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Arturiano [62]
3 years ago
12

Keystone species play a key role in A. Niche specialization B. Mutualism. C. Community stability. D. Habitat destruction

Biology
2 answers:
ArbitrLikvidat [17]3 years ago
5 0
It world by C: Community Stability
lesya [120]3 years ago
5 0

Answer: The correct answer is C) Community stability.

Keystone species is an individual species or a group of species of plants and animals, which play a significant role in maintaining the structure of community and the functioning of ecosystem.

This means that when these species are removed from an ecosystem, they will result in the dramatic changes in the structure and functioning of the ecosystem.

Example-  In terrestrial habitats, fire ants behaves like a keystone species ( a predator keystone species) that suppresses the number of individuals and species of arthropods, which could have harmful effects in the agriculture.


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2 years ago
A locus that affects susceptibility to a degenerative brain disease has two alleles, V and v. In a population, 16 people have ge
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Answer:

Yes, the population is evolving

Explanation:

Hardy-Weinberg equilibrium states that a population is not evolving if it is in hardy weinberg equilibrium as the alleles frequency do not change over time in generations.

In mathematical equation, the genotype frequency will be -P²+2Pq+q²=1 and allele frequency will be p+q=1, where p = dominant allele and q will be recessive allele.

In the question, since the observed genotype are-

P² - VV= 16

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q²- vv- 12

<u>1. Calculate allele frequency</u>

Now we will calculate if the expected value is same or close or not.

We will calculate allele frequency

Population has 120 individuals which means it has - 240 alleles.

In which V (p) will be- 32 +92= 124 allele

p- dominant allele/ total allele=  124/ 240 = 0.52

P= 0.52

q will be 1-0.52= 0.48

<u>2. Expected genotype frequency</u>

Now we will calculate expected genotypic frequency.

P²- 0.52 X 0.52= 0.27

2Pq = 2 x 0.52x 0.48 =  0.5

q²= 0.48 x 0.48 =  0.23

<u>3. Expected progeny </u>

P²= 0.27 x 120= 32 VV

2Pq= 0.5 X 120= 60 Vv

q²= 0.23 x 120= 28 vv

Since the observed value is not equal to the expected value therefore the population is not in Hardy- Weinberg equilibrium and the population is evolving.

Thus, Yes, the population is evolving is correct.

4 0
3 years ago
Read 2 more answers
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