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Doss [256]
3 years ago
13

Which statement best explains why herbivores are not considered parasites? A. They do not harm the plants they eat. B. They do n

ot live on or in the body of a host. C. They do not depend on the plants they eat for survival. D. They do not actively hunt other species for food.
Biology
2 answers:
Ymorist [56]3 years ago
8 0

The correct answer is Option B

The word parasite means that the organism must live inside(intracellular) or on the outside body of the organism(extracellular).

The herbivores do not live on the host (plants) for its survival. They just eat the plants and fruits for getting energy.

Hence, the herbivores cannot be called as parasite.

NemiM [27]3 years ago
6 0
The answer would be B due to the fact that herbivores do not live on or in the plant they eat
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Answer:

The answer is the first option. The trailing edge is composed of many filopodia containing tight parallel bundles. These filopodia push trailing edge away from the location it was previously occupying.

Explanation:

Cells migrating in tissues exhibit both filopodia and lamellipodia. These thin protrusive extensions are ideal for exploring and penetrating tissue spaces. They are also well suited for intercalating between cells, such as during the migration of leukocytes across endothelial layers.

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1 year ago
Suppose a population of flour beetles has 10,000 individuals. There are two alleles possible for the gene that determines body c
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Answer:

0.75

Explanation:

Given frequency of the two alleles i.e "B" and "b" are equal .

As per Hardy–Weinberg equilibrium, sum of frequency of the two alleles is equal to one.

i.e p+ q= 1\\

Since p=q\\

p and q will be equal to 0.5

Frequency of genotype "BB"

= p^{2} \\= 0.5^{2} \\= 0.25

Frequency of genotype "bb"

= q^{2} \\= 0.5^{2} \\= 0.25

As per second equation of Hardy–Weinberg equilibrium,

p^{2} + q^{2} + 2pq=1

Substituting the values of p and q in above equation we get -

0.25+0.25+2pq=1\\2pq = 1-0.25-0.25\\2pq = 0.5

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number of red beetles with genotype (BB)

= 0.25 * 10000\\= 2500

number of red beetles with genotype (BB)

= 0.5 * 10000\\= 5000

Total red beetles

2500+ 5000= 7500\\

Frequency of red beetles

= \frac{7500}{10000} \\= 0.75

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