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daser333 [38]
3 years ago
6

1. Describe the structure and function of the specialized cells you observed in the video. 2. Research the types of cells presen

t in the kidney.
Biology
1 answer:
Vinvika [58]3 years ago
5 0

Answer:The structure and functions of cells that found in the video include:

Explanation:

1. Nucleus:

Structure=it is a spherical body which is covered by a double membrane which contains hereditary materials.

Functions= it controls all life activities of the cell. It stores hereditary information as it contains DNA.

2. Mitochondria:

Structure: They are oval or rod shaped and bounded by a double membrane

Functions: They are site of cellular respiration.

3. Ribosomes:

Structure: They are small round bodies attached to endoplasmic reticulum.

Functions: They are responsible for protein synthesis.

4. Endoplasmic reticulum:

Structure: They are membrane-like structure that forms channel within the cytoplasm.

Functions: they aid transport of materials within the cytoplasm.

Golgi bodies:

Structure: They are series of disc-shaped sacs.

Functions: They function in synthesis, packaging and distribution of materials within the cell.

2. Types of cell present in kidney

-messangial cell

- Loop of Henle

-podocytes

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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
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<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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