Answer:
Yes, the observed results are consistent with the theoretical distribution predicted by the genetic model. The population is in equilibrium.
Explanation:
<u>Available data:</u>
- Offspring of fruit flies may have yellow or ebony bodies and normal wings or short wings.
- Expected phenotypic ratio 9:3:3:1
- 9 yellow, normal: 3 yellow, short: 3 ebony, normal: 1 ebony, short
- Sample size , N = 100
- Phenotypic distribution 59:20:11:10
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
- Chi square= ∑ ((Obs-Exp)²/Exp)
- ∑ is the sum of the terms
- Obs are the Observed individuals
- Exp are the Expected individuals
- K =genotypes number = 4
- Significance level, 5% = 0.05
- Table value = Critical value
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.
<u>yellow, normal yellow, short ebony, normal ebony, short</u>
Expected <em>56.25 18.75 18.75 6.25</em>
Observed <em> 59 20 11 10</em>
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
- X² = 5.67
- Significance level = 0.05
- Degrees of freedom = genotypes number - one = 4 - 1 = 3
- Critical value or table value = 9.348
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.
