What is the average % survival of each genotype? What is the average # offspring

Question

What is the average % survival of each genotype?

What is the average # offspring per surviving adult of each genotype?

Calculate p and q.

Scenario 2 Canadian adamantium frogs display various levels of webbing between their digits. Webbing is determined by three genotypes: Genotype Phenotype Full webbing extends entire length of digits) Rr Intermediate webbing (extends to Ythe length of the digits) rr Reduced webbing (extends less than the length of the digits The Vancouver population of Canadian adamantium frogs had originally lived in a semi mountainous region that was subject to seasonal streams and shallow lakes. However, due to climate change, their habitat has become flooded with a deep-lake persisting year-round. Initially, there were equal proportions of R and r alleles in the population's gene pool. After the formation of the permanent lake, all frogs with full webbing survived to adulthood and bore an average of 54 offspring each. 90% of frogs with intermediate webbing survived to adulthood, and those survivors bore an average of 40 offspring. 80% of frogs with reduced webbing survived to adulthood, and those survivors bore an average of 22.5 offspring.

Explanation / Answer

For genotype RR (full webbing)

Suppose 54 offspring from two frogs survived, and then the percentage of survival for this genotype will be 100%. Total candidates =54+2 = 56.

For genotype Rr (intermediate webbing):

Survivor bore an average of 40 offspring; so next generation will get 90% of survived population. Therefore, 90% of 40 = 36 will produce offspring, now total number of survived candidates = 2 (parents) + 36 = 38 from 42 (total candidates). Now percentage of survival = 38/42 x 100 = 90.48.

For genotype, rr (reduced webbing)

22.5 offspring from parents, in which 80% will survive = 22.5x 80% = 18. Now survival percentage = (18+2) / (22.5+2) x 100 = 81.63%.

P (R) and Q (r) =

Total frogs (population) = 56+42+24.5 = 122.5

Frequency P = [genotypes RR+Rr / total population] = [(56+42) /122.5] = 98/122.5 =0.8

Frequency Q = 1-P = 0.2

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