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Rom4ik [11]
3 years ago
10

hen faced with a sudden drop in environmental temperature, what will happen to an ectothermic animal? it is likely to experience

a drop in body temperature it will increase muscle activity to produce heat adjust its fats stores to be more insulating
Biology
1 answer:
sergey [27]3 years ago
6 0

Answer:in addition are formation of goose pimples and erection of hair muscles so as to trap more air thus insulation

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Which organisam is a pruduser?<br><br> A. Mouse<br> B. Snake <br> C. grass<br> D.Owl
nordsb [41]

C. grass

is the answer because it provides food for other organisms

6 0
4 years ago
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ATP and ADP both play a role in cellular reactions involving energy how are these two molecules related
Rama09 [41]
Both are used for energy to the body.
8 0
4 years ago
Which statements best describe natural selection? Check all that apply.
Vikki [24]

Answer:a:population must have genetic variation c:Natural selection occurs when the environment changes e:Organisms best suited for the environment survive and reproduce

Explanation:

3 0
3 years ago
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How did the oceans form over time?
Fantom [35]
The ocean formed<span> billions of years ago. </span>Over<span> vast periods of </span>time, our primitiveoceans formed<span>. Water remained a gas until the Earth cooled below 212 degrees Fahrenheit . At this </span>time<span>, about 3.8 billion years ago, the water condensed into rain which filled the basins that are now our </span>oceans<span>. (asteroids too)</span>
8 0
3 years ago
Cystic fibrosis is an autosomal recessive condition. The frequency of cystic fibrosis is 0.000484, or about 1 in 2000 people. Wh
mixer [17]

Answer: Allele frequency for cystic fibrosis is 0.022.

4.302% carriers (heterozygous) for cystic fibrosis in the population.

Explanation: Using F to represent the allele, homozygosity (ff) produces a disorder in autosomal recessive disorders. Therefore, the allele representation goes thus;

Cystic fibrosis (Homozygous) = ff = q²

Normal (Homozygous) = FF = p²

Carrier (Heterozygous) = Ff = 2pq

Frequency of cystic fibrosis q² is given as 0.000484

Therefore, allele frequency of cystic fibrosis, q = √0.000484 = 0.022

Recall that p+q = 1

Therefore, allele frequency for normal individual,

p = 1 - q = 1 - 0.022 = 0.978

Also, carrier (heterozygous) frequency, 2pq = 2 x 0.978 x 0.022 = 0.043032

Hence, Percentage of people who are carrier for cystic fibrosis in the population will be

0.043032 x 100 = 4.302%

6 0
3 years ago
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