The genome of the bacterium neisseria gonorrhoeae consists of one double-stranded dna molecule that contains 2220 kilobase pairs. if 85% of this dna molecule is made up of the open reading frames of genes encoding proteins, and the average protein is 300 amino acids long, how many protein- encoding genes does neisseria have? what kind of genetic information is present in the other 15% of the dna?

The length of the DNA of the Neisseria gonorrhoeae is 2220 kbps. It is to be noted that 1 kbp = 1000 bases. After converting the kbps into base pairs we get:  

2200 * 1000 = 2220000 base pairs

The entire length of the base pairs is,

2220000 * 0.34 nm = 754800 nm

After converting it into centimeters we get:  

754800 * 10^-7 = 0.7548 cm

Therefore, the total length of the molecule of DNA is 0.7548 cm.  

The 85 percent of the 2220000 base pairs = 2220000 * 85 / 100  

= 188700000 / 100 = 1887000 base pairs

Each of the amino acid is encrypted by three base pairs (triplet codon)

Thus, the number of bases needed to generate a protein with 300 amino acids is 300 * 3 = 900 base pairs

The number of protein encoding genes in N. gonorrhoeae genome is :  

= 1887000 / 900 = 2096.6 or 2097

Hence, the total number of protein encoding genes present in the genome of N. gonorrhoeae is 2097.  

The left 15 percent genes can be considered as the non-coding genes. They exhibit the information essential to monitor the expression of the 85 percent genes.  

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c

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