In a population of 100 individuals, the genotypes AA, Aa and aa have frequencies

Question

In a population of 100 individuals, the genotypes AA, Aa and aa have frequencies 0.5, 0.25, and 0.25 respectively. What is p? What is q?
Answer
A. p = 0.75, q = 0.25
B. p = 0.75, q = 0.25
C. None of these
D. p = 0.625, q = 0.375

After a single generation, what is the expected frequency of genotypes AA, Aa, and aa (assume Hardy-Weinberg in effect)?
Answer
A. 0.25, 0.5, 0.25
B. None of these
C. 0.39, 0.47, 0.14
D. 0.56, 0.38, 0.06


You can statistically test if a population is in Hardy Weinberg Equilibrium using a Chi Square Test. When compared to a probability table, this tests for differences between observed frequencies and expected frequencies.
2 value = (Observed-Expected)2/Expected
3.1) Calculate the observed number of individuals of each genotype from Question 1 and the expected number of individuals of each genotype using your answer from question 2.
3.2) Calculate the 2 value using this data.
What is the 2 value?

Answer
A. 0.6
B. 3.1
C. None of these
D. 22

Explain ALL 3 answers please and how to get it.

Explanation / Answer

Question 1: p represents the number of dominant alleles in the population (A). Remember every human cell has two sets of chromosomes and thus two alleles. One on each chromosome. If ones genotype is AA they have 2 A alleles. The Hardy-Weinberg equation is one you should memorize. p2 + 2pq + q2 =1. Here p2 represents the number of AA genotypes. 2pq homozygous Aa genotypes and q2 recessive genotypes. So in our population let's figure out the allele frequency: If AA = 0.5 then 50 individuals have the AA genotype and thus we have 100 A's! If Aa = 0.25 we have 25 more A's and 25 a's. If aa = 0.25 then we have 50 more a's So in sum, we have 125 A's and 75 a's this sum = 200 which indicates we probably did our math correct... We expect 100 individuals to have 200 alleles (we all have 2 chromosomes per cell!) So 0.625 = A (125/200) and 0.375 = a (75/100). In the next generation we expect the genotype frequency to be p2 + 2pq + q2 =1... p2 = homozygous dominant, 2pq = heterozygous and q2 homozygous recessive so just plug in the numbers 0.625x0.625 = 0.39 2x0.625x0.375 = 0.47 0.375x0.375 = 0.14 We can add these up to be sure we did our math correctly it should equal 1 and it does indeed. So x2 = (O-E)^2/E In part 1 we observed: AA = 50 Aa = 25 aa = 25 We should have had: AA = 39 Aa = 47 aa = 14 So (50-39)^2/39=3.1 + (25-47)^2/47=... and the aa problem.... add these up and you get... 22. Hopefully this helps. BB Second year medical student

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