Since the llamas will not sell well with malformed feet, you decide to select the herd against this defect. You separate the affected llamas and send them to market, allowing only the normal llamas to mate and produce the next generation, (W22 = 0, W11=1, W12 =1). What will be the allele frequency for the foot defect in the next generation? a. 1.0.279
b. 0.300
c. 0.336
d. 0.150
e. 0.228

The question is incomplete as the frequency of alleles is not given, however, the frequency and population are given below :

Frequency of a = 0.506

total population = 500, Number of aa = 128

The correct answer is - option C. 0.336.

Explanation:

Let, A = Normal allele, a = Defective allele

So, AA & Aa will develop normal phenotype & aa will develop defective phenotype.

Frequency of a = 0.506

So, Frequency of A = 1 - 0.506 = 0.494

So, frequency of AA = (0.494)2 = 0.244036

So, frequency of Aa = 2 x 0.494 x 0.506 = 0.499928

so, frequency of aa = (0.506)2 = 0.256036

total population = 500, Number of aa = 128

So, Number of AA = (0.494)2 x 500

= 122.018 which is almost 122 so considering it 122

So, Number of Aa = (2 x 0.494 x 0.506) x 500

= 249.964 which is almost 250 so considering it 250

It is given that, Relative fitness of AA (W11) & Aa (W12) is 1, and the relative fitness of aa (W22) is 0.

Now, mean fitness = (Frequency of AA x Fitness of AA) + (Frequency of Aa x Fitness of Aa) + (Frequency of aa x Fitness of aa)

= (0.244036 x 1) + (0.499928 x 1) + (0.256036 x 0) = 0.244036 + 0.499928 = 0.743964

So, after selection frequency of Aa

= (Frequency of Aa x Fitness of Aa) / mean fitness

= 0.499928 / 0.743964 = 0.67198 (Up to 5 decimal)

So, after the selection frequency of aa

= (Frequency of aa  x  Fitness of aa) / mean fitness

= 0 / 0.743964 = 0

So, frequency of a

= 1/2 of frequency of Aa + Frequency of aa

= 1/2 x 0.67198 + 0

= 0.33599 + 0

= 0.33599 which is almost 0.336

answer; i believe the correct answer is (sporophyte);