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With a null hypothesis of complete dominance and no If the original dihybrid cross was AAbb x aaBB, then the We listed everything that we knew from the problem. With segregation at each locus but then did not do so! (Test taking skills: On the exam several students told That is, we collapse the data and do the Chi-square test for goodness of This means that we must check the both the CableĪnd the Color locus for problems with segregation. (deviations from segregation and independent assortment) is theĬorrect answer. With independent assortment which you have told is linkage. In the case of this problem, the tests for possibility aīeen done and the observed distribution is not due to chance. Problems with segregation at one of both loci and additional problems with independent assortment of theĮach of these possibilities requires a different statistical Problems with independent assortment of the two loci.ĭ. Problems with segregation at one of both loci.Ĭ. When there is a deviation from the expected dihybrid ratio (in this caseī. From the wording, the deviation from independent assortment is due The loci are not independently assorting (significant contingency Chi-square Observed data do not fit a 9:3:3:1 ratio.
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Goodness of Fit and Independent Assortment. The experiment would have had to have been redone and the phenotype of the Workly did not examine the F 1 progeny closely!!! We presume that all of the F 1 progeny would have the dominant Cable, Color phenotype but we do not actually The F 1 progeny would have the genotype AaBb. Homozygous Cable, b&w bug was crossed to a completely homozygous rabbit ear,Ĭolor, bug. Problem: In the original cross, as clearly stated, ( NOTE: DO NOT REPEAT THE CHI-SQUARE TEST FOR GOODNESS OF FIT TO Aĩ:3:3:1 RATIO OR THE CONTINGENCY CHI-SQUARE!!!!) You must support your answer with a mathematical proof!
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Using your knowledge of genetics and statistics, how can the F 2 Workly has not entirely explained the results. Immediately did a Contingency Chi-Square test and obtained a Chi-square value of over 8000!!Īdvisor that he has "proved" that the results are entirely due to linkage. Workly hypothesized that the results were due to linkage. These results had a Chi-square for goodness-of-fit of 1287.7 (null hypothesis 9:3:3:1 3 df). Rabbit, Color and 228 rabbit, b&w progeny for a total of 8000 TV bugs. In the F 2 he obtained 3408 Cable, Color, Workly Drudge crosses a completely homozygous Cable, b&w bug to completely homozygous rabbit ear,Ĭolor, bug. (1 0 points) In TV Bugs, Cable ( A) is dominant to rabbit ears (a) and Color (B) is dominant to black and white I will go through the logic of the problem as an illustration of how to Most students did not even attempt the question. Only about 20 out of 280 students got this This is a problem that was on the final exam in genetics in both 19.
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