A breeder may notice that animals carrying more copies of one marker allele also tend to have different phenotypes. How can that pattern be summarised, and how much of the observed variation does it describe? Linear least squares regression fits a line through paired observations by choosing coefficients that make the squared discrepancies between the observations and the line as small as possible. The fitted association can support prediction, but it does not show that the marker is a causal QTL.
For individual \(i\), let \(y_i\) be a phenotype and \(x_i\) the dosage of the counted allele at one marker. The simple linear regression model is
\[ y_i = \beta_0 + \beta_1x_i + \varepsilon_i, \]
where \(\beta_0\) is the intercept, \(\beta_1\) is the slope, and \(\varepsilon_i\) is the unobserved model error. The intercept is the modelled mean phenotype when \(x_i=0\). The slope is the modelled difference in mean phenotype for a one-copy increase in dosage. Because a biallelic diploid marker has dosages 0, 1, and 2, the fitted line compares those coding levels. Changing which allele is counted changes the coding and therefore the coefficient signs; it does not change the underlying observations.
After the coefficients have been estimated, the fitted value for individual \(i\) is
\[ \hat y_i = \hat\beta_0 + \hat\beta_1x_i. \]
The residual is \(e_i=y_i-\hat y_i\). A positive residual means that the observed phenotype lies above the fitted line, while a negative residual means that it lies below. Residuals contain variation not represented by this one-predictor model. They should not be interpreted as purely environmental effects because they can also contain omitted genetic effects, measurement error, nonlinear structure, and other unmodelled sources of variation.
Least squares chooses \(\hat\beta_0\) and \(\hat\beta_1\) to minimise the residual sum of squares
\[ \operatorname{RSS}=\sum_{i=1}^{n}e_i^2 =\sum_{i=1}^{n}(y_i-\hat\beta_0-\hat\beta_1x_i)^2, \]
where \(n\) is the number of individuals. When the marker dosage varies, the least-squares slope can be written using the sample covariance and variance introduced on the preceding pages:
\[ \hat\beta_1=\frac{s_{xy}}{s_x^2}, \qquad \hat\beta_0=\bar y-\hat\beta_1\bar x. \]
The common \(n-1\) denominators in \(s_{xy}\) and \(s_x^2\) cancel. A positive slope means that phenotype tends to increase with the counted-allele dosage in the fitted sample; a negative slope means it tends to decrease. If all individuals have the same dosage, then \(s_x^2=0\) and the slope cannot be estimated from those data.
The coefficient of determination compares the residual sum of squares with the total sum of squares,
\[ R^2=1-\frac{\operatorname{RSS}}{\operatorname{TSS}}, \qquad \operatorname{TSS}=\sum_{i=1}^{n}(y_i-\bar y)^2. \]
For an ordinary least-squares model with an intercept, \(R^2\) is the fraction of sample phenotype variation described by the fitted predictors. It is a training-data summary, not proof of a sound model or accurate prediction for new individuals. The NIST model-validation guidance emphasises that \(R^2\) alone is not enough; residual patterns and out-of-sample evaluation can reveal failures hidden by one summary number.
The following example uses the first six animals in the simulated project dataset. The phenotype is recorded on the simulation’s arbitrary scale, and SNP007 is coded as counted-allele dosage 0, 1, or 2. These six observations make the arithmetic inspectable, but they are not an independent random sample from a breeding population.
d <- readRDS("../../demo-data/main/demo_data.rds")
animals <- data.frame(
phenotype = d$pheno$y_cont_qtl_snp[1:6],
genotype_m1 = d$snp[1:6, "SNP007"]
)
animals phenotype genotype_m1
IND001 -1.113368 0
IND002 -3.456462 0
IND003 -2.124786 2
IND004 0.760348 1
IND005 -1.656242 0
IND006 -1.778796 0For this teaching slice, the sample covariance between dosage and phenotype is approximately 0.2391 and the sample dosage variance is 0.7000. Their ratio gives
\[ \hat\beta_1=\frac{0.2391}{0.7000}=0.3416. \]
The mean dosage is 0.5 and the mean phenotype is approximately \(-1.5616\), so the intercept is
\[ \hat\beta_0=-1.5616-(0.3416)(0.5)=-1.7323. \]
The fitted line is therefore \(\hat y=-1.7323+0.3416x\). At dosage 2 it gives \(\hat y=-1.0492\). The observed phenotype for that animal is \(-2.1248\), so its residual is \(-2.1248-(-1.0492)=-1.0756\).
