Question

Your colleague has isolated a mutant bacterial strain and found the dna 5x10^8 bp in length. she has asked you to do restriction mapping. you start by determining the base composition of the bacterium, and find that the bacteria consists of 24% Thymine and 26% Guanine. you plan on using the enzyme NOTI which recognizes the sequence 5'-GCGGCCGC-3'

a) How many fragments would you expect assuming that it occurs randomly throughout the genome

b)your colleague also provided several variants of her strain of bacteria. You test all of them and found differences in their respective banding pattern, in a simple statement, what could account for this?

Answer 1

The bacteria consists of 24% Thymine and 26% Guanine.

The base-pairing rule, exists as per a consequence of stability in terms of hydrogen bond formation.
Adenine and Guanine are together referred to as purines, as they have an imidazole ring attached to their pyrimidine ring. Cytosine and Thymine are together referred to as pyrimidines, as they do not have any such imidazole ring attached to them.
A double stranded DNA has just barely enough space within its two strands to exactly accommodate only one purine and one pyrimidine. Two of the same type will simply not do, as two purines would be too large while two pyrimidines will fall too short. Thus, a purine ends up having to make a hydrogen bond with another pyrimidine.
In terms of the specificity, the molecule structures of these 4 bases are such that since one strand goes parallel and the other anti-parallel, adenine has two possible sites for hydrogen bond formation. Two exactly complement these two sites with minimum distance between polar atoms constituting the H-bond, only thymine pairs up perfectly and not cytosine. The same is true for guanine and cytosine pairing. Since one pair makes 2 hydrogen bonds and the other pair makes 3 hydrogen bonds. Hence, specificity is introduced.

Hence Number of A = Number of T = 24%
& Number of G = Number of C = 26%

We plan on using the enzyme NOT1 which recognizes the sequence 5'-GCGGCCGC-3'

Our DNA is made up of nucleotides with 4 possible bases {2 Purines labelled Adenine (A) and Guanine (G), & 2 Pyrimidines labelled Cytosine (C) and Thymine (T)}. These 4 nucleotides are arranged sequentially across two strands that come together to form a ladder with the bases as its rungs. One strand goes from 5’ to 3’ direction and the other strand is anti-parallel to this going from the 3’ to 5’ direction.
Thus there end up being sequences in the middle that are palindromic, that is, read in the same sequence of bases on both strands when reading from the 5’ to 3’ direction. Example:

5’ – AGTTAACA – 3’
3’ – TCAATTGT – 5’

which is read as 5’ – GTTAAC – 3’ on both strands.

Certain enzymes called restriction enzymes were made by bacteria to counter genetic material insertions by viruses. These restriction enzymes treated these palindromic regions in the DNA as restriction sites and were able to cut the DNA there, leading to either a blunt cut:

5’ – AGTT ||| AACA – 3’
3’ – TCAA ||| TTGT – 5’

or a staggered/cohesive cut:

5’ – AGT ||| TAACA – 3’
3’ – TCAAT ||| TGT – 5’

If a linear stretch of double-stranded DNA (dsDNA) is cut at one restriction site, it will produce two end products. Similarly, ‘n’ number of such cuts will produce ‘n+1’ number of final fragments.

Expecting that the given sequence occurs randomly throughout the genome,
we see that the chance of its occurence is equal to the chance of these exact nucleotides lining up, which is:

5'-GCGGCCGC-3'
= Octamer where the base has a 26% chance of occuring
= (26%)^8
= (0.26)^8
= 2.08827 * 10^(-5)

In a sequence of length 5x10^8 bp, we can expect this to occur:
= (5x10^8) * (2.08827 * 10^(-5)) times
= 8.1573 times ~ 8 times ANSWER

b) When the several variants of the colleague's strain of bacteria are tested and they show differences in their respective banding pattern, we can simply account for this by stating simple mutation that has rendered that restriction site useless. ANSWER