📖 ContentsChapter 11
Genetics · Chapter 11

Quantitative Genetics

Most of the traits people care about, such as height, blood pressure, milk yield, grain weight and the risk of type 2 diabetes, do not sort into neat Mendelian classes. They vary smoothly across a range because many genes and many environmental factors each nudge them a little. Quantitative genetics gives you the tools to measure that variation, divide it into genetic and environmental parts, predict how a population will respond to selection, and finally track down the individual genes hiding inside the bell curve.

By the end of this chapter you can

  • Distinguish continuous, meristic and threshold traits and explain how many additive alleles plus environmental noise produce a smooth distribution.
  • Use the multiple factor hypothesis to find the number of genes from the fraction of extreme F2 offspring and to predict phenotype class frequencies.
  • Calculate a mean, variance, standard deviation, covariance, correlation coefficient and regression slope from a small data set.
  • Partition phenotypic variance into additive, dominance, epistatic and environmental components.
  • Define and estimate broad sense and narrow sense heritability from inbred lines, twins and parent offspring regression, and state what heritability does not mean.
  • Apply the breeder's equation, R = h2S, and calculate realized heritability.
  • Explain correlated response, selection limits and why long term selection experiments keep responding.
  • Describe how QTL mapping in experimental crosses locates genes using markers and LOD scores.
  • Read a Manhattan plot, explain the genome wide significance threshold, and discuss polygenic scores and missing heritability.

Section 1Continuous traits and polygenic inheritance

Discrete and continuous variation

Mendel chose traits that fall into a few clear categories: round or wrinkled seeds, purple or white flowers. Traits like these show discontinuous (discrete) variation, and a single gene with large effect explains the categories. Most traits in nature behave differently. Measure the height of 1,000 adults, the weight of 1,000 bean seeds or the systolic blood pressure of 1,000 patients and you get a smooth, roughly bell shaped distribution with no natural breaks. These are quantitative traits, also called continuous or complex traits. Their study is quantitative genetics, and its language is the language of statistics rather than of ratios.

Quantitative traits come in three broad types, and it helps to recognize which one you are dealing with:

The multiple factor hypothesis

An early resolution of how Mendelian genes could produce continuous variation came from experiments on wheat kernel color by the Swedish geneticist Herman Nilsson-Ehle. He crossed a red kerneled line with a white kerneled line. The F1 were an intermediate pink. In some F2 families he found 1/16 as red as the red parent and 1/16 as white as the white parent, with three intermediate shades in between in the ratio 1:4:6:4:1. In other crosses only 1/64 of the F2 were as extreme as either parent.

He explained these results with the multiple factor hypothesis (the polygene hypothesis): a quantitative trait is controlled by several genes, each segregating in a Mendelian way, and each with a small, additive effect. In the two gene case, call the alleles that add red pigment A and B (the contributing or additive alleles) and the alleles that add nothing a and b (the noncontributing alleles). Every contributing allele adds the same increment of color, so the phenotype depends only on how many contributing alleles an individual carries, not on which gene they belong to. An AaBb F1 has two of four, so it is intermediate. In the F2, the number of contributing alleles follows the binomial distribution: 0, 1, 2, 3 or 4 contributing alleles occur in the ratio 1:4:6:4:1.

Three general rules follow for n genes, each with two equally effective alleles, in a cross of two heterozygotes (or in the F2 from two homozygous extreme parents):

Genes (n)Phenotypic classes (2n + 1)Fraction like one extreme parentF2 class ratio
131/41:2:1
251/161:4:6:4:1
371/641:6:15:20:15:6:1
491/2561:8:28:56:70:56:28:8:1
n2n + 1(1/4)nbinomial coefficients of (1 + 1)2n
One geneTwo genesThree genes 121 14641 1615201561 012 01234 0123456 Number of contributing (additive) alleles Numbers above bars are F2 class ratios (totals 4, 16 and 64)
Figure 11.1 As the number of additive genes rises, the number of F2 classes grows (2n + 1) and the extremes become rarer, so the histogram approaches a smooth bell curve. Add environmental variation and the classes blur together completely.
Study tip. Worked problem: Two true breeding plant lines are crossed. One averages 20 cm tall and the other 52 cm. The F1 are all about 36 cm. Among 1,024 F2 plants, 4 are about 20 cm and 4 are about 52 cm. How many genes are involved, how much does each contributing allele add, and what fraction of the F2 should be 44 cm? Step 1: Find the fraction at one extreme: 4/1,024 = 1/256. Step 2: Set (1/4)n = 1/256. Since 44 = 256, n = 4 genes. There are 2n = 8 allele positions and 2n + 1 = 9 phenotypic classes. Step 3: The difference between the extremes is 52 − 20 = 32 cm, spread over 8 contributing alleles, so each adds 32/8 = 4 cm. Step 4: A 44 cm plant has (44 − 20)/4 = 6 contributing alleles out of 8. The probability is C(8, 6) × (1/2)8 = 28/256 = 7/64. Answer: four genes, 4 cm per contributing allele, and 7/64 (about 10.9 percent) of the F2 are 44 cm. Check: the F1 has 4 of 8 contributing alleles, so 20 + 4 × 4 = 36 cm, which matches the data.

