A Burrowing owl's habitat is destroyed which is due to human activities and will come under Artificial destruction via human influence and not due to a natural destruction like cyclones, High Richter scale earthquakes, hurricanes with extremely high knot speeds, etc. Instead I'll say because of which the population of the burrowing owl will obviously decrease because they're more adapted to "their previous environment" and most likely "wouldn't adapt to a new unfamiliar environment".

To break these contradictions down simply said "they're unaware of the rules, regulations, type of soil, type of trophic levels, number of predatory organisms, etc. this makes it pretty hard to move from their once said naturally provided nature-made habitat to the burrowing owl, which got lost due to habitat annihilation by human cause. Further making the owls to adapt and change their "NATURAL TRAITS" to make it "CUSTOM" because of which these aren't going to help them instead they'd go either extinct by moving to a newly known unfamiliar habitat rather than their naturally nature gifted habitation.

So Yeah, the correct option [after the question mark ends] to be the least likely outcome would've been "the population of species of burrowing owl maybe increase as per arriving in a new habitat or introduced to newly made surroundings". This is "Highly and the most unlikely" or the "least likely predictable outcome" for burrowing owls. Introduction of species to newer habitats without any prior training, kills the species and it's progenies.

Hope this helps you and gives you the detailed analysis for this query for burrowing owls!!!!

5 0

3 years ago

Read 2 more answers

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

fenix001 [56]

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 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!

8 0

3 years ago