Will rate lifesaver:):) Thank you:):) The Rh factor of human blood is coded by t

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Will rate lifesaver:):) Thank you:):) The Rh factor of human blood is coded by the Rh gene. Thisgene has 2 alleles (Rh+ and Rh-) producing 3 possible genotypes.There are two phenotypes, Rh+ blood and Rh-blood. Homozygous Rh+and heterozygous have the same phenotype (Rh+ blood); thus havingRh- blood is a recessive trait. In certain population 80% of allpersons are Rh+ and 20% are Rh-. Assume that mating is randomfor the Rh factor. a.) Of all marriages in this population, what percentage isexpected to involve an Rh+ hunsband and Rh- wife? b.) In what percentage of all marriages is one person Rh+ andthe other Rh-? Hardy Weinberg Equalibrium:
Mating randomly
large population
free from mitation
free from mutation
free from natural selection Will rate lifesaver:):) Thank you:):) The Rh factor of human blood is coded by the Rh gene. Thisgene has 2 alleles (Rh+ and Rh-) producing 3 possible genotypes.There are two phenotypes, Rh+ blood and Rh-blood. Homozygous Rh+and heterozygous have the same phenotype (Rh+ blood); thus havingRh- blood is a recessive trait. In certain population 80% of allpersons are Rh+ and 20% are Rh-. Assume that mating is randomfor the Rh factor. a.) Of all marriages in this population, what percentage isexpected to involve an Rh+ hunsband and Rh- wife? b.) In what percentage of all marriages is one person Rh+ andthe other Rh-? Hardy Weinberg Equalibrium:
Mating randomly
large population
free from mitation
free from mutation
free from natural selection Hardy Weinberg Equalibrium: Mating randomly large population free from mitation free from mutation free from natural selection

Explanation / Answer

Rh+Rh+               Rh+ Rh-   =     80%                 Rh-Rh- =20%

UsingH.W,E.

P2 +2PQ=   80%

Q2= Rh- Rh-=20%

Q=SQUARE ROOT OF 20%= .44

P+Q= 1

Q= .44

P=.56

Percentage is expected ----Rh-wife       = Rh- Rh   =.44X.44 X 100= 19.36%

what percentage is expectedto involve an Rh+ hunsband =80.64 %

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