Why the bell curve appears

Even a few additive genes produce a distribution with a central peak and tapering tails. Two further factors smooth the steps into a continuous curve. First, real traits usually involve many more loci, often hundreds or thousands, so the steps become tiny. Second, the environment adds its own variation: nutrition, temperature, disease, maternal effects and chance events during development shift each individual a little up or down. A plant with exactly four contributing alleles might be 34 cm in poor soil and 38 cm in rich soil. When every genotypic class is spread out by environmental noise, neighboring classes overlap and the histogram looks like a single normal distribution. This is the central limit theorem at work: sums of many small independent effects tend toward a normal shape.

The overlap has an important practical consequence. In a continuous trait, you cannot read genotype from phenotype. Two plants of the same height may carry different numbers of contributing alleles, and two plants with identical genotypes may differ in height. That is why quantitative genetics works with averages and variances of populations rather than with individual genotypes.

Gene action in polygenic traits

The simple model assumes that alleles act additively: the heterozygote is exactly halfway between the two homozygotes, and the effects of different loci simply add. Real loci depart from this in two ways:

The distinction matters because, as we will see in Section 3, only the additive part of genetic variation is reliably passed from parent to offspring. Dominance and epistatic combinations are broken up each generation by segregation and recombination.

Threshold traits and liability

Many common diseases appear to run in families without following Mendelian ratios. The liability threshold model explains them. Each person has a continuous, normally distributed liability built from many genetic and environmental factors. Only those whose liability exceeds a threshold are affected. The model makes several predictions that fit clinical data well. The risk to relatives of an affected person is higher than the population risk and falls off sharply with distance of relationship, faster than for a single gene trait. The risk is higher when more relatives are affected, and higher when the affected relative has a severe form, because both indicate that the family carries many risk alleles. When the sexes have different thresholds, relatives of an affected person of the less often affected sex have higher risk. Pyloric stenosis is the classic example: it is about five times more common in boys, and the sons of affected women carry the highest risk because a woman who crossed the higher female threshold must carry an unusually heavy load of liability alleles.

Connection. Most of the conditions that fill a primary care clinic, such as hypertension, type 2 diabetes, coronary artery disease, asthma and depression, are polygenic threshold traits. A family history raises risk, but no single test result decides the outcome, and lifestyle factors such as diet, exercise and smoking shift liability just as genes do. This is why a strong family history is a reason for earlier screening and prevention rather than a diagnosis.
Key point

Quantitative traits vary continuously because many genes, each with a small and often additive effect, combine with environmental variation. With n additive genes the F2 shows 2n + 1 classes and (1/4)n at each extreme, and environmental noise blurs the classes into a normal curve. Threshold traits look discrete but rest on a continuous underlying liability.

Section 2Statistics for quantitative traits

Mean, variance and standard deviation

The mean (x̄) is the arithmetic average: add all n measurements and divide by n. It tells you where the distribution is centered, but nothing about its spread. Two populations can share a mean of 170 cm yet differ greatly in how variable they are.

The variance (s2) measures spread. For each value, find its deviation from the mean, square it so that positive and negative deviations do not cancel, add the squares, and divide by n − 1:

s2 = Σ(xi − x̄)2 / (n − 1)

Dividing by n − 1 rather than n (the degrees of freedom) corrects for the fact that the sample mean was itself calculated from the same data, which makes the raw squared deviations slightly too small on average. The variance is in squared units (cm2), which is awkward to interpret but has a great advantage: independent sources of variance add. That additivity is the foundation of heritability.

The standard deviation (s) is the square root of the variance and is in the original units. In a normal distribution, about 68 percent of values fall within 1 standard deviation of the mean, about 95 percent within 2 (more exactly 1.96) and about 99.7 percent within 3. If adult height in a population has a mean of 170 cm and a standard deviation of 7 cm, about 95 percent of adults are between 156 and 184 cm.

The standard error of the mean, s/√n, is a different quantity. It describes how precisely the sample mean estimates the true population mean, and it shrinks as sample size grows. Students often mix it up with the standard deviation, which describes variation among individuals and does not shrink with more data.

−3s−2s−1smean+1s+2s+3s 68% 68% within 1 s95% within 2 s99.7% Frequency
Figure 11.2 The normal distribution. The mean sets the center, and the standard deviation sets the width: about 68, 95 and 99.7 percent of individuals lie within 1, 2 and 3 standard deviations of the mean.
Study tip. Worked problem: Five corn ears measure 10, 12, 14, 16 and 18 cm. Find the mean, variance and standard deviation. Step 1: Mean = (10 + 12 + 14 + 16 + 18)/5 = 70/5 = 14 cm. Step 2: Deviations from the mean are −4, −2, 0, +2 and +4. Their squares are 16, 4, 0, 4 and 16, which sum to 40. Step 3: Variance = 40/(5 − 1) = 40/4 = 10 cm2. Step 4: Standard deviation = √10 = 3.16 cm. Answer: mean 14 cm, variance 10 cm2, standard deviation 3.16 cm. Quick check: the deviations must sum to zero (−4 − 2 + 0 + 2 + 4 = 0); if they do not, the mean is wrong.

Covariance and correlation

Quantitative genetics constantly asks whether two measurements vary together: the height of parents and of their children, the weight of a calf and its later milk yield, the trait values of twins. The covariance measures this joint variation. For each pair, multiply the deviation of x from its mean by the deviation of y from its mean, add the products, and divide by n − 1:

cov(x, y) = Σ(xi − x̄)(yi − ȳ) / (n − 1)

If large x values tend to go with large y values, most products are positive and the covariance is positive. If large x goes with small y, the covariance is negative. If there is no relationship, positive and negative products cancel and the covariance is near zero. The covariance between relatives is the raw material from which heritability is estimated, because relatives share genes in known proportions.

