The Offspring produced by a cross between two given types of plants can be any of the three genotypes denoted by A, B, and C. A theoretical model of any of the three genotypes denoted by A, B, and C should be in the ratio 1:2:1. For experimental verification, 100 plants are bred by crossing the two given types. Their genetic classifications are recorded. Do these data contradict the genetic model

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.

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.  

Explanation:

Available data:

Crossed genotypes: twoGenotypes among the offspring; Three → A, B, and CExpected phenotypic ratio → 1:2:1Total number of individuals, N = 100A = 18 individualsB = 55 individualsC = 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.  

 

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