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

Χ²=8.18

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

p=0.82

For the homozygote non-poisonous toads:

q = 0.18

The frequency for heterozygote poisonous toad, we can use the Hardy-Weinberg equation: , 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:

= 0.0021                   = 7.41                       = 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

i think the answer would be a dominant trait.

i think that c)gold was found there

but i am not sure