The mass of the box increased and the acceleration of the box decreased
After a few months of eating less meat and more fruits and vegetable, the liver will respond by producing more cholesterol.
<h3>What is body homeostasis?</h3>
The term body homeostasis makes reference to the internal equilibrium (homeostasis) of the body.
Body homeostasis may involve the production of certain substances when they are not consumed from the diet.
In this case, it is possible to increase cholesterol levels by having certain healthy habits (food habits).
Learn more about body homeostasis here:
brainly.com/question/13349261
They produce heat inside the core, creating a steam like substance (carbon dioxide)
Answer:
The correct answer is -
- to match the mystery food sample to a definite positive result (gelatin, oil.)
- to match the mystery food sample to a definite negative result (water)
Explanation:
Adding an indicator or any indicator to an unknown food sample or any sample and then adding the same indicator to the known sample is necessary to check and confirm the positive result and negative result that will help to match with the experiment result.
For positive result adding an indicator to get a definite positive result to mystery food for example gelatin or oil. Similarly for a definite negative result adding it to known food or water.
Answer:
<h2>
Allele frequencies for B and b.
</h2><h2>
"b"(q) allele frequency = 0.60
</h2><h2>
</h2><h2>
"B"(p) allele frequency = 0.40
</h2><h2>
</h2><h2>
Genotype frequencies;
</h2><h2>
BB = 0.16
</h2><h2>
Bb = 0.48
</h2><h2>
Bb = 0.36
</h2>
Explanation:
Given
Non-baldness (B) is dominant on baldness(b), so B is dominant over b.
Homozygous pattern baldness male (bb) = 360,
Heterozygous non- baldness (Bb)= 480,
Homozygous non-baldness (BB)= 160.
So, we can also denote then by genotypes only,
BB= 160;
Bb= 480;
bb= 360;
Total= 1000
Allele frequency q² (bb) = 360/1000=0.36
allele fequency for q( b)= √o.36=0.60.
Allele frequency for p²(BB) = 160/1000=0.16
allele frequency p(B)= √0.16 = 0.4
Expected genotype frequencies;
BB = 160/1000 = 0.16
Bb = 480/1000= 0.48
Bb = 360/1000= 0.36
