# F215 - Hardy-Weinberg

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#1
This is a post where individuals can discuss the Hardy-Weinberg equations and problems that may come up in the upcoming F215 exam.

Here is one that i need some help on as i can't seem to get the numbers to add up to 1! Help would be appreciated! Within a population of butterflies, the colour brown (B) is dominant over the colour white (b) and 40% of all butterflies are white. Given this sample of information, calculate the following:

A- The % of butterflies in a population that are heterozygous
B- The frequency of homozygous dominant individuals

http://www.k-state.edu/parasitology/.../hardwein.html
0
8 years ago
#2
(Original post by PerkyB)
This is a post where individuals can discuss the Hardy-Weinberg equations and problems that may come up in the upcoming F215 exam.

Here is one that i need some help on as i can't seem to get the numbers to add up to 1! Help would be appreciated! Within a population of butterflies, the colour brown (B) is dominant over the colour white (b) and 40% of all butterflies are white. Given this sample of information, calculate the following:

A- The % of butterflies in a population that are heterozygous
B- The frequency of homozygous dominant individuals

http://www.k-state.edu/parasitology/.../hardwein.html
Thanks for the link In response to your question, I'm not sure how you've tried to tackle it but I'll go through it in steps.

Bear in mind: BB = p2, Bb = 2pq and bb = q2

a) First, you know that q2 = 0.4 (=40/100), since q2 = bb and they can only be white if homozygous recessive.

∴ q = √q2 = 0.632 (keep all figures on calculator when working it out).

∴ p = 0.368, since p + q = 1 ∴ p = 1 - q

∴ p2 = 0.135 (simply squaring the answer for p).

∴ since you know that: p2 + 2pq + q2 = 1 then 2pq = 1 - p2 - q2.
So 2pq = 1 - 0.135 - 0.4 = 0.465 (to 3.s.f)
(You want 2pq since 2pq represents all of the heterozygous mempers of a population).

So as a % that is: 47% (to 2.s.f).

b) for this question, the answer must surely be p2 since homozygous dominant = BB, so the frequency should be 0.135 or 0.14 (to 2.s.f).

Hope that has helped or made it any clearer at all 0
8 years ago
#3
I have a strong feeling Hardy Weinberg will come up in this exam
0
8 years ago
#4
Hope it does easy marks

Posted from TSR Mobile
0
7 years ago
#5
(Original post by Rhodopsin94)
Thanks for the link In response to your question, I'm not sure how you've tried to tackle it but I'll go through it in steps.

Bear in mind: BB = p2, Bb = 2pq and bb = q2

a) First, you know that q2 = 0.4 (=40/100), since q2 = bb and they can only be white if homozygous recessive.

∴ q = √q2 = 0.632 (keep all figures on calculator when working it out).

∴ p = 0.368, since p + q = 1 ∴ p = 1 - q

∴ p2 = 0.135 (simply squaring the answer for p).

∴ since you know that: p2 + 2pq + q2 = 1 then 2pq = 1 - p2 - q2.
So 2pq = 1 - 0.135 - 0.4 = 0.465 (to 3.s.f)
(You want 2pq since 2pq represents all of the heterozygous mempers of a population).

So as a % that is: 47% (to 2.s.f).

b) for this question, the answer must surely be p2 since homozygous dominant = BB, so the frequency should be 0.135 or 0.14 (to 2.s.f).

Hope that has helped or made it any clearer at all i thought that homozygous dominant would be p2 (BB) which is not pq as that is heterozygous.

so the frequency of homoygous dominant is 0.37?
0
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