Genetics + Chi-square test

good job your getting there

Well I can't figure out... What do I fill in in that chart???

   M      N          Totals
M   29.8      
N         21.3   
Totals         100

The other number is 48.9... But I can't get it to work!

Post Merge: 11 years ago

         M             N           Totals
M        29.8           48.9       78.7
N        0???        21.3            21.3
Totals    29.8           70.2            100

I don't think this is actually right.... Do I really put 0 in there? What is wrong with this...

Post Merge: 11 years ago

Or maybe put half in under each MN???


   M                    N           Totals
M   29.8                 24.45            54.25
N   24.45           21.3           45.75
Totals   54.25   45.75   100


....?

First assume that the population is at hardy Weinberg equilibrium. The formula for that is p^2+2pq+q^2 where p is the frequency of the dominant allele, and q the frequency of the recessive allele.

Find the number of M alleles: #MM +#MN/2
Find the number of N alleles: #NN + #MN/2

Divide each by the total number of observations (2000 since the gene trait has two alleles) to find the frequencies of each genotype.

Plug those values into the formula in the first paragraph to find  the expected frequencies. Should look like 1000*M^2+2000*MN+1000N^2.

Then perform chi square with 1 degree of freedom.

I'm sorry for the slap-dash explanation, but I'm in a bit of a rush. Hopefully this helps you get on the right track. I hated these kinds of problems.

Thank you!!! That is super helpful!!!!   

I'm sorry- I still cannot figure this out.

I tried:
#MM (298) + #MN (489/2) = 542.5 # of M alleles

#NN (213) + #MN (489/2) = 457.5 # of N alleles

I really don't understand what to do with these numbers.

I also tried:
(1000 * 0.149^2) + (2000 * .2445^2) + (1000 * 0.1065^2) = 522.54325

OR

 (Didn't work at all!!!!)

(1000 * 542.5^2) + (2000 * 244.5) + (457.5^2 * 1000)

Can someone help me work through this?

I am so frustrated! :/

MM(p^2)= 298/1000= 0.298
MN(2pq)= 489/1000= 0.489
NN(q^2)= 213/1000= 0.213

Frequency of M=P=p^2+(2pq)= 0.298+(1/2)(0.489)= 0.543
Frequency of N=Q=1-P= 1-0.543= 0.457

Expected genotype frequencies
MM(p^2)= (0.543)^2= 0.295
MN(2pq)= 2(0.543)(0.457)= 0.496
NN=(q^2)= (0.457)^2= 0.210

Expected number of each genotype
MM= 0.295*1000= 295
MN= 0.496*1000= 496
NN= 0.210*1000= 210

Chi-Square= SUM(O-E)^2/E
               (298-295)^2/295+ (489-495)^2/495+ (213-210)^2/210
                  = 0.031+0.0727+0.042= 0.1457
   Using a chi-square distribution table at alpha= 0.05, with df=1, chi(table)= 3.84

   Since chi(calculated)<chi(table), do not reject null hypothesis (that is, that random mating is occurring)

Thank you!

Why are we using df=1? Wouldn't it be 2? since df=n-1? and we have 3 classes, MM, MN, NN?

We have only one degree of freedom because we have three genotypic categories and we used two numbers from the data (N and p) to calculate the expected values (3 - 2 leaves 1 degree of freedom). We did not need to use q since q = 1 - p.