My Q2 was .32 and the population was 1000, so to get the number of recessive trait you have to multiply .32*1000
Which gave me 320 as my recessive trait
The recessive phenotype is already given to us which is .32
To the the number of individual recessive allele(q) i √(.32) which gave me .57
To get the number of the dominant allele(p) i subtracted 1-.q which is 1-.57 which gave me .43
To get the number of dominat phenotype you square root your dominat allele(p) √(.43)= .66
Lastly for the heterozygoats individuals (2pq) you multiply 2(.43)(.57)= .49
Tuesday, October 27, 2015
Hardy-Weinberg.
My Q2 was .32 and the population was 1000, so to get the number of recessive trait you have to multiply .32*1000
Which gave me 320 as my recessive trait
The recessive phenotype is already given to us which is .32
To the the number of individual recessive allele(q) i √(.32) which gave me .57
To get the number of the dominant allele(p) i subtracted 1-.q which is 1-.57 which gave me .43
To get the number of dominat phenotype you square root your dominat allele(p) √(.43)= .66
Lastly for the heterozygoats individuals (2pq) you multiply 2(.43)(.57)= .49
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Hi Sylvia, I liked how you posted a picture of your work to give you a better understanding but you made a mistake in your math. So you were trying to find p^2 but instead of squaring .43 you square rooted .43 so instead of getting .66 you should have gotten .18 Also you forgot to find the number of people that are homozygous dominat individuals and heterozygous individuals in the population. Double check your work, that will always help you.
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