Chi Square Genetics Example Essay
Introduction: The chisquare test is a statistical test that can be used to determine whether observed frequencies are significantly different from expected frequencies. For example, after we calculated expected frequencies for different allozymes in the HARDYWEINBERG module we would use a chisquare test to compare the observed and expected frequencies and determine whether there is a statistically significant difference between the two. As in other statistical tests, we begin by stating a null hypothesis (H_{0}: there is no significant difference between observed and expected frequencies) and an alternative hypothesis (H_{1}: there is a significant difference). Based on the outcome of the chisquare test we will either reject or fail to reject the null hypothesis.
Importance: Chisquare tests enable us to compare observed and expected frequencies objectively, since it is not always possible to tell just by looking at them whether they are "different enough" to be considered statistically significant. Statistical significance in this case implies that the differences are not due to chance alone, but instead may be indicative of other processes at work.
Question: How is the chisquare test used to compare samples or populations? What does a comparison of observed and expected frequencies tell us about these samples?
Variables:
the chisquare test statistic  
o  observed count or frequency 
e  expected count or frequency 
n  total number of observations 
RT  row total 
CT  column total 
Methods: Shaklee et al. (1993) collected data to study genetic variation within a species of fish called the barramundi perch (Lates calcarifer). Many fish species are composed of breeding groups called stocks, which are populations that are genetically distinct from one another. One of the goals of Shaklee et al.'s study was to identify individual stocks of the barramundi perch on the basis of significant genetic differentiation. Of the 25 collections examined, those that were not significantly genetically distinct from one another were considered to be from the same stock; collections that were genetically distinct were considered to be from different stocks. Understanding species subdivision into stocks has important implications for conservation and fisheries management, since maintaining the genetic diversity of the species as a whole will require conservation of the different stocks.
We'll use some of their data here to illustrate the application of a simple chisquare test. Below are data showing allele frequencies at seven loci for eight collections of perch from different parts of the Australian coast (table adapted from Shaklee et al. 1993; all errors due to rounding are mine).
# 1  # 2  # 14  # 15  # 18  # 21  # 22  # 25  
EST2*  
*100+  249  78  97  115  101  242  128  116 
*98  26  4  0  1  2  0  2  30 
*95  126  41  60  60  52  226  125  70 
ESTD*  
*100+  390  120  155  176  171  465  335  210 
*114  15  4  0  0  0  9  2  6 
mIDHP*  
*100  387  123  152  167  152  474  333  216 
*78  0  0  5  10  4  1  0  0 
sIDHP*  
*100  354  113  111  137  143  432  310  177 
*121+  37  7  44  33  27  39  18  28 
*83  9  3  0  0  0  1  1  3 
LDHC*  
*100  373  115  156  175  154  400  245  208 
*90+  29  9  1  1  1  75  25  5 
PGDH*  
*100  382  122  130  145  153  378  240  199 
*88+  5  2  21  18  16  95  89  3 
PROT*  
*100+  399  120  149  168  147  453  326  207 
*97  8  4  8  9  9  22  5  9 
We can use the chisquare test to compare collections # 1 and # 25 at the EST2* locus. The expected values are the allele frequencies we would expect if there were no difference between the two collections at this locus. We can calculate the expected allele frequencies using the row and column totals from a table of the observed frequencies for these two collections.
For the first cell (collection #1, allele *100+) we begin by calculating the probability of an observation being in the first row, regardless of column. To do this, take the row total (365) and divide it by n (617) (note that n changes depending on which locus and which pair of populations is being compared). Based on these two collections, the probability of a barramundi perch having the *100+ allele at the EST2* locus is 0.5916 (365/617). Next, we calculate the probability of an observation being in the first column, regardless of row, by taking the column total (401) and dividing it by n (617). The probability of an observation coming from collection #1 as opposed to collection #25 is 0.6499 (401/617).
We have now determined the probability of a perch having a given allele at this locus, and the probability of being in a given collection. But what is the probability that an individual observation will have the *100+ allele at the EST2* locus and be from collection #1? The probability of two outcomes occurring together is called the joint probability, and is calculated by multiplying the two separate probabilities: 0.5916 x 0.6499 = 0.3845. It follows that in a sample of 617 fish we would expect 617 x 0.3845 = 237 individuals to be from collection #1 and have the *100+ allele, and we have now calculated our expected value for the first cell in the table. This calculation can be simplified with the following formula:
e = (RT/n)(CT/n)*n
Verify that the other expected frequencies have been calculated correctly.
