The following DNA sequence is present towards the 5’ end of the open reading fra
ID: 18676 • Letter: T
Question
The following DNA sequence is present towards the 5’ end of the open reading frame (i.e. the entire stretch of DNA will be transcribed).
5’-AAGACTTATTAACGGTGTTTT-3’
3’-TTCTGAATAATTGCCACAAAA-5’
A) If DNA replication initiates from the left side of this DNA sequence, which strand will be the template for leading strand synthesis and which strand will be the template for lagging strand synthesis?
Template for leading strand synthesis =
Template for lagging strand synthesis =
B) If the top strand is used as the template for transcription, what will the sequence of the RNA product be? Make sure to label the ends of the molecule and place the end at which translation will begin towards the left side of the page.
C) Take the mRNA that you have written in part B and translate it into protein sequence (label the ends of the protein). Start at the first 3 nucleotides as if the 3 nucleotides just before this sequence were AUG and translation started there. Don’t let this example confuse how translation begins in your mind. It requires an AUG and doesn’t start at the very end of the mRNA (remember the 5’-UTR).
Answer = -fMet-
Explanation / Answer
(a) Template for leading strand synthesis = bottom
Template for lagging strand synthesis = top
(b)
3'-UUCUGAAUAAUUGCCACAAAA-5' (RNA sequence prod.) [read from 5' to 3']
5’-AAGACTTATTAACGGTGTTTT-3’ (template)
(c) there is no start codon (AUG) in the RNA sequence product so it's hard to know where to start the nucleotides