15

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In genetics, we employ a statistical technique known as the chi-squared test for linkage to investigate whether two genes behave independently during inheritance or if they are connected on the same chromosome. The underlying assumption of the null hypothesis is that these genes segregate independently, indicating that the presence or absence of one gene doesn't influence the presence of the other.
In simpler terms, this test helps us decipher whether these genes are intertwined or act as if they are situated on entirely separate chromosomes. Our evaluation hinges on a comparison between the actual phenotypic ratios observed in the F2 generation and the expected ratios assuming independent assortment. A noteworthy disparity between the observed and expected ratios signifies potential linkage between the genes. To quantify this difference, we employ the chi-squared statistic, which gauges the extent of deviation between observed and expected outcomes, ultimately guiding our determination regarding the linkage of these genes.
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