Observed frequencies Expected frequencies
allele  # 1  # 25  allele  # 1  
*100+  249  116  365  *100+  237  128  365 
*98  26  30  56  *98  36  20  56 
*95  126  70  196  *95  127  69  196 
CT  401  216  n=617  CT  401  216  n=617 
Note also that the row and column totals remain the same. Now we can use the chisquare test to compare the observed and expected frequencies. The chisquare test statistic is calculated with the following formula:
For each cell, the expected frequency is subtracted from the observed frequency, the difference is squared, and the total is divided by the expected frequency. The values are then summed across all cells. This sum is the chisquare test statistic. For the example here,
= 0.608 + 2.778 + 0.008 + 1.125 + 5.000 + 0.014 = 9.533.
Interpretation: The critical value for the chisquare in this case () is 5.991; if the calculated chisquare value is equal to or greater than this critical value, we can conclude that the probability of the null hypothesis being correct is 0.05 or less a very small probability indeed! Our calculated value of 9.533 is greater than the critical value of 5.991. We therefore reject the null hypothesis, and conclude that there is a significant difference between the observed and expected frequencies of alleles at the EST2* locus for these two collections of barramundi perch. (Critical values for the chisquare are determined from a statistical table based on the significance level at which the test is being performed [0.05 in our case] and a number called degrees of freedom [2 in this example], but the details are beyond the scope of this module).
Conclusions: Our rejection of the null hypothesis allows us to conclude that the two collections of barramundi perch compared here are genetically distinct at the EST2* locus. In other words, the frequencies of the three alleles at this locus are significantly different between the two populations. Using somewhat more complicated applications of the chisquare test, the authors concluded that the 25 collections they analyzed came from seven genetically distinct stocks, or populations, from adjacent stretches of the northeastern Australian coast. One of the goals of conservation and/or management is the preservation of genetic diversity within a species. Management decisions based on the assumption that a species' genetic variation is distributed across populations could have disastrous consequences for the future of the species if the populations are indeed genetically distinct. Techniques for identifying amounts and patterns of genetic variation within a species are critical tools for biologists.
Additional Questions:
1) Are the allele frequencies at the other six loci also significantly different between collections #1 and #25? (**For loci with two alleles instead of three, the critical value of the chisquare is 3.841, but otherwise the procedure is the same).
2) Use the chisquare test to compare allele frequencies for collections #14 and #15. Can you determine whether or not these two collections are from the same stock?
Sources: Rohlf, F. J. and R. R. Sokal. 1995. Biometry, 3rd ed. W. H. Freeman and Company, New York, NY.
Rohlf, F. J. and R. R. Sokal. 1995. Statistical Tables, 3rd ed. W. H. Freeman and Company, New York, NY.
Shaklee, J. B., J. Salini, and R. N. Garrett. 1993. Electrophoretic characterization of multiple genetic stocks of barramundi perch in Queensland, Australia. Transactions of the American Fisheries Society 122:685701.
copyright 1999 by M. Beals, L. Gross, and S. Harrell
ChiSquare (X2) Test for Independence
How to Calculate a ChiSquare Test of Independence
 The number of quadrants with both species present
 The number of quadrants with Species A but not Species B
 The number of quadrants with Species B but not Species A
 The number of quadrants with neither species
 X2 = 0.0 + 0.1 = 0.1 + 0.2 = 0.4
Degrees of Freedom = (number of rows  1) X (number of columns  1)
In our example, DF = (21) X (21) = 1 X 1 = 1
*the row and column for the total in the contingency table are not included
The X2 distribution table is organized by the Level of Significance. The level of significance is the maximum tolerable probability of accepting a false null hypothesis. We use 0.05.
 If the calculated value is lower than the 0.05 level of significance, accept the null hypothesis and conclude that there is NO significant association between the variables.
 If the calculated value is higher than the 0.05 level of significance, reject the null hypothesis and conclude that there IS a significant association between the variables.
For example, with a DF=1, a value greater than 3.841 is required to be considered statistically significant (at p = 0.05). Since the X2 we calculated (0.4) is less than 3.841, there is NOT a significant association between Species A and Species B. The location of Species A has no significant effect on the location of Species B, any association between species is likely due to chance and sampling error.
Null Hypothesis: "There is not a significant association between variables, the variables are independent of each other; any association between variables is likely due to chance and sampling error." For example:
 Alternative Hypothesis: "There is a significant (positive or negative) association between variables; the association between variables is likely not due to chance or sampling error." For example:

0 thoughts on “Chi Square Genetics Example Essay”
>