4.4: Chi-square tests how well genetic data fit a hypothesis
- Page ID
- 142556
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Explanation
For a variety of reasons, the phenotypic ratios observed from real crosses rarely match the exact ratios expected based on a Punnett Square or other prediction techniques. There are many possible explanations for deviations from expected ratios. Sometimes these deviations are due to sampling effects, in other words, the random selection of a non-representative subset of individuals for observation. On the other hand, it may be because certain genotypes have a less than 100% survival rate. For example, Drosophila crosses sometimes give unexpected results because the more mutant alleles a zygote has the less likely it is to survive to become an adult. Genotypes that cause death for embryos or larvae are underrepresented when adult flies are counted.
The χ2 Test For Goodness-of-fit
A statistical procedure called the chi-square (χ2) test can be used to help a geneticist decide whether the deviation between observed and expected ratios is due to sampling effects, or whether the difference is so large that some other explanation must be sought by re-examining the assumptions used to calculate the expected ratio. The procedure for performing a chi-square test is covered in the labs.
Learning Objectives
- Study the use of chi-square test of goodness-of-fit when you have one nominal variable
- To see if the number of observations in each category fits a theoretical expectation, and the sample size is large
When to use it
Use the chi-square test of goodness-of-fit when you have one nominal variable with two or more values (such as red, pink and white flowers). You compare the observed counts of observations in each category with the expected counts, which you calculate using some kind of theoretical expectation (such as a \(1:1\) sex ratio or a \(1:2:1\) ratio in a genetic cross).
Null hypothesis
The statistical null hypothesis is that the number of observations in each category is equal to that predicted by a biological theory, and the alternative hypothesis is that the observed numbers are different from the expected. The null hypothesis is usually an extrinsic hypothesis, where you knew the expected proportions before doing the experiment. Examples include a \(1:1\) sex ratio or a \(1:2:1\) ratio in a genetic cross. Another example would be looking at an area of shore that had 59% of the area covered in sand, \(28\%\) mud and \(13\%\) rocks; if you were investigating where seagulls like to stand, your null hypothesis would be that \(59\%\) of the seagulls were standing on sand, \(28\%\) on mud and \(13\%\) on rocks.
In some situations, you have an intrinsic hypothesis. This is a null hypothesis where you calculate the expected proportions after you do the experiment, using some of the information from the data. The best-known example of an intrinsic hypothesis is the Hardy-Weinberg proportions of population genetics: if the frequency of one allele in a population is \(p\) and the other allele is \(q\), the null hypothesis is that expected frequencies of the three genotypes are \(p^2\), \(2pq\), and \(q^2\). This is an intrinsic hypothesis, because you estimate \(p\) and \(q\) from the data after you collect the data, you can't predict \(p\) and \(q\) before the experiment.
How the test works
Unlike the exact test of goodness-of-fit, the chi-square test does not directly calculate the probability of obtaining the observed results or something more extreme. Instead, like almost all statistical tests, the chi-square test has an intermediate step; it uses the data to calculate a test statistic that measures how far the observed data are from the null expectation. You then use a mathematical relationship, in this case the chi-square distribution, to estimate the probability of obtaining that value of the test statistic.
You calculate the test statistic by taking an observed number (\(O\)), subtracting the expected number (\(E\)), then squaring this difference. The larger the deviation from the null hypothesis, the larger the difference is between observed and expected. Squaring the differences makes them all positive. You then divide each difference by the expected number, and you add up these standardized differences. It is conventionally called a "chi-square" statistic, although this is somewhat confusing because it's just one of many test statistics that follows the theoretical chi-square distribution. The equation is:
\[\text{chi}^{2}=\sum \frac{(O-E)^2}{E}\]
As with most test statistics, the larger the difference between observed and expected, the larger the test statistic becomes. To give an example, let's say your null hypothesis is a \(3:1\) ratio of smooth wings to wrinkled wings in offspring from a bunch of Drosophila crosses. You observe \(770\) flies with smooth wings and \(230\) flies with wrinkled wings; the expected values are \(750\) smooth-winged and \(250\) wrinkled-winged flies. Entering these numbers into the equation, the chi-square value is \(2.13\). If you had observed \(760\) smooth-winged flies and \(240\) wrinkled-wing flies, which is closer to the null hypothesis, your chi-square value would have been smaller, at \(0.53\); if you'd observed \(800\) smooth-winged and \(200\) wrinkled-wing flies, which is further from the null hypothesis, your chi-square value would have been \(13.33\).
The distribution of the test statistic under the null hypothesis is approximately the same as the theoretical chi-square distribution. This means that once you know the chi-square value and the number of degrees of freedom, you can calculate the probability of getting that value of chi-square using the chi-square distribution. The number of degrees of freedom is the number of categories minus one, so for our example there is one degree of freedom. Using the CHIDIST function in a spreadsheet, you enter =CHIDIST(2.13, 1) and calculate that the probability of getting a chi-square value of \(2.13\) with one degree of freedom is \(P=0.144\).