Covariance depends on the units of measurement, so it is hard to compare across traits. The correlation coefficient (r) standardizes it by dividing by both standard deviations:

r = cov(x, y) / (sx sy)

The correlation ranges from −1 (perfect negative linear relationship) through 0 (no linear relationship) to +1 (perfect positive). Two cautions apply. First, correlation does not imply causation: ice cream sales and drowning deaths are positively correlated because both rise in summer. Second, r measures only linear association. The square of r, r2, gives the proportion of the variation in y that is accounted for by its linear relationship with x.

Regression

Correlation tells you how tightly points cluster around a line. Regression tells you the line itself, so that you can predict y from x. The least squares regression line y = a + bx has slope

b = cov(x, y) / sx2

and intercept a = ȳ − b x̄. The slope says how much y changes, on average, for each one unit change in x. In quantitative genetics the regression of offspring on parents is especially useful: as Section 3 shows, the slope of offspring values regressed on the average of the two parents (the midparent value) estimates narrow sense heritability directly.

Francis Galton noticed the key pattern in the late 1800s while studying human height. Children of very tall parents were tall, but on average not as tall as their parents; children of very short parents were short, but not as short. He called this regression toward the mean. It happens because part of what made the parents extreme was environment and nonadditive gene combinations that are not transmitted, so the children drift back toward the population mean. The slope of the regression line measures how much of the parental deviation is passed on.

Study tip. Worked problem: Five midparent heights of a plant (x) and the mean heights of their offspring (y), in cm: (10, 12), (12, 12), (14, 15), (16, 15), (18, 16). Find the covariance, the correlation and the regression slope. Step 1: x̄ = 14 (from the previous problem, with sx2 = 10). ȳ = (12 + 12 + 15 + 15 + 16)/5 = 70/5 = 14. Step 2: Deviations of y are −2, −2, +1, +1, +2. Cross products with the x deviations (−4, −2, 0, +2, +4) are 8, 4, 0, 2 and 8. Sum = 22, so cov(x, y) = 22/4 = 5.5. Step 3: Squared y deviations are 4, 4, 1, 1, 4, sum 14, so sy2 = 14/4 = 3.5 and sy = 1.871. sx = 3.162. Step 4: r = 5.5/(3.162 × 1.871) = 5.5/5.916 = 0.93. Step 5: b = 5.5/10 = 0.55; intercept a = 14 − 0.55 × 14 = 6.3. The line is y = 6.3 + 0.55x. Answer: covariance 5.5 cm2, r = 0.93, slope 0.55. Because x is the midparent value, the slope estimates a narrow sense heritability of about 0.55.
StatisticFormulaUnitsWhat it tells you
MeanΣx / nSame as dataCenter of the distribution
VarianceΣ(x − x̄)2 / (n − 1)Squared unitsSpread; independent sources add
Standard errors / √nSame as dataPrecision of the sample mean
CovarianceΣ(x − x̄)(y − ȳ) / (n − 1)Product of unitsDirection and size of joint variation
Correlation (r)cov / (sxsy)None, −1 to +1Strength of linear association
Regression slope (b)cov / sx2y units per x unitPredicted change in y per unit x
Key point

The mean locates a distribution and the variance measures its spread; variances from independent sources add, which is why quantitative genetics partitions variance. Covariance and correlation measure how two variables move together, and the regression slope (covariance divided by the variance of x) predicts one from the other. Offspring on midparent regression ties these statistics directly to heritability.

Section 3Heritability: broad and narrow sense

Partitioning phenotypic variance

The central idea of quantitative genetics is that the total variation we see in a trait, the phenotypic variance (VP), can be divided into parts that have different causes. In the simplest model, an individual's phenotype is the sum of a genetic value and an environmental deviation, P = G + E. If genes and environment are independent, their variances add:

VP = VG + VE

Real populations add two complications. If particular genotypes tend to find themselves in particular environments (for example, children with high musical aptitude are more likely to receive music lessons), there is a genotype environment covariance that adds 2cov(G, E). If genotypes respond differently to the same environmental change (one wheat variety yields best in drought, another in wet years), there is a genotype by environment interaction, VG×E. A fuller version is therefore VP = VG + VE + VG×E + 2cov(G, E). Most problems assume these two terms are zero.

The genetic variance itself has three components that correspond to the kinds of gene action described in Section 1:

Putting the pieces together: VP = VA + VD + VI + VE. The environmental variance can also be split into a common or shared part (VEc, effects shared by members of a family, such as household diet) and a special or unique part (effects specific to each individual, including random developmental noise and measurement error).

Phenotypic variance, VP VA additiveVDVIVE environment VG genetic: broad sense H2 = VG / VP narrow sense h2 = VA / VP VP = VA + VD + VI + VE Only VA makes offspring resemble parents and drives response to selection
Figure 11.3 Partitioning phenotypic variance. Broad sense heritability counts every genetic component; narrow sense heritability counts only the additive component, which is what parents transmit to offspring.

Two kinds of heritability

Heritability is the proportion of phenotypic variance in a population that is due to genetic differences among individuals. There are two versions, and confusing them is a classic exam error.

