01:447:384 Lecture Notes - Lecture 22: Chi-Squared Test, Chi-Squared Distribution, Null Hypothesis

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coffeegoat901

23 Apr 2020

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Chi is from the greek letter (pronounced ki ). The chi square test is symbolized as 2. The chi square test is also referred to as the goodness of fit test" (e. g. , how well the observed and expected data match). This is a statistical test used to compare the observed values to predicted (or expected) values. Usually these values are the numbers of offspring resulting from a genetic cross mating. The easiest way to calculate a chi square value is to complete the table below. Phenotypes column: the difference in one feature being studies that varied among the offspring. Observed column: contains the numbers of offspring seen for each phenotype. The total is the sum of all the offspring observed. Expected column: contains the number of offspring you would predict based on (1) the genotypes of the parents in the cross, (2) the type of inheritance relationship between the alleles (for example: complete dominance, co- dominance, incomplete dominance).

Related Questions

Assume that a cross is made between a heterozygous tall pea plant and a homozygous short pea plant. Fifty offspring are produced in the following frequency: 30 = tall 20 = short

(a) What frequency of tall and short plants is expected?

(b) To test the goodness of fit between the observed and expected values, provide the needed statement of the null hypothesis.

(d) How many degrees of freedom are associated with this test of significance?

bronzetrout874

Introduction: A Chi-square test is used to compare observed data with expected data according to a hypothesis. For instance, if you were crossbreeding 2 heterozygous pea plants, you would expect to see a 3:1 phenotypic ratio in the offspring. In this case, if you were to breed 400 pea plants, you would expect to see 300 plants showing the dominant trait and 100 showing the recessive trait. But what happens if you observe only 260 plants with the dominant trait and 140 plants with the recessive trait? Does this mean something is wrong with Mendelian genetics or is this difference in expected results just due to chance (random sampling error)? These are the questions that can be answered using Chi-square statistics. The results of this statistical test is used to either reject or accept (fail to reject) the null hypothesis. The null hypothesis states there is no significant difference between the observed results and the expected results. This means that if the null hypothesis is accepted, the difference in observed and expected results was just a matter of chance and so the observed results basically "fit" with what was expected. Degrees of freedom (df) = number of independent outcomes (Y) being compared less 1 df = Y-1 At the 95% confidence interval we are 95% confident that there is a significant difference between the observed and expected results, therefore rejecting the null hypothesis. Probability Value - Is the decimal value determined from the X2 table and is the probability of accepting the null hypothesis. A 0.05 probability value equates to a 95% confidence interval.

The Chi-squared test formula is: Example: If we cross two pea plants that are heterozygous yellow pods, we would expect a 3:1 phenotypic ratio. So let's say we actually did the cross and got 280 plants with green pods and 120 plants with yellow pods. Question: Is this a 3:1 phenotypic ratio? This is the value of Chi-squared Test. We have a total of 400 plants and we expect a 300 green:100 yellow phenotypic ratio If the calculated Chi-squared value is less than the critical value listed in the Chi-squared table, then we accept the null hypothesis. This means that there is no significant difference between the observed and the expected values. Our degrees of freedom (df) = 2 outcomes - 1, or df = 1. Now we go the X2 table below and using the df = 1 and probability value of 0.05, our critical value is 3.84. Since our calculated X2 value is 5.33, and is larger than the critical value, we reject the null hypothesis and can say (at 95% confidence) that there is a significant difference between our observed and expected values.

The parent generation is yellowed podded and green podded pea plants. You cross a yellow podded pea plant with a green podded pea plant and you get 100% yellow podded plants in the F1 Generation (Phenotypic ratio 4 : 0, yellow to green). What will be the expected phenotypic ratio when you allow the F1 generation to reproduce?

Fill out the Punnett square.

If we actually did the cross and got 1150 yellow and 350 green. Would this be a consistent with what was expected?

Learning Outcomes Questions

1. Why would you run a Chi-squared test?

To determine if our data is consistent with expected results.
		a To determine if our data is consistent with expected results. b To determine if our data exactly matches the expected results.
		c To determine the expected results.
	d	To compare the phenotypic ratios to the genotypic ratios.

2. Determine the degrees of Freedom of the phenotypic ratio for this genetic cross.

a. 1

b. 2

c. 3

d. 4

e. 5

3. Using the data given, what is the result of your Chi-squared analysis? x2= ___.

	a.	2.22
	b	2.71
	c	4.36
	d	187.78
	e	448.27

4. Using the results of your Chi-squared analysis, do we fail to reject or reject the null hypothesis?

a.		Fail to reject the null
b.		Reject the null
c.		It cannot be determined from the data given

silversalmon27

The Mononoke lab studies tree sprites, small woodland creates with large white heads that rattle (I recommend that movie Princess Mononoke if you want to see some cool tree sprites â I watched it last weekend with my son and it inspired this question!).

A genetecist in the Monokoe lab performs a mutant screen in which she identifies a mutant tree sprite with a small, rattleless head. She crosses her mutant tree sprite to a wild-type tree sprite and finds that all the tree sprites in the F1 generation are normal. She crosses F1 sibling tree sprites together and anlyszes the prgoeny from the F2 generation. She finds 81 tree sprites with small, rattleless heads and 219 tree sprites with large heads that rattle.

A. Is this mutant most likely dominant or recessive?

B. Given your answer in (A), what is the expected numbers of small, rattleless (mutant) and large, rattling (normal) headed tree sprites int eh F2 generation? (Show your work for full credit)

C. Use the chi-square test to determine the probability the differeence between the expected and observed values is due to chance. (Show your work for full credit).

D. If two small-headed, rattleless (mutant) tree sprites from the F2 generation were mated, what would the be the expected phenotypic ratio of the progeny?

E. A second tree sprite geneteicst at a different university performed a similar genetic screen and also discovered a tree sprite with small, rattleless heads. The two geneticist meet at the âMythical Woodland Creaturesâ meeting and decide to exchange mutant tree sprites. They both find that, when they cross the two small, rattleless headed tree sprites from different laboratories, all the offspring are normal. However, when they cross two small, rattleless headed tree sprites from the same laboratory, all the progeny have two small, rattleless heads. What is the most likely explanation for this result? (It is NOT experimental error or sloppy record keeping!)

ochreraven956

01:447:384 Lecture Notes - Lecture 22: Chi-Squared Test, Chi-Squared Distribution, Null Hypothesis

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