Data are provided below for two different populations. For each of these populat
ID: 163150 • Letter: D
Question
Data are provided below for two different populations. For each of these populations, calculate the frequencies for the other three haploid genotypes (Ab, aB, and ab) and the linkage disequilibrium coefficient (D). Also, for each of the four haploid genotypes in the population, state whether it occurs with more, less, or the same frequency as you would expect at random given the allele frequencies [no further explanation is required]: i. F(AB) - 0.49; F(A) = 0.5; F(B) = 0.7 ii. F(AB) = 0.16; F(A) = 0.2; F(B) = 0.8Explanation / Answer
Population 1
Give
F(AB) =0.49
=========
CONSIDERIN ONE LOCUS
Sum of alleles at a locus is equal to one
F(A)=0.5 ---------p1
------So F(a)=0.5---------p2
===================
CONSIDERIN second LOCUS
F(B)=0.7----------q1
---- F(b)=0.3-----------q2
genotype
Frequency
genotype Ab
=0.5 x 0.3 =0.15
Genotype aB
=0.5 x 0.7 =0.35
Genotype ab
=0.5 x 0.3 =0.15
P11 = p1q1= 0.5 x 0.5 = 0.25
P22 = p2q2= 0.5 x 0.3 = 0.15
P12 = p1q2= 0.5 x 0.3= 0.15
P21 = p2q1= 0.5 x 0.7= 0.35
D = (P11)(P22) -(P12)(P21)
D = (0.25) (0.15) - (0.15) (0.35) = 0.015
Population 2
Give
F(AB) =0.16
F(A=0.2
F(B)=0.8
=========
CONSIDERIN ONE LOCUS
Sum of alleles at a locus is equal to one
F(A)=0.2 ---------p1
------So F(a)=0.8---------p2
===================
CONSIDERIN second LOCUS
F(B)=0.8----------q1
---- F(b)=0.2-----------q2
genotype
Frequency
genotype Ab
=0.2 x 0.2 =0.04
Genotype aB
=0.8 x 0.8 =0.64
Genotype ab
=0.8x 0.2 =0.16
P11 = p1q1= 0.2 x 0.8 = 0.16
P22 = p2q2= 0.8 x 0.2 = 0.16
P12 = p1q2= 0.2 x 0.2= 0.04
P21 = p2q1= 0.8 x 0.8= 0.64
D = (P11)(P22) -(P12)(P21)
D = (0.16) (0.16) - (0.04) (0.64) = 0
genotype
Frequency
genotype Ab
=0.5 x 0.3 =0.15
Genotype aB
=0.5 x 0.7 =0.35
Genotype ab
=0.5 x 0.3 =0.15
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