The shape of the chi-square distribution depends on the number of degrees of freedom. For an extrinsic null hypothesis (the much more common situation, where you know the proportions predicted by the null hypothesis before collecting the data), the number of degrees of freedom is simply the number of values of the variable, minus one. Thus if you are testing a null hypothesis of a \(1:1\) sex ratio, there are two possible values (male and female), and therefore one degree of freedom. This is because once you know how many of the total are females (a number which is "free" to vary from \(0\) to the sample size), the number of males is determined. If there are three values of the variable (such as red, pink, and white), there are two degrees of freedom, and so on.
An intrinsic null hypothesis is one where you estimate one or more parameters from the data in order to get the numbers for your null hypothesis. As described above, one example is Hardy-Weinberg proportions. For an intrinsic null hypothesis, the number of degrees of freedom is calculated by taking the number of values of the variable, subtracting \(1\) for each parameter estimated from the data, then subtracting \(1\) more. Thus for Hardy-Weinberg proportions with two alleles and three genotypes, there are three values of the variable (the three genotypes); you subtract one for the parameter estimated from the data (the allele frequency, \(p\)); and then you subtract one more, yielding one degree of freedom. There are other statistical issues involved in testing fit to Hardy-Weinberg expectations, so if you need to do this, see Engels (2009) and the older references he cites.
Examples
Extrinsic Hypothesis examples
Example
European crossbills (Loxia curvirostra) have the tip of the upper bill either right or left of the lower bill, which helps them extract seeds from pine cones. Some have hypothesized that frequency-dependent selection would keep the number of right and left-billed birds at a \(1:1\) ratio. Groth (1992) observed \(1752\) right-billed and \(1895\) left-billed crossbills.
Calculate the expected frequency of right-billed birds by multiplying the total sample size (\(3647\)) by the expected proportion (\(0.5\)) to yield \(1823.5\). Do the same for left-billed birds. The number of degrees of freedom when an for an extrinsic hypothesis is the number of classes minus one. In this case, there are two classes (right and left), so there is one degree of freedom.
The result is chi-square=\(5.61\), \(1d.f.\), \(P=0.018\), indicating that you can reject the null hypothesis; there are significantly more left-billed crossbills than right-billed.
Example
Shivrain et al. (2006) crossed clearfield rice, which are resistant to the herbicide imazethapyr, with red rice, which are susceptible to imazethapyr. They then crossed the hybrid offspring and examined the \(F_2\) generation, where they found \(772\) resistant plants, \(1611\) moderately resistant plants, and \(737\) susceptible plants. If resistance is controlled by a single gene with two co-dominant alleles, you would expect a \(1:2:1\) ratio. Comparing the observed numbers with the \(1:2:1\) ratio, the chi-square value is \(4.12\). There are two degrees of freedom (the three categories, minus one), so the \(P\) value is \(0.127\); there is no significant difference from a \(1:2:1\) ratio.
Example
Mannan and Meslow (1984) studied bird foraging behavior in a forest in Oregon. In a managed forest, \(54\%\) of the canopy volume was Douglas fir, \(28\%\) was ponderosa pine, \(5\%\) was grand fir, and \(1\%\) was western larch. They made \(156\) observations of foraging by red-breasted nuthatches; \(70\) observations (\(45\%\) of the total) in Douglas fir, \(79\) (\(51\%\)) in ponderosa pine, \(3\) (\(2\%\)) in grand fir, and \(4\) (\(3\%\)) in western larch. The biological null hypothesis is that the birds forage randomly, without regard to what species of tree they're in; the statistical null hypothesis is that the proportions of foraging events are equal to the proportions of canopy volume. The difference in proportions is significant (chi-square=\(13.59\), \(3d.f.\), \(P=0.0035\)).
The expected numbers in this example are pretty small, so it would be better to analyze it with an exact test. I'm leaving it here because it's a good example of an extrinsic hypothesis that comes from measuring something (canopy volume, in this case), not a mathematical theory; I've had a hard time finding good examples of this.
References
- Engels, W.R. 2009. Exact tests for Hardy-Weinberg proportions. Genetics 183: 1431-1441.
- Groth, J.G. 1992. Further information on the genetics of bill crossing in crossbills. Auk 109:383–385.
- Mannan, R.W., and E.C. Meslow. 1984. Bird populations and vegetation characteristics in managed and old-growth forests, northeastern Oregon. Journal of Wildlife Management 48: 1219-1238.
- McDonald, J.H. 1989. Selection component analysis of the Mpi locus in the amphipod Platorchestia platensis. Heredity 62: 243-249.
- Shivrain, V.K., N.R. Burgos, K.A.K. Moldenhauer, R.W. McNew, and T.L. Baldwin. 2006. Characterization of spontaneous crosses between Clearfield rice (Oryza sativa) and red rice (Oryza sativa). Weed Technology 20: 576-584.