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Alja [10]
3 years ago
6

1. The cells in our bodies are surrounded by these types of solutions.

Biology
1 answer:
slamgirl [31]3 years ago
7 0

Answer:

1. The cells in our bodies are surrounded by these types of solutions. → Isotonic solution.

3. When an animal cell is places in this solution, it will burst (get layer)  → Hypotonic solution.

4. When an animal cell is placed in this solution, it will shrivel or shrink (get smaller) → Hypertonic solution.

Explanation:

The cells in the body are in a balance of substances —concentration of solutes— between their cytoplasm and the extracellular space. This balance is dynamic in living beings, due to the constant exchange of ions and substances between the intracellular and extracellular space.  For this reason, the extracellular medium is isotonic with the cytoplasm.

<u>A cell can lose or gain water depending on the amount of solutes that a medium has in which it is found</u>, with respect to the cytoplasm. This difference in solutes concentrations produces an osmotic gradient that drags water from the least concentrated solution to the most concentrated, through the process of osmosis, which seeks to achieve an equilibrium of concentrations.

  • <em>When a animal cell is exposed to a </em><em>hypertonic solution</em><em> </em>—<em>with a higher concentration of solutes</em>— <em>it loses water and tends to </em><em>dehydrate and become smaller</em><em>.</em>
  • <em>An animal cell in a </em><em>hypotonic solution</em><em> receives water, so it can </em><em>expand and even burst</em><em>.</em>

In practice, the concentrations of intracellular and extracellular solutes depend not only on the osmotic gradient, but also on the concentration gradient of substances.

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3 years ago
In a population of cats, the phenotypic frequency of black cats is 91%, and the phenotypic frequency of white cats is 9%. Assumi
Talja [164]

Answer:

1. Allele frequency of b = 0.09 (or 9%)

2. Allele frequency of B = 0.91 (0.91%)

3. Genotype frequency of BB = 0.8281 (or 82.81%)

4. Genotype frequency of Bb = 0.1638 (or 16.38%)

Explanation:

Given that:

p = the frequency of the dominant allele (represented here by B)  = 0.91

q = the frequency of the recessive allele (represented here by b)  = 0.09

For a population in genetic equilibrium:

p + q = 1.0 (The sum of the frequencies of both alleles is 100%.)

(p + q)^2 = 1

Therefore:

p^2 + 2pq + q^2 = 1

in which:  

p^2 = frequency of BB (homozygous dominant)

2pq = frequency of Bb (heterozygous)

q^2 = frequency of bb (homozygous recessive)

p^2 = 0.91^2 = 0.8281

2pq = 2(0.91)(0.9) = 0.1638

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Answer:

The lack of ATP would slow nerve responses to generate an action potential

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