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Mademuasel [1]
3 years ago
13

The Pacific sea otter was hunted to near extinction in the early 1900s. The loss of this predator resulted in several changes to

the kelp forest ecosystem. Populations of sea urchins, which are one of the sea otters' favorite prey, grew to historically high numbers. Sea urchins graze on kelp, and the extremely large populations of sea urchins significantly reduced kelp biomass. Researchers have hypothesized that overgrazing of kelp by urchins may have hastened the extinction of the Stellar's sea cow, which also depended on kelp as a food source. Identify the ecosystem relationships that were disrupted in the kelp forests by the loss of sea otters.
Biology
1 answer:
BaLLatris [955]3 years ago
7 0

Answer:

  • Interspecific competition between sea urchins and Stellar's sea cows  
  • Predation fo sea urchins by sea otters

Explanation:

As urchins and Stellar's sea cows feed on seaweed, since these algae are low in quantity, these two organisms end up competing for food, creating an inter-specific competition, which occurs when two organisms compete for the same resource within a same ecological niche. However, as the hedgehog has advantages in this niche, this competition was interrupted.

The otters' predatorism in relation to sea urchins was also interrupted, as the otters reached a very low population level, failing to control the hedgehog population and generating an environmental imbalance.

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small group of 100 people decide to isolate themselves from the world and move to a small and remote deserted island. Out of thi
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Answer:

Frequency of p = 0.684

Frequency of p = 0.316

Number of individuals with homozygous dominant (AA) = 47

Number of individuals with heterozygous (Aa)= 43

Number of individuals with homozygous recessive (aa) = 10

Explanation:

Out of 100 people, 10 have albino skin (aa)

So, the frequency of homozygous recessive individuals (q^{2}) is \frac{10}{100} = 0.1

Now, q will be

= \sqrt{q^{2} } = \sqrt{0.1} \\= 0.316

As per Hardy Weinberg's equation -

p + q = 1

Substituting the value of q in above equation, we get -

p + 0.316 = 1p = 1 -0.316\\p = 0.684

Now the frequency of homozygous dominant (AA) will be

p^{2} = 0.684^{2} \\= 0.467

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Like wise out of 100 people 0.1 * 100 = 10 people are homozygous recessive (aa)

As per As per Hardy Weinberg's equation-

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