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erastovalidia [21]
2 years ago
11

A beaker of water is separated by a selectively permeable membrane that allows the passage of water but not ions. A solution of

4% sodium chloride (NaCl) is placed on one side of the membrane and pure water is placed on the other side of the membrane. The solutions are allowed to equilibrate. At this point, which process is going on
Biology
1 answer:
postnew [5]2 years ago
3 0

Answer:

osmosis

Explanation:

The process that was going is <u>osmosis</u>.

<em>Osmosis is the movement of water molecules from regions of high water potential to regions of low water potentials through a selectively permeable membrane.</em>

In this case, the membrane is only permeable to water movement and not ions and water molecules move from the side with pure water (which happens to have higher water potential) to the side with 4% sodium chloride until equilibrium in water potential is established between the two sides.

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, cell, Synapse,  Neuron,

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In the African savanna, more wildebeests are born than the environment can support. Which is most likely to happen over a period
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Eventually, the environment will be unable to support the production of wildebeests, this will cause there to be a loss of vegitation that then causes a food shortage, and then the wildebeests will starve.  

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Hope this helps! :)

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2 years ago
The term "net energy” refers to the amount of energy available from a resource minus the amount of
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What are the dated fossils called that help scientists to identify or date rocks or fossils?
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Explanation:

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5 0
3 years ago
Genes A and B are 10 map units apart and genes B and C are 20 map units apart, with gene B located between genes A and C. What w
Marizza181 [45]

Answer:

2% of the progeny will be double crossovers for the trihybrid test cross

Explanation:  

By knowing the positions of genes, we can estimate the distances in MU between them per region.

  • Genes A and B are 10 map units apart (Region I)
  • Genes B and C are 20 map units apart (Region II)
  • Genes A and C are 30 map units apart

----A-------10MU--------B-------------20MU-------------C---

             Region I                      Region II  

We can estimate the recombination frequencies by dividing each distance by 100.

• recombination frequency of A-B region = 10MU / 100 = 0.10

• recombination frequency of B-C region = 20MU / 100 = 0.20

Now that we know the recombination frequencies in each region, we can calculate the expected double recombinant frequency, EDRF, like this:

EDRF = recombination frequency in region I x recombination frequency in region II.

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2% of the progeny will be double crossovers for the trihybrid test cross

   

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