Question

In a population of 150 toads, poisonous (P) is dominant over non-poisonous (p). According to the Hardy-Weinberg equation, the expected proportion of PP to Pp to pp is 16:8:1. The actual population had 100 homozygote poisonous toads, 45 heterozygote poisonous toads, and 5 homozygote non-poisonous toads. Compute the chi-square test statistic. a. 0.19 b. 0.48 c. 0.53 d. 0.87

Answer 1

Answer: Χ²=8.18

Explanation: In this population, the frequency of homozygote poisonous toads is:

$p^{2} =\frac{100}{150}$

$p^{2} = 0.667$

$p = \sqrt{0.667}$

p=0.82

For the homozygote non-poisonous toads:

$q^{2} = \frac{5}{150}$

$q^{2} =0.034$

$q = \sqrt{0.034}$

q = 0.18

The frequency for heterozygote poisonous toad, we can use the Hardy-Weinberg equation: $p^{2} + 2pq + q^{2} = 1$, in which, heterozygote frequency is given by 2pq

2pq = 2*0.82*0.18 = 0.3

Now, to compute the chi-square test, follow the instructions:

1) Find the observed values: in this case, they are the found frequency:

0.82       0.3         0.18

2) Find the expected values: As the question mentioned, the expected proportion is:

16            8             1

3) Subtract the observed value from the expected value:

0.82 - 16 = - 0.184          0.3 - 8 = -7.7             0.18 - 1 = - 0.818

4) Square each value from above:

(-0.184)² = 0.034            (-7.7)² = 59.3                 (-0.818)² = 0.77

5) Divide each value by expected value:

$\frac{0.034}{16}$ = 0.0021                   $\frac{59.3}{8}$ = 7.41                      $\frac{0.77}{1}$ = 0.77

6) Add all the values and we will have the chi-square test:

Χ² = 0.0021 + 7.41 + 0.77 = 8.18

The chi-square test is Χ² = 8.18