No they are two different environments with different factors
Answer:
Plant cells deal with osmosis by being enclosed in a cell wall.
Animal cells use active transport systems to deal with the problem of osmosis.
Fresh water protists have contractile vacuoles to deal with osmosis.
Many bacteria have cell wall to protect them from osmosis.
Explanation:
Plant cells have a rigid cell wall. If a plant cell is places in a place where the conditions are hypotonic, then the cell will tale up water by osmosis but the cell wall will prevent it from bursting. This condition is termed as the cell being 'turgid'.
As animal cells do not have the rigid cell wall, they use the mechanism of active transport system to stop the cell from bursting during osmosis. In this process, ions are moved out of the cell so that the pressure in the cell due to osmosis can be reduced.
Fresh water protists have a structure present in them called as the contractile vacuole. The contractile vacuole has the capability to remove any excess water from the cell as well storing water if there is not enough water.
Bacteria have peptidoglycan cell walls to prevent osmosis.
<span>Nucleus. Would be your answer</span>
In fact, adding salt does<span> the very opposite of </span>making<span> water </span>boil faster. Instead, itmakes<span> it take longer for the water to </span>boil<span>! The </span>salt<span> actually increases the boiling point of the water, which is when the tendency for the water to evaporate is greater than the tendency for it to remain a liquid on a molecular level.</span>
Answer:
The fraction of heterozygous individuals in the population is 32/100 that equals 0.32 which is the genotipic proportion for these endividuals.
Explanation:
According to Hardy-Weinberg, the allelic frequencies in a locus are represented as p and q, referring to the alleles. The genotypic frequencies after one generation are p² (Homozygous for allele p), 2pq (Heterozygous), q² (Homozygous for the allele q). Populations in H-W equilibrium will get the same allelic frequencies generation after generation. The sum of these allelic frequencies equals 1, this is p + q = 1.
In the exposed example, the r-6 allelic frequency is 0,2. This means that if r-6=0.2, then the other allele frequency (R) is=0.8, and the sum of both the allelic frequencies equals one. This is:
p + q = 1
r-6 + R = 1
0.2 + 0.8 = 1
Then, the genotypic proportion for the homozygous individuals RR is 0.8 ² = 0.64
The genotypic proportion for the homozygous individuals r-6r-6 is 0.2² = 0.04
And the genotypic proportion for heterozygous individuals Rr-6 is 2xRxr-6 = 2 x 0.8 x 0.2 = 0.32
