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An insertion mutation occurs, leaving 31 nucleotides in a DNA sequence.

What is the maximum number of codons that could be produced by this sequence?

User Siame
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2 Answers

4 votes

Answer:

10

Step-by-step explanation:

Every three nucleotides is a codon. Each codon codes for an amino acid. If we have 31 divided by, 3, that gives us 10 codons, with one nucleotide left over.

User Study
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3 votes

Answer:

10 codons

Step-by-step explanation:

During gene expression, the order of the nucleotide sequence in the DNA determines the order of the nucleotide sequence in the mRNA (transcription). Transcription, which is the first stage of gene expression, involves the synthesis of a mRNA molecule using a DNA template. The nucleotide sequence of a DNA molecule is used to produce the nucleotide sequence in a single stranded mRNA molecule.

During translation, which is the second stage of gene expression, the mRNA sequence is read in a group of three (3) nucleotides called CODON. This means three successive nucleotide bases in the mRNA sequence represents one CODON. Each codon specifies a particular amino acid.

However, if an Insertion mutation (addition of one or more nucleotide bases) occur in the DNA sequence that is transcibed into a mRNA sequence leaving 31 nucleotides in the DNA sequence, a maximum of 10 codons will be produced by this sequence.

If 3 nucleotides represent 1 codon

31 nucleotides will represent 31/3 = 10 codons remaining 1 nucleotide

Hence, the maximum codon that can be produced by a DNA sequence of 31 nucleotides is 10.

User TomByrer
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