Question

The human genome is 3×10⁹ bp in length. Calculate how many fragments would result from the complete digestion of the human genome with specific enzymes. Sau3A (˘GATC), BamHI (G˘GATCC), EcoRI (G˘AATTC), NotI (GC˘GGCCGC). How would your initial answer for BamHI change if you knew that the average GC content of the human genome was 45

1) 7.32×10⁵
2) 1.15×10⁷
3) 5.81×10⁵
4) 8.69×10⁵
5) 1.97×10⁴

Answer 1

The number of fragments resulting from the complete digestion of the human genome depends on the length of the recognition sequence for each specific enzyme. None of the option is correct.

Step-by-step explanation:

The human genome is 3x109 base pairs (bp) in length. To calculate the number of fragments resulting from the complete digestion of the human genome with each specific enzyme, we need to divide the length of the genome by the length of the recognition sequence for each enzyme. Sau3A has a recognition sequence of GATC, which is 4 bp long, so the number of fragments would be 3x109 / 4 = 7.5x108.

For BamHI (GATCC), the recognition sequence is 5 bp long. If the average GC content of the human genome is 45%, it means that 45% of the nucleotides are G or C. Therefore, each G or A base has a 55% chance of being followed by a C or T base. So, the probability of the BamHI recognition sequence occurring is (0.45)x(0.55)x(0.55)x(0.45)x(0.55) = 0.048. Hence, the number of fragments would be 3x109 / (5x0.048) = 1.25x1011 when considering the average GC content of the human genome.

EcoRI (AATTC) and NotI (GCGGCCGC) have recognition sequences that are 5 and 8 bp long, respectively. Therefore, the number of fragments resulting from complete digestion with these enzymes would be 3x109 / (5x0.048) = 1.25x1011 for EcoRI and 3x109 / (8x0.048) = 6.25x1010 for NotI.