2 Name Ameera Sose Section Genetics Lab 6- Restriction Mapping Pre-lab Below is

Question

2 Name Ameera Sose Section Genetics Lab 6- Restriction Mapping Pre-lab Below is a restriction map of pBLU. What is the size (in bp) of pBLU plasmid? 1. pBLU Miu! 632 ori Mlul 1057 pBLU (5437 bp) HindIl 3190 1837 The sites at which each of the 2 different restriction enzymes will cut pBLU are shown in the map above (for example, HindIll cuts at base-pair 3190). The location of the base numbered 1 is shown, but neither of the enzymes cuts at this position. Imagine that the plasmid is a piece of string that you have tied into a loop and you cut it at the indicated positions. For each enzyme, Hindlll and Mlul, determine how many fragments you would expect to see, and the size (in bp) of each fragment. (4) 2. a. HindIlI b. Mlul

Explanation / Answer

Restriction enzymes are endonucleases that cleave fragment of DNA.

If the linear DNA fragment contains a restriction site for enzyme A, then the enzyme cleaves it and thus we obtain 2 fragments DNA. The sizes of these two fragments add up to the size of original uncleaved DNA fragment.

However, if the DNA fragment it is circular, then the enzyme A cleaves it once and thus linearizes it.

Now in the given question, the plasmid DNA provided to us is circular and it contains sites for the restriction enzymes HindIII and MluI.

(A) The plasmid contains only one restriction site at position 3190 for the enzyme HindIII and thus as explained previously cleaveage with this enzyme results in a single fragment of the size 5437bp.

(B) The plasmid contains 3 MluI sites. Therefore, the activity of the enzyme at one site linearizes the DNA and the activity of enzyme at the remaining two sites cleave the linear DNA into 3 fragments.

Therefore, if the plasmid DNA is cut with MluI enzyme, 3 fragments of DNA will be obtained.

Now lets look at the sizes of the fragment.

Lets assume MluI site at 632 linearises the DNA. This results in a Linear DNA of size 5437 bp with MluI ite at position 0.

In the cicular DNA the other 2 sites are at position 1057 and 1837 respectively. However this is not the case in the linear DNA. On the linear DNA these sites are (1057-632 =) 425 bp and (1837 - 632 =) 1205 bp respectively.

Therefore, on cutting this linear DNA with the enzyme, we get 3 fragments of sizes 425 bp, (1205 - 425 =) 780 bp and (5437 - 1205 =) 4232 bp

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.