HARDY weinberg equi.

[predentgal]

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A recessive trait appears in 81% of the individuals in a population that is in Hardy-Weinberg equilibrium. What percent of the population in the next generation is expected to be homozygous dominant?

Answer is 1%

In solving the problem, the explanation assumes that .81 is q^2 but how does having a recessive trait equate to a homozygous recessive trait(q^2). Can someone explain, I'm soo confused.

Thankss

 

[Baylor2011]

Typically for a recessive trait to be expressed, it has to be homozygous recessive. If it were heterozygous, the dominant trait will mask the recessive, displaying the dominant phenotype.

 

[Kilerfornia]

How would you solve this problem if you started from scratch?

 

[RCT PC CRN]

First thing, If population is in HW equilibrium allele freq (p and q) stay constant.

Recessive trait (i.e. Phenotype) 81% => q^2 = 0.81 => q=0.9
p+q=1 => p=0.1

Homozygous dominant = p^2 = 0.01 or 1%

 

[Kilerfornia]

Thanks Alot😀

 

[predentgal]

Ok now I understand that a recessive trait has to be q^2 but in the equation p + q=1 what's q?

 

[Baylor2011]

http://www.k-state.edu/parasitology/biology198/hardwein.html

 

[predentgal]

Thanks 🙂

 

[LiteBear2011]

Ok now I understand that a recessive trait has to be q^2 but in the equation p + q=1 what's q?

In the "p+q=1" relationship, p and q refer to the allele frequency in a population. Consistent with the letters used in the original problem "p" represents the allele frequency of the dominant allele and "q" the recessive allele; the sum of the two equals, of course, 100% or 1.