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ale4655 [162]
3 years ago
14

When were the two major extinction periods that amphibians have survived?

Biology
1 answer:
faltersainse [42]3 years ago
8 0

in the 1950s and in 2007

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The production of red blood cells is regulated by a feedback loop involving erythropoietin. In non-disease states, the productio
LuckyWell [14K]

Answer:

Negative feedback.

Explanation:

The production of erythropoietin by the liver and kidneys is a negative feedback because with the production of erythropoietin, our body tends to move to become more stable state. The erythropoietin is produced by the liver and kidneys to increase the production of red blood cells if the quantity of red blood cells are lower than the normal range. If the number of red blood cells are decreases the oxygen level in our body is low that leads to unstable condition of the body but when the concentration of red blood cells increases, the body move towards stability.

5 0
3 years ago
Water at the top of the slope has potential energy. <br> true or false?
Vinil7 [7]
True. Because its staying still
7 0
2 years ago
1. What season would it be in the northern
Mekhanik [1.2K]
Northern hemisphere : Summer because it is tilted towards the sun.

Southern Hemisphere : Winter because it is tilted away from the sun.
8 0
3 years ago
Which structure is found in all animals
tester [92]

Answer:

cells

Explanation:

These include structures such as the plasma membrane, cytoplasm, nucleus, mitochondria, and ribosomes

3 0
3 years ago
Read 2 more answers
A new kind of tulip is produced that develops only purple or pink flowers. Assume that flower color is controlled by a single-ge
Sophie [7]

Answer:

a. purple allele (C) = 0.609, pink allele (c) = 0.391

b. purple homozygotes = 371, pink homozygotes = 153, heterozygotes = 476

Explanation:

Given -

Purple flowers - 847

Pink flowers - 153

The frequency of recessive genotype i.e

q^ 2 = \frac{153}{1000} \\q^ 2 = 0.153\\

Frequency of recessive allele i.e q is equal to

q = \sqrt{0.153} \\q = 0.391

As per hardy Weinberg's first equilibrium equation -

p + q = 1\\p = 1-q\\p = 1-0.391\\p = 0.609

Frequency of purple homozygous species

= p^2\\= 0.609^2\\= 0.371

Number of purple homozygous species = 0.371 * 1000= 371

Number of pink homozygous species = 0.153 * 1000= 153

Heterozygous species is equal to

(1-0.371-0.153)* 1000\\= 0.476 * 1000\\= 476

5 0
3 years ago
Read 2 more answers
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