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Extra credit question The genes (loci) that influence quantitative traits are known as: _________________________ Uses of heritability • The degree to which offspring resemble their parents is determined by the narrow-sense heritability h2 • The efficacy of natural and artificial selection is also determined by h2 h2 = 1 h2 = 0 VA/VP = 1 VA/VP = 0 Efficacy of artificial selection: size of Labradors Breeder’s Question Q: A horse breeder wants to win the Kentucky Derby. If she breeds her mare to a really fast stallion, how likely is it that the colt will be faster than all the other three-year-olds when it runs in the Derby? A: It depends on the heritability of running speed! Breeder’s Equation • R = h2 S • S = Selection differential difference between selected parents and the population as a whole (within a generation) • R = response to selection difference between selected offspring and the unselected population (across generations) Breeder’s Equation R = h2 S A dog breeder chooses his largest dogs to breed together. The average height of the breed is 60 cm (at the shoulder), and the dogs he chooses to breed average 70 cm tall. He knows from previous work that the heritability of height is 0.5. How big can he expect the offspring to be? R = h2 S = 0.5 * 10cm = 5cm Breeder’s Equation R = h2 S = 0.5 * 10 cm = 5 cm If the response to selection is 5 cm, he can expect his puppies to grow to be 60 cm + 5 cm = 65 cm tall Exactly the same equation can be used to understand natural selection! Efficacy of natural selection: Darwin’s finches h2 = 0.8 If large bills are favored in drought years, what effect will an El Nino year have on the population? R = h2 S Before El Nino mean bill depth = 10 cm. Birds that survived the drought had bills that were 2 mm deeper (on average) than the population mean. Q: What happened to the average bill depth in the next generation? A: R = 0.8 * 2mm = 1.6 mm. Bill depth next generation = 10 + 1.6 = 11.6 mm. How do you measure the heritability? h2 = 1 h2 = 0 Another way to measure h2 if R = h2 S, then h2 = R/S A corn breeder chooses to breed from plants that have large cobs. The average cob length in his crop is 15 cm, but he breeds from plants that average 18 cm cobs. Next year, he measures cob length in the offspring of the selected plants and discovers that the mean is 16 cm What is the heritability of cob length in this population? h2 = R/S = 1 cm/3 cm = 0.33 Most quantitative traits have substantial random environmental variance 2 h = VA/VP = VA/(VA+VD+VE) 2 h =0.5, If then 50% of the phenotypic variance is additive genetic. What’s the other 50%? Hydrangeas Environmental effects on Quantitative Traits: Monozygotic Twins Heritability measured in one population does not tell you anything about differences between populations Environmental variation (VE) is not the same within and between populations Jones’s Farm h2 1.0 milk yield 3 qts/day Smith’s Farm 1.0 2 qts/day Genetic and Environmental Effects on Quantitative Traits Can these methods be applied to humans? Environmental variation cannot be controlled Twin Studies • Identical twins share 100% of alleles • Fraternal twins share 50% of alleles • Assume both kinds of twins share VE to the same extent Correlations between identical co-twins Complete genetic determination No genetic determination 176 174 r=1.0 172 r=0.0 185 180 Trait in Twin 2 170 Trait in Twin 2 190 168 166 164 162 160 158 175 170 165 160 155 150 145 156 140 154 150 155 160 165 Trait in Twin 1 170 175 150 155 160 165 Trait in Twin 1 170 175 Correlations between co-twins for adult height Monozygotic (identical) Dizygotic (fraternal) 180 190 r=0.91 r=0.46 185 180 Twin 2 Height Twin 2 Height 175 170 165 160 175 170 165 160 155 150 155 145 150 140 150 155 160 165 Twin 1 Height 170 175 150 155 160 165 Twin 1 Height 170 175 Heritability estimates from correlations between co-twins • Identical twins share 100% of alleles • Fraternal twins share 50% of alleles H2 ≈ 2* (ri - rf) Height example: ri = 0.91, rf = 0.46 H2 = ? H2 = 0.90 ri Identical Fingerprint ridges 0.96 Height 0.90 IQ score 0.83 Social maturity 0.97 rf Fraternal 0.47 0.57 0.66 0.89 H2 0.98 0.66 0.34 0.16 Heritabilty estimates from human twin studies are biased 1. H2≈2* (ri - rf) overestimates true H2 by 0.5* VD/VP 2. H2 includes all genetic effects, not just additive genetic 3. GxE interaction increases variability among fraternal twins (it lowers rf) but does not affect ri, so H2 is overestimated Heritabilty estimates from human twin studies are biased 4. Estimates assume identical and fraternal twins share environments to the same extent. Do you think this is true? a) Identical twins share embryonic membrane. b) Identical twins share more similar social environment. Heritabilty estimates from human twin studies are biased 5. Studies often based on small sample sizes, and therefore estimates are not very precise (large standard errors) 6. Some studies include male-female fraternal twins, whereas identical twins are always the same sex. Conclusions Heritabiltiy estimates derived from human twin studies should be considered very approximate, and probably too high. Molecular Biology and Quantitative Genetics QTL Mapping • Use DNA-based markers • Marker is anything that differs among strains (RFLPs, microsatellites, single-base pair differnces) • Markers close to genes will tend to be inherited along with genes (recombination rare) Small m s Large m s M S M S F1 m s M S Markers that are close to the QTL are inherited with it (recombination is rare) m s F1: M S Markers that are close to the QTL are inherited with it (recombination is rare) F2: m s m s Small m s M S Intermediate M S Large M S If a marker is far from any gene affecting fruit weight, there will be recombination between the fruitweight and the marker A marker far from a gene will not show a statistical association with the phenotype Many Markers Results • 28 QTLs responsible for fruit-weight variation in tomato • One gene has been cloned and transferred between plants • When “small” allele is transferred to normally large-fruited variety, the fruit weight is reduced by 30% • Proof that the gene is really a QTL