Broad sense heritability (H2) includes all genetic variance:

H2 = VG / VP = (VA + VD + VI) / VP

It tells you how much of the variation reflects genotype in general, which matters, for example, when clones or inbred lines can be propagated, because a clone transmits its entire genotype.

Narrow sense heritability (h2) includes only the additive part:

h2 = VA / VP

It is the quantity that predicts how closely offspring resemble parents and how a sexually reproducing population will respond to selection. Plant and animal breeders care most about h2. Both heritabilities range from 0 (no variation is genetic) to 1 (all variation is genetic), and h2 can never exceed H2.

Trait (typical population)Approximate h2Comment
Adult height, humans (well nourished populations)0.8Among the most heritable human traits
Body mass index, humans0.4 to 0.7Twin estimates higher than family estimates
Milk yield, dairy cattle0.3Moderate; steady genetic progress in breeding programs
Back fat thickness, pigs0.5Responds quickly to selection
Egg production, chickens0.1 to 0.3Lower, as for many fitness related traits
Litter size, pigs and miceAbout 0.1Reproductive traits tend to have low h2
Abdominal bristle number, fruit flies0.5Classic laboratory selection trait

Estimating heritability

Heritability cannot be measured directly. It is estimated by comparing groups whose genetic or environmental relationships are known. Four approaches are commonly taught.

1. Genetically uniform populations. Members of an inbred line, or the F1 from a cross of two inbred lines, are genetically identical, so all their variance is environmental: VP = VE. The F2 from the same cross contains both genetic and environmental variance. Grown in the same environment, VG = VP(F2) − VE. This yields H2.

2. Parent offspring regression. Plot offspring phenotypes against parental phenotypes. When offspring are regressed on the midparent value, the slope b estimates h2 directly. When they are regressed on one parent only, the slope estimates h2/2, because a single parent contributes only half of the offspring's alleles; double the slope to get h2. The method works because the covariance between a parent and its offspring equals ½VA, while dominance and epistatic deviations are not shared.

3. Twin studies. Monozygotic (identical) twins share all their genes; dizygotic (fraternal) twins share on average half, like ordinary siblings. If both kinds of twins share their environments to a similar degree, the extra similarity of identical twins reflects genes. Falconer's formula gives a heritability estimate from the twin correlations:

h2 ≈ 2(rMZ − rDZ)

Because the difference in genetic sharing is one half, doubling the difference in correlation recovers the additive genetic proportion. Twins reared apart provide another approach: the correlation of identical twins raised in separate homes estimates H2 directly, provided the homes are not systematically similar. Twin studies have limits: identical twins may be treated more alike, and adoptive homes are not random environments, biases that tend to inflate estimates.

4. Response to selection. The amount a population changes after selection gives the realized heritability, discussed in Section 4.

Study tip. Worked problem: Two inbred tomato lines and their F1 and F2 are grown together. The phenotypic variances for fruit weight (in g2) are: line 1, 4.0; line 2, 6.0; F1, 5.0; F2, 20.0. A separate analysis shows VA = 9.0 in the F2. Find VE, VG, H2, VD + VI and h2. Step 1: The inbred lines and F1 are genetically uniform, so their variance is environmental. Average them: VE = (4.0 + 6.0 + 5.0)/3 = 5.0. Step 2: VG = VP(F2) − VE = 20.0 − 5.0 = 15.0. Step 3: H2 = 15.0/20.0 = 0.75. Step 4: Nonadditive genetic variance = VG − VA = 15.0 − 9.0 = 6.0. Step 5: h2 = VA/VP = 9.0/20.0 = 0.45. Answer: VE = 5.0, VG = 15.0, H2 = 0.75, VD + VI = 6.0 and h2 = 0.45. Check that h2 ≤ H2: 0.45 ≤ 0.75, as it must be.
Study tip. Worked problem: For a personality score, identical twins correlate at rMZ = 0.50 and fraternal twins at rDZ = 0.30. Estimate h2, and estimate the shared (common) environmental proportion c2 = 2rDZ − rMZ. Step 1: h2 ≈ 2(0.50 − 0.30) = 2(0.20) = 0.40. Step 2: c2 = 2(0.30) − 0.50 = 0.60 − 0.50 = 0.10. Step 3: The remainder, 1 − 0.40 − 0.10 = 0.50, is the unique environment plus measurement error. Answer: about 40 percent additive genetic, 10 percent shared environment and 50 percent unique environment. Note that identical twins correlate at only 0.50: even with identical genes, half of the variation is not shared.

What heritability does not mean

Heritability is one of the most misunderstood numbers in biology. Keep these points straight:

Connection. In medicine, heritability is often confused with risk. A disease can be highly heritable and still rare, or common and only modestly heritable. Knowing that blood pressure has a heritability of roughly 0.3 to 0.5 does not tell a patient how much of their own hypertension is genetic; it tells researchers that searching for blood pressure genes is worthwhile, and it reminds clinicians that diet, salt intake, exercise and weight remain powerful levers for every patient.
Key point

Phenotypic variance divides into additive, dominance, epistatic and environmental parts. Broad sense heritability (H2 = VG/VP) includes all genetic variance; narrow sense heritability (h2 = VA/VP) includes only the transmissible additive part. Heritability is estimated from inbred lines, parent offspring regression, twins and selection, and it describes variation in one population in one environment, never the genetic share of an individual's trait.

