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Bond [772]
3 years ago
10

The regulation of body temperature is an example of which type of homeostatic control?

Biology
1 answer:
Arturiano [62]3 years ago
6 0
Its our warm bloodlessness we stay at one temp because our blood has to be that warm at all times 
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Name the form of nitrogen which higher plants use​
Tresset [83]

Answer:

Nitrate

Nitrate is the form of nitrogen most used by plants for growth and development. Nitrate is the form that can most easily be lost to groundwater. Ammonium taken in by plants is used directly in proteins. This form is not lost as easily from the soil.

Explanation:

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What trait most likely evolved in ferns that was not
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HI Mate! The right answer is vascular tissue!

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Taxonomy seeks to _____.
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Taxonomy seeks to organize all living things in a logical manner.
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Read 2 more answers
- Why are cars and trucks today powered by petroleum products?
AysviL [449]

Answer:

Due to structure of engine.

Explanation:

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6 0
3 years ago
A population of 150 individuals has an allele frequency of 0.2 for the dominant allele (H) and a frequency of 0.8 for the recess
Amanda [17]
Hardy-Weinberg Equation (HW) states that following certain biological tenets or requirements, the total frequency of all homozygous dominant alleles (p) and the total frequency of all homozygous recessive alleles (q) for a gene, account for the total # of alleles for that gene in that HW population, which is 100% or 1.00 as a decimel. So in short: p + q = 1, and additionally (p+q)^2 = 1^2, or 1
So (p+q)(p+q) algebraically works out to p^2 + 2pq + q^2 = 1, where p^2 = genotype frequency of homozygous dominant individuals, 2pq = genotype frequency of heterozygous individuals, and q^2 = genotype frequency of homozygous recessive individuals.
The problem states that Ptotal = 150 individuals, H frequency (p) = 0.2, and h frequency (q) = 0.8.
So homozygous dominant individuals (HH) = p^2 = (0.2)^2 = 0.04 or 4% of 150 --> 6 people
Heterozygous individuals (Hh) = 2pq = 2(0.2)(0.8) = 0.32 or 32% of 150
--> 48 people
And homozygous recessive individuals (hh) = q^2 = (0.8)^2 = 0.64 = 64% of 150 --> 96 people
Hope that helps you to understand how to solve these types of population genetics problems!
6 0
3 years ago
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