To measure evolution of a continuous trait

Determine whether its distribution changes across generations

e.g. Determine whether mean changes from one generation to next

To study evolution of a discrete trait

Must determine whether its distribution changes across generations

Thus, need a way to describe the distribution of phenotypes of the trait

Discrete trait

Can be classified into discrete categories

Examples

e.g. coat colors of cats (black, white, spotted)

e.g. chromosome morphs of Drosophila

e.g. allozymes (determined by starch gel electrophoresis)

Describing distribution of phenotypes

Every individual can be placed into a category

Can compare populations by looking at the distribution of individuals among the categories

e.g. a sample from the population of cats in my neighborhood in 1980

Spotting phenotype	Number of individuals = Np
No spots	80
Some spots	100
Many spots	20
Total	200 = N

and a sample from the same population of cats in 2000

Spotting phenotype	Number of individuals = Np
No spots	65
Some spots	52
Many spots	13
Total	130 = N

Can ask: Is the distribution of phenotypes the same in the two samples?

It’s kind of difficult to tell, because the sample size is different

Frequencies

Can describe distribution of individuals among categories by their frequencies

A way of standardizing counts

For camparison across samples of different size

Proceedure

Sum all the categories to get the total population (or sample) size

Frequency = number in category/number in all categories, combined

Thus

e.g. the sample from the population of cats in my neighborhood in 1980

Spotting phenotype	Np	Freq(phenotype)
No spots	80	80/200 = 0.4
Some spots	100	100/200 = 0.5
Many spots	20	20/200 = 0.1
Total	200	200/200 = 1.0

and the sample from 2000

Spotting phenotype	Np	Freq(phenotype)
No spots	65	65/130 = 0.5
Some spots	52	52/130 = 0.4
Many spots	13	13/130 = 0.1
Total	130	130/130 = 1.0

How to describe the distribution of genetic variation in a population?

Just as with the continuously varying trait

To differentiate evolution from some phenotypically plastic response to an environmental change, must determine whether there has been a genetic change in population

Genotype frequencies

If we can ascertain the genotypes of these phenotypes

Then we can estimate the genotype frequency in the population in 1980

Spotting phenotype	Np	Freq(phenotype)	Genotype	Ng	Freq(genotype)
No spots	80	80/200 = 0.4	ss	80	0.4
Some spots	100	100/200 = 0.5	Ss	100	0.5
Many spots	20	20/200 = 0.1	SS	20	0.1
Total	200	200/200 = 1.0		200	1.0

And, in 2000

Spotting phenotype	Np	Freq(phenotype)	Genotype	Ng	Freq(genotype)
No spots	65	65/130 = 0.5	ss	65	0.5
Some spots	52	52/130 = 0.4	Ss	52	0.4
Many spots	13	13/130 = 0.1	SS	13	0.1
Total	130	130/130 = 1.0		130	1.0

So, did this cat population evolve?

It appears that it did – the genotype frequencies have changed across generations

However,

The formal signature of evolution is change in gene frequencies across generations

We could get a change in genotype frequencies without seeing a change in allele frequencies

For example, there could be a decrease in the numbers of heterozygotes and a compensating increase in the numbers of homozygotes

In this case, there would actually be no evolution, but we would think that there had been

Thus, to determine whether evolutionary really occurred, we must determine whether the allele frequencies have changed

From the genotypic data, we can easily estimate the distribution of alleles in the population at each date

In 1980

Allele	Numberallele	Nallele	Freq(allele)
S	(# SS x 2) + (#Ss x 1)	(20 x 2) + (100 x 1) = 140	140/400 = 0.35
s	(# ss x 2) + (#Ss x 1)	(80 x 2) + (100 x 1) = 260	260/400 = 0.65
total	(# cats x 2)	(200 x 2) = 400	400/400 = 1.00

And, in 2000

Allele	Numberallele	Nallele	Freq(allele)
S	(# SS x 2) + (#Ss x 1)	(13 x 2) + (52 x 1) = 78	78/260 = 0.3
s	(# ss x 2) + (#Ss x 1)	(65 x 2) + (52 x 1) = 182	182/260 = 0.7
total	(# cats x 2)	(130 x 2) = 260	260/260 = 1.00

Let’s make freq(S) = p and freq(s) = q

Then, can state that p + q = 1.0

Note on codominant genes

In this example, I used co-dominant genes

Becase they allow me to ascertain genotype frequencies directly from phenotype frequencies

Which allows me to directly calculate allele frequencies

What happens when I am working with genes having dominant alleles?