Section 4Response to selection

The breeder's equation

For thousands of years farmers improved crops and livestock by saving seed from the best plants and breeding the best animals. Quantitative genetics turns that practice into a prediction. Suppose a breeder measures a trait in a population, picks the individuals with the highest values to be parents, and measures their offspring. Three quantities are involved:

The breeder's equation states:

R = h2 S

The logic is the same as for regression toward the mean. The selected parents are superior partly because of their additive genes and partly because of good luck in environment and favorable dominance or epistatic combinations. Only the additive part is transmitted, so offspring recover only the fraction h2 of their parents' advantage. If h2 = 1, the offspring would match their selected parents exactly. If h2 = 0, the offspring would revert completely to the original population mean, no matter how carefully the parents were chosen.

Parental generation selected parents population mean S Offspring of selected parents R offspring mean R = h2 S
Figure 11.4 The breeder's equation. Truncation selection of the top of the parental distribution gives a selection differential S. Their offspring shift by a smaller response R; the ratio R/S is the realized heritability.
Study tip. Worked problem: A flock of hens averages 240 eggs per year. The hens chosen as breeders average 260 eggs, and the roosters used are sons of hens averaging 260 as well, so S = 20 for both sexes. If h2 = 0.25, what is the expected mean of the next generation? How many generations of equally strong selection would raise the mean to 260 eggs, assuming h2 stays constant? Step 1: S = 260 − 240 = 20 eggs. Step 2: R = h2S = 0.25 × 20 = 5 eggs. Step 3: New mean = 240 + 5 = 245 eggs. Step 4: Total gain needed = 260 − 240 = 20 eggs; at 5 eggs per generation, 20/5 = 4 generations. Answer: 245 eggs after one generation and about 4 generations to reach 260. In practice h2 often declines as favorable alleles approach fixation, so progress slows.

Realized heritability

Rearranging the breeder's equation gives a direct way to estimate heritability from a selection experiment:

h2 = R / S

This is the realized heritability. For example, if mice average 40 g, the selected parents average 50 g (S = 10 g) and their offspring average 43 g (R = 3 g), then h2 = 3/10 = 0.30. Realized heritability is valuable because it is measured from the very process it is meant to predict, rather than from correlations among relatives.

Selection intensity

The size of S depends on two things: how strictly the breeder selects and how variable the trait is. Breeders separate these by writing S = i σP, where σP is the phenotypic standard deviation and i is the selection intensity, the selection differential in standard deviation units. For a normally distributed trait, keeping the top 50 percent gives i ≈ 0.80, the top 20 percent gives i ≈ 1.40, the top 10 percent gives i ≈ 1.76, and the top 1 percent gives i ≈ 2.67. The breeder's equation then becomes R = i h2 σP. Dividing by the generation interval gives gain per year, which is why genomic selection of young bulls has sped dairy improvement.

FactorEffect on response per yearHow breeders use it
Heritability (h2)Higher h2, larger responseReduce environmental noise with uniform management and repeated measurement
Selection intensity (i)Stricter selection, larger responseSelect fewer, more extreme parents (limited by inbreeding risk)
Phenotypic standard deviation (σP)More variation, more room to selectIntroduce new germplasm to restore variation
Generation intervalShorter interval, faster yearly gainSelect young animals using genomic predictions

Correlated response

Selecting on one trait often changes others. A correlated response occurs when two traits share a genetic correlation, usually because some genes affect both (pleiotropy) or because genes for the two traits are tightly linked. Broiler chickens selected intensely for rapid growth also developed leg problems and heart failure; dairy cows selected for milk yield showed declining fertility; and in many species, selection for large body size lengthens development time. Correlated responses can also be exploited: if a trait is hard or expensive to measure, a breeder can select on an easily measured trait that is genetically correlated with it.

Long term selection and selection limits

Two famous experiments show what sustained selection can do. In the Illinois long term corn selection experiment, begun in the late 1800s, separate lines were selected for high and low kernel oil content. After more than 100 generations, the high oil line had risen from about 5 percent oil to more than 20 percent, and the low oil line had fallen to near 1 percent, values far outside the range of the starting population.

Why does response continue so long? First, a polygenic trait has many loci, and a starting population is very unlikely to contain an individual with the favorable allele at every one; selection gradually assembles combinations that never existed before. Second, new mutations continually add a small amount of additive variance each generation. Eventually, however, response may slow or stop, reaching a selection plateau. Causes include fixation of favorable alleles (VA exhausted), opposing natural selection (very high growth lines may become infertile), and negative genetic correlations with fitness.

Study tip. Worked problem: Fruit fly abdominal bristle number has a mean of 36 and a phenotypic standard deviation of 4. A researcher keeps the top 10 percent (i = 1.76) as parents, and h2 = 0.5. Predict the response and the next generation mean. Step 1: S = i σP = 1.76 × 4 = 7.04 bristles. Step 2: R = h2 S = 0.5 × 7.04 = 3.52 bristles. Step 3: New mean = 36 + 3.52 = 39.52, about 39.5 bristles. Answer: a response of about 3.5 bristles, to a mean near 39.5. Quick check: R should always be smaller than S when h2 < 1, and here 3.52 < 7.04.
Connection. The same equation that guides cattle breeding describes evolution in the wild. Peter and Rosemary Grant measured beak depth in Galápagos medium ground finches before and after a severe drought, when only birds with deeper beaks could crack the remaining large, hard seeds. With a high heritability of beak depth, the next generation's beaks were measurably deeper, close to the change predicted by R = h2S. Natural selection, in other words, is the breeder's equation with nature choosing the parents.
Key point

The breeder's equation R = h2S predicts how far the next generation moves when parents are selected; rearranged, R/S gives realized heritability. Response increases with heritability, selection intensity and phenotypic variation and decreases with longer generation intervals. Genetic correlations cause correlated responses, and long term selection keeps working until additive variance is used up or opposed by natural selection.

