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Alexxx [7]
2 years ago
15

A research team studied a large lizard population in the Sonoran Desert of Arizona. In these lizards, gene A is one gene that af

fects the color pattern on the dorsal surface of the lizard’s body. In 2010, the research team sampled a large number of lizards and determined the genotypes of the lizards for the A gene. They found the following genotype frequencies: A1A1 = 0.17, A1A2 = 0.15 A2A2 = 0.68. What is the frequency of the A1 allele? 0.49 0.5 0.34 0.17 0.245 0.755 0.15
Biology
1 answer:
BabaBlast [244]2 years ago
6 0

Answer:

Option A, 0.49

Explanation:

As per Hardy Weinberg equation -

p^2 is the frequency of occurrence of genotype A_1A_1

The frequency of allele A_1 will be equal to square root of frequency of \sqrt{p_2}

Substituting the given value in above equation, we get -

Frequency of allele A_1= \sqrt{p^2}

Frequency of allele A_1 = \sqrt{0.17}

Frequency of allele A_1= 0.412

So the nearest option is option 0.49, i.e option A

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Malthus is the correct answer

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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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