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rivs9 rivs9
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12 years ago
I'm not sure how to answer this as I still do not fully understand the Hardy Weinberg Principle.

In a population of 1000 individuals, 190 individuals demonstrate the dominant trait.  Calculate the following:
Frequency of the dominant allele.
Frequency of the recessive allele.
Number of homozygous dominant individuals.
Number of heterozygous dominant individuals.
Number of homozygous recessive individuals.
 
If you could please help with answering this & explaining how you got the answer it would be greatly appreciated. Thank you Slight Smile
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wrote...
12 years ago
Hardy Weinberg just looks at the change in allele frequencies in a population.  If the population is in equilibrium the p and q will stay the same through generations.

1=p^2+2pq+q^2 (thats just probability of homo-dom, hetero, homo-rec and since these are the only possiblities the total equals one) the first 2 terms in the equation are the dominant phenotype and the q^ 2 are the phenotype/genotype homo recessive.
p+q=1 (the formuals only works with 2 allele populations since p and q each stand for an allele)
So from the above info
1000 population
190 are Aa or AA (show dom trait)
this means that 810 show the recessive trait.
We know that q^2 is 0.810 (810/1000)
so q=0.9 so p=1-q =0.1 (take the square root of recessice individuals frequency to give recessive allele frequency)
NOTE always start with individuals who display recessive trait since you know the genotype and phenotype of these individuals
Dom allele =0.1=p
Rece allele =0.9=q
Number of Homozygous dom = 0.01 (which is P^2 = 0.1^2)
Number of homozygous dominant individuals.= P^2= 0.1^2= 0.01
Number of heterozygous dominant individuals.=2pq=2*0.1*0.9=0.18
Number of homozygous recessive individuals=q^2=0.81
to check add these up and you get 1.
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