PleaseHelp me to solve this question I had received this ans

PleaseHelp me to solve this question!

I had received this answer , but I Need Explain How to solve this question step by step. this question is Restiriction mapping problem, but quite difficult to solve for me.

pUWL22 is a circular double stranded, 10 kb plasmid. It contains a gene encoding an enzyme that confers ampicillin resistance in the host bacterium.

Cloning into the Kpn1 and Sst1 sites abolishes ampicillin resistance, whereas cloning into other sites on the plasmid does not.

Digestion with the following restriction enzymes singly and in combination results in the fragments listed;

Enzymes Fragments produced (kb)

Hind3 10.5

kpn1 10.5

Not1 10.5

Pvu2 10.5

Sst1 10.5

Stu1 10.5

Hind3+Kpn1 8.5 2.0

Hind3+Not1 8.0 2.5

Hind3+Pvu2 6.0 4.5

Hind 3+Sst1 7.5 3.0

Hind3+Stu1 7.5 3.0

Stu1 +Not1 10.0 0.5

Stu 1+Pvu2 7.5+3.0

Questions;

a: Map the cleavage sites for these enzymes on the circular map of PUWL22. Designate the Hind 3 site as 12o\'clock on the map.

b:Propose two other double digests using the enzymes listed in the table above that could be done to confrim your map. Predict the sizes of the fragments generated from these digests.

Solution

Digestion with the restriction enzymes singly produced following same size fragments i.e. :

Hind3 10.5

kpn1 10.5

Not1 10.5

Pvu2 10.5

Sst1 10.5

Stu1 10.5

This mean the plasmid has single corresponding sites for each enzyme and liniarizes when nicked in any point.

Combination or double digestion will cut the plasmid in two points and will produce 2 fragments according to positions of the enzymes in the circular DNA as follows:

Hind3+Kpn1 8.5 2.0

Hind3+Not1 8.0 2.5

Hind3+Pvu2 6.0 4.5

Hind 3+Sst1 7.5 3.0

Hind3+Stu1 7.5 3.0

Stu1 +Not1 10.0 0.5

Stu 1+Pvu2 7.5+3.0

Designating the Hind3 site as 12o\'clock on the map among Kpn1, Not1, Sst1, Stu1, and Pvu2 the nearest site to Hind 3 is Kpn1 as it produces the shortest fragment (a complimentary biggest fragment is also formed) then Not1 then Sst1/ Stu1 in equal distance but in opposite side and the Pvu2 is farthest.

So the initial liniar mapping would be as following:

Hind3-2-Kpn1-0.5-Not1-0.5-Stu1/Sst1-1.5-Pvu2-3-
As Stu1 +Not1 10.0 0.5 and Stu 1+Pvu2 7.5+3.0

The order will be as Not1-0.5-Stu1 or

Sst1-3-Hind3-2-Kpn1-0.5-Not1-0.5-Stu1-1.5-Pvu2-3-
Or
Hind3-2-Kpn1-0.5-Not1-0.5-Stu1-1.5-Pvu2-3-Sst1-3-

a:The above liniar map can be mapped in a circular fashion designating Hind3 site as 12o\'clock on the map as Hind3 will come after Sst1-3- to complete the circle of PUWL22.

b: Double digestion with Pvu2-and-Sst1 (Pvu2-3-Sst1) will produce 7.5+3.0.
Double digestion with Kpn1 and Stu1 (Kpn1-0.5--0.5-Stu1) will produce 9.5+1. (Digest in NEBuffer 1.1 at 37°C.)

KpnI & PvuII=2.5+8 (Digest in NEBuffer 2.1 at 37°C)