Question

1. click on enter or edit dna and copy this code then click :

atgccaggcggcgagagctaa
click the \unfold button\ to see the protein sequence. click on each individual amino acid and write the sequence:

1. how many dna triplets(codons) were in the original sequence?______

how many amino acids are in the final protein?______

Explanation:

Step1: Transcribe DNA to mRNA

DNA: ATG CCAGGC GCGAGAGCTAA
mRNA (replace T with U, complement): AUG CCA GGC GCG AGA GCU AAU

Step2: Split mRNA into codons

Codons: AUG, CCA, GGC, GCG, AGA, GCU, UAA

Step3: Count total DNA triplets

DNA length = 21, triplets = $\frac{21}{3}=7$

Step4: Translate codons to amino acids

AUG = Met, CCA = Pro, GGC = Gly, GCG = Ala, AGA = Arg, GCU = Ala, UAA = Stop (no amino acid)

Step5: Count amino acids

Count non-stop codons: 6

Answer:

Protein sequence: Met-Pro-Gly-Ala-Arg-Ala

Number of DNA triplets: 7

Number of amino acids: 6