BIOL 240 1st Edition Lecture 5Outline of Last Lecture I. Dihybrid CrossesII. Mendel’s Law of Independent AssortmentIII. Chromosome Theory of HeredityOutline of Current Lecture I. Chi-Square TestingCurrent LectureI. Chi-Square Test (Χ2)- Tests if there is a significant difference in the data and if the difference is too extreme to be explained by chance. - Example: Test Mendel’s hypothesis that dominant and recessive alleles segregate in a 1:1 ratio. To experiment, cross two heterozygous parents.For this case, we can predict that ¾ will be purple (D) and ¼ will be white (d).- Let’s say 500 flowers were produced: 367 were purple and 133 were white. This is how you would calculate Χ2……….Purple flowers: Observed=367Expected= ¾ x 500 = 375Observed – Expected = 367-375= -8(Observed-Expected)2= (-8)2=64(Observed-Expected)2/Expected= 64/375= 0.17White flowers: Observed= 133Expected= ¼ x 500 = 125Observed – Expected = 133-125 = 8(Observed-Expected)2=(8)2= 64(Observed-Expected)2/Expected= 64/125= 0.51These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Χ2 = 0.17 + 0.51 = 0.68- Once you have found Χ2, you must consult the table. To do this, find your degrees of freedom (# of categories – 1), in this case it would be 1. Find the range of probability in which your Χ2 value lies.Our calculated Χ2 value of 0.68 lies between 0.455 and 2.706, which corresponds to 0.1<p<0.5. Rejecting the hypothesisrequires p<0.05, so we cannot reject the
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