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 proportionof individuals with the different blood types (Type A, B, AB and O) if the population is composed of thefollowing:Туре A - 80Турe АB - 35 Туре В - 90 Туре О- 195(r) Type 0 = 195/400 = 0.488(p) Type A = 80/400 = 0.200Type AB = 35/400 = 0.087(q) Type B = 90/400 = 0.225Gene frequency:Genotypic frequencyfO)= r=r2 =0.4882f(AA) = p2=0.22 f(AA)=0.04f0)=r= 0.488f(AO) =2pr =2(0.2)(0.488) =0.195ELA)=p=1-(g+)f(BB) = q2 =0.4882-D0.238=1-(0.225+0.488)f(BO) =2gr=2(0.225)(0.488)=0.220HA)=p =0.287f(AB)=2pq=2(0.200)(0.488)=0.195FL00)=r2=0.4882=0.488UB)=q=1-{g+)UB)=q=1-(0.200+0.488)EB)=q=0.312Proportion:TYPE A = AA+AO(p2+2pr) x totale(0.04+0.195) x 400 = 94TYPE B = BB+BO(q2+2gr) x total=(0.238+0.220) x 400 = 183.2TYPE AB = AB(2pq) x total=(0.195) x 400 = 78TYPE O = 00(r2) x total=(0.488) x 400 = 195.2

In the absence of disrupting events, the Hardy-Weinberg equilibrium states that genetic changes in a…

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

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

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter19: Population Genetics And Human Evolution

Section: Chapter Questions

Problem 11QP: Using the HardyWeinberg Law in Human Genetics In a given population, the frequencies of the four...

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Genetic Variation

Genetic variation refers to the variation in the genome sequences between individual organisms of a species. Individual differences or population differences can both be referred to as genetic variations. It is primarily caused by mutation, but other factors such as genetic drift and sexual reproduction also play a major role.

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Quantitative genetics is the part of genetics that deals with the continuous trait, where the expression of various genes influences the phenotypes. Thus genes are expressed together to produce a trait with continuous variability. This is unlike the classical traits or qualitative traits, where each trait is controlled by the expression of a single or very few genes to produce a discontinuous variation.

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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

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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
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