How many possible open reading frames (frames without stop codons) are there that extend through the following sequence? 5'... CTTACAGTTTATTGATACGGAGAAGG... 3' 3'... GAATGTCAAATAACTATGCCTCTTCC... 5'

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 TGA 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 TAA G

4)Now you need to find start (AGT in terms of DNA) and stop codons(TAA, TAG, TGA, also in terms of DNA), in 5->3 and de reverse complementary.

As you can see, you have for open reading frames that lack of stop codon, unfortunately, they also lack of AGT, the start codon.

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!

annyksl annyksl · Answer 2 · 2021-11-01T11:36:40+01:00

According to the DNA sequence shown in the question above, we can see that there are 3 reading frames without stop codons.

You can find this answer as follows:

Isolate the 5'-3' Sequence: You will not need the 3'-5' Sequence, so you will need to isolate the 5'-3' Sequence. In this, you should observe the sequence of nitrogenous bases, identify the codons (set of three nitrogenous bases) and create three sequences. The first sequence will exclude the first nucleotide, the second sequence will exclude the second nucleotide, and the third sequence will exclude nothing. Based on this, you will have the three sequences below:

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 TGA TAC GGA GAA GG... 3' (with stop codons)

Create complementary sequences: You should do this based on the complementarity of the nitrogenous bases. In this case, it is necessary to remember that Adenine (A) is complemented by Thymine (T), Guanine (G) is complemented by cytosine (C), and vice versa. In this case, you will have the following sequences:

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'

Identify the inverse complement: Still taking into account the complementarity of the nitrogenous bases, you should find the inverse complementary sequences. In this case, you will find the sequences:

CCT TCT CCG TAT CAA TAA ACT GTA AG (with stop codon)

CC TTC TCC GTA TCA ATA AAC TGT AAG (no stop codon)

C CTT CTC CGT ATC AAT AAA CTG TAA G (with stop codons)

From this, we can see which sequences have one of the stop codons, which are TAA, TAG, TGA. As you can see, only the first two frames of the 5'-3' sequences and the first two frames of the inverse sequences do not have these codons, so it is possible to observe 3 open reading frames without stop codons.

More information:

https://brainly.com/question/25253880?referrer=searchResults