You want to make a genomic library of the human genome. So, you will need to cho

Question

You want to make a genomic library of the human genome. So, you will need to choose an enzyme that 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 x103 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. (hint:Calculate about how many times will each of the following restriction sites be present in the entire genome? Think about the frequency of each base)

You will need the three enzymes and the restriction sequences to answer this. Obviously you will need to to do a little math for this problem. The enzymes and sequences are:

(a) BamHI (restriction site = 5'—GGATCC—3')

(b) EcoRI (restriction site = 5'—GAATTC—3')

(c) HaeIII (restriction site = 5'—GGCC—3')

Explanation / Answer

To make a genomic library, the vector and the gene of interest is very important. The library consist of a collection of total genomic DNA, from the specific species and then the collected DNA to be stored in a population of known vectors, each vector would be having different insert of DNA.

Here in this example we have the DNA to be cloned having 9 x103 bp = 9000 bp. We also have the information as having AT rich sequences more in the human genome, so here the 60% of 9000 would be 5400 bp.

Let us look at the pattern of genomic DNA fragments that would be generated if we use the restriction enzymes (RE). If a RE is 40mer enzyme, for that he average fragment size would be approx 256bp. Here in our example we have Hae III as such

5'—GGCC—3'

3'—CCGG—5'

Similarly if we take a 6-mer sequence we can generate an average DNA size approximately 4000 bp. On the other hand if we have an 8-mer to work with we will be getting approximately 64,000 bp. Here we also have to know that the fragments larger and smaller than the averages are also produced.

In case of the vectors, their size of construct and the capacity to a DNA insert has to be lesser than a 10Kb construct. Whereas lambda vectors we can take up to 10-20 Kb. In case we use a 6-mer or 8-mer DNA sequence enzymes, they would be producing larger DNA fragments, which might not be represented in a library using some vector system like that of a bacterium.

The probability we need to calculate for the specific random cuts in the genomic DNA sequence. We can use the RE as BamHI, where the sequence is

5'— G GATCC —3'

3'—CCTAG G—5'

5'— G               GATCC —3'

3'—CCTAG              G—5'

Vector Ends (BamHI)

5'— G GATC------------/ /------------------ GATCC —3'

3'—CCTAG -------------/ /---------------- CTAGG—5'

                        Insert + Vector Ligation

Now how many would be generated to calculate,

Given here the human genome size is 3 x 103 bp, now let us take an standard example where using an average plasmid insert size of 6 kb,
1 genome equivalent we get = (3 x 109 bp) / (6 x 103 bp) = 5 x 105 plasmids.

While same applies to our Phage substitution vector insert of 20 Kb

1 genome equivalent = (3 x 109 / (2 x 104 bp) = 1.5 x 105 phage.

Using the genome equivalence we assume to have a formula called ‘rule of thumb’,

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