Suppose that as long bills were evolving in pelicans, bill length was controlled

Question

Suppose that as long bills were evolving in pelicans, bill length was controlled by a single gene with two alleles, B and X, such that BB individuals had short bills and XX individuals had long bills. Further, suppose that B was dominant to X so that BX individuals had short bills just like BB individuals. Since long bills allowed pelicans to obtain more food, pelicans with long bills survived best and produced, on average, 2 surviving offspring during their lives, while individuals with short bills produced, on average, 0.75 surviving offspring during their lives.

a) Suppose that in a virtually infinite population of pelicans that mate randomly with regard to bill length, do not experience mutation, and do not move among populations, the frequency of the X allele in the gametes at the start of a generation is 0.3. What will be the frequency of the XX genotype in the zygotes produced by union of these gametes?

b) Determine the frequency of the XX genotype in the surviving adults that develop from the zygote population.

Explanation / Answer

The above problem can be solved by using the formulas;

1. p+q=1

2. p2+q2+2pq=1

where p is the frequency of the dominant allele in the population

q is the frequency of the recessive allele in the population

p2 is the percentage of the homozygous dominant individuals

q2 is the percentage of thehoomozygous recessive individuals

2pq is the percentage of the heterozygous individuals

given:

Bill length: BB-short bills XX-long bills Bx-short bills

p(X) frequency of X allele is = 0.3

then the frequency of XX genotype will be square root of 0.3 which is =0.55

Answer a)=0.55

Given that 50% of long beaks survive the Answer b) =0.55 /2=0.275

answer b)=0.275

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