Suppose ecah of nucleotides....distnace between two stop
codons?
The answer is d.
about 21 codons.
The explanation is
The total number of nucleotides in DNA = 4, so the probability of A
or G or C or T nucleodies = 1/4.
The probability of stop codons is (ATT, ATC, ACT) 1/4 * 1/4 *
1/4 = 1/64.
which means for every 64 nucleotides there is a chance of one stop
codons. As each codon has 3 nucleotides, the propbability for
reperation is 64/3 = 21.3, so the answer is 21.
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