Phenotypic and Genetic Variation in Rapid Cycling Brassica Parts III & IV

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1 1 Phenotypic and Genetic Variation in Rapid Cycling Brassica Parts III & IV Objective: During this part of the Brassica lab, you will be preparing to breed two populations of plants. Both will be considered the Parentals. One population will be your selective breeding experiment. Those plants that are not used in the selective process will be randomly pollinated so that you can collect data to calculate the heritability of trichome density. The pollination of both groups of plants will need to be conducted over several days during which the plants are flowering. This will require you to come in on your own time. 1) HERITABILITY Variance Components The variance in trichome density that you have calculated is the phenotypic variance in the population (V P ). Phenotypic variance is comprised of genotypic variation (V G ) and environmental variation (V E ). V P = V G + V E The proportion of the total variation in the phenotype that is due to genes is the heritable portion. Unlike Lamarck s inheritance of acquired characteristics, a parent can only pass on those characteristics of the phenotype that are coded for by the genes. For example, consider Siamese cats. What makes the points (ears, nose, tail tip and feet) on a seal point Siamese cat dark is temperature sensitive tyrosinase, which initiates a sequence leading from tyrosine to melanin. How dark those points are (and whether the cat is a blue point, seal point or chocolate point) is determined by the genes. However, the environment (temperature) also plays a role in determining the activity of the enzyme produced by those genes. The colder the environment, the greater the melanin production, and the darker the points. Heritability The heritability, h 2, of a character is the proportion of phenotypic variance due to variance in the genes. Its value ranges between and 1. Note that the symbol for heritability is squared. This is to help remind us that it is a relationship of variances, which are also symbolized by squared variables (σ 2, s 2 ). h 2 VG = VP You have estimated phenotypic variance (V P ) already, now you need to estimate the genetic variance (V G ). Recall that the fur color of a Siamese cat is the result of genes inherited from the mother, genes inherited from the father, and the influence of the environment. If we know the phenotype of the kitten s parents, we can determine the relationship between its phenotype and theirs. If the kitten s phenotype is exactly the average between the phenotype of each parent, then the environment had no detectable effect on the kitten s fur color. If the kitten s phenotype is not the average of the two parents (we call this average the midparent value ), then the environment also influenced it. This can be demonstrated graphically. Below are data from two populations of kittens that were produced by an assortment of parents. In the first population (left), nearly all of the variation in

2 2 kitten phenotypes is due to variation among their parents. In other words, the environment played little measurable role in modifying the phenotype of the kittens. The slope of the relationship is.9, which is the percentage of the phenotypic variation that can be attributed to genetic influences. In the second population (right), the environment modified the kitten phenotypes, and the slope of the relationship is.64. This means that nearly two thirds of the variation in kitten phenotype is due to genetic influences. The other third cannot be explained by variation in parental genotype, so it must be due to environmental variation. In other words, the slope of the regression line in these graphs tell us what proportion of the variation in kitten phenotype that is inherited. The slope is the heritability, h 2. High Heritability Low heritability 1 1 kitten phenotype kitten phenotype midparent phenotype midparent phenotype Measuring heritability To measure the heritability, you will need to create graphs like the ones on above. You will have the phenotype measurements for one generation, but not two. You will obtain data for the second generation by collecting the seed from population of plants not used in your selection experiment (see below in part 2). 2) SELECTION Background Case 1: No fitness differences among individuals We need more than variation and heritability for evolution to occur by natural selection. If all individuals in a population with heritable variation in a character contribute equally to the next generation (with a constant environment), then the next generation will have the same distribution of phenotypes that the parent generation had. (Recall assumption of H-W equilibrium.) Example 1: Imagine a population of 1 individuals that vary in a highly heritable character. Each individual mates, and each mated pair (5 of them) produce two offspring. The parent generation then dies. The next generation thus has 1 individuals that have the same mean phenotype as their parents, and likely the same amount of variation in that heritable character.

3 3 Case 2: Fitness differences among individuals based on phenotype. If all parents do not contribute equally to the next generation, then the phenotype of the next generation may or may not be the same as the parents, depending on who produced offspring. Example 2:Truncation selection. Imagine drawing 1 individuals from that population of 1, and letting that small pool mate (5 pairs) to produce the next generation of 1 individuals (2 offspring each). These 1 parents were the individuals with the most extreme version of a phenotype that you measured. If the phenotypic character has a heritability of 1 (h 2 =1) then the average phenotype of the offspring should be equal to the average of the phenotype of the 1 parents. This kind of selection is called truncation selection because the probability of survival is truncated abruptly. Only individuals in the top 1% range of the phenotype survive and reproduce. Fitness Function The relationship between phenotype and the survival probability is called the fitness function (also called a fitness surface). This function can vary in shape. In example 2 above, the fitness function looks something like this: 1 Fitness (w) gene frequency/genotype frequency/phenotype frequency Fitness functions may also take other shapes. For example, the graph may be humped, indicating that intermediates have greater fitness than the either of the extremes. 1 Fitness (w) gene frequency/genotype frequency/phenotype frequency

4 4 What you will do Today Let us assume that the fitness function for trichome density is flat: all individuals (all trichome densities) have the same probability of survival. Now, imagine that you introduce an herbivore to your population. Assume that the herbivore attacks the plants late in their lives late enough that the flowers can be pollinated, but before the plants can produce seeds. This way, you do not have to worry about whether a father lived to fertilize the seeds that you are going to let live or die. Questions to consider 1) How is this herbivore affected by trichome density? 2) What effect does the herbivore have on plant survival? Choose a fitness function for your plant population. Based on your answers to the two questions above, what is the new relationship between phenotype and fitness in your populations? Draw this relationship as a fitness function. Label the axes in more detail. What is the type of units on your x-axis? fitness (w) Apply selection to your population Determining the fate of each plant. What the fitness function describes is the probability of survival of each individual based on phenotype. Now that you have a fitness function you need to determine what the probability of survival is for each parent in your population. How do you determine whether an individual contributes to the next generation or not with the probabilities proposed by your fitness function? For example, an individual with a phenotype with a survival probability of.67 doesn t live 67%; it either lives 1% or it dies (% ). We can use either dice or a random number table to determine the fate of each plant. If the fitness of a phenotype is, then it is simply eliminated from the breeding population. It s survival probability is zero. Dice - But if a phenotype has a fitness of.5, then it has a 5/5 chance of survival. You could roll a die (6 sides) that could produce values from 1-6. You could assign values 1,3,5 to the fate of death, and the values of 2,4,6 to the fate of survival.

5 5 Random number table - In a similar way, you could assign numbers, 1,3,5,7,9.. to the fate of death and the numbers of 2,4,6,8.. to the fate of life and draw these numbers from a random number table. Or you could select a range of values (-5). All values above 25 would be assigned a fate to live, and all values between -25 would be assigned the fate of death. Parent ID # trichomes Survival Probability Ultimate fate Once you have listed all of your plants and determined their fates- Plants that have been assigned the fate to live will be used in controlled breeding for your selection experiment. Plants that have been assigned the fate of death will not be put to death, but will be used in another controlled breeding experiment so that you can estimate heritability h 2. This lab was derived from: J. Winterer, Franklin and Marshall College Wisconsin Fast Plants Projct (1999) Quantitative Variation, Selection and Inheritance with Fast Plants. (1993) Hairy s Inheritance: Getting a handle on Variation

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