Question

This DNA sequence comes from part of a gene on a chromosome of a eukaryotic organism.

 i) If the bottom strand of the sequence is the template strand and the top strand is the coding strand, circle the answer below that best represents the direction that RNA polymerase would move along the template DNA strand. 

Left to Right    Right to Left 

Lagging to Leading    A site to P site

 ii) Given the information in 6 i, write the sequence of the mRNA that is transcribed from this region. Label the 5' and 3' end of this molecule. 

iii) This portion of the mRNA contains the region where release factor binds. What is the role of release factor during translation? 

iv) Given the information in 6 ii, what protein sequence is encoded by this portion of the mRNA? Show your work for full credit. Be sure to label the N and C terminus of the protein sequence. 

v) If a tRNA has an anticodon that is 5'-CCA-3', write the codon below. Label the 5' and 3'ends for full credit. 

vi) What amino acid would the tRNA in 6 v be bringing to the ribosome? Write out the full name of the amino acid for full credit.  

Answer 1

i) The RNA polymerase will move along Left to right in the template strand, i.e from the 3' end to the 5' end. The reason behind this is the fact that Transcription will always occur towards the same direction in which the coding strand moves. Since RNA polymerase cannot bind to the coding strand (obviously due to misalignment of 3' and 5' ends), hence in order to replicate the coding strand the template strand needs to be transcribed in 3' to 5' direction only for getting the exact replica of coding strand.

ii) The sequence of mRNA will be as follows -

5'-ACAUUGCAGGGAUGCAUAAAUCU-3'

Reason : Obviously the mRNA sequence will resemble to that of coding strand, except T being replaced with U. Nothing to elaborate in this. mRNA is the replica of Coding strand.

iii) Release factor is also called the stop codon. It codes for the specific protein which signals or indicates the termination for the entire process of translation. The stop codons (UGA, UAA, UAG) enter the ribosomal subunit and trigger the abrupt stoppage of the Translation process. Presence of stop codons in the mRNA sequence will signify that the mRNA needs to be translated upto here only.

iv) The protein sequence will be as follows :

N---TLQGCIN---C

Reason: We will firstly make a sequential triad (group of 3) of all the nucleotides, called Codons. The codons and their respective amino acids will be determined by comparing them to the corresponding entry in the table mentioned in the question. The results will be as follows --

T-ACA

L-UUG

Q-CAG

G-GGA

C-UGC

I-AUA

N-AAU

v) The codon will be 5'--UGG--3'

Reason: The anticodon on tRNA has exactly reverse and complementary sequence of the codon, since it runs from the opposite direction. To get the codon from tRNA anticodon, we need to reverse the sequence and then write the complementary nucleotide. For CCA, the complementary will be GGU, and its reverse will be UGG.

vi) Since the codon sequence is UGG, so it will bring W amino acid to the ribosome (refer to the table given in question). Here, W stands for Tryptophan (Trp).

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