626/2018 Take lest: Assignment 2 (Mendelian Genetics) Summer 2018 Question 7 x Q
ID: 281526 • Letter: 6
Question
626/2018 Take lest: Assignment 2 (Mendelian Genetics) Summer 2018 Question 7 x Question Completion Status: 8. What phenotypic ratio do you expect in the F2 generation? (2 marks) Enter phenotypes in the following order: 1. owls with short ears and solid-coloured chest feathers 2. owls with long ears and solid-coloured chest feathers 3. owls with short ears and speckled chest feathers 4. owls with long ears and speckled chest feathers 4 pointsSave Answer QUESTION 9 Part ll: Dihybrid Cross- Question 9 This question is based on the information provided above in Questions 7 and 8 9. Looking at the F2 generation, answer the following probability questions (a) What is the probability of having an owl with short ears? (1 mark) (b) What is the probability of having an owl with long ears? (1 mark) (c) What is the probability of having an owl with long ears and solid coloured chest feathers? (1 mark) (d) What is the pobability of having an owl with long ears and (1 mark) speckled chest feathers? 2 points Save Answer QUESTION 10 Qu lowing situation Please nnt that nn an Y minatinn un" will NOT h niven the formula for the Chi- ClickSave and Submit to save ad submit. Click Save Ail Answers to sve all answers. Save All Answers CloseExplanation / Answer
Chi square test represent difference in two things thats why two hypothesis designed
Null hypothesis (H0) there is no difference beteween observed and expected frequencies
Alternative hypothesis (H1) there is a significant difference between observed and expected frequencies
In the dihybrid cross , phenotypic ratio is 9:3:3:1 so to calculate expected value we use total observed number
Observed. Expected
382. 650 *9/16 = 366
111. 650*3/16 = 122
124. 650*3/16= 122
33. 650*1/16= 40
So overall value of both expected and observed equal to 650
Chi square calculated by £ (O-E)^2/E
O. E. O-E. (O-E)^2. (O-E)^2/E
382. 366. 16. 256. 0.69
111. 122. -11. 121. 0.99
124. 122. 2. 4. 0.032
33. 40. -7. 49. 1.225
Total. 650. 2.937
Value of chisquare = 2.937
Degree of freedom = ( no. Of phenotype-1)
4-1 = 3
When you observe on chi square table, you will find a point where our value is less than the 0.05 . If the value is less than 0.05 means null hypoyhesis is accepted means there is no significant difference between observed and expected value in this dihybrid cross.
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