Snouters are small critters that live on both sides of the Ohio River. A populat

Question

Snouters are small critters that live on both sides of the Ohio River. A population on the north side of the Ohio River is fixed for an allele for blue feet. The population on the south side of the river is fixed for an allele for pink feet. Forty blue-footed snouters from the north and 60 pinkfooted snouters from the south colonize a newly formed island in the middle of the river. Please calculate the following.
(1 pt) What are the genotype frequencies at the time of colonization (Parental Generation, P1)? (Just the colonizers before reproduction)
(1 pt) What are the allele frequencies at the time of colonization (P1)? (Just the colonizers before reproduction, N = 100)
(2 pts) What are the genotype frequencies of the offspring of the next generation (First generation of offspring, F1 ), assuming random mating?
(1 pt) What are the allele frequencies of the offspring of the next generation (F1 )?
Heterozygotes have purple feet that are susceptible to snouter purple-foot fungus that sterilizes purple-footed snouters (They cannot reproduce). Assume that the F1 generation above is the parents for the next set of questions.
(3 pts) Calculate the allele frequencies of the parents that can breed (F1 with no purple footed individuals). Calculate the genotype and allele frequencies of the offspring (F2) in the second generation assuming random mating (Hint: the sum of the allele frequencies represented by the homozygotes are equal to 1, keep this in mind when you calculate the allele frequency of F1 as parents. It may help if you assume a population size for F1 prior to reproduction.)   
(2 pt) Has evolution occurred? Why? If the purple foot fungus were to continue on this island for several generations, what color feet would most likely become fixed on the island and why?

Explanation / Answer

1. The genotype for the blue feet snouters can be represented by BB and for the pink feet snouters by PP. Thus the genotypic ratio before reproduction would be:

Blue: Pink = 20: 80 = 1:4

Genotypic frequencies:

BB = 0.2

PP = 0.8

2. Allelic frequencies (before reproduction)

(Frequency of B allele) = 2 x 20 /(200) = 0.2

Frequency of P allele = 2 x 80 / (200) = 0.8

3. Upon crossing the two parental types i.e. BB and PP and other same genotype crosses i.e. BB x BB and PP x PP, we get:

the F1 progeny as ---> BP, BB, PP

Let us assume a population size of 200 for the F1 generation, then the genotypic frequencies can be represencted as follows:

Acc. to Hardy-Weinberg equilibrium the allelic frequencies remain the same in a population.

Frequency (BP) = 2 x 0.2 x 0.8 = 0.32

Frequency (BB) = 0.2 x 0.2 = 0.04

Frequency (PP) = 0.8 x 0.8 = 0.64

4.  The allelic frequencies in this case can be calculated as follows:

No. of BP snouters = 64

No. of BB snouters = 8

No. of PP snouters = 128

Allelic frequencies are:

B = (2 x 8) + 64 / (400) = 0.2

P = (2 x 128) + 64 / (400) = 0.8

5. The parents that can breed will be those without the BP genotype, and thus the BB or PP genotype:

BB = 8

PP = 128

Thus the total number of individuals that can breed are 136.

Allelic frequency (B) = (2 x 8) / (2 x 136) = 0.058

Allelic frequency (P) = (2 x 128) / (2 x 136) = 0.942

thus, upon random mating we might get any of the genotypes BB,PP and BP again and we can calculate the genotypic and allelic frequencies as follows:

Genotypic frequencies:

BB = 0.058 x 0.058 = 0.003 (No. of BB individuals = 0)

PP = 0.942 x 0.942 = 0.887 (No. of PP individuals = 121)

BP = 2 x 0.058 x 0.942 = 0.110 (No. of BP individuals = 15)

Allelic frequencies remain the same as per Hardy-Weinberg equilibrium and can be shown to be:

Allelic frequency (B) = (2 x 0) + 15 / (2 x 136) = 0.058

Allelic frequency (P) = (2 x 121) + 15 / (2 x 136) = 0.942

6. Evolution has clearly occurred as shown in the above calculations, the frequency of the PP genotype was dominant from the beginning over the BB genotype, and over the next 2 generations, the PP genotype frequency and P allele frequency has increased from 0.8 to 0.942. This shows that as the heterozygotes BP gave sterile phenotypes thus it was not a favourable trait and hence had to be eliminated.

Thus the P allele being more dominant in frequency is selected more often than the B allele during reproduction so as to change the trend towards a dominant PP genotype population i.e.. a pink feet snouter population. Thus as can be seen the no. of BB genotype snouters reduces to almost zero by the end of the 2nd generation. As a result of this, as the heterozygotes BP are sterile, thus the PP individuals will mate amidst each other to give a purely PP population which is superior to both the other genotypes.

If the purple snouter fungus continues on the island then it will infect the individuals with the purple feet only thus promoting the purple colour feet to become dominant on the island. However, as the individuals with purple colour feet cannot reproduce, thus they will not be able to survive on the island. Hence, even though the purple snouter fungus persists on the island and the purple feet will be dominant, the other two types i.e. snouters with blue and pink feet should also survive on the island.

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