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3 years ago

To protect itself from herbivores, plants

Biology

2 answers:

svetoff [14.1K]3 years ago

6 0

Answer: B) both secrete toxins and have thorns

Plants undergoes several adaptations in order to survive in their environment and to defend themselves from the herbivores. The presence of thorns on the surface of the plants is a structural adaptation, these thorns will get inserted in the herbivores and will discourage their feeding on the plants. They secrete toxins to the external surface, it is physiological adaptation, which can kill the herbivores and also discourage other members from feeding.

makkiz [27]3 years ago

5 0

Plants secrete toxins and have thorns to protect themselves from herbivores.

Further Explanation:

When a plant is attacked by the pathogen, sometimes a hypersensitive response (HR) is initiated. HR response leads to rapid cell death of the infected tissue. This process uses the wounded tissue as a barrier that prevents pests from attacking the rest of the plant tissue. When a leaf is chewed by the herbivore, proteinase inhibitors are produced not only at the wounded site but throughout the plant. They bind to the digestive enzymes in the gut of herbivores.

Plants also secrete secondary metabolite. They are organic compounds that are not directly involved in the growth, reproduction or development of plants.These compounds are generally involved in the defense mechanism of the plants. Also, plant undergo in many adaptation to protect themselves from herbivores and other pathogens. Thorns are the structural adaptation of plants that insert into the herbivores and protect themselves from plants.

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Answer Details:

Grade: High school

Subject: Biology

Chapter: Plant Defense Mechanism

Keywords:

Plant, herbivores, algae, rapid cell death, proteinase, hypersensitive, response, pathogen, insert, protect, response, attacked.

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Answer:

See the answer below

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Using the formula for calculating Chi square ( $X^2$ ):

$X^2$ = $\frac{(O - E)^2}{E}$ where O = observed frequency and E = expected frequency.

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Hence, expected frequencies are calculated as:

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phenotype (3) = 3/16 x 293 = 54.94

phenotype (4) = 1/16 x 293 = 18.31

The $X^2$ is calculated thus:

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1 88 164.81 $\frac{(88 - 164.81)^2}{164.81}$ = 35.80

2 62 54.94 $\frac{(62 - 54.94)^2}{54.94}$ = 0.91

3 62 54.94 $\frac{(62 - 54.94)^2}{54.94}$ = 0.91

4 81 18.31 $\frac{(81 - 18.31)^2}{18.31}$ = 214.64

Total $X^2$ = 35.80 + 0.91 + 0.91 + 214.64 = 252.26

To the nearest tenth = 252

Degree of freedom = n - 1 = 4- 1 = 3

$X^2$ tabulated = 7.815

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Can someone plz help me? :(

Brums [2.3K]

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