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Fudgin [204]
3 years ago
6

What is the most likely plant response to an attack by herbivores?

Biology
2 answers:
Damm [24]3 years ago
6 0

The options are:

A) leaf abscission to prevent further loss

B) early flowering to try to reproduce before being eaten

C) production of chemical compounds for defense or to attract predators

D) production of physical defenses, such as thorns

Answer:

The correct answer of this question is- C) production of chemical compounds for defense or to attract predators

Explanation:

In response to attacks by herbivores, plants produce chemicals that are toxic to the herbivores which can even kill them or some chemicals can spread in the air and can attract the predators of those herbivores that attack that plant species.

Plants produce secondary metabolites that play an important role in protecting the plant from predators because they can be toxic and antimicrobial for example flavonoids, alkaloids, saponins. Plants do not get harmed by these metabolites because they have their own detoxification method. So the correct answer is C.

Sav [38]3 years ago
6 0

The question is incomplete as it does not have the options which are:

A) leaf abscission to prevent further loss

B) early flowering to try and reproduce before being eaten

C) production of chemical compounds for defense or to attract predators

D) production of physical defenses, such as thorns

E) production of thicker bark and cuticle to make it more difficult to eat

Answer:

D) production of physical defenses, such as thorns

Explanation:

The plants are autotrophic organism which can produce food on their own therefore they become the producers of the food chain that is they are eaten by the primary consumers.

So to protect themselves the plants adapted to protect themselves and formed the needle-like structures called thorns by modifying their stem. The needle-like structure can injure or harm the herbivore when they try to eat them.

Thus, Option-D is correct.

You might be interested in
Imagine that a newly discovered, recessively inherited disease is expressed only in individuals with type O blood, although the
erastova [34]

Complete question:

Imagine that a newly discovered, recessively inherited disease is expressed only in individuals with type O blood, although the disease and blood group are independently inherited.

A normal man with type A blood and a normal woman with type B blood have already had one child with the disease. The woman is now pregnant for a second time.

Assuming that both parents are heterozygous for the gene that causes the disease, what is the probability that the second child will also have the disease? Express your answer as a fraction using the slash symbol and no spaces (for example, 1/2).

Answer:

The probability that the second child will also have the disease is 1/16.

Explanation:

<u>Available data:</u>

  • Two genes independently inherited: one for blood type, the other for disease
  • Man with type A blood  x  Woman with type B blood
  • Both parents are heterozygous for the gene that causes the disease; Dd

If the man has A blood, and the woman has B blood, and they already have an affected child, this means that they must be heterozygous for blood type too.

Cross:  

Parentals)           AiDd         x      BiDd

Gametes)  AD  Ad  iD  id       BD  Bd  iD  id

Punnett square)      AD       Ad       iD       id

                     BD   ABDD  ABDd  BiDD BiDd

                     Bd   ABDd   ABdd   BiDd Bidd

                     iD    AiDD    AiDd    iiDD   iiDd

                     id    AiDd     Aidd     iiDd    iidd

F1) <u>Genotype</u>:

1/16 ABDD

2/16 ABDd

1/16 ABdd

1/16 AiDD

1/16 BiDD

2/16 AiDd

2/16 BiDd

1/16 Aidd

1/16 Bidd

1/16 iiDD

2/16 iiDd

1/16 iidd

<u>Phenotype:</u>

3/16 A/B normal

4/16 A normal

4/16 B normal

3/16 0 normal

1/16 0 affected by the disease.

3 0
3 years ago
terry physics pushes a 75kg lab cart so that is accelerates from rest with a velocity of 4.4m/s in 3.1 secounds. what is the for
svetlana [45]

106.5N

Explanation:

Given parameters:

Mass of cart = 75kg

Velocity = 4.4m/s

time taken = 3.1s

Unknown:

Force = ?

Solution:

To solve this problem, we are going to apply newton's second law of motion. The law states that "the acceleration of a body is directly proportional to the force on the body and inversely proportional to the mass".

      F = m  x   a

F = force on body

m = mass of body

a = acceleration of the body

 But we know that acceleration is the change in velocity with time of a body;

     a  = \frac{change in velocity }{time}

  F = m x \frac{V}{t}

V is the velocity

t is the time taken

   F = \frac{75  x  4.4}{3.1}   = 106.5N

learn more:

Newton's law brainly.com/question/11411375

#learnwithBrainly

6 0
3 years ago
Which statement describes adaptions necessary for plants to live on land? Check all that apply​
Alex777 [14]
Hm what’s the all that apply?
8 0
3 years ago
Morning! I just need some help with this question! Ex. If it is A and it is mitosis then you would put A-Mitosis
polet [3.4K]

In cell biology, mitosis (/maɪˈtoʊsɪs/) is a part of the cell cycle in which replicated chromosomes are separated into two new nuclei. Cell division gives rise to genetically identical cells in which the total number of chromosomes is maintained. Therefore, mitosis is also known as equational division. In general, mitosis is preceded by S phase of interphase (during which DNA replication occurs) and is often followed by telophase and cytokinesis; which divides the cytoplasm, organelles and cell membrane of one cell into two new cells containing roughly equal shares of these cellular components. The different stages of mitosis altogether define the mitotic (M) phase of an animal cell cycle—the division of the mother cell into two daughter cells genetically identical to each other. So yes i would put A-Mitosis

Closed intranuclear pleuromitosis is typical of Foraminifera, some Prasinomonadida, some Kinetoplastida, the Oxymonadida, the Haplosporidia, many fungi ( chytrids, oomycetes, zygomycetes, ascomycetes ), and some Radiolaria ( Spumellaria and Acantharia ...

Closed extranuclear pleuromitosis occurs in Trichomonadida and Dinoflagellata.

Closed orthomitosis is found among diatoms, ciliates, some Microsporidia, unicellular yeasts and some multicellular fungi.

Semiopen pleuromitosis is typical of most Apicomplexa.

Semiopen orthomitosis occurs with different variants in some amoebae ( Lobosa) and some green flagellates (e.g., Raphidophyta or Volvox ).

5 0
2 years ago
Assume you inoculated 100 cells, with a generation time of 20 minutes, into 100 ml of nutrient broth. You then inoculated 100 ce
Marina86 [1]

Answer:

a) the same number of cells in both

Explanation:

In cells reproduction, in both cases we consider the same specie, the same generation time, and we assume the same broth.

The only advantage of the container with more milliliters of nutrients is that when the population increases, it will need more nutrients, so maybe the reproduction rate in the container with 100 ml will be lower.

But if the temperature, the quantity of nutrients are the same in both containers, so the volume is not a variable to affect the speed or low the reproduction rate.

6 0
3 years ago
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