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

Step-by-step explanation: In this population, the frequency of homozygote poisonous toads is:

`p^(2) =(100)/(150)`

`p^(2) = 0.667`

`p = √(0.667)`

p=0.82

For the homozygote non-poisonous toads:

`q^(2) = (5)/(150)`

`q^(2) =0.034`

`q = √(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:

`(0.034)/(16)` = 0.0021
`(59.3)/(8)` = 7.41
`(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