The normal wild type anticodon recognizes UAC on mRNA which codes Tyr. Since the anticodon is mutated and has the sequence UUA, it would now recognize the codon AAU on the mRNA which codes for Aspargine Asn. The mutated tRNA is charged with Tyr as in the normal tRNA the amino acid Tyrosine would replace all Aspargines in the protein chain.
Let me know if you have questions
Best
Sanjay
Sonia B.
- tRNA with the anticodon sequence of 5’-UCA-3’ recognises the codon UGA, which codes for the stop codon. However, the mutated tRNA is already charged with Tyr, hence the amino acid Tyr will be added to the sequence of amino acids at the codon UGA. The translation will pass the codon UGA and continue until the ribosome reaches a different stop codon or finish translating the mRNA strand.
- However, in the case where the ribosome comes to a codon UCA, the translation process will continue only if there is tRNA that can fill in the space for tRNA recognising codon UCA. If there is no tRNA filling the space of the UCA recognising tRNA, the translation process will stop.
- Hence, the protein produced will be shorter than the wild type protein if the ribosome reaches the UCA codon and no tRNA can fill in the space. The protein produced will be longer if no UCA codon is present and the ribosome pass the codon UGA.
- However, in the case where the ribosome comes to a codon UCA, the translation process will continue only if there is tRNA that can fill in the space for tRNA recognising codon UCA. If there is no tRNA filling the space of the UCA recognising tRNA, the translation process will stop.
- Hence, the protein produced will be shorter than the wild type protein if the ribosome reaches the UCA codon and no tRNA can fill in the space. The protein produced will be longer if no UCA codon is present and the ribosome pass the codon UGA.
is this right?
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09/12/17
Sanjay B.
tutor
Hi Sonia,
So sorry, I made a mistake. I did not take into account the directions 5' to 3'. Sorry, my bad! Lets try this again
As per the initial question - WT tRNA recognizes 5' UAC 3' (codon for tyr) (not UGA as in your comment?)
WT - mRNA 5' UAC 3'
so - tRNA 3' AUG 5'
Mutant - tRNA 3' AUU 5' +tyr
mRNA 5' UAA 3' (stop codon)
This would cause insertion of tyrosine where the UAA stop codon appears causing extension in place of termination. This could result in longer proteins. Termination from other codons will occur normally.
With respect to tyrosine, the codon UAU UAC should allow normal tyr insertion, due to wobble. Thus with respect to tyrosine the protein chain would not be shorter.
This is correct now, unless I am missing something?!!
Thanks
Sanjay
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09/12/17
Sonia B.
hello, thnks for that, could u explain this in a paragraph
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09/12/17
Sonia B.
Hello could u help me with this one?
What are the similarities and differences between the G2 and the M phase checkpoints? What are the phenotypic outcomes for the cell when either of the two checkpoints fails but the cell continues to divide?
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09/12/17
Sonia B.
09/12/17