Section 5QTL mapping and genome wide association

From variance components to genes

Heritability tells us that genes matter for a trait, but not which genes, where they are, or how large their effects are. A quantitative trait locus (QTL) is a region of the genome containing one or more genes that influence a quantitative trait. Two families of methods find them. QTL mapping uses controlled crosses between lines that differ in the trait; genome wide association studies (GWAS) use natural populations, including humans, where controlled crosses are impossible. Both rely on the same principle: if a genetic marker sits near a gene that affects the trait, individuals carrying different marker alleles will differ, on average, in the trait.

QTL mapping in experimental crosses

The classic design starts with two inbred lines that differ strongly in the trait, for example a high yielding and a low yielding tomato. The lines also differ at thousands of molecular markers, such as single nucleotide polymorphisms (SNPs) or microsatellites, scattered across the genome. They are crossed to produce an F1, and the F1 is either intercrossed to make an F2, backcrossed to one parent, or selfed for several generations to make recombinant inbred lines (RILs), each a homozygous mosaic of the two parental genomes. Every individual in the mapping population is then genotyped at all markers and measured for the trait.

At each marker, the population is sorted by marker genotype and the trait means are compared. At a marker unlinked to any QTL, the genotype classes have the same mean, because independent assortment randomizes the trait genes relative to the marker. At a marker tightly linked to a QTL, marker genotype predicts QTL genotype, so the classes differ. The size and pattern of the differences estimate the QTL's effects:

Because recombination between the marker and the QTL weakens the association, the effect is strongest at the marker closest to the QTL. Interval mapping refines this by testing positions between markers, using the known map distances to calculate the probability of each QTL genotype at each position. The evidence at each position is expressed as a LOD score, the logarithm (base 10) of the odds that the data arose with a QTL at that position versus with no QTL. A LOD of 3 means the data are 103 = 1,000 times more likely under the QTL hypothesis. Plotting LOD along each chromosome produces peaks at likely QTL positions; thresholds of about 3 (often set more exactly by permutation testing) declare significance, and the region within about one LOD unit of the peak gives a confidence interval.

Study tip. Worked problem: In an F2 from two inbred lines of mice, body weight is sorted by genotype at marker M. Mean weights are: M1M1, 52 g; M1M2, 47 g; M2M2, 42 g. Estimate the additive and dominance effects. Then do the same if the heterozygote mean had been 51 g. Step 1: Midpoint of the homozygotes = (52 + 42)/2 = 47 g. Step 2: Additive effect a = (52 − 42)/2 = 5 g per allele substitution (each M1 allele adds about 5 g relative to the midpoint). Step 3: Dominance effect d = heterozygote mean − midpoint = 47 − 47 = 0. The QTL near M acts additively. Step 4: If the heterozygote were 51 g, d = 51 − 47 = 4 g, and d/a = 4/5 = 0.8, so the M1 linked allele would be nearly dominant. Answer: a = 5 g and d = 0 for the original data; with a 51 g heterozygote, a = 5 g, d = 4 g and the high weight allele is partially dominant. A marker with no linked QTL would show three equal means.

Limits of QTL mapping

QTL mapping in crosses is powerful but coarse. An F2 has had only one round of meiosis to break up parental chromosomes, so a QTL peak often spans 10 to 30 centimorgans and hundreds of genes. Only variants that differ between the two parental lines can be detected, and small effect QTL are easily missed unless the mapping population is large. Intervals are narrowed with further crosses and candidate genes tested by gene editing. Success stories include the tomato gene fw2.2, which accounts for a large share of the difference in fruit size between wild and cultivated tomatoes, and the maize gene tb1, whose altered regulation helped turn bushy teosinte into single stalked corn.

Genome wide association studies

In humans and in many natural populations, experimental crosses are not possible. Instead, a GWAS genotypes hundreds of thousands to millions of SNPs in thousands to hundreds of thousands of unrelated people and tests each SNP for association with the trait. For a continuous trait like height, the test asks whether the trait mean differs among SNP genotypes, usually by regressing the trait on the number of copies of one allele (0, 1 or 2). For a disease, it compares allele frequencies in cases and controls and expresses the effect as an odds ratio.

GWAS depends on linkage disequilibrium (LD), the nonrandom association of alleles at nearby loci. In a population, recombination has broken up most chromosome segments over many generations, but alleles very close together are still inherited as blocks called haplotypes. A genotyped SNP therefore acts as a tag for untyped variants in the same LD block. A significant SNP is usually not the causal variant itself, only a neighbor in LD with it. Because LD blocks in human populations span only tens of kilobases, GWAS localizes signals far more precisely than F2 QTL mapping, often to a handful of genes.

