A population of 400 plants is genotyped for a flower-color locus. Observed counts are RR = 90, Rr = 220, rr = 90. The student calculates p = 0.5, q = 0.5 and computes a chi-square value of 10.0 (df = 1, critical = 3.84) to test for Hardy-Weinberg equilibrium.
What conclusion is appropriate?
- A
Fail to reject the null; the population is in Hardy-Weinberg equilibrium.
- Bcheck_circle
Reject the null; observed genotypes deviate significantly from Hardy-Weinberg expectations.
- C
Sample size is too small to compute chi-square.
- D
p and q must be recalculated because they sum to 1.
Explanation
Wait - chi-square sums to 1+2+1=4. With df=1, critical=3.84, 4.0 > 3.84, reject. The stem says chi=10.0 which also exceeds 3.84, so the conclusion is the same: significant deviation.