Question

+/+	+/-	-/-
7	9	2

=18 students

1. An allelic frequency is a ratio of the number of copies of an allele over the total number of alleles present (2 x homozygous+heterozygous/total number of alleles). Based on the class data, calculate the allelic frequencies of the Alu insertion of your lab section.
2. If the population is genetically stable, the allelic frequencies will remain constant from one generation to the next. If we expect the allelic frequency of the Alu insertion to be 45% (P=0.45) and the lack on insertion is 55% (Q=0.55), how do these values compare with your class genotypic ratios? Show the expected genotypic ratios and Class ratios?

Answer 1

Answer-

Allelic frequency of Alu insertion (+) = p = 0.639; allelic frequency of lack of insertion (-) = q = 0.361

After one generation genotypic numbers will remain same as they are in Hardy Weinberg equilibrium. So genotypic ratio in class will be 7:9:2 (in favors of ++, +-, -- respectively). Expected Genotypic ratio will be 4:9:5 (if p = 0.45 and q = 0.55 and if we consider total expected population to be 18).

Explanation is provided below:

++ (1) Total students = 18 ++ +- -- -7 = 9 = 2 N N N- Allele frequency of (t) =p=2N+ + + N +- Total alleles (2N) extes 23=0.639 QN +Nt- Allele frequency of (-) =9=- - (2x32 + 9 3 -0.361 = (@x7)+9 2N 36 Now , expected allelic frequency of (+) =P = 0.45 (-) = 4 = 0.55 we know, q=1 (P+Q) ² = 1 p² + 204 + 2 = 1 ++ +- -- Eupected frequency of genotype(++) = P2=(0-45)% = 0.2025 genotype (+-) = 2P4 = 2X0-4 5 X0.55-0.495 genotype (--) =92 = 0.55) = 0-3025 It we consider 18 students Expected number of genotype (++) = 0.2025 X18 = 3.645 ~4 , (+-) = 0.495 X18 = 8.91 a 9 " . .. (--) = 0.3025X18 - 5.445 ~ 5 . Expected genotypic ratio = 4:9:5 (In number) class ratio = 7:9:2 (In number as gi