Assume that the DNA sequence below represents exon 3 of a ge

Assume that the DNA sequence below represents exon 3 of a gene. How many possible open reading frames (i.e., frames without any stop codons) extend through this sequence (remember that transcription can use either strand as a template)? The stop codons are UAA, UAG, and UGA. 5\'-AAAGGTATGTTTCGAATGATACTAACATAACAAAGAACATTTTCAGGAGGACCCTTGG GGAGGACCCTTGGAGGGTACCGG-3\' 3\'-TTTCCATACAAAGCTTACTATGATTGTATTGTTTCTTGTAAAA GTCCTCCTGGGAACCTCCCATGGCC-5\'

Solution

There is NO stop codon in this given given DNA sequence.but the total nucleotides present in this sequence is 68 insted of 69 so the last set is not coded any aminoacid in this DNA fragment.the last set 67 & 68 th Nucleotide represent CC so the set is coded amimoacid proline (bcz CCU,CCC,CCA,CCG are representing Proline).

LINE 1

DNA :   5 \'- AAA GGT ATG TTT CGA ATG ATA CTA ACA TAA CAA AGA ACA TTT TCA

mRNA: 3 \'- UUU CCA UAC AAA GCU UAC UAU GAV UGU AUU GUU UCU UGU AAA AGU

A.a : phe pro tyr lys ala tyr tyr asp cys ile val ser cys lys ser

LINE 2 (continution of above sequence )

DNA :   5\' - GGA GGA CCC TTG GAG GGT ACC GG - 3\'

mRNA: 3\' - CCU CCU GGG AAC CUC CCA UGG CC - 5\'

A.a :    pro pro gly asn leu pro trp -

[A.a= AMINO ACID]