Chi-Square Genetics Practice

University:
Eastern Kentucky University
Course:
BIO 315 | Genetics
Academic year:

2023
Views:

416

Pages:

1
Author:

customer-8742184

Chi-Square Genetics Practice Name: _________________________________ Per: _____ AP Biology | Mr. Austin 1. In peas, green pods (G) are dominant over yellow pods (g). In the P generation, a plant homozygous for green pods is crossed with a plant with yellow pods. The resulting F1 plants were then crossed. The seeds in the observed F2 generation were: Green pods = 720 Yellow pods = 260 Null Hypothesis (H0): Write a null hypothesis for this cross. Perform a chi square test to determine if the data fits the predicted phenotypic ratio. Ho is that the observed results will fit the expected Mendelian phenotypic ratio For a monohyboid cross between two I heterozygotes GgxGg the expected phenotypic ratio is 3 green pods : I yellow pod The · ㅣ Conclusion: The chi square value is [ greater | less ] than the critical value for ___ degrees of freedom and a P-value of 0.05. The null hypothesis [ can | cannot ] be rejected. The data [ does | does not ] fit the predicted phenotypic ratio. 2. Color blindness is a sex-linked trait in wombats. A female who is a carrier of the color-blind allele mates with a male who is color blind. The phenotypes of their offspring are: Phenotype Observed (o) Expected (e) (o-e)2/e Normal female 133 (133-13033/130 0 069 130 129 130)// 3) 0 2 zF ( Color-blind female 124 130 = - - Normal male Color-blind male 127 136 . . (127 -13//30 0069 (136 130) /130 0 277 30 - O = - - . Total XE 792 320 520 Write a null hypothesis for this cross. Perform a chi square test by completing the table above to determine if the data fits the predicted phenotypic ratio. Null Hypothesis (H0): the He is that no significant difference between the observed and phenotypes do not change significantly from the : : ratio Mendelian 1 1 1 expected 1 - 0 . offspring : Conclusion: The chi square value is [ greater | less ] than the critical value for ___ 3 degrees of freedom and a P-value of 0.05. The null hypothesis [ can | cannot ] be rejected. The data [ does | does not ] fit the predicted phenotypic ratio. 3. In corn, purple kernels (P) are dominant over yellow (p), and inflated kernels (I) are dominant over shrunken (i). An ear of corn was produced by two dihybrid parents. The ear of corn has 400 kernels total with the following phenotypes: · · · and the expected t ㄱ Write a null hypothesis for this cross. Perform a chi square test to determine if the data fits the predicted phenotypic ratio. : 2 observed Expected 10-e)/2 225 X E Phenotype X 3 S7 purple smooth 1x400 purple shrunken: = = = yellow shrunken tx400 23 = = E gx400 7s yellow smooth: x400 75 · :3thereis nosignificant difference betweentheobservedkenelphe en Null Hypothesis (H0): Purple, smooth = 213 Purple, shrunken = 83 Yellow, smooth = 72 Yellow, shrunken = 32 = purple smooth puople 213 shiunken 3 yellow smooth a 225 s s ( て 13 - 225) / 225- 0 (83 75)2/75 = 0 833 . - と (F2 YG) 1F5 : - 0 . . .6φ BF 4-1 = = 12 (32 25) 25 : -96 total 3573 yoo 300 3 degrees of freedom and a P-value of 0.05. Conclusion: The chi square value is [ greater | less ] than the critical value for ___ The null hypothesis [ can | cannot ] be rejected. The data [ does | does not ] fit the predicted phenotypic ratio. : yellow shrunken 2 5 / . 1