Jeffrey Pines from Mt. Rose were genotyped at a resistance gene for a type of fungi that affects pines. Genotypes of 1000 of these Jeffrey Pines were: 300 R1R1, 500 R1R2, and 200 R2R2. Use a chi- square (χ2) test to determine if this population is in Hardy- Weinberg equilibrium? The critical chi- square value at P=.05 for 1 degree of freedom is 3.84.
A) 0.10; fail to reject
B) 10; fail to reject
C) 0.10; reject
D) 10; reject
E) 37.5; reject

Question

Quizplus · Accepted Answer

To determine if the population is in Hardy-Weinberg equilibrium, we first calculate the allele frequencies. Let p be the frequency of R1 and q be the frequency of R2. From the genotypes, we have:- 300 R1R1: 600 R1 alleles- 500 R1R2: 500 R1 alleles and 500 R2 alleles- 200 R2R2: 400 R2 allelesTotal R1 alleles = 600 + 500 = 1100Total R2 alleles = 500 + 400 = 900Total alleles = 1100 + 900 = 2000p = 1100 / 2000 = 0.55q = 900 / 2000 = 0.45Expected frequencies under Hardy-Weinberg equilibrium are:- R1R1: p^2 * 1000 = (0.55)^2 * 1000 = 302.5- R1R2: 2pq * 1000 = 2 * 0.55 * 0.45 * 1000 = 495- R2R2: q^2 * 1000 = (0.45)^2 * 1000 = 202.5Chi-square (χ2) calculation:χ2 = [(observed - expected)^2 / expected] for each genotype= [(300 - 302.5)^2 / 302.5] + [(500 - 495)^2 / 495] + [(200 - 202.5)^2 / 202.5]= 0.021 + 0.051 + 0.031= 0.103Since the calculated χ2 value (0.10) is less than the critical value of 3.84, we fail to reject the null hypothesis, indicating that the population is in Hardy-Weinberg equilibrium.