Variation and its inheritance

1. Variation and its inheritance The foundation of inheritance is the laws of Mendelian genetics. Mendel succeeded and understood particulate (discontinuous or discrete) inheritance. Darwin also undertook experiments on plant inheritance, but failed to recognize the significance of his results. Darwin studied inheritance of flower form in Antirrhinum (snapdragon). Most flowers are asymmetric, but occasion- ally symmetric flowers occur. This is discontinuous variation.

2. Darwin did the same thing as Mendel – he crossed snapdragons that produced asymmetrical flowers (normal, wild type) with plants that produced peloric (symmetrical) flowers. • In the F1 generation, all plants had asymmetrical flowers. • He produced an F2 generation by self-pollination of F1 flowers. Among the F2s he found the following: • 88/127: asymmetrical • 2/127: intermediate • 37/127: peloric • What would you say about this pattern of inheritance? • Darwin did not recognize what these numbers suggested.

3. Do we need to go over Mendelian one and two factor inheritance? The Law of Segregation? The Law of Independent Assortment? What Darwin had found was Mendelian inheritance of flower shape. Such patterns are called heteromorphism. The wild type parents were SS. The peloric parent had the genoptype ss. All F1s were heterozygous Ss. Within tolerances, the F2s showed a 3:1 ratio of phenotypes, and, most likely, a 1:2:1 ratio of genotypes.

4. Situations involving discontinuous traits like Mendel’s flower colour (and other traits) in peas and Darwin’s flower shape in snapdragons involve what are called major gene inheritance. An only slightly more complicated example is a polymorphism in white clover, Trifolium repens, with respect to the production of cyanogenic glycosides and the release of cyanide when leaves are damaged (as a defense). The production of the glycoside is under the control of one locus, and the presence of allele Ac.

5. The release of cyanide is under the control of a second locus. Recessive homozygotes (li/li) do not produce the enzyme linamarase required to release cyanide. Only plants with the Li allele are cyanogenic. Genes with discrete patterns of inheritance that are easily scored, are often used as “marker genes” to track populations. However, it was remarkable and statistically unlikely that Mendel stumbled on 7 unlinked traits with discrete inheritance. Most traits are polygenic, and are inherited with continuous distributions of quantitative characteristics. Discontinuous traits tend not to be strongly influenced by environmental conditions. Continuous traits are much more frequently affected by environmental conditions.

7. The multiple loci involved in regulation of continuous traits are not individually different from the loci Mendel studied. To distinguish them, they are collectively and individually called Quantitative Trait Loci (or QTLs). Given that quantitative traits tend to respond to environmental conditions, the way in which a quantitative trait responds to a range of conditions is called the reaction norm, and the variation in phenotype is called phenotypic plasticity. The plasticity itself may have a heritable component. Reaction norms of plant size for maternal families of Abutilon theofrasti across a range of soil nutrient conditions.

8. Note that the reaction norms are not identical. When phenotypic differences between genotypes vary (differ) between environments, we say there is a genotype x environment interaction. One example of this is the response of the plant world’s “lab rat”, Arabidopsis thaliana, to cold treatment. Either seeds were vernalized for 4 weeks, or rosettes of leaves were cold-treated for the same time.

9. The time to bolting (growing a flowering stalk) was measured for different genotypic families. There was clear evidence of considerable difference, ranging from little response to large response, across the range of families studied…

10. In the last example there is clear evidence of both genetic and environmental effect on the ‘bolting phenotype’. In it we know the genotypes. In most situations involving quantitative inheritance, we don’t know as much. How do you separate the effects of genetic and environmental variability on phenotypic variation? One method is to remove environmental variation by growing plants in a common garden. This was the approach of Clausen, Keck and Hiesey (1948) in studying variation in yarrow from the Pacific coast near Stanford University up the Sierra Nevada mountains. In their native environments the plants varied in size and stature. All grown in a common garden at Stanford…

12. In the absence of environmental variation, the in plant architecture observed among these populations of Achillea lanulosa must be genetic. The common garden is not a native habitat for any of those populations. Another approach is a reciprocal transplant study. Plants are moved between habitats where they differ (and as control removed and replanted in the same habitat).

