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Readme [11.4K]
3 years ago
9

Examine the list of organisms.

Biology
1 answer:
astraxan [27]3 years ago
4 0

Answer:

amoeba bacteria

Explanation:

amoeba is a bacteria

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Cellular respiration involves two phases. The anaerobic phase does not involve oxygen, while the aerobic phase does. Where does
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While the majority of anaerobic (without oxygen) and aerobic (with oxygen) respiration occurs in the cytoplasm and mitochondria, respectively, of the cell.

<h3>What is cellular respiration?</h3>

The process through which food, in the form of sugar (glucose), is converted into energy within cells is known as cellular respiration. All kinds of cellular operations are then powered by the energy that is stored in ATP molecules.

There are three types of cellular respiration

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Chromium (Cr) can combine with chlorine (Cl) to form chromium chloride (CrCl3).
Viktor [21]
The right answer is going to be D


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3 years ago
Most black bears (Ursus americanus) are black or brown in color. However, occasional white bears of this species appear in some
Sphinxa [80]

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.

5 0
3 years ago
A. a biosphere <br> B. a community <br> C. a population <br> D. a biome
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there is no question here

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3 years ago
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Answer:

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