A.) Assuming there is random mating, find the gene frequencies, genotypic frequencies, and the proportion of individuals with the different blood types (Type A, B, AB and O) if the population is composed of the following: Туре А - 80 Туре АВ - 35 Туре В - 90 Туре О - 195 Type O = 195/400 = 0.488 Type A = 80/400 = 0.200 Type AB = 35/400 = 0.087 %3D

Biology (MindTap Course List)
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Chapter19: Evolutionary Change In Populations
Section: Chapter Questions
Problem 3TYU: The MN blood group is of interest to population geneticists because (a) people with genotype MN...
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A.) Assuming there is random mating, find the gene frequencies, genotypic frequencies, and the
proportion of individuals with the different blood types (Type A, B, AB and O) if the population is
composed of the following:
Туре А - 80
Туре АВ - 35 Туре В - 90
Type o - 195
Type O = 195/400 = 0.488
Type A = 80/400 = 0.200
Type AB = 35/400 = 0.087
Type B = 90/400 = 0.225
Gene frequency:
Genotypic frequency:
f(0)= r=r2 =0.4882
f(AA) = p2=0.22 f(AA)=0.04
f(0)=r= 0.488
f(AO) =2pr =2(0.2)(0.488) =0.195
E(A)=p=1-(g+r)
f(BB) = q2 =0.4882=0.238
=1-(0.225+0.488)
f(BO) =2qr=2(0.225)(0.488)=0.220
f(A)=p =0.287
f(AB)=2pq=2(0.200)(0.488)=0.195
f(o0)=r2=0.4882=0.488
f(B)=q=1-(p+r)
f(B)=q=1-(0.200+0.488)
f(B)=q=0.312
Proportion:
1.) TYPE A = AA+AO
4.) TYPE O = 00
(p2+2pr) x total=(0.04+0.195) x 400 = 94
(r2) x total=(0.488) x 400 = 195.2
2.) TYPE B = BB+BO
(q2+2qr) x total=(0.238+0.220) x 400 = 183.2
3.) TYPE AB = AB
(2pq) x total=(0.195) x 400 = 78
Transcribed Image Text:A.) Assuming there is random mating, find the gene frequencies, genotypic frequencies, and the proportion of individuals with the different blood types (Type A, B, AB and O) if the population is composed of the following: Туре А - 80 Туре АВ - 35 Туре В - 90 Type o - 195 Type O = 195/400 = 0.488 Type A = 80/400 = 0.200 Type AB = 35/400 = 0.087 Type B = 90/400 = 0.225 Gene frequency: Genotypic frequency: f(0)= r=r2 =0.4882 f(AA) = p2=0.22 f(AA)=0.04 f(0)=r= 0.488 f(AO) =2pr =2(0.2)(0.488) =0.195 E(A)=p=1-(g+r) f(BB) = q2 =0.4882=0.238 =1-(0.225+0.488) f(BO) =2qr=2(0.225)(0.488)=0.220 f(A)=p =0.287 f(AB)=2pq=2(0.200)(0.488)=0.195 f(o0)=r2=0.4882=0.488 f(B)=q=1-(p+r) f(B)=q=1-(0.200+0.488) f(B)=q=0.312 Proportion: 1.) TYPE A = AA+AO 4.) TYPE O = 00 (p2+2pr) x total=(0.04+0.195) x 400 = 94 (r2) x total=(0.488) x 400 = 195.2 2.) TYPE B = BB+BO (q2+2qr) x total=(0.238+0.220) x 400 = 183.2 3.) TYPE AB = AB (2pq) x total=(0.195) x 400 = 78
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