3. Transcribe the DNA into mRNA
Sequence A: AGA AGG GAG GAU UUG CAA GGU GGC CAA GAA UUA GGC GGC GGU CCC GGG GCG GGG AGU CUU CAA CCA Sequence B: AGU CUG CAA AAA CGG GGC AUU GUU GAA CAA UGU UGU ACC AGU AUU UGC AGU CUC UAC CAG UUC GAG AAU UAC UGA Sequence C: AUG UUU GUA CAU UUG UGU GGG AGU CAC CUG GUU GAG CGU UGU AUU UGG UUU GUG GCG AGC GCG GCU UUU UCU AUA
4. Beginning sequence: Sequence C since it starts with AUG (the starting codon) Middle sequence: Sequence A since it does not end with an ending codon or start with a starting codon. End sequence: Sequence B since it ends with UGA (the stop codon)
5. Remove codons 24 to 66, including codon 66
AUG UUU GUA CAU UUG UGU GGG AGU CAC CUG GUU GAG CGU UGU AUU UGG UUU GUG GCG AGC GCG GCU UUU CAG UUC GAG AAU UAC UGA
6. Translate the mRNA into protein using the genetic code.
Methionine (start), Phenylalamine, Valine, Histidine, Leucine, Cysteine, Glycine, Serine, Histidine, Levcine, Valine, Glutamic acid, Arginine, Cysteine, Isoleucine, Tryptophan, Phenylalamine, Valine, Alanine, Serine, Alanine, Alanine, Phenylalamine, Glutamine, Phenylalamine, Glutamic acid, Asparagine, Tryosine, Stop.
a) Sequence C was the beginning fragment since it began with the starting codon, AUG.
b) Sequence B was the ending fragment since it ends with the stop codon, UGA.
c) The introns are removed before protein translation. Only the exons are part of the mRNA that is used to make the final functional protein. The enzyme that removes the introns is spliceosomes.
d) This protein contains 28 amino acids. (29, if the stop codon is included.)
e) This is a prokaryotic genetic sequence since only prokaryotic genetic sequence contains introns.
f) It would not be possible to obtain the same DNA nucleotide sequence because the introns that have been removed would have been part of the DNA sequence when translated backwards. So if a protein were to be translated...