Statistics: Chi-squared 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.

Problem:
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 1st generation [F1 generation] (Phenotypic ratio 4 yellow : 0 green).
What will be the expected phenotypic ratio when you allow the F1 generation to reproduce? Work a Punnett square.
We observed 1150 yellow and 350 green when actually crossing F1 generation.
Would this be a consistent with what was expected?


Do your work and answer the questions on a separate paper/ document and upload it here.
Why would you run a Chi-squared test?
To determine if our data exactly matches the expected results.
To determine if our data is consistent with expected results.
To determine the expected results.
To compare the phenotypic ratios to the genotypic ratios.
Determine the degrees of Freedom of the phenotypic ratio for this genetic cross.
1
2
3
4
5
Using the data given, what is the result of your Chi-squared analysis? x2= ___.
2.22
2.71
4.36
187.78
448.27
Using the results of your Chi-squared analysis, do we fail to reject or reject the null (i.e. no connection) hypothesis?
Fail to reject the null
Reject the null
It cannot be determined from the data given
If we flipped a coin 100 times. Which of the following is the minimum number necessary to achieve significance?
55
60
63
70
75

We perform a Chi-square analysis to find out if the difference between observed and expected is due to chance or not. In this example, 1) FD=1 / 2) X² = 2.22 / 3) Fail to reject the null / 4) 75.

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Since a complete introduction to Chi-square was provided, we will proceed with the problem analysis.

1st Cross: Yellow   x   green

Parentals)  YY    x      yy

F1) 100% Yy, yellow

2nd Cross:  Yellow   x   yellow (From F1)

Parentals)    Yy    x     Yy

F2) 3/4 = 75% of the progeny is expected to be yellow

     1/4 = 25% of the progeny is expected to be green

We know from the data, that there are 1500 individuals in F2. So we can get the expected number of individuals from this data.

100 % of the progeny ----------------------- 1500 individuals

75 % yellow ------------------------------------X = 1125 individuals

25% green -------------------------------------X = 375 individuals

We assume the population is in H-W equilibrium, but we observe differences in what we expect to see and what we actually see. We want to know why.

Now we will make a table resuming information

                               Yellow                       Green

Observed                 1150                            350

Expected                  1125                            375

(Obs-Exp)²/Exp        0.555                         1.666

We can see that the table value is greater than the X² value, so there is not enough evidence to reject the null hypothesis.

The genotypes might be in equilibrium, and there might be independent assortment.

This results suggest that the difference between the observed individuals and the expected individuals is by random chances.

So now let us answer the questions,

1) Determine the degrees of Freedom of the phenotypic ratio for this genetic cross ⇒ DF = 2 - 1 = 1

2) What is the result of your Chi-squared analysis? ⇒ X² = 2.22

3) Do we fail to reject or reject the null hypothesis? ⇒

We Fail to reject the null hypothesis. There is not enough evidence to reject it.

4) If we flipped a coin 100 times. Which of the following is the minimum number necessary to achieve significance? 75 is the minimum number necessary to achieve significance

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