Multiple testing and the significance threshold

With a million tests, the usual P < 0.05 would produce about 50,000 false positives by chance alone. GWAS therefore uses a Bonferroni style correction, dividing 0.05 by the approximate number of independent common variants in the human genome, about one million: 0.05/1,000,000 = 5 × 10−8. This is the conventional genome wide significance threshold. Results are displayed on a Manhattan plot, with −log10(P) on the vertical axis so that small P values become tall points. The threshold 5 × 10−8 sits at −log10(5 × 10−8) = 8 − log10(5) = 8 − 0.70 = 7.3. Significant associations must then be replicated in an independent sample.

0246810 −log10(P) genome wide significance, P = 5 × 10−8 (7.3) 123456789101112 Chromosome (each dot is one SNP) significant locus
Figure 11.5 A Manhattan plot. Each dot is the association test for one SNP, plotted as −log10(P) by genome position, so stronger associations sit higher. Peaks that cross the dashed line at P = 5 × 10−8 are genome wide significant; the tower shape comes from neighboring SNPs in linkage disequilibrium with the causal variant.
FeatureQTL mapping in crossesGWAS in populations
Study populationF2, backcross or recombinant inbred lines from two parentsUnrelated individuals, or cases and controls
Source of marker trait associationLinkage created by the crossHistorical linkage disequilibrium
ResolutionCoarse, often 10 to 30 cMFine, often tens of kilobases
Variants detectedOnly those differing between the two parentsCommon variants segregating in the population
Typical sample sizeHundreds of individualsThousands to over a million
Significance measureLOD score, threshold about 3P value, threshold 5 × 10−8
Main pitfallLarge intervals, missed small effectsPopulation stratification, tiny effect sizes

Pitfalls: population stratification

If cases and controls come from populations with different ancestries, any SNP whose allele frequency differs between those ancestries will appear associated with the disease, whether or not it has anything to do with it. This population stratification produces false positives. Modern GWAS correct for stratification by including principal components of genome wide genotypes, or relatedness matrices, in the analysis, and by checking that the bulk of the P values follow the null expectation on a quantile quantile (Q-Q) plot.

What GWAS has revealed

Two decades of GWAS have transformed our picture of complex traits. Most traits are highly polygenic: adult height is associated with more than 10,000 independent SNPs, each shifting height by about 1 mm or less. Most associated variants lie in noncoding DNA, especially in enhancers active in the relevant tissue, implying that they change gene regulation rather than protein structure. Many loci are shared among traits, which is pleiotropy on a genomic scale. Some findings have pointed to unexpected biology: GWAS of Crohn disease highlighted autophagy genes, and GWAS of age related macular degeneration identified complement factor H, opening new treatment directions.

Early GWAS posed the puzzle of missing heritability. Twin studies put the heritability of height near 0.8, yet the first significant SNPs together explained only a few percent of the variance. The gap has largely closed for height as samples grew into the millions, showing that much of the missing heritability was spread across thousands of variants too small to detect individually. Remaining gaps are attributed to rare variants poorly tagged by SNP arrays, structural variants, imperfect LD tagging, gene by environment interaction and inflated twin estimates.

Polygenic scores

A polygenic score (polygenic risk score) adds up an individual's alleles across many associated SNPs, each weighted by its GWAS effect size: score = Σ (number of effect alleles × effect size). For coronary artery disease, breast cancer and type 2 diabetes, people in the top few percent of the score distribution can have risk comparable to carriers of single gene mutations. Polygenic scores also have important limits. They predict poorly in populations whose ancestry differs from the GWAS sample, because LD patterns and allele frequencies differ; most large GWAS have been done in people of European ancestry, so scores currently work best for them, an equity problem the field is working to correct. A score gives probabilities for a population of people with similar scores, not a diagnosis for an individual, and its predictive power is capped by the trait's heritability.

Study tip. Worked problem: A simplified polygenic score uses three SNPs with effect sizes (in mm of height per effect allele) of +2.0, +1.5 and −1.0. A person carries 2, 1 and 2 effect alleles at the three SNPs. Calculate the score. Then calculate −log10(P) for a SNP with P = 2 × 10−9 and decide whether it is genome wide significant. Step 1: Multiply and add: (2 × 2.0) + (1 × 1.5) + (2 × −1.0) = 4.0 + 1.5 − 2.0 = 3.5 mm. Step 2: −log10(2 × 10−9) = 9 − log10(2) = 9 − 0.30 = 8.7. Step 3: Compare with the threshold of 7.3 (P = 5 × 10−8). Since 8.7 > 7.3 (equivalently 2 × 10−9 < 5 × 10−8), the SNP is significant. Answer: a score of +3.5 mm and a significant SNP at −log10(P) = 8.7. Remember that a taller point on a Manhattan plot means a smaller P value.
Key point

QTL mapping finds trait genes by testing whether marker genotypes in a cross differ in trait means, summarizing evidence as LOD scores; it has coarse resolution. GWAS tests millions of SNPs in natural populations, relies on linkage disequilibrium, uses the 5 × 10−8 threshold to control false positives, and must correct for population stratification. It has shown that complex traits are highly polygenic, mostly regulatory, and that polygenic scores are useful but ancestry dependent.

