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ivanzaharov [21]
1 year ago
13

In a population of flowers growing in a meadow, C1 and C2 are autosomal codominant alleles that control flower color. The allele

s are polymorphic in the population, with f (C1) = 0.8 and f (C2) = 0.2. Flowers that are C1C1 are yellow, orange flowers are C1C2, and C2C2 flowers are red. A storm blows a new species of hungry insects into the meadow, and they begin to eat yellow and orange flowers but not red flowers. The predation exerts strong natural selection on the flower population, resulting in relative fitness values of C1C1 = 0.30, C1C2 = 0.60, and C2C2 = 1.0.
Required:
a. What is the C1C2 genotype frequency among the progeny of predation survivors?
b. What is the C2C2 genotype frequency among the progeny of predation survivors?
c. What is the C2C2 genotype frequency among the progeny of predation survivors?
d. What is the equilibrium C2 allele frequency in the predation environment?
Biology
1 answer:
Sergeu [11.5K]1 year ago
8 0
B is yours answer have a nice day
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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:

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  • j k / j k = 20%
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  • 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:

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  • 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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