Chapter 11: Q 11. (page 718)

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 $α = 0.05$ significance level of the proposed genetic model.

Short Answer

Expert verified

$P$ exceeds the significance level, indicating that $H_{0}$ is not rejected. As a result, the phenotypes occur in the following proportions: $9 : 3 : 3 : 1$

Step by step solution

Given information

Significance level: $5 % = 0.05$

Ratio: $R_{1} : R_{2} : W_{1} : W_{2} = 9 : 3 : 3 : 1$

$n = 4$

Concept

Null hypothesis: $H_{0}$

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

Calculation

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:

$E 1 = T \times \frac{R_{1}}{R} = 200 \times \frac{9}{16} = 112.5 E_{2} = T \times \frac{R_{2}}{R} = 200 \times \frac{3}{16} = 37.5 E_{3} = T \times \frac{W_{1}}{R} = 200 \times \frac{3}{16} = 37.5 E_{4} = T \times \frac{W_{2}}{R} = 200 \times \frac{1}{16} = 12.5$

Now, use the test −statistics:

$x^{2} = \sum_{i = 1}^{4} \frac{{(O_{i} - E_{i})}^{2}}{E_{i}} x^{2} = \frac{{(99 - 112.5)}^{2}}{112.5} + \frac{{(42 - 37.5)}^{2}}{37.5} + \frac{{(49 - 37.5)}^{2}}{37.5} + \frac{{(10 - 12.5)}^{2}}{12.5} x^{2} = 6.1867$