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trasher [3.6K]
2 years ago
12

Path-breaking medical procedures have begun to pay attention to how microscopic entities aid or hinder bodily health. For exampl

e, fecal transplants have helped cure chronic digestive ailments of people who have suffered for years. These new procedures treat the body not as a separate, contained biological entity but as a complex: ___________
a. biomedicine.
b. anthrobiotic environment.
c. personal biosystem.
d. human microbiome.
Biology
1 answer:
Norma-Jean [14]2 years ago
6 0

Human microbiome that is as complex entity of all the microorganism found in the entire body structure of all living organism and the exact location where they are found.

Fecal transplants is idea for replenishing of good bacteria in the body due to resistance from prolong use of antibiotics. It is specifically for treatment of resistance of antibiotic therapy due <u>C. difficile colitis. </u>

In patient  above 65, infected with  C. difficile colitis and  with resistance to antibiotic especially oral vancomycin, fecal transplant is applied to  effectively cure this.

The transplant is conducted through the Colonoscopy. Where the fecal is discharge into the colon  as the Colonoscope is withdrawal from the colon

.

Since natural  good bacteria are present  in fecal and have no effect on the good bacteria in the body, this procedure increased the total number of bacterial(microbiota),by stabilizing the the microbiome and curing the patients.

more:brainly.com/question/25839750

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References:

Bates, BC et al., Climate Change and Water. Technical Paper of the Intergovernmental Panel on Climate Change. Geneva: IPCC Secretariat, 2008. 224 p.

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Most black bears (Ursus americanus) are black or brown in color. However, occasional white bears of this species appear in some
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Answer and Explanation:

<em><u>The number of observed individuals</u></em>:

  • AA 42
  • AG 24
  • GG 21

<u><em>Total number of individuals, N</em></u>= 87 = 42 + 24 + 21

<u><em>Allelic frequencies</em></u>:

  • f(p) = (2 x AA + AG)/ 2 x N

       f (p)= (2 x 42 + 24) /2 x 87

       f (p) = (84 + 24) / 174

       f (p)= 108 / 174

      f (p) = 0.62

  • f (q) = (2 x GG + AG)/2 x N

        f (q) = (2 x 21 + 24 )/2 x 87

        f (q) = (42 + 24)/ 174

        f (q) = 66/174

        f (q) = 0.38

p + q = 1

0.62 + 0.38 = 1

<em><u>The expected genotypic frequency:</u></em>

  • F (AA)= 0.62 ² = 0.3844
  • F (AG) = 2 x A x G = 2 x 0.62 x 0.38 = 0.4712
  • F (GG) = 0.38 ² = 0.1444

AA + AG + GG = 0.3844 + 0.4712 + 0.1444 = 1

<u><em>The number of expected individuals</em></u>:

AA= (0.62)² x 87 = 0.3844 x 87 = 33.44

AG= (0.4712) x 87 = 40.99

GG= (0.38)² x 87 = 12.563

<u><em>Total number of expected individuals</em></u> = 33.44 + 40.99 + 12.563 = 87

<u><em>Chi square</em></u>= sum (O-E)²/E

  • AA= (O-E)² /E

        AA=(42 - 33.44) ² / 33.44

        AA= 2.2

  • AB= (O-E)² /E      

        AB= (24 - 40.99)²/ 40.99

        AB=7.04

  • BB=(O-E)² /E

        BB= (21-12.563)²/12.563

        BB= 5.66

<u><em>Chi square</em></u>= sum ((O-E)²/E) = 2.2 + 7.04 + 5.66 = 14.9

<u><em>Degrees of freedom</em></u> = genotypes - alleles = 3 - 1 = 2

p value less than 0.05

There is enough evidence to reject the nule hypothesis. The genotype frequencies are not in equilibrium.

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