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Kisachek [45]
2 years ago
3

Summarize, using water potential, why water leaves the slugs tissues and moves into the surround environment.​

Biology
1 answer:
scoundrel [369]2 years ago
3 0

Answer:

The salt pulls water out of the slug in light of the fact that the NaCl breaks separated into two particles (negative and positive charge) and this declines the water potential on the outside of the slug.

Explanation:

The proportion of the overall propensity of water to move starting with one zone then onto the next, and is regularly spoken to by the Greek letter Ψ (Psi).  The salt pulls water out of the slug in light of the fact that the NaCl breaks separated into two particles (negative and positive charge) and this declines the water potential on the outside of the slug.

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The statement that organisms that have a physical trait that makes it easier for them to access food supplies are unlikely to pass on that trait to their offspring is false. Organisms pass their traits and adaptation abilities to their offspring. These organisms have also characteristics that are advantageous for reproduction in a specific environment and leave more offspring in the next generation. The process is called natural selection.


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In one row in the table, the data for the three trials is inconsistent. Which table cell contains data that are inconsistent for
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Answer: A

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The cell for trial 2 with a lot of water. Trust me

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On your first attempt to run a PCR, you realize you forgot to add one of the two primers. The graduate student you work with sug
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2 years ago
Two autosomal genes, J and K, are 60 map units apart. You perform the following testcross: J K / j k x j k / j k. At what freque
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Answer:

  • J K / j k = 20%
  • j k / j k = 20%
  • J k / j k = 30%
  • j K / j k = 30%                

Explanation:

To calculate the recombination frequency, we have to know that 1% of recombinations = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.

The map unit is the distance between the pair of genes for which every 100 meiotic products, one of them results in a recombinant one.

So, en the exposed example:

  • J and K are autosomal genes
  • J and K are separated by 60 M.U.
  • 60 M.U. means that there is 60% of recombination.

Cross)             J K / j k                    x                  j k / j k

Gametes) JK  Parental                                     jk, jk, jk, jk

                jk   Parental                                  

                Jk   Recombinant                          

                 jK   Recombinant

One map unit equals 1% of recombination frequency. This means that every 100 meiotic products, one of them is a recombinant one.

1 M.U. -------------- 1% recombination

60 M.U. ------------ 60% recombination

                              30% Jk  +  30% jK

100 M.U. - 60 M.U. = 40 M.U.

40M.U.--------------40 % Parental (Not recombinant)

                            20% JK   +   20% jk

Punnet Square)           JK       jk      Jk      jK

                          jk     JK/jk   jk/jk   Jk/jk   jK/jk

J K / j k = 20%

j k / j k = 20%

J k / j k = 30%

j K / j k = 30%                                

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