# F215 - Hardy-Weinberg

#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

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)

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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