You want to make a genomic library of the human genome. So, you will need to cho
ID: 100930 • Letter: Y
Question
You want to make a genomic library of the human genome. So, you will need to choose an enzyme at will digest the DNA that will extensively cut the genome. It is known that approximately 60% of human DNA is made up of AT. Also, known is that the human genome has 9 times 10^3 bp of DNA. Which of the following three enzymes would you want to use to digest the DNA as many times as possible to clone the DNA fragments into vectors? Show all your work. (a) BamHI (restriction site = 5'-GGATCC-3'.) (b) EcoRI (restriction site = 5'-GAA TTC-3') (c) Haell (restriction site 5'-GGCC-3')Explanation / Answer
The AT content of human genome is approximately 60% which means A base constitutes 30% (0.30) and T base constitutes 30% (0.30) of total human genome. The remaining 40% of human genome has GC content which means 20% (0.20) of G base and 20% (0.20) of C base is present in human genome.
The human genome has 3 X 109 bp of DNA but not 9 X 103 bp of DNA.
a. Bam HI recognizes restriction site 5’ – GGATCC – 3’
The probability of the presence of the sequence in the DNA is as follows:
(0.20) (0.20) (.30) (.30) (.20) (.20) = 1.4 e-4 = 0.00014
The probability of finding one Bam HI restriction site in 60% AT content human genome is
1/0.00014 = 7142.8= 7143 bp.
So, for every 7143 bp distance of DNA, a Bam HI restriction site is present.
The size of human genome is 3 X 109 bp of DNA.
If we assume that Bam HI enzyme will be able to cut every single restriction site available in the human genome, the number of cuts the enzyme can make in the human genome is
3 X 109/7143 = 419991.6 = 419992 cuts.
b. Eco RI recognizes restriction site 5’ – GAATTC – 3’
The probability of the presence of the sequence in the DNA is as follows:
(0.20) (0.30) (.30) (.30) (.30) (.20) = 3.2 e-4 = 0.00032
The probability of finding one Eco RI restriction site in 60% AT content human genome is
1/0.00032 = 3125 bp.
So, for every 3125 bp distance of DNA, a Eco RI restriction site is present.
The size of human genome is 3 X 109 bp of DNA.
If we assume that Eco RI enzyme will be able to cut every single restriction site available in the human genome, the number of cuts the enzyme can make in the human genome is
3 X 109/3125 = 960000 cuts.
c. Hae III recognizes restriction site 5’ – GGCC – 3’
The probability of the presence of the sequence in the DNA is as follows:
(0.20) (0.20) (.20) (.20) = 1.6 e-3 = 0.0016
The probability of finding one Hae III restriction site in 60% AT content human genome is
1/0.0016 = 625 bp.
So, for every 625 bp distance of DNA, a Hae III restriction site is present.
The size of human genome is 3 X 109 bp of DNA.
If we assume that Hae III enzyme will be able to cut every single restriction site available in the human genome, the number of cuts the enzyme can make in the human genome is
3 X 109/625 = 4800000 cuts.
Hae III enzyme can be used to digest the DNA as many times as possible to clone the DNA fragments into vectors as restriction digestion with Hae III produces more DNA fragments compared to Bam HI and Eco RI digestion of human genome.
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