If the genetic code is a triplet,how many cod | vedclass

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(D) The genetic code is based on $4$ nitrogenous bases: Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Uracil $(U)$.Since the genetic code is a triple

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$64$

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(D) The genetic code is based on $4$ nitrogenous bases: Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Uracil $(U)$.Since the genetic code is a triplet,each codon consists of $3$ nitrogenous bases.The total number of possible codons is calculated by the formula $4^n$,where $n$ is the number of bases per codon.Here,$n = 3$,so the total number of codons = $4^3 = 4 	imes 4 	imes 4 = 64$.

If the genetic code is a triplet,how many codons will be formed?

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