can some kind soul please explain this Q, thanks - rsvr

[ArchivalUser]

this is a kaplan qbank q, but rsvr presented the q differently. the original q says the start codon of the disease gene is near the END OF 1st exon of gene X, therefore the answer for that is AAAA-- 1st intron of gene X.

[ArchivalUser]

so if we draw picture..

(intron1 ) (exon 1) (5'UTR)
GENE X 5' --------/----------/--------------/---------------------3'

DIS. GENE 3'--------/----------/--------------/-------------/-----5'
(TATA) (5' UTR) * (exon1) (intron1)

* (start codon) = at the end of exon1 of GENE X

since the mutation is in 5'UTR of dis gene, the polymorphism is complementary to 1st intron of gene x.. So A is the answer
I hope its correct.