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slava [35]
3 years ago
13

How are microorganisms involved in the movement of nitrogen from the atmosphere to the bodies of other organisms? Why can’t othe

r organisms just get nitrogen from air?
Biology
1 answer:
Maru [420]3 years ago
3 0

Microorganisms that are involved in the nitrogen cycle are able to break the triple of elemental nitrogen using special enzymes that they have developed and make ammonium. Other organisms like plants are unable to break the triple bond and therefore rely on these microorganisms so that they can absorb the ammonia

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Which three of the following entities are acellular?.
arsen [322]

Answer:

Prions archea, viroids bacteria, viruses protists are acellular while fungi is cellular.

While viruses have indeed been closely researched for more than a century for infections of people and animals, most viruses, particularly those that infect microorganisms, are still poorly understood. 

Pathogenic RNA molecules are known as viroids. Contagious protein particles are identified as prions. Viruses are larger than that of viroids. Viroids are larger than prions in size.

Molds, wild yeasts, as well as mushrooms are examples of fungi. Numerous fungi are multicellular and also have nuclei in their cells. A protein-encased DNA or RNA core defines viruses being non - cellular organisms. Some scientists disagree with the notion that viruses are living things.

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brainly.com/question/28475730cellular

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3 years ago
Where did the energy to power the light-independent reactions come from?
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6 0
2 years ago
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Alu element at the PV92 locus on Chromosome 16. In this population, 436 people have a heterozygous genotype and 102 people have
Lesechka [4]

Answer:

Explanation:

According to the exercise we can infer that for the Alu insert:

p: positive allelic frequency

q: negative allelic frequency

maintaining that the population is in equilibrium we can carry out the following formula

p + q = 1 and pp + 2pq + qq = 1

looking for the genotype frequency we clear and obtain the following data

genotype frequency of 2pq = 436/1000 = 0.436

The genotype frequency of qq = 102/1000 = 0.102

This is how we look now:

number of positive people for Alu = 1000- (436 + 102) = 1000- 538 = 46

In this way it is resolved that:

genotypic frequency of pp = 462/1000 = 0.462

  p = 0.462 + (0.436 / 2) = 0.462 + 0.218 = 0.680

q = 0.102+ (0.436 / 2) = 0.102 + 0.218 = 0.320

According to the exercise carried out it is deduced:

The value of p in the population is = 0.68

We conclude that our prognosis showing a homozygous positive genotype is: 0.462

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