Download Phenotypic and Genetic Variation in Rapid Cycling Brassica Parts III

BIO 370 Evol Bio
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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 (VP). Phenotypic variance is comprised of genotypic variation (VG) and
environmental variation (VE).
VP = VG + VE
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, h2, of a character is the proportion of phenotypic variance due to variance in the
genes. Its value ranges between 0 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 (σ2, s2).
V
h2 = G
VP
You have estimated phenotypic variance (VP) already, now you need to estimate the genetic
variance (VG). 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
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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
0.9, 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 0.64. 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, h2.
Low heritability
10
10
8
8
kitten phenotype
kitten phenotype
High Heritability
6
4
2
6
4
2
0
0
0
2
4
6
8
midparent phenotype
10
0
2
4
6
8
10
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 1000 individuals that vary in a highly heritable character.
Each individual mates, and each mated pair (500 of them) produce two offspring. The parent
generation then dies. The next generation thus has 1000 individuals that have the same mean
phenotype as their parents, and likely the same amount of variation in that heritable character.
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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 100 individuals from that population of 1000, and letting that small pool mate
(50 pairs) to produce the next generation of 1000 individuals (20 offspring each). These 100
parents were the individuals with the most extreme version of a phenotype that you measured. If
the phenotypic character has a heritability of 1 (h2=1) then the average phenotype of the
offspring should be equal to the average of the phenotype of the 100 parents.
This kind of selection is called truncation selection because the probability of survival is
truncated abruptly. Only individuals in the top 10% 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)
0
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)
0
gene frequency/genotype frequency/phenotype frequency
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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 0.67 doesn’t live 67%; it either lives 100% or it dies (0% ). We can
use either dice or a random number table to determine the fate of each plant.
•
•
If the fitness of a phenotype is 0, then it is simply eliminated from the breeding population.
It’s survival probability is zero.
Dice - But if a phenotype has a fitness of 0.50, then it has a 50/50 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.
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•
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 (0-50). All values above 25 would be assigned a
fate to live, and all values between 0-25 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 fatesPlants 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 h2.
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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