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The\({\chi ^2}\)value means nothing on its own- it is used to find the probability that, assuming the hypothesis is true, the observed data set could have resulted from random fluctuations. A low probability suggests that the observed data are consistent with the hypothesis, and thus the hypothesis should be rejected, A standard cutoff point used by biologists is a probability of 0.05(5%). If the probability corresponding to the\({\chi ^2}\)value is 0.05or considered statistically significant, the hypothesis (that the genes are unlinked) should be rejected. If the probability is above 0.05, the results are not statistically significant: the observed data are consistent with the hypothesis.

To find the probability, locate your\({\chi ^2}\)value in the\({\chi ^2}\)Distribution table in Appendix F. The 鈥渄egree of freedom鈥� (pdf) of your data set is the number of categories (here,4 phenotypes), minus 1, so df=3.

(a). Determines which values on the df =3 line of the table your calculated\({\chi ^2}\)value lies between.

(b). The column headings for these values show the probability range for your\({\chi ^2}\)number. Based on whether there is non-significant (p\( \le \)0.05) or significant (p>0.05) difference between the observed and expected values, are the data consistent with the hypothesis that the two genes are unlinked and assorting independently, or is there enough evidence to reject this hypothesis?

Step by step solution

01

Chi-square

The chi-square value is calculated to compare with the degree of freedom distribution to find if the hypothesis is supported or not. The standard value of probability is 0.05.If the probability value is greater than 0.05, then the hypothesis is significant. If the probability value is less than or equal to 0.05, then the hypothesis is not significant.

02

Explanation of part “a”

The sum value is calculated as 2.14 from the previous subdivision.

The degree of freedom given in the question is 3. By comparing the degrees of freedom, the\({\chi ^2}\)value lies in between 1.42 and 2.37. The value 2.14 lies between the value of 1.42 and 2.37 by appendix F.

03

Step 3: Explanation of part “b”

The genes that possess unlinked alleles get assorted independently. The unlinked alleles are present in different chromosomes. The phenotypic ratio of the F₁dihybrid cross is 1:1:1:1.

The hypothesis in the given question is supported since the p-value lies between 0.5 and 0.70. The p value is greater than 0.05, so the hypothesis is supported, and the p value is significant by the given condition.

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