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olya-2409 [2.1K]
3 years ago
7

Which of the following is an example of the endocrine system maintaining homeostasis? (3 points)

Biology
1 answer:
motikmotik3 years ago
4 0

Answer:

C. Sending a message to the pituitary gland to start producing a hormone when the levels in the body are too low

Explanation:

It's not A, because that's the nervous system, not endocrine.

It's not B, because homeostasis = balance, and not stopping production isn't balance

It's not D, because that's the nervous system again.

The endocrine system governs hormones.

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You are studying a population of ground squirrels. You trap 100 individuals and find 9 that have a light colored coat, and the r
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The question is incomplete. Here is the complete question.

You are studying a population of ground squirrels. You trap 100 individuals and find 9 that have a light colored coat and the rest have a dark coat. You know that light coat color is recessive in ground squirrels. What is the allele frequency for the light coat allele  and the dark coat allele? How many of the dark coated individuals do you expect to be heterozygotes at the coat color gene? Assume the population is at H-W equilibrium.

Answer: Allele frequency for light coat: q = 0.3;

              Allele frequency for dark coat: p = 0.7;

             There will be 42 heterozygote;

Explanation: A population in H-W equilibrium is a population which doesn't change from one generation to the next if there are no disturbing factors, as migration or mutation.

It follows the equation: p^{2}+2pq+q^{2}=1.

p represents the frequency of the dominant allele;

q represents the frequency of the recessive allele;

2pq represents heterozygote individuals;

The sum of frequencies of both alleles must be 1, i.e.: p+q=1

For the population of squirrels:

  • Frequency of recessive allele

q^{2} = \frac{9}{100}

q = \sqrt{\frac{9}{100}}

q = 0.3

  • Frequency of dominant allele

p+q=1

p=1-q

p=1-0.3

p = 0.7

  • Frequency of heterozygote

2pq = 2*0.7*0.3 = 0.42

In a population of 100 individuals: 0.42*100 = 42

The allele frequency for the light coat is 0.3; for the dark coat is 0.7.

It is expected to have 42 heterozygote individuals in this population.

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