Genetic/statistics help needed..

[Syringer]

I don't get how to calculate the 2nd/3rd gens for models..how does one do it?

"Suppose that the prob of the genotypes, AA, Aa, aa are p, 2q, and r, respectively, in the first generation. Find the prob in the 2nd and 3rd gen, and show that they are the same."

 

[Zee]

depends, are we examining the genes of a normal animal, plant or a spider?

 

[Scouzer]

lots of punnets squares is the only way i learned...

BTW: What class is this for?

 

[UNCjigga]

Ask the fruitflies...they're smarter than they let on!

 

[MySoS]

I think you use Chi(pronounced Ki) Square.

 

[nmcglennon]

use the hardy weinburg equation.
p^2 +2pq+q^2 = 1

Your probabilities will always equal 1.

p^2 is the probability of genotype AA
pq is the probability of genotype Aa
q^2 is the probability of genotype aa

Note Hardy Weinberg does not apply to cases of Genetic Drift, Nat. Selection, Mutation, Mieotic Drive, Gene Flow, etc.
try Googling for the Hardy Weinberg Equilibrium

 

[nmcglennon]

heres one:
http://users.rcn.com/jkimball..../H/Hardy_Weinberg.html

 

[jagec]

Originally posted by: MySoS
I think you use Chi(pronounced Ki) Square.

I thought it was S^(pi*d)+e^r

 

[nmcglennon]

I think you are supposed to show change in the gene pool.

"A population in Hardy-Weinberg equilibrium shows no change. What the law tells us is that populations are able to maintain a reservoir of variability so that if future conditions require it, the gene pool can change. If recessive alleles were continually tending to disappear, the population would soon become homozygous. Under Hardy-Weinberg conditions, genes that have no present selective value will nonetheless be retained."

You still have the same number of genes (same number of a and same number of A), but you have a diff proportion of AA, Aa, aa. These change through generations, and you can calculate them... I had to do it in my genetics class last semester.

 

[nmcglennon]

Another link:
http://en.wikipedia.org/wiki/Hardy-Weinberg_ratio

 

[Syringer]

Originally posted by: nmcglennon
use the hardy weinburg equation.
p^2 +2pq+q^2 = 1

Your probabilities will always equal 1.

p^2 is the probability of genotype AA
pq is the probability of genotype Aa
q^2 is the probability of genotype aa

Note Hardy Weinberg does not apply to cases of Genetic Drift, Nat. Selection, Mutation, Mieotic Drive, Gene Flow, etc.
try Googling for the Hardy Weinberg Equilibrium

How does that apply with 3 variables?

 

[Syringer]

.

 

[Syringer]

Originally posted by: Syringer

Originally posted by: nmcglennon
use the hardy weinburg equation.
p^2 +2pq+q^2 = 1

Your probabilities will always equal 1.

p^2 is the probability of genotype AA
pq is the probability of genotype Aa
q^2 is the probability of genotype aa

Note Hardy Weinberg does not apply to cases of Genetic Drift, Nat. Selection, Mutation, Mieotic Drive, Gene Flow, etc.
try Googling for the Hardy Weinberg Equilibrium

How does that apply with 3 variables?

Does that mean I can subsititute p for p^2, q for pq, and q for q^2?