Two Sanger sequencing reactions were carried out using the same DNA template and primer. The sequencing gels are shown below.

Lane A of reaction 2 gel shows the absence of lower bands and accumulation of higher bands (represented by a black thick band). Which of the following represents the correct reason?
Options given in the image:

Correct option is C

Correct Answer:
(c) Only
Explanation:
In Sanger sequencing, bands in lane A arise due to incorporation of ddATP causing chain termination. A very low concentration of ddATP results in fewer termination events, allowing DNA synthesis to continue and producing longer fragments. This leads to disappearance of lower bands and accumulation of higher bands in lane A.
Information Booster:
· High processivity of DNA polymerase (option b) would lead to longer fragments, but it doesn’t explain the absence of lower bands completely.
· Absence of the base A near the 5’ region (option a) would just result in missing termination at those positions but not a complete accumulation of longer products.
· High concentration of ddTTP (option d) would affect T termination, not A.
Additional Knowledge (Incorrect Options Explained):
(a) Incorrect – Even if there’s no A near 5’ end, there would still be termination at other positions; lower bands would appear.
(b) Incorrect – High processivity alone doesn’t prevent formation of shorter fragments; ddNTP concentration is the limiting factor.
(d) Incorrect – ddTTP affects termination at T sites; here the thick band is at A, so irrelevant.