Answer: Χ²=8.18
Explanation: In this population, the frequency of homozygote poisonous toads is:
![p^{2} =\frac{100}{150}](https://tex.z-dn.net/?f=p%5E%7B2%7D%20%3D%5Cfrac%7B100%7D%7B150%7D)
![p^{2} = 0.667](https://tex.z-dn.net/?f=p%5E%7B2%7D%20%3D%200.667)
![p = \sqrt{0.667}](https://tex.z-dn.net/?f=p%20%3D%20%5Csqrt%7B0.667%7D)
p=0.82
For the homozygote non-poisonous toads:
![q^{2} = \frac{5}{150}](https://tex.z-dn.net/?f=q%5E%7B2%7D%20%3D%20%5Cfrac%7B5%7D%7B150%7D)
![q^{2} =0.034](https://tex.z-dn.net/?f=q%5E%7B2%7D%20%3D0.034)
![q = \sqrt{0.034}](https://tex.z-dn.net/?f=q%20%3D%20%5Csqrt%7B0.034%7D)
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