Question

The yeast genome is about 1.3 X 10^7 bp. The average gene codes for a protein of 42,000 molecular weight (Da). The intergene distance (the average distance between adjacent ORFs) is about 500 bp. How many protein-coding genes could be accommodated by the yeast genome?

Answer 1

One protein molecule is formed by several numbers of different amino acids. Hence, the molecular weight of a protein = number of amino acids present × average molecular weight of amino acid

So, the average number of amino acids in each protein

= average molecular weight of a protein ÷ average molecular weight of one amino acid

= 42000 Da ÷ 110Da(average molecular weight of one an amino acid) = 381.8181 = 382 (the fraction number is rounded to the nearest whole number as the number of amino acids can't be a fraction)

One amino acid is translated from one codon. One codon contains three base pairs. Hence, 382 amino acids are translated from (382×3)=1146 number of base pairs (bp).

One protein is translated from the ORF (open reading frame, i.e., the coding region of a gene which is translated to form a protein) of a gene. So, for the above question, it can be said that one gene = one protein = 382 amino acids = 1146 bp

One gene and one intergenic distance contain total 1146+500 = 1646 base pairs.

The total size of the genome = number of genes × total number of base pairs in one gene and one intergenic distance

So, 1.3×10^7 bp = number of genes × 1646 bp

Hence, the number of genes present in the yeast genome = 1.3×10^7 ÷ 1646 = 7897.934 = 7898 (the fraction number is rounded to the nearest whole number as the number of genes can't be a fraction).

Hence, according to the above question, the yeast genome contains approximately 7898 protein-coding genes.