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ddd [48]
2 years ago
6

Using complete sentences, compare conditions in the Belgian Congo with conditions in the Congo Free State under King Leopold II.

Biology
2 answers:
Alex_Xolod [135]2 years ago
6 0
<span>Employee of the Elder-Dempster shipping company based out of Antwerp. Responsible for basically starting the international human rights movement in the Congo Free State. Created the Congo Reform Association and was a constant thorn in the side of Leopold II. He formed his own newspaper 'The West African Mail' and wrote 'Red Rubber' to publicise the atrocities committed by King Leopold II and his officers in the Congo.</span>
elena-s [515]2 years ago
3 0

King Leopold II treated the Congo as his personal property and exploited the country for slave labor, rubber, and ivory. The Belgian Congo was established in 1908 after the international community pressured the Belgian government to annex the country and take it out of the king's hands. Conditions in Belgian Congo improved as the Belgian government supported education and established trade. Nevertheless, the people of the Congo increased their demands for independence.

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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
Dmitry [639]

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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