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Environment: Polluted Forest Moths Released G1 G2 G3 G4 G5 Typica 125 88 83 76 29 Carbonaria 510 735 885 1042 1406 Total 635 823 968 1118 1435 Phenotype Frequency Color Initial Frequency Frequency G5 Typica Light 9 Carbonaria Dark Allele Frequency Allele Initial Allele Frequency G5 Allele Frequency q d p D Genotype Frequency Moths Genotype Color Moths Released Initial Frequency Frequency G5 Number of Moths G5 q2 Typica dd Light 2pq Carbonaria Dd Dark p2 Carbonaria DD Dark

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marianaegarciaperedo

The recessive allele frequency decreased (q = 0.14) and the dominant one increased (p = 0.86). Genotypic frequencies followed this tendency too (q² =0.02, 2pq = 0.24 and p² = 0.74). The Carbonaria phenotype increased to 0.98, while Typica showed a frequency of 0.02.  

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Available data:

  • Some moths were released in the forest (N=1000). 250 were white, and 750 were black.

  • The color is defined by a single diallelic gene. The dominant allele -D- codes for black color (carbonaria), while the recessive allele -d- codes for white (Typica).

  • These individuals produced five new generations since they were released, G1, G2, G3, G4, G5.

                     

                     Moths realesed         G1          G2         G3         G4         G5        

Typica                     250                  125         88          83          76          29

Carbonaria             750                   510         735       885      1042      1406

Total                       1000                  635        823        968      1118       1435

Phenotype frequencies

                       Color             Initial Frequency               G5 Frequency                      

 

Typica             white                       0.25                                                

Carbonaria      Black                       0.75                                                

Allele Frequencies

                         Allele        Initial Allele Frequency     G5 Allele Frequency        

q                           d                            0.5                                                      

p                          D                            0.5                                                        

Genotype Frequency

        Moths   Genotype Color  Released  Initi.Freq.   G5 Freq.   Nº F5 moths

q²      Typica        dd       White      250           0.25                                        

2pq  Carbon.      Dd       Black       500           0.50                                        

p²     Carbon.      DD       Black       250           0.25                                        

This information is guiding you to know how to calculate frequencies and total numbers. Information about released individuals is an example of how you need to proceed.

Firts, we will assume that this population is under Hardy-Weinberg equilibrium. So let us review some theoretical framework.

The allelic frequencies in a locus ⇒ p and q ⇒ dominant and  

                                                                               recessive alleles.

The genotypic frequencies after one generation are p² (H0m0zyg0us dominant), 2pq (Heterozyg0us), q² (H0m0zyg0us recessive).

If a populations is in H-W equilibrium, it will get the same allelic frequencies generation after generation.  

  • The sum of allelic frequencies equals 1 ⇒ p + q = 1.

  • The sum of genotypic frequencies equals 1 ⇒ p² + 2pq + q² = 1

Now let us analyze the problem.

We need to get information on G5 generation. We will do it step by step.

1) Phenotype frequencies

To get the Phenotype frequencies, we just need to divide the number of individuals with each phenotype by the total number of individuals in this generation. So,

  • Total number of individuals in G5  →  1435
  • White = Typica = 29 individuals
  • Black = Carbonaria = 1406 individuals

⇒ F(Typica) = 29 / 1435 = 0.02  

⇒ F(Carbonaria) = 1406 / 1435 = 0.979 ≅ 0.98    

2) Allelic Frequencies

We can use the phenotypic frequencies to get the allelic frequencies.

Remember that carbonaria (black) moths include h0m0zyg0us dominant (DD) and heter0zyg0us (Dd) individuals. So we can not get the allelic frequencies from this data.

We can only use the allelic frequency of Typica (White) individuals. Typica phenotypic frequency only includes h0m0zyg0us recessive individuals, dd.

We know that,

H0m0zyg0us recessive genotype  →  dd

Genotypic frequency →  F(dd) → Represented as q²            

F(Typica) = F(dd) = q² = 0.02

Recessive allele → d

Recessive allelic frequency →  f(d) → Represented as q

f(d) = q = ??

q² = 0.02

q = √ 0.02

q = 0.1414 ≅ 0.14

0.14 is the recessive allelic frequency. Now we should calculate the dominant allelic frequency. To do this, we will clear the following formula,

p + q = 1

p + 0.14 = 1

p = 1 - 0.14

p = 0.86

So, now we also know that

⇒ f(D) = p = 0.86

⇒ f(d) = q = 0.14

3) Genotypic Frequencies

Now, we need to get the genotypic frequencies, F(xx)

⇒ F(DD) = p² = 0.86² = 0.7396 ≅ 0.74

⇒ F(Dd) = 2pq = 2 x 0.86 x 0.14 = 0.24

⇒ F(dd) = q² = 0.14² = 0.0196 ≅ 0.02    

4) Number of individuals

Finally, we need to tell the number of individuals with each genotype. We just need to multiply each frequency by the total number of individuals in G5.

F(DD) = p² = 0.74

F(Dd) = 2pq = 0.24

F(dd) = q² = 0.02    

Total number of individuals = 1435

⇒ DD Black -Carbonaria- individuals → 0.74 x 1435 = 1,061.9 ≅ 1062

⇒ Dd Black -Carbonaria- individuals → 0.24 x 1435 = 344.4 ≅ 344

⇒ dd White -Typica- individuals → 0.02 x 1435 = 28.7 ≅ 29

From this results, we can conclude that the moths population is not in H-W equilibrium, because their allelic and genotypic frequencies changed through generations.

It seems that Natural selection is favoring the dominant phenotype by increasing the frequency of the dominant allele over the recessive one. Probably directional selection is acting on this population.

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Related link: brainly.com/question/12724120?referrer=searchResults                          

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