34. The CP locus in clover plants controls leaf cyanide production, which helps
ID: 268903 • Letter: 3
Question
34. The CP locus in clover plants controls leaf cyanide production, which helps the plant defend itself against insects and grazing animal herbivory. There are two alleles at the CP locus: A and a. Individual plants that are homozygous AA or aa produce very low, but slightly different levels of leaf cyanides. Aa heterozygous plants produce very large amounts of leaf cyanides. In a clover population that experiences high levels of deer grazing the relative fitness values among genotypes are as follows: Aa 1.0 Waa 0.4 Which of the following most closely describes the expected equilibrium frequency of the A allele in this clover population? a. The expected equilibrium frequency of the A allele is 1.0 b. The expected equilibrium frequency of the A allele is 0.43 c. The expected equilibrium frequency of the A allele is 0.57 d. The expected equilibrium frequency of the A allele is 0.67Explanation / Answer
Accoring to HW law (p+q)^2 = p^2+ 2pq+ q^2 = 1
p and q are the alleles.
p^2 - represents the p homozygous frequency
2pq - represents the heterozygote frequency
q^2 - represents the q homozygous frequency.
p = 0.5 and q = 0.5,
p^2 - represents the p homozygous frequency = 0.5*0.5 = 0.25
2pq - represents the heterozygote frequency = 2* 0.5*0.5 = 0.50
q^2 - represents the q homozygous frequency. = 0.5*0.5 = 0.25
After fitness the frequencies =
Genotype frequencies before fitness
Fitness values
Genotype frequencies after fitness
0.25
* 0.2
= 0.05
0.5
* 1
= 0.5
0.25
* 0.4
= 0.1
So, the frequencies of each allele after fitness –
Genotype
Freequency
Allele A
Allele a
Total
AA
0.05
0.1
0
0.1
Aa
0.5
0.5
0.5
1
aa
0.1
0
0.2
0.2
Total
0.65
0.6
0.7
1.3
Allele frequencies: Frequency of a allele/Frequency of all alleles
Allele A
= 0.6/1.3
= 0.46
Allele a
=0.7/1.3
= 0.54
So, the answer is 0.46.
As the exact answer is not found and the near by answer 0.43,
the answer is B. The expected equilibrium of the A allele = 0.43
Genotype frequencies before fitness
Fitness values
Genotype frequencies after fitness
0.25
* 0.2
= 0.05
0.5
* 1
= 0.5
0.25
* 0.4
= 0.1
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