Answer:
Species A produced only 2 amino acids. Since there are 5 bases, each base can specigy one amino acid, with redundancy. The minimum codon length for species A is 1.
Species B: Produces 56 amino acids. Since there are 4 bases, to code for 56 amino acids, the minimum codon length should be 4^3 = 4 x 4 x 4 = 64. A triplet codon is required to code the 56 amino acids of this species.
Species C: Produce 162 amino acids. Since a triplet codon can code for only 64 amino acids, a codon of four bases is required for this species. The four base codon gives 4^4= 4 x 4 x 4 x 4 = 254 different codons. Hence, the answer is 4 base codon.
The minimum codon length for each species are ;
Given data
Three different bacterial species ( Species A, B and C )
Each contains genetic materials not DNA
Each contains different number of amino acids;
Each genetic material contains four ( 4 ) bases
Calculate the minimum codon length of each species using the total number of amino acids
For Specie A ;
number of base = 4
amino acids = 2
∴ minimum codon length = 1 ( due to redundancy )
For Specie B
number of bases = 4
amino acids = 56
∴minimum codon length = 3
Because 3 base codon is needed to code 56 amino acids ( i.e. 4^3 > 56 amino acids )
For Specie C
number of bases = 4
amino acids = 162
∴ minimum codon length = 4
Because 4 codon is needed to code 162 amino acids ( i.e 4^4 > 162 amino acids )
Hence we can conclude that the minimum codon length for each species are ; Specie A = 1 codon length For Specie B = 3 codon length For specie C = 4 codon length.
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