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sequence of nucleotides
Anonymous
Annonn
wrote...
3 months ago
3 months ago
sequence of nucleotides
The following sequence of nucleotides is found in a single-stranded DNA template:
AGCATACAGCAGACCGTTGGTCTGAAAAA
AGCATACA
A) Assume that RNA polymerase proceeds along this template from left to right. Which end of the DNA template is 5’ and which end is 3’?
Explain your answer.
B) Assume this DNA template sequence comes from the end of a bacterial transcription unit. Where will transcription terminate?
Explain your answer, including the structure-function relationship.
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Anonymous
wrote...
#1
3 months ago
AGCA
TAC
AGCAGACCGTTGGTCTGAAAAA AGCATACA
TAC is the antisense DNA equivalent of the start codon "AUG".
During transcription, the RNA polymerase read the template DNA strand in the 3′→5′ direction, but the mRNA is formed in the 5′ to 3′ direction.
Hence:
3'-AGCA
TAC
AGCAGACCGTTGGTCTGAAAAA AGCATACA-5'
Quote
B) Assume this DNA template sequence comes from the end of a bacterial transcription unit. Where will transcription terminate? Explain your answer, including the structure-function relationship.
It is known that the termination of the transcription in the bacteria happens by two different mechanisms:
1. Rho-dependent
2. Rho-independent
The Rho-dependent transcription termination happens because of the presence of the Rho protein which dissociates the RNA polymerase whereas the Rho-independent mechanism works by making a hairpin loop in the transcript(mRNA) which sends a signal to stop the transcription.
To make a hairpin loop and show how that can be made in the mRNA we first translate this given sequence into the mRNA. We know that following are the complementary nucleotide to the DNA in the RNA as shown below.
A T C G = DNA
U A G C = RNA
the complementary binding is essential to make the mRNA from the DNA. the T is replaced by U(Uracil) in the RNA when it transcribes. keeping these rules in the mind we will transcribe the given sequence to the mRNA as follow
AGCATACAGCAGACCGTTGGTCTGAAAAA
AGCATACA= DNA Sequence
AGCAUACAGCAGACC
GUU
GGUCUGAAAAAAGCAUACA = mRNA Sequence
To show that Rho-independent transcription termination is possible in this mRNA we have to find the ability of this mRNA sequence to form a hairpin loop
If we take
GUU
which is highlighted as the head of the hairpin we can make a hairpin.
If we took 5 nucleotides upstream of the G of GUU and 5 nucleotides downstream to the second U in
GU
U these sequences are palindromic in nature and can form a hairpin.
We have used an online tool known as VECTORBUILDER to find out if this sequence have any ability to form hairpin loop as shown in the picture:
{attached}
Once the hairpin is formed it will send signal to the RNA polymerase and this will stop the transcription
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