Offspring of certain fruit flies may have yellow or ebony bodies and normal wings or short wings. Genetic theory predicts that these traits will appear in the ratio 9:3:3:1 (9 yellow, normal: 3 yellow, short: 3 ebony, normal: 1 ebony, short). A researcher checks 100 randomly selected fruit flies and finds the distribution of the traits to be 59, 20, 11, and 10, respectively. Are the results this researcher observed consistent with the theoretical distribution predicted by the genetic model

Answer:

Yes, the observed results are consistent with the theoretical distribution predicted by the genetic model. The population is in equilibrium.

Explanation:

Available data:

To know if these results are consistent with the expected ones, we need to develop a chi-square analysis. To do it we need the following information

- ∑ is the sum of the terms

- Obs are the Observed individuals

- Exp are the Expected individuals

- K =genotypes number = 4  

First, define the hypothesis:

Hypothesis: The allele of this population will assort independently. The population is in equilibrium

H₀= Individuals will be equally distributed.  

H₁ = Individuals will not be equally distributed.  

Second, we need to get the expected number of individuals with each phenotype. To do it, we will use the expected phenotypic ratio and the total number of individuals in the sample. We can just perform a three simple rule, as follows:

  16 ------------------------------------ 100 individuals in the sample ------- 100%

  9 yellow, normal ---------------- X = 56.25 individuals -------------------56.25%

  3 yellow, short ------------------- X = 18.75 individuals --------------------18.75%

  3 ebony, normal -----------------X = 18.75 individuals ---------------------18.75%

  1 ebony, short ---------------------X = 6.25 individuals ----------------------6.25%

Now that we know the expected numbers of individuals with each genotype, we can compare them with the observed ones.

                  yellow, normal      yellow, short     ebony, normal    ebony, short

Expected            56.25                    18.75                   18.75                    6.25

Observed            59                           20                      11                          10

The chi-square value = Σ(Obs-Exp)²/Exp.

So now we need to calculate (Obs-Exp)²/Exp

(Obs-Exp)²/Exp = (59-56.25)²/56.25 = 0.134

(Obs-Exp)²/Exp = (20 - 18.75)²/18.75 = 0.083

(Obs-Exp)²/Exp = (11 - 18.75 )²/18.75 = 3.203

(Obs-Exp)²/Exp = (10 - 6.25)²/6.25 = 2.25

X² = Σ(Obs-Exp)²/Exp =  0.134 + 0.083 + 3.203 + 2.25 = 5.67

P₀.₀₅ > X2

9.348 > 5.67

There is not enough evidence to reject the null hypothesis. The genotypes might be in equilibrium, and there might be an independent assortment.

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