Question

6. crossed a female worm homozygous for the dpy-17 e164 allele with a wild- type male worm heterozygous for dpy-17 el64. He then scored all of the developing eggs/larva for the Dpy phenotype. He found 88 with the Dpy phenotype and 98 wild type. Do a Chi square test on phenotypic data to determine if your data supports the hypothesis that the mutation segregates as a single-gene mutation. Show your work in a table and include 2 pts a. Null hypothesis b. X2 value c. dof d. p value e. Conclusion

Answer 1

Answer:

• Observed number of Dpy phenotype = 88
• Observed number of wild type phenotype = 98
• Total number of worms = 88+98 = 186
• Expected number of Dpy phenotype = 186/2 = 93
• Expected number of wild type phenotype = 186/2 =93
• Chi square analysis can be performed as below:

	Observed (O)	Expected ('E)	(O-E)	(O-E)^2	(O-E)^2/E
Dpy Phenotype	88	93	-5	25	0.269
Wild Type Phenotype	98	93	5	25	0.269
Total	186	186		Chi square (sum)	0.538

Null Hypothesis: $H_0:$ Mutation segregates as a single gene mutation and there is no difference between the observed wild type and Dpy phenotypes compared to the expected values.

Degree of freedom = number of categories - 1 = 2 - 1 = 1

Significance level used alpha = 0.05

Probability value corresponding to a chi square value of 0.538 for 1 degree of freedom is 0.463263. The result is not significant at p < 0.05.

Since p value 0f 0.538 is > significance level (0.05), therefore null hypothesis is not rejected. Data supports the hypothesis that the mutation segregates as a single-gene mutation.