Fruit flies Biologists wish to mate pairs of fruit flies having genetic makeup RrCc, indicating that each has one dominant gene (R) and one recessive gene (r) for eye color, along with one dominant (C) and one recessive (c) gene for wing type. Each offspring will receive one gene for each of the two traits from each parent, so the biologists predict that the following phenotypes should occur in a ratio of $9:3:3:1:$

Assume that the conditions for inference are met. Carry out a test at the $\alpha =0.05$ significance level of the proposed genetic model.

Significance level: $5\%=0.05$

Ratio: $R_1:R_2:W_1:W_2=9:3:3:1$

$n=4$

Null hypothesis: $H_0$

The null hypothesis is rejected when the value of $P$ is less or equal to the significance level.

The null hypothesis: $H_0$

The phenotype occur in a ratio of $9:3:3:1$

Alternative hypothesis: $H_1$

The phenotype does not occur in a ratio of $9:3:3:1$

Now,

Take the sum: $S=9+3+3+1=16$

Create a table with the observed and expected values:

The expected value is calculated as:

$$\begin{gathered} E1=T\times \frac{R_1}{R}=200\times \frac{9}{16}=112.5 \\ {E}_2=T\times \frac{R_2}{R}=200\times \frac{3}{16}=37.5 \\ {E}_3=T\times \frac{W_1}{R}=200\times \frac{3}{16}=37.5 \\ {E}_4=T\times \frac{W_2}{R}=200\times \frac{1}{16}=12.5 \end{gathered}$$

Now, use the test −statistics:

$$\begin{gathered} x^2=\sum _{i=1}^4\frac{{(O_i-E_i)}^2}{E_i} \\ {x}^2=\frac{{(99-112.5)}^2}{112.5}+\frac{{(42-37.5)}^2}{37.5}+\frac{{(49-37.5)}^2}{37.5}+\frac{{(10-12.5)}^2}{12.5} \\ {x}^2=6.1867 \end{gathered}$$

Degree of freedom:

$df=n-1=4-1=3$

For $x^2=6.1867$and $df=3$the $P-$value is $0.1029$

So, the null hypothesis is not rejected when the value of P is not less or equal to the significance level.

$P>\propto :0.1029>0.05$: fail to reject $H_0$

Hence,

The phenotypes occurs in the given ratio $9:3:3:1$

Therefore, the$P>\propto :0.1029>0.05$