In a broad range of terrestrial environments, microorganisms are the key decomposers of organic matter and release nutrients in the soil for plant growth as well as CO2 and CH4 into the atmosphere.
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All organisms arose from a single common ancestor.
An endosymbiont or endobiont is any organism that lives inside the frame or cells of some other organism most often, though no longer continually, in a mutualistic relationship.
Bacterial endosymbionts result in dramatic phenotypes in their arthropod hosts, including cytoplasmic incompatibility, feminization, parthenogenesis, male killing, parasitoid protection, and pathogen blocking.
Endosymbionts, such as Wolbachia, Rickettsia, and Cardinium, are a type of bacteria generally located in arthropod species inclusive of bugs, spiders, crustaceans, and mites, in addition to other invertebrates along with filarial nematodesEndosymbiosis is a form of symbiosis in which the symbiont lives in the body of its host and the symbiont in an endosymbiosis is known as an endosymbiont.
Learn more about endosymbiotic here;-brainly.com/question/1698852
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Interspecific competition
A type of competition in which organisms belong to different species compete with each others for same resources.
Example
Leopard and lion.
intraspecific competition
A type of competition in which organism belong to same species compete with each others for same resources.
Example
birds for breeding or shelter compete with each others
Answer:
6,25%
Explanation:
Considering that the couple has a trait of sickle cell anemia, we know that both are heterozygous for the disease (Aa) and therefore can have children with the following genotypes:
Parents: Aa X Aa
Children: AA(A x A), Aa(A x a), Aa (a x A) and aa(a x a)
Knowing that sickle cell anemia only occurs in homozygous individuals, the probability for children to have the disease according to each crossing is:
A x A = 1/4 = 25%
A x a = 1/4 = 25%
a x A = 1/4 = 25%
a x a = 1/4 = 25%
The probability of forming each homozygous child (aa) is 1/4 or 25%. Since they are two children, the probability of both having sickle cell anemia is calculated by multiplying the probability of each, so:
1/4 × 1/4 = 1/16 = 0.0625 = 6.25%
It is concluded that the probability of a heterozygous couple for sickle cell anemia to have two children with the disease is 6.25%.
