Hardy-Weinberg

Just a quick question,

When you are doing a hardy weinberg exam question there will always be a figure in the question whether it's 16% are recessive or 1 in 1000 have the condition. What I want to know is how do you work out whether or not to square root that number. For example say with the 16%. You convert it to 0.16 then sometimes you either...
1) Square root the 0.16=0.4
2) Do 1-0.4=0.6
3) 0.6*0.4*2 to find the heterozygous

OR in some questions you do
1) 1-0.16=0.84
2) 2*0.84*0.16

So basically I want to know when do you have to square root the number in the question before you do 1-__?

Reply 1

acciolucy

9

If they tell you the number of organisms who are homozygous recessive, that is a P^2 or Q^2 value ( so 0.16).

In order the find P orQ, you must square root that to find the number of the recessive allele, e.g 0.4 have that allele.

To find the number of heterozygous you are finding 2 x P x Q, (with P and Q representing each allele), so you must find P or Q, which may be directly given or you may have to calculate.

Hope this helps!

Reply 2

acciolucy

9

Basically any 'homozygous' number represents a P squared or Q squared value, any 'allele' frequency/number represents a P or Q value.

Reply 3

daniel1005

OP

4

Original post by acciolucy

Basically any 'homozygous' number represents a P squared or Q squared value, any 'allele' frequency/number represents a P or Q value.

Thanks so much! That clears it up

Reply 4

daniel1005

OP

4

Original post by acciolucy

Basically any 'homozygous' number represents a P squared or Q squared value, any 'allele' frequency/number represents a P or Q value.

There is one question which screwed with my head a bit which I did recently. It was this:
The shells of this snail may be unbanded or banded. The absence or presence of bands is controlled by a single gene with two alleles. The allele for unbanded, B, is dominant to the allele for banded, b.

A population of snails contained 51% unbanded snails. Use H-W equation to calculate the % of population you would expect to be heterozygous.

So my working was that
BB= unbanded
Bb= unbanded
bb= banded

So if there is 51% unbanded snails, this means 51% of the population is BB and Bb? And I just don't get why you have to square root 0.51 because the square root of BB and Bb doesn't really tell you anything.

Reply 5

lerjj

13

Original post by daniel1005

There is one question which screwed with my head a bit which I did recently. It was this:
The shells of this snail may be unbanded or banded. The absence or presence of bands is controlled by a single gene with two alleles. The allele for unbanded, B, is dominant to the allele for banded, b.

A population of snails contained 51% unbanded snails. Use H-W equation to calculate the % of population you would expect to be heterozygous.

So my working was that
BB= unbanded
Bb= unbanded
bb= banded

So if there is 51% unbanded snails, this means 51% of the population is BB and Bb? And I just don't get why you have to square root 0.51 because the square root of BB and Bb doesn't really tell you anything.

Yo don't- (Bb + BB)^=51% tells you that (bb)^2=49%
this in turn gives you the frequency of b at 70% and B at 30%
From there you can work out all the genotypic frequencies.

Reply 6

acciolucy

9

See attached working -

image.jpg505.3KB

Hardy-Weinberg

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