R’s lm() fits the same least-squares model and provides the fitted values and residuals.
phenotype genotype_m1 fitted residual
IND001 -1.113368 0 -1.732326 0.61895832
IND002 -3.456462 0 -1.732326 -1.72413553
IND003 -2.124786 2 -1.049224 -1.07556164
IND004 0.760348 1 -1.390775 2.15112329
IND005 -1.656242 0 -1.732326 0.07608478
IND006 -1.778796 0 -1.732326 -0.04646921coef(fit)(Intercept) genotype_m1
-1.732326 0.341551 summary(fit)$r.squared[1] 0.04272633The fitted coefficients agree with the manual calculation. The residual sum of squares is approximately 9.1479 and the total sum of squares is 9.5562, giving \(R^2=0.0427\). Thus the fitted line describes about 4.3% of phenotype variation among these six records. That small value does not prove that SNP007 is biologically unimportant, just as a large value would not prove causality. Sampling variation, family structure, linkage disequilibrium, and other loci all affect a one-marker association.
Least squares can always produce coefficients when the design has sufficient rank, but statistical interpretation needs assumptions. The model assumes that the conditional mean of the phenotype is linear in the predictors and that the unobserved model error has conditional mean zero. For the usual standard errors and tests, errors from independent individuals are assumed to be independent and to have constant conditional variance. A normal-error assumption is additionally used for exact small-sample tests and intervals; it is not required merely to calculate the least-squares line. Fitted residuals estimate aspects of those unobserved errors for model checking, but they are not the errors themselves. With an intercept, their sum is zero by the least-squares calculation, and fitted residuals need not be statistically independent.
These assumptions are conditional on the variables included in the model. Relatives can have correlated model errors, management groups can shift the phenotype mean, and a curved relationship can leave a pattern in residuals. The six project animals come from the same simulated full-sib family, so this example is a calculation rather than evidence for independent-population inference. Residual plots, study design, and biological knowledge are needed before the line is used beyond this slice.
Prediction also requires care. A fitted value describes the modelled mean at a supplied predictor value. It does not predict every individual’s phenotype exactly, and extrapolation beyond the dosage range used for fitting has no support from these observations. Prediction for new breeding candidates must eventually be assessed on records that were not used to fit the model.
The intercept is the fitted mean phenotype at dosage 0. The slope is the fitted phenotype difference for a one-copy increase in the counted allele, so dosage 1 is fitted 0.3416 units above dosage 0. The coefficient summarises an association in six simulated records. LD, family structure, and omitted loci can generate that association, so the coefficient does not establish that SNP007 causes the trait difference.
The fitted value is \(-1.7323+0.3416(2)=-1.0491\), with the difference from the unrounded page result caused by rounding the coefficients. The residual is \(-2.1248-(-1.0491)=-1.0757\). Using the full coefficients gives fitted value \(-1.0492\) and residual \(-1.0756\).
The slope is \(s_{xy}/s_x^2=0.36/0.80=0.45\) phenotype units per additional copy of the counted allele. This calculation assumes that the marker dosage varies; if its variance were zero, the slope would be undefined.
\(R^2=1-18/24=0.25\), so the fitted predictors describe 25% of the sample variation around the sample mean. This value alone cannot establish causality, validate the residual assumptions, or show that prediction will be accurate for new individuals.
The constant conditional error-variance assumption is doubtful. Ordinary least-squares coefficients can still be calculated, but standard errors and intervals derived under constant variance may be inappropriate. The pattern also indicates that a single summary such as \(R^2\) is not a complete model check.
The stated dosage counts copies of one allele in a diploid individual, so its possible values are only 0, 1, and 2. Dosage 3 is outside both the biological coding and the fitted data range. Substituting it into the equation produces a number, but not a supported biological prediction.
One marker gives a readable line, but genomic datasets contain many correlated markers. The next page extends the same model to a design matrix and explains why regularisation is needed when ordinary least squares becomes unstable or has no unique coefficient solution.