Background for Problems 2-6. In flies (Drosophila), the curly-wing phenotype is
ID: 273711 • Letter: B
Question
Background for Problems 2-6. In flies (Drosophila), the curly-wing phenotype is dominant to straight, and red eye color is dominant to scarlet. When "dihybrid" flies that were heterozygous for both genes (and therefore had curly wings and red eyes) were mated (to each other), the following phenotypes were observed among a total of 432 progeny 219 curly wings & red eyes 75 curly wings & scarlet eyes 103 straight wings & red eyes 35 straight wings & scarlet eyes 2) Assuming Mendelian behavior, what is the "expected" number of progeny in each of the four phenotypic 3) Calculate the Chi-square value 4) Which of the four categories made the largest contribution to your Chi-square value. 5) Use the table in your texthook(Chapter 3) to determine the two closest probabilty/P values Gie,the range into o) Based on you analysis, would you reject or fail to reject the hypothesis that the traits/genes behave in a classes? (Be sure your numbers add up to a total of 432.) which the actual P must fall). Mendelian fashion?Explanation / Answer
Answer:
Based on the given information:
Curly wing (C) is dominant to straight wing (c)
Red eye color (R) is dominant to scarlet eye color (r)
Dihybrid flies heterozygous for both genes (Curly wings and red eye) were mated to each other: CcRr x CcRr
Dihybrid cross can be represented as Punnett square:
Punnett Square:
Different possible Phenotypes are:
Frequency: Curly, Red = 9; Curly, Scarlet = 3; Straight, Red = 3; Straight, Scarlet = 1
Therefore, expected Phenotypic Ratio is: 9:3:3:1
2) Expected Number for different Phenotypes can be determined as below:
Chi Square Test:
Degree of Freedom = Number of categories - 1 = 4 - 1 = 3
3) P-value corresponding to Chi Square value of 11.160 with 3 degree of freedom is: P-Value = 0.010891. The result is significant at p < 0.05
4) The largest contribution to the Chi square value is made by Straight wing and red eye category.
5) P-Value = 0.010891. Critical Chi-Square Value for probability level 0.05 = 7.815
Chi Square value obtained (11.160) > Critical Chi Square value (7.815) --> Null Hypothesis is rejected.
6) Since P-value (0.010891) is less than the significance level (0.05), therefore, Null Hypothesis is rejected. Based on the experimental evidence, their is a statistically significant difference between expected phenotypic ratio (Mendelian fashion) and observed phenotypic ratio).
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