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Title: For autosomal recessive inheritance what is the chi-square value for these results? Post by: lenam on Mar 14, 2013 A cross of two heterozygous individuals produces 80 dominants and 20 recessives. For autosomal recessive inheritance what is the chi-square value for these results (to two decimal points)?
Also, for the chi-square above, do you accept or reject the chance hypothesis? Please explain how you arrived at the answer. I know the basic equation needed but I don't know where to begin with the information provided. Thank you! Title: For autosomal recessive inheritance what is the chi-square value for these results? Post by: Ilovebios on Mar 14, 2013 Aa X Aa should yield 75% dominant and 25% recessive.
Thus, out of 100 individuals, we should obtain 75 as dominant and 25 as recessive. ? = degree of freedom (n-1) = 1 for this case. ?² = ? [(o-e)²/e] = ? [d²/e], where d=deviation; o= observed; e=expected First of all, we calculate for the dominant character. d = o-e = 80-75=5 d²= 5²=25 d²/e = 25/75 = 0.33 Now for the recessive character, d = o-e = 20-25= -5 d²= (-5)² = 25 d²/e = 25/25 = 1.0 Now, ?² = ? [d²/e] = 0.33 + 1.0 ?² = 1.33 Now we have chi-square value (1.33) and degree of freedom (1). Now use the chart to determine the p-value. http://passel.unl.edu/Image/Namuth-CovertDeana956176274/chi-sqaure%20distribution%20table.PNG Look in the first horizontal line (for ?=1), you are getting chi square as 1.32 at p value = 0.25. Since the p value (0.25) is greater than 0.05, the deviation is not of statistical significance and impies that chance deviations are occuring. Thus, we accept the chance hypothesis..!! This was my first chi-square solving question!! Hope i made it clear to you! I need to learn it even more. |