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UNO [17]
3 years ago
5

A geneticist analyzes the four products of a meiosis for an Aa Bb plant and finds that two of the products have the AB alleles a

nd the other two products have the ab alleles. Is this consistent with the principle of independent assortment? Assume that crossing over does not occur.
Biology
1 answer:
slega [8]3 years ago
3 0

Answer:

No, this is not consistent with the principle of independent assortment.

Explanation:

The principle of independent assortment states that alleles from different genes assort independently. This means that if a plant has a genotype Aa Bb, all four alleles (A, a, B, and b) are going to segregate equally, so we will have the following four gametes after meiosis:

- AB

- Ab

- aB

- ab

If the researcher finds that two of the four products are AB, probably there would be a deviation of Mendel's laws.

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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
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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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