Hardy-Weinberg Equilibrium Question

[DDS-King]

Hey guys I was hoping someone could help me out. I don't know why but I can never understand which variable I am given in the question.

I know that p and q are ALLELE frequencies and p2 and q2 are GENOTYPE frequencies, but how do you know which is given by the information in the question?

A population of sheep is in Hardy-Weinberg equilibrium. The trait for black wool is dominant, and the trait for white wool is recessive. Out of 100 sheep, 84 were found to have black wool. How many sheep are heterozygous?

• A. 24
• B. 36
• C. 48
• D. 60
• E. 72

Any help is GREATLY appreciated!

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[510586]

Is this correct? Was the answer C?

 

[DDS-King]

View attachment 195274 Is this correct? Was the answer C?

Thanks. I understand that, but how did you know that 16/100 was p^2 and not p?

 

[510586]

Thanks. I understand that, but how did you know that 16/100 was p^2 and not p?

It's actually q^2. for alleles it's p+q=1 (p being dominant, q being recessive). Since they only gave the sheep population (which is phenotypes), we use q^2 (which we know is 16 because recessive have to be aa not Aa/AA). q would be just a not aa (and we don't know a because they gave us phenotypes NOT genotypes). Just do p^2+2pq+q^2=1 for anything when given phenotypes and p or q if given genotypes/allele frequencies. Hope this helps

 

[nornton]

The first thing i like to do on these types of problems is write out all the equations I know. So, p^2 + 2pq + q^2 =1 and p + q = 1. 84 had black wool so they are either homozygous dominant or heterozygous. Therefore, p^2 + 2pq = .84. Then you know that q^2 = .16 and if you take the square root of that q=.4. Now you can figure out p by plugging q into p + q = 1 and you get p=.6. You are trying to find out how many are heterozygous so plug your p and q values into 2pq and you get .48 or 48 sheep.

 

[DDS-King]

It's actually q^2. for alleles it's p+q=1 (p being dominant, q being recessive). Since they only gave the sheep population (which is phenotypes), we use q^2 (which we know is 16 because recessive have to be aa not Aa/AA). q would be just a not aa (and we don't know a because they gave us phenotypes NOT genotypes). Just do p^2+2pq+q^2=1 for anything when given phenotypes and p or q if given genotypes/allele frequencies. Hope this helps

Got it! Thanks