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sweet-ann [11.9K]
2 years ago
14

In Holstein cattle the allele for black hair color (B) is dominant over the allele for red hair color (b), and the allele for po

lled (P), or lacking horns, is dominant over the allele for having horns (p). What is the expected phenotypic ration of the offspring of a BbPp x BbPp cross if these alleles sort independently?
Biology
1 answer:
torisob [31]2 years ago
7 0

Answer:

A dominant trait can be described as the one which masks the effect of a recessive trait. The alleles for a recessive trait gets suppressed by the dominant one.

A punnet square can be described as a diagram which depicts the outcomes of a cross. For single traits, the cross is monohybrid. For studying two traits at a time, a dihybrid cross can be made.

The punnet square for the following Holstein cattle can be shown as:

           BP           Bp          bP             bp

BP     BBPP     BBPp     BbPP         BbPp

Bp     BBPp     BBpp     BbPp          Bbpp

bP      BbPP     BbPp     bbPP          bbPp

bp      BbPp      Bbpp     bbPp          bbpp

The results show the phenotypic ratio to be 9 black/polled : 3 black/horned : 3 red/polled : 1 red/horned

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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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  • 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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