Which is more probable to interpret a hybridization band of 4.5 Kb in Southern blot analysis in a transgenic event?
A- Two copies of the T-DNA were inserted at the same locus where they joined head to tail
B- One Copy of the T-DNA was inserted in a locus where there is a BamHI recognition site in the plant genome that is 2.2 Kb far from the right border
N.B. The probe is marked in red in the blow image and the genomic DNA is digested with BamHI
The BamHI has a 6 bp recognition site which means that It has a probability to occur every 4096 bases across the genome
Hi Mohamed ,
1. According to your diagram and probe location:
(a) When you get RBRB joins (2 copies) in the same locus: you should get 2 bands
(b) When you get LB>RBRBRBRB
My bad! It is head to tail. Thanks a lot. But when two copies joined at that order LB>RBLB>RB it would give a 4.5 Kb band. Also, one copy of the T-DNA inserted in a locus where there is a BamHI recognition site in the plant genome that is 2.2 Kb far from the right border would have the same result?
which is more probable to happen?
If the two T-DNA cassettes are in head-to-tail arrangement: LB>RBLB>RB. You should have one cassette internal cut band (4.5 kb) [first band] + one band (> or = 2.3 kb) [second band]. There is a chance that the second band is also 4.5 kb, if the BamHI in the plant genome that is 2.2 kb away from the righ border.
- Badges
- Science topic
- Similar topics
- Biological Science
- Genetics