Imagine discovering three different bacterial species on a meteorite. Each species contains genetic material that is not DNA, but the genetic material of each species contains four bases. Each species has a different number of amino acids. Use the total number of amino acids per species to determine the minimum codon length for each species.

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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 ;

  • Specie A = 1 codon length
  • For Specie B = 3 codon length
  • For specie C = 4 codon length

Given data

Three different bacterial species ( Species A, B and C )

Each contains genetic materials not DNA

Each contains different number of amino acids;

  • Specie A = 2 amino acids ,
  • Specie B = 56 amino acids,     ( missing  values )
  • Specie C = 162 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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