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Chi-Square Analysis Goodness of Fit "Linkage Studies of the Tomato" (Trans. Royal Canad. Inst. (1931))

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We will develop the use of the distribution through an example from biology. Consider two different characteristics of tomatoes, leaf shape and plant size. The leaf shape may be potato-leafed or cut- leafed, and the plant may be tall or dwarf. If we cross a tall cut-leaf tomato with a dwarf potato-leaf tomato and examine the progeny we will discover a uniform F 1 generation. The traits tall and cut-leaf are each dominant, while dwarf and potato-leaf are recessive. We use the letter T for height, and C for leaf shape, so the alleles are T, t, C, and c.

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Tall cut-leaf tomato Dwarf potato-leaf tomato We will examine a Punnett square to illustrate this dihybrid cross. Notice the uniformity among the offspring, all are TtCc.

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Now we cross the F 1 among themselves to produce the F 2 : gametes TCTctCtc TCTTCCTTCcTtCCTtCc TcTTCcTTccTtCcTtcc tCTtCCTtCcttCCttCc tcTtCcTtccttCcttcc

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Now we identify the tall cut-leaf tomatoes: gametes TCTctCtc TCTTCCTTCcTtCCTtCc TcTTCcTTccTtCcTtcc tCTtCCTtCcttCCttCc tcTtCcTtccttCcttcc ttCc

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Now we identify the tall potato-leaf tomatoes: gametes TCTctCtc TCTTCCTTCcTtCCTtCc TcTTCcTTccTtCcTtcc tCTtCCTtCcttCCttCc tcTtCcTtccttCcttcc

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Next we identify the dwarf cut-leaf tomatoes: gametes TCTctCtc TCTTCCTTCcTtCCTtCc TcTTCcTTccTtCcTtcc tCTtCCTtCcttCCttCc tcTtCcTtccttCcttcc

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gametes TCTctCtc TCTTCCTTCcTtCCTtCc TcTTCcTTccTtCcTtcc tCTtCCTtCcttCCttCc tcTtCcTtccttCcttcc Finally, the last type of tomato is dwarf potato-leaf:

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So now we have four phenotypes (different physical forms) of tomatoes originating from the single phenotype of the F 1 generation. They are, along with their genotypes and expected frequencies: Tall cut-leaf TTCC, TTCc, TtCC, TtCc Tall potato- leaf TTcc, Ttcc Dwarf cut-leafttCC, ttCc Dwarf potato- leaf ttcc

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If our understanding of genetics is correct and we have constructed the crosses we believe we have, we expect the proportions of the four phenotypes to fit our calculations. With the distribution, we are able to test to see if groups of individuals are present in the same proportions as we expect. This is rather like conducting multiple Z-tests for proportions, all at once. In this example we carry out the dihybrid cross to produce an F 1 generation, and, as expected, the F 1 are all of the same phenotype, tall and cut-leaf. Further, the F 1 are crossed among themselves to produce the F 2 generation. We record the numbers of individuals in each category.

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The following table gives the observed numbers of each category. Phenotype Observed Expected frequency Tall cut-leaf926 Tall potato- leaf 288 Dwarf cut- leaf 293 Dwarf potato-leaf 104

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To make a test for “goodness of fit” we start as with all other tests of significance, with a null hypothesis. Step 1: H 0 : The F 2 generation is comprised of four phenotypes in the proportions predicted by our calculations (based on Mendelian genetics). H a : The F 2 generation is not comprised of four phenotypes in the proportions predicted by our calculations. Another way of saying this is that for the null hypothesis the population fits our expected pattern, and for the alternate hypothesis, it does not fit our pattern.

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Step 2: Assumptions: Our first assumption is that our data are counts. (We cannot use proportions or means.) With, we do not always have a sample of a population, and sometimes examine an entire population, as with this example. We must ensure that we have a representative sample, when we work from a sample. 1. All expected counts must be one or more. In order to check assumptions for this goodness of fit test we must calculate the expected counts for each category. Then we must meet two criteria: 2. No more than 20% of the counts may be less than 5.

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We calculate the expected counts by finding the total number of observations and multiplying that by each expected frequency. Phenotype Observed counts Expected frequency Expected counts Tall cut-leaf926 Tall potato-leaf288 Dwarf cut-leaf293 Dwarf potato- leaf 104

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As you can see, all expected counts are greater than 5, so all assumptions are met. Step 3: The formula for the test statistic is: where o = observed counts, and e = expected counts This calculation needs to be made in the graphing calculator. Enter the observed counts in L 1. Enter the expected frequencies in L 2, as exact numbers. (Enter numbers like 1/3, directly, as fractions, never round to just.3 or.33.)

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In L 3 multiply L 2 by 1611. This will give the expected counts. The sum of L 1 can be found using 1-Var Stats. Now in L 4, enter, this will give you the contribution for each category. Finally, is the sum of L 4. For this problem, the statistic is 1.4687. In, we always need to know and report the degrees of freedom. The degrees of freedom are the number of categories minus one. Here we have 3 degrees of freedom.

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Step 4:

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Step 5:Step 6: Fail to reject H 0, a test statistic this large may occur by chance alone almost 70% of the time. Step 7: We lack strong evidence that the pattern of tomato phenotypes is different from the expected. That is, the F 2 generation are present in the expected proportions. The area can also be found with cdf(1.4687,10^99,3).

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THE END

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