there are three questions under number 13 about this senario You notice a change
ID: 79416 • Letter: T
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there are three questions under number 13 about this senario
You notice a change in one species of the fish population in a large pond you are studying. The initial species population had mostly Speckled (Black and White) fish with smaller but equal numbers of pure Black fish or pure White fish. You counted 50 speckled fish, 25 pure White fish, and 25 pure Black fish in the population six months age. What are the gene frequencies for the Black allele and the White allele in this original population? What kind of gene expression do you think is occurring in the Speckled fish? Since your last count black gravel was spread on the bottom of the pond. Your recent count showed 49 pure Black fish, 42 speckled fish, and only 9 pure White fish. What are the new gene frequencies for this population? What do you think is happening to this species of fish in the pond? Will the White allele ever disappear?Explanation / Answer
Ans. 13. Part A. Let the trait of color be regulated by following two alleles-
Allele W – gives pure white phenotype in homozygous (WW) condition
Allele B – gives pure black phenotype in homozygous (BB) condition
So, the genotypes of possible color phenotype are as follow-
Specked = WB
Pure black = BB
Pure white = WW
Given Population size = 50 (WB, speckled) + 25 (WW, pure white) + 25 (BB, pure black)
= 100
Total number of alleles for the trait in population = 2 x population size
= 2 x 100 = 200
Now,
(Allelic frequency (No. of given allele / Total number of alleles in population)
Being diploid each individual has 2 alleles of the gene. The homozygotes have two copies of the respective allele, whereas the heterozygotes have 1 copy of both the alleles. That is, WW individuals have 2 copies of allele W, whereas WB individuals have one copy each of allele W and B.
Allelic frequency of allele W, f(W) =
(No. of Allele W in population / Total no. of alleles for the trait)
= (2 x no. of WW fish + 1 x no. of BW fish) / (200)
= (2 x 25 + 1 x 50) / 200
= 100/ 200
= 0.5
Therefore, f(W) = 0.5
Also, f(B) = (No. of Allele B in population / Total no. of alleles for the trait)
= (2 x no. of BB fish + 1 x no. of BW fish) / (200)
= (2 x 25 + 1 x 50) / 200
= 0.5
Therefore, f(B) = 0.5
# Inheritance pattern: It is an example of co-dominance, where both allele W and B are simultaneously and independently expressed in the heterozygote (WB) giving random patches of black and white color.
If it were case of complete dominance, the BW fish would have been either totally white or totally black depending on which allele was dominant.
If it were case of incomplete dominance, the heterozygote WB would have given an uniform phenotype between pure white and pure- say brown color fish.
Part B. Population size =
42 (WB, speckled) + 9 (WW, pure white) + 49 (BB, pure black)
= 100
Total number of alleles for the trait in population = 2 x population size
= 2 x 100 = 200
Now, f(W) = (2 x 9 + 1 x 42) / 200 = 0.3
f(W) = (2 x 49 + 1 x 42) / 200 = 0.7
Part C: Note that the frequency of allele B in final population (0.7) is greater than the same in initial population (0.3). Change in allelic frequency of an allele indicates that the population is evolving and allele B is preferably selected in the population. It might be because the resultant phenotype may increase reproductive fitness, say by making black color invisible to predators- so, more black fish survive and they reproduce more.
# No, the allelic W for white color would not disappear. Note that the number of sparkled fish (WB) is 42 in a total population of 100. So, allele W is allele would be not disappeared.
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