1. CHI-SQUARE ANALYSIS

2. In genetic studies the chi-square test is used to evaluate a genetic theory or hypothesis by comparing actual breeding results to theoretically expected results. The test is designed to convert the differences (or deviations) between the two into the probability of their occurring by chance, taking into account both the size of the sample and the number of variables (degrees of freedom).

3. There are two steps to perform the test: 1.Calculating the chi-square value from the test result figures using a standard formula. (convert the differences (or deviations) between the two into the probability of their occurring by chance) 2.Comparing the chi-square value with a scale of values given by a standard probability. (look at the calculated value on a chart)

4. The chi-square test is used to evaluate the difference between experimental (observed) data and expected (hypothetical) data; it is also called the “goodness of fit” test. This is sometimes set up using a “modified null” hypothesis, predicting an outcome and then evaluating how closely the data fit. X 2 =  (O-E) 2 E O is the observed value E is the expected value

5. Sample data and calculation: The F 2 generation of a cross of Drosophila melanogaster was examined with the following results: Red-eyed females= 78 Red-eyed males = 69 White-eyed females = 62 White-eyed males = 73

6. Given the parents used for this cross, a 1:1:1:1 ratio is expected. The total number of flies observed was 78 + 69 + 62 + 73, or 282. Thus, if this is divided by 4 (because of the 4 equal groups expected), 70.5 flies should have been observed with each phenotype.

7. X 2 =  (78-70.5) 2 + (69-70.5) 2 + (62-70.5) 2 + (73-70.5) 2 70.5 70.5 70.570.5 X 2 = .80 +.03 + 1.02 +.09 X 2 = 1.94

8. Obtaining a precise value is not the purpose of this test. What is important is to determine whether the deviations from theoretically expected results are of such significance as to invalidate the theory being examined. The significant level is customarily set at 5% so that, for instance, where there are two phenotypic classes (degrees of freedom, n = 1), the figure of 3.84 marks the division between deviations which are significant and those which are not significant.

9. n P=.99 P =.95P =.80 P =.50 P =.20 P =.05 P =.01 1.00015 7.00393.0642.4551.6423.8416.635 2.0201.103.4461.3863.2195.9919.210 3.115.3521.0052.3664.6427.81511.345 4.297.7111.6493.3575.9899.48813.277 5.5541.1452.3434.3517.28911.07015.086 6.8721.6353.0705.3488.55812.59216.812 71.2392.1673.8226.3469.80314.06718.475 81.6462.7334.5947.34411.03015.05720.090 92.0883.3255.3808.34312.24216.91921.666 102.5583.9406.1799.34213.44218.30723.209 155.2297.26110.30714.33919.31124.99630.578 208.26010.85114.57819.33725.03831.41037.566 2511.52414.61118.94024.33730.67537.65244.314 3014.95318.49323.36429.33636.25043.77350.892

10. **Using the sample data, for 4 phenotypic classes (n = 3), a chi-square value of 1.94 is obtained which lies between values indicating probabilities of 50% and 80%; let’s say 60%. This indicates that in a large number of similar tests deviations as great as, or greater than that observed, would occur in about 60% of those tests by chance alone. This is greater than the significant 5% level and we can conclude that this test does NOT invalidate theoretical expectations.

11. Problem: Using Mendel’s methods in crossing pea plants, the following results were collected in a cross between two plants heterozygous for seed shape and seed color (RrYy X RrYy). (R = round, r = wrinkled; Y = yellow, y = green) RESULTS: 547 yellow-round; 193 green-round; 195 yellow-wrinkled; 65 green-wrinkled. a) What are the expected phenotype ratios in the offspring? b) In 1000 offspring, how many are expected of each phenotype? c) Perform a Chi-Squared test with the results obtained. Do these results invalidate theoretical expectations?

12. a) What are the expected phenotype ratios in the offspring? RRYYRRYyRrYYRrYy RRYyRryyRrYyRryy RrYYRrYyrrYYrrYy RrYyRryyrrYyrryy RyrYryRY Ry rY ry EXPECTED: 9/16 yellow-round; 3/16 green-round; 3/16 yellow-wrinkled; 1/16 green-wrinkled

13. b) In 1000 offspring, how many are expected of each phenotype? In 1000 offspring: 9/16 yellow-round = 562.5 3/16 green-round = 187.5 3/16 yellow-wrinkled = 187.5 1/16 green-wrinkled = 62.5

14. c) Perform a Chi-Square test with the results obtained. Do these results invalidate theoretical expectations?

15. X 2 =  (547-562.5) 2 + (193-187.5) 2 + (195-187.5) 2 + (65-62.5) 2 562.5 187.5 187.5 62.5

16. c) Perform a Chi-Square test with the results obtained. Do these results invalidate theoretical expectations? X 2 =  (547-562.5) 2 + (193-187.5) 2 + (195-187.5) 2 + (65-62.5) 2 562.5 187.5 187.5 62.5 X 2 =  0.427 + 0.16 + 0.3 + 0.1

17. c) Perform a Chi-Square test with the results obtained. Do these results invalidate theoretical expectations? X 2 =  (547-562.5) 2 + (193-187.5) 2 + (195-187.5) 2 + (65-62.5) 2 562.5 187.5 187.5 62.5 X 2 =  0.427 + 0.16 + 0.3 + 0.1 X 2 = 0.987

18. n P=.99 P =.95P =.80 P =.50 P =.20 P =.05 P =.01 1.00015 7.00393.0642.4551.6423.8416.635 2.0201.103.4461.3863.2195.9919.210 3.115.3521.0052.3664.6427.81511.345 4.297.7111.6493.3575.9899.48813.277 5.5541.1452.3434.3517.28911.07015.086 6.8721.6353.0705.3488.55812.59216.812 71.2392.1673.8226.3469.80314.06718.475 81.6462.7334.5947.34411.03015.05720.090 92.0883.3255.3808.34312.24216.91921.666 102.5583.9406.1799.34213.44218.30723.209 155.2297.26110.30714.33919.31124.99630.578 208.26010.85114.57819.33725.03831.41037.566 2511.52414.61118.94024.33730.67537.65244.314 3014.95318.49323.36429.33636.25043.77350.892

19. So, can we accept these results or do they invalidate theoretical expectations? **Using the sample data, for 4 phenotypic classes (n = 3), a chi-squared value of 0.987 is obtained which lies between values indicating probabilities of 80% and 95%; let’s say 82%. This indicates that in a large number of similar tests deviations as great as, or greater than that observed, would occur in about 82% of those tests by chance alone. This is greater than the significant 5% level and we can conclude that this test does NOT invalidate theoretical expectations. (We CAN ACCEPT the results!)