Can’t determine genotypes directly from phenotypes

So, can’t calculate allele frequencies

What to do?

To estimate genotype frequencies

Make two assumptions about the population

Assume that it’s at evolutionary equilibrium, and

Assume that individuals are mating randomly with respect to their chromosomal type

A population that meets these two assumptions is at what’s called Hardy-Weinberg equilibrium

A population at H-W equilibrium will display a characteristic relationship between the genotype frequencies and the allele frequencies

Called the H-W ratios

For a locus with two alleles

If p = freq(A)

And q = freq(a)

Then, freq(AA) = p²

Freq(Aa) = 2pq

Freq(aa) = q²

Note that, since p = 1-q,

Then p² + 2pq + q² = 1

Which is true of all good frequencies

Can use this relationship to estimate allele frequencies from frequencies of known genotypes

First, what do we mean when we assume that a population is at evolutionary equilibrium?

More general question: What’s an equilibrium?

Where things don’t change (e.g. downtown Pollocksville, NC, is the center of the universe, because the whole universe revolves around it and nothing at the center moves)

Equilibrium – a point at which the system remains the same from one time point to another

Types of equilibria (digression with diagrams)

Stable (marble in bottom of bowl)

When unperturbed, system does not change

Perturbing system changes things, but then they come right back to the same place

Unstable (marble at top of hill)

As long as it’s not perturbed, system stays where it is

Once system is perturbed, however, it leaves that equilibrium point for some other place

Neutral (marble on flat tabletop)

When unperturbed, system does not change

Perturbing system changes where it is, but then it stays at that point until further perturbation

How do we know when population is at evolutionary equilibrium?

When it doesn’t evolve

How do we know it isn’t evolving?

Allele frequencies are not changing

How to determine that?

One way is to watch it for a long time

Another way is to determine if there is some characteristic, defining aspect of allele and genotype frequencies in populations that aren’t evolving

It turns out that there is

They are at H-W equilibrium

To ascertain what that characteristic is, we ask what are all the factors that could make a population evolve

Then model a population in which none of those factors is operating

Hardy-Weinberg Principle

Based on a model population

Just as chemists and physicists find it convenient to picture gas molecules as inelastic spheres

From which to derive the Ideal Gas Law for describing the behavior of real gaseous systems

Population geneticists make certain simplifying assumptions about populations

From which to derive certain generalizing principles with which to describe the behavior of real populations

The H-W Principle is just such a generalizing principle

Helps us understand, and hence describe, the behavior of alleles in real populations

But, is based on a certain type of ideal population

The ideal population is a population not experiencing any evolutionary forces

What are the evolutionary forces that could act on allele frequency at a locus?

Small populations can experience changes in allele frequencies at all loci, due to genetic drift

Mutation at a locus creates new classes of individuals, which causes allele frequency changes

Gene flow (either immigration of individuals into population or emigration of individual out of population) can change allele frequencies, if the migrants have a different allele frequency than the residents

Selection on a locus can change its allele frequencies

In addition, must also assume that individuals in population mate at random with respect to their genotype at the locus

Non-random mating doesn’t actually cause changes in allele frequencies, but it can cause changes in genotype frequencies, which is inconvenient for our model

H-W principle states that

Even though the population exhibits phenotypic variability for the trait coded by this locus, and

Even though that variability is heritable (and, thus, there is genetic variability for the trait)

