Please check if my answer is correct. Thank you. 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: Type A - 80          Type AB – 35       Type B - 90         Type O - 195

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Please check if my answer is correct. Thank you.

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:

Type A - 80          Type AB – 35       Type B - 90         Type O - 195

 

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 A - 80
Турe АB - 35 Туре В - 90 Туре О- 195
(r) Type 0 = 195/400 = 0.488
(p) Type A = 80/400 = 0.200
Type AB = 35/400 = 0.087
(q) Type B = 90/400 = 0.225
Gene frequency:
Genotypic frequency
fO)= r=r2 =0.4882
f(AA) = p2=0.22 f(AA)=0.04
f0)=r= 0.488
f(AO) =2pr =2(0.2)(0.488) =0.195
ELA)=p=1-(g+)
f(BB) = q2 =0.4882-D0.238
=1-(0.225+0.488)
f(BO) =2gr=2(0.225)(0.488)=0.220
HA)=p =0.287
f(AB)=2pq=2(0.200)(0.488)=0.195
FL00)=r2=0.4882=0.488
UB)=q=1-{g+)
UB)=q=1-(0.200+0.488)
EB)=q=0.312
Proportion:
TYPE A = AA+AO
(p2+2pr) x totale(0.04+0.195) x 400 = 94
TYPE B = BB+BO
(q2+2gr) x total=(0.238+0.220) x 400 = 183.2
TYPE AB = AB
(2pq) x total=(0.195) x 400 = 78
TYPE O = 00
(r2) x total=(0.488) x 400 = 195.2
Transcribed Image Text: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 A - 80 Турe АB - 35 Туре В - 90 Туре О- 195 (r) Type 0 = 195/400 = 0.488 (p) Type A = 80/400 = 0.200 Type AB = 35/400 = 0.087 (q) Type B = 90/400 = 0.225 Gene frequency: Genotypic frequency fO)= r=r2 =0.4882 f(AA) = p2=0.22 f(AA)=0.04 f0)=r= 0.488 f(AO) =2pr =2(0.2)(0.488) =0.195 ELA)=p=1-(g+) f(BB) = q2 =0.4882-D0.238 =1-(0.225+0.488) f(BO) =2gr=2(0.225)(0.488)=0.220 HA)=p =0.287 f(AB)=2pq=2(0.200)(0.488)=0.195 FL00)=r2=0.4882=0.488 UB)=q=1-{g+) UB)=q=1-(0.200+0.488) EB)=q=0.312 Proportion: TYPE A = AA+AO (p2+2pr) x totale(0.04+0.195) x 400 = 94 TYPE B = BB+BO (q2+2gr) x total=(0.238+0.220) x 400 = 183.2 TYPE AB = AB (2pq) x total=(0.195) x 400 = 78 TYPE O = 00 (r2) x total=(0.488) x 400 = 195.2
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