Question

1. The mature mRNA molecule for the EDS1 transcript is 1903 nucleotides long, starting with an initiation codon at one end and a termination codon at the other end. The number of amino acids in the protein translated from this mRNA is:

2. You’ve just determined the first four amino acids in the amino-terminal end of a the EDS1L Arabidopsis protein. The peptide sequence is: (NH4)- met-ala- phe-glu- (COOH). What is the nucleotide sequence in the portion of the mRNA encoding this protein?

Answer 1

Given that a mature mRNA molecule of EDS1 transcript is 1903 nucleotides.
There is one initiation codon = methionine= 1 amino acid
There is one termination codon= 1903-3= 1900
So there are 1900 nucleotides and each amino acid constitutes 3 nucleotides.
So number of amino acids formed is = 1900/3= 633.3
I think there is a problem with number of nucloetides.
The nucleotides should be either 1902 or 1905, then aminoacids translated would be either 634 or 635.

2.
The peptide sequence is: (NH4)- met-ala- phe-glu- (COOH)
Nucleotide sequence is: auggcguuugaa (most likely because amino acids codons are redundant)