Question 15 (3 points) Deduce the amino acid sequence provided by the following DNA sequence. 5' CCAGATGCAGGAGTGACCG 3' Second letter First letter U C A G U UUU Phenylalanine (Phe) UUC UUA Leucine (Leu) UUG UCU Serine (Ser) UCC UCA UCG UAU Tyrosine (Tyr) UAC UAA Stop UAG Stop UGU Cysteine (Cys) UGC UGA Stop UGG Tryptophan (Trp) U C A G C CUU Leucine (Leu) CUC CUA CUG CCU Proline (Pro) CCC CCA CCG CAU Histidine (His) CAC CAA Glutamine (Gln) CAG CGU Arginine (Arg) CGC CGA CGG U C A G A AUU Isoleucine (Ile) AUC AUA AUG Start Methionine (Met) ACU Threonine (Thr) ACC ACA ACG AAU Asparagine (Asn) AAC AAA Lysine (Lys) AAG AGU Serine (Ser) AGC AGA Arginine (Arg) AGG U C A G G GUU Valine (Val) GUC GUA GUG GCU Alanine (Ala) GCC GCA GCG GAU Aspartic acid (Asp) GAC GAA Glutamic acid (Glu) GAG GGU Glycine (Gly) GGC GGA GGG U C A G Third letter GLY - LEU - ARG - PRO - HIS - TRP - ... None of the options are correct PRO - ASP - ALA - GLY - VAL - THR - .... MET - GLN - GLU - STOP MET - PRO - ASP - ALA - GLY - VAL - THR - ....
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The DNA sequence is: 5' CCAGATGCAGGAGTGACCG 3' The mRNA sequence will be: 5' GGUCUACGUCCUCACUGGC 3' Now, we need to translate the mRNA sequence into an amino acid sequence using the genetic code. The mRNA sequence can be divided into codons as follows: Show more…
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