8. (10 points) A particular dihybrid cross resulted in 38 plants with smooth lea

Question

8. (10 points) A particular dihybrid cross resulted in 38 plants with smooth leaves, 82 plants with slightly rough leaves, and 44 plants with very rough leaves. Use Chi-square analysis to determine if this data supports incomplete dominance. Table: Chi-Square Probabilities r 095 0.99 0.97 .950,90 0.10 005 0,025 0.0 0.005 0.95 0.90 0.100.050.025 0.01 0.005 -i 0.001 i 0.004 -0.016 -2.706 L 3, 841r 5, 0241 6 ssl. 7879 6055997.378 9.21010.597 6.25 7.815 9.348 345 12.838 4 0.207 0.297 0.4840.064.7799.488 143 13.277 14.860 0.412 0.554 0.8345 1.6109.236 11.070O 12.833 15.086 16.750 0.0010.004 0.016 2.7 3.841 5.024 6.635 7.879 2/ 0.020 0.01 0103 0.2 4 0.01? 0.0720 0.1151 0.21? 0.35211 0.5 0.6760 0.8721 1.237|| 1.635| 2.2? 10.645| 12.592| 14.449 16.812| 18.548 7| 2.833| 12.01? 14.06? 16.0131 18.475 20.278 | 1.344/ 1.646 2.18? 2.733| 3.4901 13.3621 15.50? 17.53512009? 21.955 3.325 4.168 14.684 16.91919.023 21.666 23.589 101 2.1Sol 2.5581 3.247| 3.940| 4.865| 15.98? 18.30? 20.483] 23.20? 25.188 0.98? 1.23? 1.69? 2.16 91.735 2.088 2.7

Explanation / Answer

We expect a proportion of 1:2:1 phenotypic ratios for incomplete dominance.

So

38+82+44=164

2+1+1=4

Getting the expected number of individuals

Chi-square test

We have three phenotypic clases, so the degrees of freedom are

Df=3-1=2

Using 2 degree of freedom and 0.05 like level of significance we got a chi-square probability of 5.991 which is so much bigger than the chi-square calculated (0.44).

So we accepted the hypothesis than the cross is under incomplete dominance.

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