ebonyediva
contestada

Environment: Clean Forest
Moths
Released
G1
G2
G3
G4
G5
Typica
810
405
468
569
691
857
Carbonaria
190
72
66
64
61
56
Total
1000
477
534
633
752
913
Phenotype Frequency
Color
Initial Frequency
Frequency G5
(Round to 2 decimal
places)
Typica
Light
0.81
0
Carbonaria
Dark
0.19
0
Allele Frequency
Allele
Initial Allele
Frequency
G5 Allele Frequency
(Round to 2 decimal
places)
9
d
0.90
Ñ€
0.10

Respuesta :

marianaegarciaperedo

The recessive allele frequency increased (q = 0.97) and the dominant one decreased (p = 0.03). Genotypic frequencies followed this tendency too (q² =0.94, 2pq = 0.058 and p² = 0.001). The Carbonaria phenotype decreased to 0.06, while Typica showed a frequency of 0.94.  

------------------------------------------

Available data:

  • Some moths were released in the forest (N=1000). 810 were white, and 190 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                     810                  405        468        569       691        857

Carbonaria             190                   72           66           64         61          56

Total                       1000                477         534         633       752       913

Phenotype frequencies

                        Color             Initial Frequency               G5 Frequency

Typica             white                       0.81                                                

Carbonaria      Black                       0.19                                                

Allele Frequencies

                          Allele        Initial Allele Frequency     G5 Allele Frequency

q                           d                       ��    0.9                                                      

p                          D                            0.1                                                        

Genotype Frequency

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

q²      Typica        dd       White      810           0.81                                        

2pq  Carbon.      Dd       Black       180           0.18                                        

p²     Carbon.      DD       Black       10              0.01                                        

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  →  913
  • White = Typica = 857 individuals
  • Black = Carbonaria = 56 individuals

⇒ F(Typica) = 857 / 913 = 0.938 ≅ 0.94

⇒ F(Carbonaria) = 56 / 913 = 0.061 ≅ 0.06    

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 = 0.94
  • Recessive allele → d
  • Recessive allelic frequency →  f(d) → Represented as q
  • f(d) = q = ??

q² = 0.94

q = √ 0.94

q = 0.969 ≅ 0.97

0.97 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 + 0.97 = 1

p = 1 - 0.97

p = 0.03

So, now we also know that

⇒ f(D) = p = 0.03

⇒ f(d) = q = 0.97

3) Genotypic Frequencies

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

⇒ F(DD) = p² = 0.03² = 0.0009 ≅ 0.001

⇒ F(Dd) = 2pq = 2 x 0.03 x 0.97 = 0.058

⇒ F(dd) = q² = 0.97² = 0.9409 ≅ 0.94    

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 = 0.001
  • F(Dd) = 2pq = 0.058
  • F(dd) = q² = 0.94    
  • Total number of individuals = 913

⇒ DD Black -Carbonaria- individuals → 0.001 x 913 = 0.913 ≅ 0.91

⇒ Dd Black -Carbonaria- individuals → 0.058 x 913 = 52.954 ≅ 53

⇒ dd White -Typica- individuals → 0.94 x 913 = 858.22 ≅ 858

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 recessive phenotype by increasing the frequency of the recessive allele over the dominat one. Probably directional selection is acting on this population.

----------------------------------

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

Otras preguntas