D Question 1 1 pts A population contains 100,000 individuals, 16,000 with yellow
ID: 197154 • Letter: D
Question
D Question 1 1 pts A population contains 100,000 individuals, 16,000 with yellow fruit and 84,000 with red fruit. You know that fruit color segregates as a single Mendelian locus with two alleles, with red dominant to yellow. Assuming that the population is in Hardy- Weinberg equilibrium, how many of the red-fruited individuals are homozygous at this locus? Question 2 1 pts The position of petal spots in Clarkia gracilis is controlled by two codominant alleles (Pb and Pc) at a single locus. pbpb Basal pbpc Double PcPC Central A population of C.gracilis contains 600 individuals with a single basal spot, 260 individuals with a single central spot, and 140 individuals with both a basal and central spot. What is the frequency of the Pc allele in this population?Explanation / Answer
As per Hardy-Weinberg Equilibrium, let
p = Frequency of Dominant Allele
q = Frequency of Recessive Allele
Also p + q = 1
Now,
p2 = Frequency of Homozygous Dominant Genotype
2pq = Frequency of Heterozygous Dominant Genotype
q2 = Frequency of Homozygous Recessive Genotype
ANSWER 1
Total Number = 100,000
Number of Dominant Phenotype = 84,000
Number of Recessive Phenotype = 16,000
As per Hardy-Weinberg Equilibrium:
Number of Recessive Phenotype:
(Frequency of Recessive Allele*Frequency of Recessive Allele)*100000 = 16,000
Therefore, Frequency of Recessive Allele = 0.4
Therefore, Frequency of Dominant Allele = 1-0.4 = 0.6
Number of Homozygous Dominant Genotype
= (Frequency of Dominant Allele*Frequency of Dominant Allele)*100000
= (0.6 * 0.6) * 100000
= 0.36 * 100000
= 36000 ANSWER
ANSWER 2
Total Number = 1000
Number of Homozygous Basal Genotype = 600
Number of Codominant Double Genotype = 140
Number of Homozygous Central Genotype = 260
Frequency of Pc allele
= (2*Number of Homozygous Central Genotype + Number of Codominant Double Genotype) / Total Allele
= (2*260 + 140) / 1000*2
= 660 / 2000
= 0.33 ANSWER
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