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20 May 2019 Chi2 Test For Genetics Help sheet.

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Presentation on theme: "20 May 2019 Chi2 Test For Genetics Help sheet."— Presentation transcript:

1 20 May 2019 Chi2 Test For Genetics Help sheet

2 Chi2 Test For Genetics Objective
Use the chi-squared test to determine if there is a significant difference between observed and expected frequency of distribution

3 Information The Chi2 test is a way of determining if there is a difference between the expected number of offspring with a particular characteristic and the actual number of offspring (observed) We first need to suggest a null hypothesis, this is to say that ‘There is no difference between the observed values and the xpected values’ The Chi2 formula is used to obtain a value which is the compared with a probability value (p Value) from a table. If the calculated value is lower than the value from the table then we accept the null hypothesis, if the calculated value is higher than the value from the table then we reject the null hypothesis

4 Question 1 Two purple crested flowers were bred together, the resulting seeds were grown and the colours of flowers for the offspring recorded. There were 21 white coloured flowers and 39 purple flowers Which of the two alleles was the dominant one? Can you determine the genotypes of the parent plants? Is there a difference between the phenotypes obtained and the expected results?

5 Calculating the expected value
If we are expecting a 3:1 ratio and have 39 purple flowers and 21 white flowers we need to perform a calculation: Expected value = Expected ratio Sum of ratios X Total sample size Expected value = (3 + 1) X ( ) Expected value for the purple flowers = ¾ X 60 = 45

6 Degrees of freedom = Total options - 1
Formula Remember Form a null hypothesis- There is no difference between the observed and expected values If the calculated value is less than the critical value in the chi2 table accept the null hypothesis If the calculated value is more than the critical value in the chi2 table rejectthe null hypothesis Degrees of freedom = Total options - 1

7 Chi-Square Table

8 Question 2 Two fruit flies were bred together, one with red eyes and curved antennae, the other with white eyes and straight antennae. The offspring consisted of: 8 Red Curved antennae 5 Red Straight antennae 2 White curved antennae 1 White straight antennae Which of the two alleles was the dominant one? Are the two characteristics on the same chromosome or separate chromosomes? (if separate there would be a 9:3:3:1 ratio)

9 Chi-Square Table

10 Question 3 Two fruit flies were bred together, one with a grey body and long wings, the other with an ebony body and short wings. The offspring consisted of: 20 Grey short 30 Grey Long 63 Ebony short 80 Ebony long Which of the two alleles was the dominant one? Are the two characteristics on the same chromosome or separate chromosomes? (if separate there would be a 9:3:3:1 ratio)

11 Question 3 Two fruit flies were bred together, one with a grey body and long wings, the other with an ebony body and short wings. The offspring consisted of: 20 Grey short 30 Grey Long 63 Ebony short 80 Ebony long Hint to convert the ratio to the actual value: (Ratio value/ratio total) X Sample total Eg (9/16) X 193 = Which of the two alleles was the dominant one? Are the two characteristics on the same chromosome or separate chromosomes? (if separate there would be a 9:3:3:1 ratio)

12 Chi-Square Table

13 Learning outcome Use the information on the next slide to determine if there is a significant difference between observed and expected frequency of distribution for a pea plant population

14 Another Example: from Mendel
Expected 9:3:3:1 Ratio phenotype observed expected proportion expected number round yellow 315 round green 101 wrinkled yellow 108 wrinkled green 32 total 556

15 Chi-Square Table


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