13. Differences in some phenotypic characteristic that persist across habitats have a genetic component, and differences in the variation between habitats demonstrates an environmental component. In general, we can, in theory, specify a phenotypic value (size, height, leaf area, or whatever) as the sum of a genotypic value and environmental deviation: P = G + E

14. Maybe someday we’ll be able to specify the genotypic value in an individual from DNA sequences. However, we can use the population level variability to assess the contributions of genes and environment to phenotypes: The total phenotypic variation is the sum of contributions from genetic and environmental causes ( and a covariance of genes and environment): VP = VG + VE + covGE What fraction of the total variation is due to genetics? VG/VP This ratio is called the broad-sense heritability.

15. Genetic variation, to be thorough, should be subdivided. Only a fraction of it is ‘available’ to selection. That portion is called the additive component, VA. There is also a component commonly called dominance deviation. This refers to differences in phenotypic expression due to the character of the other allele at the same locus. And there is a third component due to epistasis, which is the effect on the phenotypic expression of a gene due to the characteristics of alleles at other loci. The examples in the text show how dominance (box 6A) and epistasis effect phenotypes: Assume plant height is determined by a single diploid locus. There are no environmental effects on height. Genotype AA produces plants 100cm tall, aa plants 20cm tall.

16. If individuals with genotype Aa all have the intermediate height (60cm), then the effects of alleles A and a are strictly additive; there is no evidence of dominance or epistasis. However, what would you find if the A allele is dominant to a? (Remember that we don’t know what the parental or offspring genotypes are; all we know is the measured height of parents and offspring)

17. If both parents are AA, then all offspring will be 100cm tall. If one parent is AA and one Aa, all offspring will be 100cm tall. If both parents are aa, then all offspring will be 20cm tall. If both are Aa, then ¾ of the offspring will be 100cm tall, and ¼ will be 20cm. The average among offspring will be 80cm. Difference from the parental average indicates dominance.

18. Epistasis: assume that these plants have two phenotypes for leaf shape – lanceolate and ‘heart-shaped’ (cordate).

19. How would we explain differences in height that are effected by leaf shape? Plants with the AA genotype are 110cm tall with lanceolate leaves, and 90cm with cordate leaves. Plants with the aa genotype are 15cm with lanceolate leaves and 25cm with cordate leaves. The effects need not be opposite on the two genotypes, but the only explanation is that there is an epistatic interaction between the height determining locus and the leaf shape locus. So, the complete equation is: VP = VA + VD + VI + VE + VGxE + covGE

20. A quick reminder of the statistics you may have forgotten. The variance of a characteristic is: xiis a single measure of the characteristic is the mean of measurements of the characteristic n is the number of measurements made (if only a sample of individuals is measured. If all are measured then the proper denominator is n.

21. When a plant can be cloned, you can sort out VE and get at the components that way, having removed VG. Another frequent question is how many genes are involved in determining the quantitative trait under study? You can determine that if you can create two different inbred (homozygous) lines. The two parental lines are genetically uniform, so all variation in each line must be environmental. Now cross the two lines. All offspring will be heterozygous at all involved loci. But variance among individuals is still all environmental. Now cross the F1 heterozygotes. The F2s will vary genetically as well. The difference between variances in the F1s and F2s estimates VG.

22. Here’s what the experiment looks like genetically:

23. F2 generation Gene 2: B1/B1 Or B1/B2 Or B2/B2 Gene 3: C1/C1 Or C1/C2 Or C2/C2 Gene 4: D1/D1 Or D1/D2 Or D2/D2 Gene 1: A1/A1 Or A1/A2 Or A2/A2 • For these 4 loci, 3 X 3 X 3 X 3 = 81 different genotypes are possible The variation observed in these F2s includes both genetic variation among the 81 different possible genotypes and the environmental variance which should still be equal to that seen in the parental and F1 generations. Now we can estimate VG. VG = variance in F2 – variance in F1

24. If we assume each locus has equal impact on the observed phenotype, then the number of loci involved is estimated as: n = (mean of line 1 – mean of line 2)2 4VG