 102 bytes or I byle - to ore 10-9 GB So (368x10* * 2) types (9.4 x10*2 ) as = 2:175 x 10-3 GB

Step=4 ( Conversion of GB to iOS cwsite) 2 GB = 1 i Os esit 1 GB = (1) Sunet 50, 2.175 x103 GB = (2175 x10?) iOS unit = 2 1.0875 x 103 iOS cenit / To.00 10875 ios unit Answer to Part 6): bas of According 7,250 360 aa Step 1 to the question, the genome ORES that encode preoleins average length. (Calculation of no of nucleotides in ORFs) involved. 1 ORF = 360aa 7250 ORFS - (7250 x360) aa 1 aa = 3bl nucleotides So (7250 x360) aa = (7250 x 360 x 3) by nucleotidan = 783x10h nucleotides

Since only one streand encodes fore a gene, | So no.& nucleotides encoding ORES is equal to no. of base pairs in ORFS. Step-2 (Calculation of percentage of bacterial genome that encodes cales complete n % age of bacterial genome =>/783x10" x 100 7.1 118.76x1067 189.38 %