If a population meets the conditions stated above

Allele frequency at this locus will not change, and hence

The population will not evolve at this locus

And, the population will exhibit H-W ratios of genotype and allele frequencies

Deriving H-W principle

Imagine a large, ideal population with a particular life history

e.g. a sea urchin

At time t, population consists entirely of adults about to breed

A breeding event in which all individuals release gametes into the water simultaneously

Gametes diffuse out into the water column and act somewhat like a population of gas molecules

That is, meeting of two particular gametes is result of random process

Gametes that meet then fuse together, yielding a population of zygotes

Zygotes then grow and develop into mature adults at time t+1

At which time, life cycle begins again

To detect whether allele frequency changes between generations

Census adult population at time t and determine genotype of each

Again census adult population at time t+1

Compare allele frequencies from the two censuses

So, our model population meets the following assumptions

Non-overlapping generations

Random meeting of gametes

Hence, random mating

No differential mortality or fecundity

Hence, no selection

Large population

Hence, no genetic drift

No mutation

No net migration

Initial parameter values in adults at time t = 1

Population size = N

Initial genotype frequencies: p_AA,0, p_Aa,0, p_aa,0

Initial allele frequencies then are:

Shedding gametes: Let each individual produce gametes

Assume that each individual produces m gametes

Gamete type	From AA	From Aa	From aa
A	NmpAA,0
a			Nmpaa,0
Total	NmpAA,0	NmpAa,0	Nmpaa,0

Total number of gametes is simply

Nm(p_AA,0 + p_Aa,0 + p_aa,0) = Nm

Because (p_AA,0 + p_Aa,0 + p_aa,0) = 1 (it’s a frequency)

Hence, the allele frequency of A alleles in gametes is

And, q_a,gametes = 1 – p_A,gametes = q_a,0

Thus, the shedding of gametes does not change the allele frequency

Fusing of gametes

Random mating means that any individual gamete has an equal probability of fusing with all other gametes

Consequently, probability that a particular gamete will fuse with a gamete of genotype A is equal to the fraction of genotype A gametes in the gamete pool

That probability = p_A,0

Similarly, the probability that a particular gamete will fuse with a gamete of genotype a is equal to the fraction of gametes in the gamete pool that are of genotype a

That probability = q_a,0

We can model the process of fusion between two gametes as one of drawing two gametes at random from the gamete pool

Then, the product law comes into play

Product Law

Statement: The chance that a number of independent events will occur together equals the product of the chances that each event will occur separately

A mathematical law

Works for pennies

The chance of flipping a head is ½

If you flip a penny 100 times, on average you will expect to get 50 heads

It’s a probability event, so won’t always get 50 heads

Just expect to do so

Each flip of the penny is independent of all other flips

Thus, the chance of getting three heads in a row is

½ x ½ x ½ = 1/8

Applies to biological entities such as gametes

Reach into gamete pool (like a drawer of socks of different colors)

Chance of getting A allele = p_A,0

Chance of getting another A allele = p_A,0

So, by product law, chance of getting two A alleles to create an AA zygote = p_A,0 x p_A,0 = p_A,0²

Similarly for aa zygote

By product law, chance of getting two A alleles to create an aa zygote = q_a,0 x q_a,0 = q_a,0²

A little trickier for Aa zygote

Can get it in two different ways

Can first draw A allele = p_A,0 and then draw a allele = q_a,0

So, chance of getting Aa zygote by drawing one A allele and one a allele = p_A,0 x q_a,0

But, can also first draw a allele and then draw A allele

So, chance of getting aA zygote by drawing one a allele and one A allele = q_a,0 x p_A,0

So, summing over these two ways = 2(q_a,0 x p_A,0 )

Note that these probabilities all sum to 1

p_A,0² + 2(q_a,0 x p_A,0 ) + q_a,0² = (p_A,0 + q_a,0)² = 1² = 1

Zygote number

Because the total number of gametes produced is Nm, and

Becaue two gametes are needed for each zygote,

We know the number of zygotes is ½(Nm)

