Complete question:
The offspring produced between two given types of plants can be any of the three genotypes, denoted by A, B and C. A theoretical model of gene inheritance suggests that the offspring of types A, B and C should be in a 1:2:1 ratio (meaning 25% A, 50% B, and 25% C). For experimental verification, 100 plants are bred by crossing the two given types. Their genetic classifications are recorded in the table below.
Genotype Observed frequency
A → 18 individuals
B → 55 individuals
C → 27 individuals
Do these contradict the genetic model?
Use a 0.05 level of significance.
Determine the chi-square test statistic.
Answer:
Do these contradict the genetic model? No, according to the chi-square test, there is not enough evidence to reject the null hypothesis of the population being in equilibrium.
Step-by-step explanation:
Available data:
- Crossed genotypes: two
- Genotypes among the offspring; Three → A, B, and C
- Expected phenotypic ratio → 1:2:1
- Total number of individuals, N = 100
- A = 18 individuals
- B = 55 individuals
- C = 27 individuals
So, let us first state the hypothesis:
- H₀= the population is equilibrium for this locus → F(A) = 25%, F(B) = 50%, F(C) = 25%
- H₁ = the population is not in equilibrium
Now, let us calculate the number of expected individuals, according to their expected ratio.
4 -------------- 100% -------------100 individuals
1 --------------- 25% -------------X = 25 individuals A
2 -------------- 50% -------------X = 50 individuals B
1----------------- 25%--------------X = 25 individuals C
A B C
- Observed 18 55 27
- Expected 25 50 25
- (Obs-Exp)²/Exp 1.96 0.5 0.16
(Obs-Exp)²/Exp
A) (18 - 25)²/25 = 49/25 = 1.96
B) (55 - 50)² / 50 = 25/50 = 0.5
C) (27 - 25)²/25 = 4/25 = 0.16
Chi square = X² = Σ(Obs-Exp)²/Exp
- ∑ is the sum of the terms
- O are the Observed individuals: 2 in chamber B, and 18 in chamber A.
- E are the Expected individuals: 10 in each chamber
X² = ∑ ((O-E)²/E) = 1.96 + 0.5 + 0.16 = 2.62
Freedom degrees = 2
Significance level, 5% = 0.05
Table value/Critical value = 5.99
X² < Critical value
2.62 < 5.99
These results suggest that there is not enough evidence to reject the null hypothesis. We can assume that the locus under study in this population is in equilibrium H-W.