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LUCKY_DIMON [66]
3 years ago
7

Cquired immunity involves both a humoral immune response and a cell-mediated immune response. To review the definitions of these

immune responses, see Hint 1.
Sort the items into the appropriate bins depending on whether they are involved in the humoral response, the cell-mediated response, or both the humoral and the cell-mediated responses.
antigen presenting cells
plasma cells
memory cells
antibodies
b cells
helper T cells and cytokinesis
cytotoxic T cells
The categories are humoral response, cell mediated response, and both.
Biology
2 answers:
Ivenika [448]3 years ago
4 0

Answer:

Antigen presenting cells - cell-mediated and humoral response

Plasma cells - Humoral response

Memory cells - Humoral response

Antibodies - Humoral response

B cells - Humoral response

Helper T cells and cytokines - Cell-mediated response

Cytotoxic T cells - Cell mediated response

Explanation:

Humoral secretes antibodies to fight against antigens. They are B cells which are also referred to as antibody-mediated response. Cell-mediated immunity works inside the infected cells where it destroys the pathogens by the process of lysis or by releasing cytokines. Plasma cells, memory cells, antibodies and B cells are humoral because they secretes antibodies which fight against antigens and are found outside of the cell. Helper T cells , cytokines and cytotoxic T cells are cell-mediated immunity because they secrete antigens which fight against pathogens. Antigen presenting cells are both cell mediated and humoral because they can function as both B cells and T cells

Rudiy273 years ago
3 0

T cells identify pathogens based on antigens on the surface of pathogens. The T cells then multiply and stimulate B cells to produce antibodies. The antibodies attach to the antigens, making the pathogens a target for phagocytes.

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The Offspring produced by a cross between two given types of plants can be any of the three genotypes denoted by A, B, and C. A
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Complete question:

The offspring produced between two given types of plants can be any of the three genotypes, denoted by A, B and C. A theoretical model of gene inheritance suggests that the offspring of types A, B and C should be in a 1:2:1 ratio (meaning 25% A, 50% B, and 25% C). For experimental verification, 100 plants are bred by crossing the two given types. Their genetic classifications are recorded in the table below.

<em><u>Genotype       Observed frequency</u></em>

    A             →        18 individuals

    B             →        55 individuals

    C             →        27 individuals

Do these contradict the genetic model?  

Use a 0.05 level of significance.

Determine the chi-square test statistic.

Answer:

Do these contradict the genetic model? No, according to the chi-square test, there is not enough evidence to reject the null hypothesis of the population being in equilibrium.  

Explanation:

<u>Available data</u>:

  • Crossed genotypes: two
  • Genotypes among the offspring; Three → A, B, and C
  • Expected phenotypic ratio → 1:2:1
  • Total number of individuals, N = 100
  • A = 18 individuals
  • B = 55 individuals
  • C = 27 individuals

So, let us first state the hypothesis:

  • H₀= the population is equilibrium for this locus → F(A) = 25%,  F(B) = 50%, F(C) = 25%  
  • H₁ = the population is not in equilibrium

Now, let us calculate the number of expected individuals, according to their expected ratio.

4 -------------- 100% -------------100 individuals

1 ---------------  25% -------------X = 25 individuals A

2 --------------  50% -------------X = 50 individuals B

1----------------- 25%--------------X = 25 individuals C

<u>                                                   A                             B                           C</u>

  • Observed                         18                            55                         27
  • Expected                         25                           50                         25
  • (Obs-Exp)²/Exp                1.96                        0.5                        0.16

<u>(Obs-Exp)²/Exp</u>

A)  (18 - 25)²/25 = 49/25 = 1.96

B)  (55 - 50)² / 50 = 25/50 = 0.5

C)  (27 - 25)²/25 = 4/25 = 0.16

Chi square = X² = Σ(Obs-Exp)²/Exp  

  • ∑ is the sum of the terms
  • O are the Observed individuals: 2 in chamber B, and 18 in chamber A.  
  • E are the Expected individuals: 10 in each chamber  

X² = ∑ ((O-E)²/E) = 1.96 + 0.5 + 0.16 = 2.62

Freedom degrees = 2

Significance level, 5% = 0.05  

Table value/Critical value = 5.99

X² < Critical value

2.62 < 5.99    

<em>These results suggest that there is </em><u><em>not enough evidence to reject</em></u><em> the null hypothesis. We can assume that </em><u><em>the locus under study in this population is in equilibrium H-W.  </em></u>

 

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ANSWER
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explanation: what i found on google
8 0
3 years ago
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