Hardy Weinberg question

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Hardy Weinberg question

Post by Alla on Sat Mar 15, 2008 12:54 am

First of all, your population is quite small and you are assuming that both parents are heterozygous (or is it given in your problem?). In this case, the expected phenotypic ratio of dominant: recessive is 3:1.

Your total number of individuals in this experiment is 34, so your

expected numbers would be 25.5 Brown: 8.5 Blue and
observed numbers are 24 Brown: 10 Blue

if you do a chi-square test with these numbers, you get:

--------BR-----BL
Obs____24____10
Exp____25.5__8.5
Dev ___-1.5__1.5
d2 ____2.25__2.25
d2 /e__0.09__0.26

chi2 =0.35; df=1 and p is close to 0.9 (therefore, hypothesis is correct)

Your data is missing one important point. You need to know in advance one of two things:

1. If parents are known to be heterozygous, you can predict from the ratios which traits are dominant or recessive;
2. If you know which traits are dominant and which recessive, you can figure out the genotype of parents from this data.

To give you an example: if you are given information that Blue is dominant, then parents of your generation would be one homozygous brown and other heterozygous blue. The chi squared for this will be the same as above.

Hardy–Weinberg equation is better suited for large populations where you can assume that most parents are heterozygous. In the small population size, you cannot assume that.

To answer your second question, the only ways you would know which ones and how many of your individuals with dominant phenotype are heterozygous are

1. collect data of subsequent generations or
2. do DNA analysis.

I hope this helps.

I would appreciate it if you registered and asked further questions through this site.

Good luck!

Alla
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Re: Hardy Weinberg question

Post by curious1234 on Sat Mar 15, 2008 2:57 am

Thanks, I think that really helped.

Just one more question. How did you determine the expected values of 25.5 and 8.5 ?

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Re: Hardy Weinberg question

Post by Alla on Sat Mar 15, 2008 3:04 am

You take the total of experimental group (34 in this case), divide by the total from empirical ratio and multiply it by trait coefficient:

3 Brown: 1 Blue (total 4)

Expected Brown = (34/4)x3
Expected Blue = (34/4)x1

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Re: Hardy Weinberg question

Post by curious1234 on Sat Mar 15, 2008 3:07 am

Thanks !

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Re: Hardy Weinberg question

Post by Alla on Sat Mar 15, 2008 3:07 am

You are welcome!

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