So, total number of zygotes of each genotype is

Genotype	# zygotes	# adults at time t + 1
AA		Save room for this column
Aa
aa

Finally, the zygotes must grow up to form the next generation of adults

During this period of growth, many will die; only a fraction will survive to reproduce

Let this fraction of survivors be l

Since we want to hold all variables constant except the effects of random mating, l will be the same for all genotypes

Thus, the number of adult individuals at time t + 1 will be

The number of individuals of each genotype is simply l times the number of zygotes of that genotype

Genotype	# zygotes	# adults at time t + 1
AA
Aa
aa

Now, let’s determine the allele frequencies in the adults of the new generation

Cancel out the

Cancel out

So, the frequency of the A allele does not change from time t to time t + 1

And, of course, since q_a,1 = 1 – p_A,0 = q_a,0

No evolution because no change in allele frequences!

A problem

Levy and Levin (1975) used electrophoresis to study the phosphoglucose isomerase-2 locus in the evening primrose Oenothera biennis. . . They observed two alleles affecting electrophoretic mobility of the enzyme, PGI-2a and PGI-2b. In 57 strains, they observed 35 PGI-2a/PGI-2a, 19 PGI-2a/PGI-2b, and 3 PGI-2b/PGI-2b. . .

Calculate the expected numbers of the three genotypes . . . assuming that the genotypes occur in Hardy-Weinberg proportions.

Answer

Make p the frequency of PGI-2a and q the frequency of PGI-2b.

The frequency of PGI-2a is equal to 35/57 + 0.5*19/57 = 0.78.

Thus, p = 0.78 and q = 1 - 0.78 = 0.22.

Given Hardy-Weinberg proportions, the expected frequency of PGI-2a/PGI-2a = p2 = 0.61, of PGI-2a/PGI-2b = 2pq = 0.34, and of PGI-2b/PGI-2b = q2 = 0.05.

As a check, 0.61 + 0.34 + 0.5 do indeed equal 1.0. 0.61 * 57 = 35,

0.34 * 57 is a little greater than 19, and

0.05 * 57 is a little less than 3.

Though they didn't ask, you probably would, to a first approximation (i.e., it's always good form to do the statistics even on the obvious), assume that this population is in Hardy-Weinberg equilibrium.

Another problem

You have been studying a population of annual plants in which a flower color polymorphism (white, pale, or dark) has been in Hardy-Weinberg equilibrium at the chalcone synthetase gene, in which the codominant white (w) allele occurs at a frequency of 0.2. This year, however, a new pollinator has arrived in the area. This pollinator has perfect color fidelity, meaning that when it starts seeking flowers on any particular day, it randomly finds the first flower but after that initial visit, visits only flowers of that color intensity. This pollinator is very effective and at very high density. Due to its aggressive behavior, it prevents the native pollinators from visiting flowers.

What is the term that population geneticists use to describe the type of mating that this situation produces?

What were the genotype frequencies before the new pollinator arrived?

What do you predict the genotype frequencies will be in the seeds that will be produced this fall?

How about the adults that will be flowering next year?

Answer

First calculate the H-W genotype frequencies:

WW (dark) plants are 64% of population and produce only WW progeny;

ww (white) plants are 4% of the population and produce only ww progeny;

Ww (light) parents are 32% of the population and produce offspring that are 25% WW, 50% Ww, and 25% ww.

Since these latter progeny make up only 32% of the population, the final frequencies are:

WW = 0.64 + (0.25 * 0.32) = 0.64 + .08 = 0.72

ww = 0.04 +(0.25 * 0.32) = 0.04 + 0.08 = 0.12

Ww = 0.32 * 0.5 = 0.16

Check for summation to 1: 0.72 + 0.12 + 0.16 = 1.0

Note that assortative mating has decreased the frequency of heterozygotes in the population.

What assumptions did you make in this situation?

No selection (all adults equally likely to mate, all progeny equally fit)

Large population (already knew that from previous HW equil)

No change in migration rates