Consider a locus with two alleles - A and a. These alleles are codominant, meaning that the fitness of the heterozygote is halfway between either homozygote. Consider further a population of randomly mating green frogs where the genotype counts are AA = 500, Aa = 250, and aa = 250. In this population the relative fitnesses of each genotype are AA = 1.00, Aa = 0.80, and aa = 0.60. What is the mean realtive fitness within this population? Please give your answer to two decimal places.

Consider a locus with two alleles - A and a. These alleles are codominant, meaning that the fitness of the heterozygote is halfway between either homozygote. Consider further a population of randomly mating green frogs where the genotype counts are AA = 500, Aa = 250, and aa = 250. In this population the relative fitnesses of each genotype are AA = 1.00, Aa = 0.80, and aa = 0.60. What is the expected allele frequency change for A after one generation with selection? Please give your answer to two decimal places.

Question 1.

The mean relative fitness is 0.85

Question 2.

The expected allele frequency change of A, p = 0.63: AA, p2 =0.63 x 0.63 = 0.39.

Explanation:

Mean relative fitness is given as

p2W(AA) + 2pqW(Aa) + q2W(aa).

Where W(AA)= 1.00, W(Aa) = 0.80, W(aa) = 0.60

Firstly, find p2, 2pq, q2 which are the frequencies of AA, Aa and aa respectively.

AA= 500,Aa=250 ,aa=250. Total is 1000

Frequency of A, p=

250/2)+500= 625.

625/1000= 0.625.

Therefore A, p is 0.625

p2 = 0.625x0.625= 0.39

For q, since p+q =1

q= 1 - 0.625 = 0.375

q2= 0.375x0.375=0.14.

2pq= 2x0.625x0.375= 0.47.

Mean relative fitness is

p2W(AA) + 2pqW(Aa) + q2W(aa).

0.39x1+ (0.47x0.80) + (0.14x0.60)=0.85.

Question 2.

From Hardy-Weinberg equilibrium

F = p2 + 2pq + q2.

Frequency of A, p= 0.625 and AA, p2 is 0.625 x 0. 0625= 0.39.

answer: 45

explanationyou have to add as well to be it could be

it has to be understood that the sea water contains a huge amount of sodium and chlorine ions and if they enter the cell of the atlantic cod fish, the cell will shrivel and the fish will die. to avoid such a condition, the fish uses active transport for the purpose of removing the ions against the gradient.

explanation:

the atlantic cod is not a maritime osmoconformer. its native application of salts is less than that of the seawater it lives in. it is continually falling water from the gills to the ocean via osmosis.

countering that, the fish is going to be swallowing seawater. but since its internal collection of salts is lower than seawater, the fish will want to go free of excess salt, and since that's going upon a gradient, then active transport, requiring energy, must be used to pump out the salt. much marine fish use the organs to filter out the salt and these fish discharge very straight urine.