How many amino acids will be coded by the mRN | vedclass

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(C) The original sequence is $5'-CCC-UCA-UAG-UCA-UAC-3'$.The $12^{th}$ nucleotide is the '$A$' in the second '$UCA$' codon (sequence: $C_1 C_2 C_3 U_4

Vedclass · Accepted Answer

Two amino acids

Vedclass · Answer

(C) The original sequence is $5'-CCC-UCA-UAG-UCA-UAC-3'$.The $12^{th}$ nucleotide is the '$A$' in the second '$UCA$' codon (sequence: $C_1 C_2 C_3 U_4 C_5 A_6 U_7 A_8 G_9 U_{10} C_{11} A_{12} U_{13} A_{14} C_{15}$).Inserting an adenosine $(A)$ after the $12^{th}$ nucleotide results in the sequence: $5'-CCC-UCA-UAG-UCA-AUAC-3'$.Now,let's read the codons:$1$. $CCC$ (Proline)$2$. $UCA$ (Serine)$3$. $UAG$ (Stop codon)The translation stops at the $UAG$ stop codon.Therefore,only two amino acids are coded before the translation is terminated.

How many amino acids will be coded by the $mRNA$ sequence $5'-CCCUCAUAGUCAUAC-3'$ if an adenosine residue is inserted after the $12^{th}$ nucleotide?

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