The chapter in one breath

  • Quantitative traits (continuous, meristic and threshold) vary smoothly because many genes and the environment each contribute small effects.
  • With n additive genes, an F2 shows 2n + 1 classes and (1/4)n at each extreme; environmental noise blurs the classes into a normal curve.
  • The variance measures spread and adds across independent sources; covariance, correlation and the regression slope describe how two variables move together.
  • VP = VA + VD + VI + VE, with possible genotype by environment terms.
  • Broad sense heritability H2 = VG/VP; narrow sense heritability h2 = VA/VP, the transmissible part.
  • Heritability is estimated from inbred lines, offspring on midparent regression (slope = h2), twins (h2 ≈ 2[rMZ − rDZ]) and selection experiments.
  • Heritability describes one population in one environment; it says nothing about individuals or differences between groups.
  • The breeder's equation R = h2S predicts response; realized heritability is R/S, and S = iσP.
  • Correlated responses arise from pleiotropy and linkage; long term selection continues until additive variance runs out or natural selection opposes it.
  • QTL mapping uses markers and LOD scores in crosses; GWAS uses linkage disequilibrium, the 5 × 10−8 threshold and Manhattan plots, and yields polygenic scores.

Words worth knowing

TermWhat it means
Quantitative traitA trait that varies continuously across a range and is influenced by many genes and the environment.
Threshold traitA present or absent trait that appears when an underlying continuous liability exceeds a threshold.
Multiple factor hypothesisThe idea that quantitative traits are controlled by several Mendelian genes with small, additive effects.
VarianceThe average squared deviation from the mean (dividing by n − 1); a measure of spread.
Standard deviationThe square root of the variance, in the same units as the data.
CovarianceA measure of how two variables vary together, positive when they rise together.
Correlation coefficientCovariance standardized by both standard deviations, ranging from −1 to +1.
Regression slopeThe predicted change in y per unit change in x, equal to covariance divided by the variance of x.
Additive variance (VA)Genetic variance from the average effects of alleles; the part transmitted from parent to offspring.
Dominance variance (VD)Genetic variance from interactions between alleles at the same locus.
Epistatic variance (VI)Genetic variance from interactions between alleles at different loci.
Genotype by environment interactionWhen genotypes respond differently to the same change in environment.
Broad sense heritability (H2)The proportion of phenotypic variance due to all genetic variance.
Narrow sense heritability (h2)The proportion of phenotypic variance due to additive genetic variance.
Selection differential (S)Mean of the selected parents minus the mean of the population.
Response to selection (R)Mean of the offspring minus the mean of the original population.
Realized heritabilityHeritability estimated from a selection experiment as R/S.
Selection intensity (i)The selection differential expressed in phenotypic standard deviations.
Correlated responseChange in an unselected trait caused by its genetic correlation with a selected trait.
Quantitative trait locus (QTL)A genomic region containing a gene or genes that affect a quantitative trait.
LOD scoreThe base 10 logarithm of the odds favoring linkage or a QTL over no linkage.
Linkage disequilibriumNonrandom association of alleles at nearby loci in a population.
GWASA genome wide association study that tests many SNPs across the genome for association with a trait.
Population stratificationAncestry differences between study groups that create false associations.
Polygenic scoreA weighted sum of an individual's trait associated alleles used to estimate genetic predisposition.

Check yourself

In an F2, 1/64 of plants are as short as the short parent. How many additive genes are segregating, and how many phenotypic classes do you expect?
Answer. (1/4)n = 1/64 gives n = 3 genes, so 2n + 1 = 7 classes in the ratio 1:6:15:20:15:6:1.
Four measurements are 6, 8, 10 and 12. What are the variance and standard deviation?
Answer. Mean = 9. Squared deviations are 9, 1, 1, 9, sum 20. Variance = 20/3 = 6.67, and the standard deviation is √6.67 = 2.58.
Why does a strong curved relationship between two variables sometimes give a correlation near zero?
Answer. The correlation coefficient measures only linear association. In a U shaped relationship, positive and negative cross products cancel, so r is small even though y depends strongly on x.
VA = 30, VD = 10, VI = 0 and VE = 60. Calculate H2 and h2.
Answer. VP = 100. H2 = 40/100 = 0.40 and h2 = 30/100 = 0.30.
The slope of offspring regressed on one parent (not the midparent) is 0.2. What is h2?
Answer. A single parent transmits half of the offspring's alleles, so the slope estimates h2/2. h2 = 2 × 0.2 = 0.4.
Identical twins correlate at 0.86 for a trait and fraternal twins at 0.47. Estimate h2.
Answer. h2 ≈ 2(0.86 − 0.47) = 2(0.39) = 0.78.
A trait has h2 = 0.9 in a town. A neighboring town's mean is 10 percent lower. Can you conclude the difference is genetic?
Answer. No. Heritability describes variation within a population and says nothing about the cause of differences between populations, which could be entirely environmental.
A herd averages 500 kg. Selected parents average 560 kg, and offspring average 518 kg. What is the realized heritability?
Answer. S = 60 kg and R = 18 kg, so h2 = R/S = 18/60 = 0.30.
At marker Q in an F2, genotype means are 30, 30 and 30 for Q1Q1, Q1Q2 and Q2Q2. What does this suggest?
Answer. No QTL is linked to marker Q: the genotype classes do not differ, so a = 0 and d = 0 at that position.
Why is the genome wide significance threshold 5 × 10−8, and why does a GWAS hit usually not identify the causal variant itself?
Answer. It is 0.05 divided by about one million independent common variants tested, a Bonferroni correction. The significant SNP is usually only in linkage disequilibrium with the causal variant, so fine mapping and functional studies are needed.