What type of cell must be in osmotic equilibrium with there surrounding environment

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

ratelena [41]

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.

Genotype Observed frequency

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:

Available data:

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

A B C

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

(Obs-Exp)²/Exp

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

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

6 0

3 years ago

The vitamin necessary for fibrinogen and prothrombin formation (found in green vegetables and released by intestinal bacteria) i

Volgvan

It is vitamin B, vitamin D, vitamin A, vitamin E, and vitamin K

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Which bird(s) was/were most successful after 1,000,000 years? Why were they more successful?

GalinKa [24]

Answer:

•The southwest population of birds

•The tiny birds with a medium beak and a plumage of yellow

• The northeast population of birds where most succesful after 1,000,000 years this could be because their beaks where around medium length and where curved, Their body sixe was also medium.

Adaptations are formed by the random mutations of the cell of an organism by which natural selection has forced them to. This enables the organism to survive and thrive.

8 0

2 years ago

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What is produced if a cell divides by mitosis but does not undergo cytokinesis?

skelet666 [1.2K]

Answer:

If it does not undergo cytokinesis, it means that the cell membrane does not divide to give two daughter cells, hence the two nucleus of daughter cells remain in the parent cell. Over time, this may trigger checkpoint responses in the cell and if cytokinesis cannot be activated to divide the parent cell into two daughter cells, the cell will be killed via apoptosis.

Explanation:

4 0

3 years ago

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Science-

Blizzard [7]

Answer:

The answer would be "plant-like protists".

Explanation:

These protists make their energy on sunlight. This would make them a protist and a plant-like cell.

6 0

2 years ago