Question

Two pea plants with purple flowers are crossed. Among the offspring, 63 have purple flowers, and 17 have white flowers. With a chi-square test, compare the observed numbers with a 3:1 ratio and determine if the the difference between observed and expected is statistically significant. Note: Chi-square critical value with 1 degree of freedom is 3.84 with alpha equal to 0.05

Answer 1

Chi-square value = 0.6

The difference between the observed individuals and the expected individuals is not statistically significant. There is not enough evidence to reject the null hypothesis.

Step-by-step explanation:

H₀= Individuals will be equally distributed

H₁ = Individuals will not be equally distributed.

• Chi square= ∑ ((O-E)²/E)

∑ is the sum of the terms

O are the Observed individuals

E are the Expected individuals

• Freedom degrees = 1

• Significance level, 5% = 0.05

• Table value/Critical value = 3.84

The number of observed individuals:

• 63 have purple flowers,
• 17 have white flowers

Total number of individuals, N = 80 = 63 + 17

Observed Genotypic frequencies:

• F(PF) = 63/80 = 0.79
• F(pf) = 17/80 = 0.21

p² + 2pq + q² = 1

p² + 2pq = F(PF)

0.79 + 0.21 = 1

The expected genotypic frequency:

F (PF)= 0.75

F (pf) = 0.25

The number of expected individuals:

• PF= 0.75 x 80 = 60
• wf = 0.25 x 80 = 20

Chi square = sum (O-E)²/E

• PF = (O-E)² /E

PF =(63 - 60) ² / 60

PF = 0.15

• pf = (O-E)² /E

pf= (17 - 20)²/20

pf= 0.45

Chi square= sum ((O-E)²/E) = 0.15 + 0.45 = 0.60

• Chi square = 0.6
• Degrees of freedom = 1
• alpha 0.05
• Table value/Critical value = 3.84

0.6 < 3.84 meaning that the difference between the observed individuals and the expected individuals is not statistically significant. There is not enough evidence to reject the null hypothesis. Probably both parentals were heterozygous for the trait.