Chapter 4, Problem 1AQ

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?

Summary Introduction

To discuss:

The Neisseria gonorrhoeae bacterial genome consists of one double stranded DNA and that contains 2220 Kbp. If 85% of this DNA sequence is made up of open reading frames of genes encoding proteins, and the average protein size is 300 amino acids long, Neisseria contains how many protein-encoding genes. Other 15% of the DNA sequence contains what kind of genetic information.

Concept introduction:

Open reading frames or ORF is a specific segment of DNA or RNA molecule and that part can be translated into a protein sequence. An open reading frame (ORF) (sequence of nucleotides) contains a start codon (AUG) followed by a stretch of various codons and end with a stop codon (UAA, UGA, or UAG). An ORF region in the mRNA is essential for its translation.

Explanation of Solution

The Neisseria gonorrhoeae bacterial genome contains 2220 Kbp or 2,220,000 bp.

Each base pair size is 0.34 nm.

Therefore 2220 Kbp x 0.34 nm is 754, 800 nm.

The length of DNA is 754, 800 nm or 0.07548 cm

If 85% of the bacterial genome is composed of open reading frames, 1,887 Kbp of DNA sequence could be the open reading frames.

($2,220,000\times 85÷100=1,887,000\text{ bp or 1887 kbp}$

Average size of the protein is 300 amino acids. Each amino acid is encoded by three nucleotides (one codon). Therefore, 900 bp of DNA sequence in the open reading frame encode proteins.

$(300\times 3=900)$

Number of protein coding bacterial gene is 2097. The remaining 15 % of the bacterial DNA can be non-coding genes, which may regulate gene expression. 

$(1,887,000÷900=2097)$