Please help I really dont understand this at all The rTAA AC

The rTAA ACC was cleaved using the restriction site at AIAGCTT \"hich has a cleavage a. sequence of the shortest fragment age results in two fragments, write the b. Design a 3 nt primer. c. In smaller fragment was incubated with DNA the presence of design 3 nt primer, the reaction was then split into presence dATP, dTTP, dGTP, and polymerase in the of sequence of was further each with a different The oligonucleotide determined using Predict the band pattern on the gel er separation. el ddATP ddTTP ddGTP ddCTP

Solution

1.(a) The shortest segment will be 5\'-AGCTTACC-3\' because the enzyme cleave at A/AGCTT.

(b) As the primer is desined against 3\' end of of the DNA strand and according to the sequesnce of DNA the three nt at 3\' end are ACC so the primer of three nt will be TGG.

(c) The ddATP,ddTTP,ddCTP,ddGTP terminate the DNA strand elongation.Hence they will bind to their complementary nt and will stop the elongation so the largest and the uppermost band will be of ddCTP because it will travel longer on DNA strand in comparison to others to find its complimentary i.e \'G\'. Next band will be of ddATP then it will be of ddTTP and then the last one will be of ddGTP as it will bind to its complimentary in the beginging only and remember that DNA synthesis take place always in 5\' - 3\' direction.(5\'-AGCTTACC-3\')

2. The sequence will be 5\'-TGCTGAGCCT 3\' because the uppermost band is in case of ddATP it means that it has travel the longest on DNA strand to find its complimentary (5\' T) and same is the case with other.