Answer: 20 fragments
Explanation:
This particular restriction enzyme has a 4 bp (base pair) recognition sequence therefore it will cleave once every 4ⁿ bases where n indicates the number pf bases in the recognition sequence.
- The recognition sequence is 4 bp therefore this restriction enzyme cleaves once every 4⁴ = 256 bases. This is the length of the restriction fragment.
- Calculate the number of fragments by dividing the total number of bases ( where you assume equal frequency of bases ie. A+T = C+G) by the length of the restriction base.
- therefore 5000bp/256bp = <u>19,53</u> now this number ca n be rounded up to 20.
- The expected number of fragments on the electrophoresis gel will be 20 fragments.
Your answer would be
H) A parallel decline in non-edible species.
Hope I could help!!
~TheAperture~
The process likely to be occurring in the plant's roots will be nitrogen fixation.
<h3>What is nitrogen fixation?</h3>
It is a process whereby atmospheric nitrogen is converted to usable forms of nitrogen in the soil.
Nitrogen fixation takes place in the root nodules of leguminous plants in association with some symbiotic bacteria.
Thus, the only process that could be happening in the plant's root is nitrogen fixation.
More on nitrogen fixation can be found here: brainly.com/question/19938608
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According to the observation, I would conclude that the fungus has asexual form of reproduction. This is because asexual spores germinate and produce new hyphae anytime and anywhere as long as the conditions are favorable. Sexual pores on the other hand need a time of dormancy after the are formed before they produce more hyphae.
