Under which condition(s) would genetic testing for predisposition to an inherited disorder in a minor child be considered reason
1 answer:
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Answer:
There are 4 reading frames without stop codons.
Explanation:
1) To find the possible reading frames you need to separate your sequence in codons (trios) for the 5'-> 3' sense, first, excluding the first nucleotide, then excluding 2 nucleotides and then not excluding nucleotides.
5'... C TTA CAG TTT ATT GAT ACG GAG AAG G... 3' NO STOP CODON
5'... CT TAC AGT TTA TTG ATA CGG AGA AGG... 3' NO STOP CODON
5'... CTT ACA GTT TAT <u>TGA</u> TAC GGA GAA GG... 3'
2)Then you find the complementary sequence (which you already have)
3'... GAA TGT CAA ATA ACT ATG CCT CTT CC... 5'
3'... GA ATG TCA AAT AAC TAT GCC TCT TCC... 5'
3'... G AAT GTC AAA TAA CTA TGC CTC TTC C... 5'
3) Next, you have to obtain the reverse complementary
CCT TCT CCG TAT CAA TAA ACT GTA AG NO STOP CODON
CC TTC TCC GTA TCA ATA AAC TGT AAG NO STOP CODON
C CTT CTC CGT ATC AAT AAA CTG <u>TAA</u> G
4)Now you need to find start (AGT in terms of DNA) and stop codons(<u>TAA</u>, <u>TAG,</u> <u>TGA</u>, also in terms of DNA), in 5->3 and de reverse complementary.
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<em>As you can see, you have for open reading frames that lack of stop codon, unfortunately, they also lack of AGT, the start codon.</em>
I added a table where you can find thestart and stop codons as well as the proteins that they translate to in terms of RNA.
I hope you find this information interesting and useful, good luck!
