3. Before a power plant went on-line in New York Harbor, 1000 killifish ( Fundul

Question

3. Before a power plant went on-line in New York Harbor, 1000 killifish (Fundulus heteroclitis) of known electrophoresis genotypes were placed in an experimental enclosure: 40 FF, 320 FS and 640 SS individuals (F stands for fast migrating protein allele, S = slow migrating allele in gel electrophoresis). Two weeks after the power plant began operation and used harbor water in the cooling towers, the 680 fish that were still alive were collected and analyzed for genotype, with the following results: 40 FF, 256 FS and 384 SS. (i) What are the allele frequencies after selection? (ii) Determine the relative fitnesses of each genotype and the selection coefficient against each genotype. (hints: Think about fitness as the percentage of each genotype that survives the heat stress (e.g. “freq. after”/”freq. before”); maximal relative fitness = 1.0 (100%); you do not need to use Hardy-Weinberg, or allele frequency to answer question

Explanation / Answer

1.what are the allelic frequencies ?

answer; it is caluculated by using the formula ; After selection

1. frequency for FF allele number of FF allele / total number of allele which is equal to 40/ 40+256+384 equal to 0.05

2. frequency of allele for SS allele number of SS allele / total number of allele which is equal to 384 /40+256+384 equal to 0.56

2. Determine the relative fitnesses of each genotype and the selection coefficient against each genotype.

answer ;

Relative fitness ;Average number of surviving progeny of a perticular genotype compared with the average number of progeny of competing genotypes after a single generation .

1. W for FF allele equal to 2* number of FF alelle +number of FS allele / 2* total number of allele s which is equal to 2 [ 40 +256] / 2*[680] equal to 592/1360 equal to 0.43

relative frequencies for allele FF is 0.43 ,W is 0.43 for FF

2. W for SS allele equal to 2* number of FF alelle +number of FS allele / 2* total number of allele s which is equal to

2* 384 / 2*680 equal to 768/1360 equal to 0.56

Relative frequencies for SS allele is 0.56 , w for SS is 0.56

selection coefficient; it is a measure of relative reduction in contribution that a perticular geneotype makes to gamates as compared to another population .

S = 1-W ,

so for FF allele S = 1- 0.43 which is equal to 0.57 , so selection coefficient for SS allele is 0.57 .

for SS allele S = 1-0.56 which is equal to 0.44 , so selection coefficient for SS is 0.44

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