Presentation on theme: " Read Chapter 6 of text We saw in chapter 5 that a cross between two individuals heterozygous for a dominant allele produces a 3:1 ratio of individuals."— Presentation transcript:
We saw in chapter 5 that a cross between two individuals heterozygous for a dominant allele produces a 3:1 ratio of individuals expressing the dominant phenotype: to those expressing the recessive phenotype. For example brachydachtyly (shortening of the digits) displays this pattern of inheritance.
In the early 1900’s when Mendel’s work was rediscovered there was confusion about how these simple patterns of inheritance affected populations. Why, for example, was not 3 of every 4 people a person with brachdactyly? Why did not dominant alleles replace recessive alleles?
The confusion stemmed from confusing what was happening at the level of the individual with what occurs at the population level. Individual-level thinking enables us to figure out the result of particular crosses.
Population level thinking however is needed to figure out how the genetic characteristics of populations change over time. It enables us to figure out quantitatively what is happening in a population as a result of evolution. Remember, evolution occurs when genotype frequencies change over time.
Study of the distribution of alleles in populations and causes of allele frequency changes
Diploid individuals carry two alleles at every locus › Homozygous: alleles are the same › Heterozygous: alleles are different Remember Evolution: change in allele frequencies from one generation to the next
Hardy-Weinberg serves as the fundamental null model in population genetics
Null models provide us with a baseline. They tell us what we expect to be the case if certain forces are not operating. The Hardy-Weinberg equilibrium tells us what we expect to happen to genotype frequencies when forces such as natural selection are not operating on a population.
The Hardy-Weinberg model enables us to determine what allele and genotype frequencies we would expect to find in a population if all that is happening is alleles are being randomly assigned to gametes when gametes are made (during meiosis) and those gametes meet up at random.
The Hardy-Weinberg model examines a situation in which there is one gene with two alleles A 1 and A 2. Recall that alleles are different versions of a gene. There are three possible genotypes A 1 A 1, A 2 A 2,and A 1 A 2
Hardy and Weinberg used their model to predict what would happen to allele frequencies and genotype frequencies in a population in the absence of any evolutionary forces. Their model produced three important conclusions
The three conclusions of the H-W model. In the absence of evolutionary processes acting on them: 1. The frequencies of the alleles A 1 and A 2 do not change over time. 2. If we know the allele frequencies in a population we can predict the equilibrium genotype frequencies (frequencies of A 1 A 1, A 2 A 2,and A 1 A 2 ).
3. A gene not initially at H-W equilibrium will reach H-W equilibrium in one generation.
1. No selection. › If individuals with certain genotypes survived better than others, allele frequencies would change from one generation to the next.
2. No mutation › If new alleles were produced by mutation or alleles mutated at different rates, allele frequencies would change from one generation to the next.
3. No migration › Movement of individuals in or out of a population would alter allele and genotype frequencies.
4. Large population size. › Population is large enough that chance plays no role. Eggs and sperm collide at same frequencies as the actual frequencies of p and q. › If assumption was violated and by chance some individuals contributed more alleles than others to next generation allele frequencies might change. This mechanism of allele frequency change is called Genetic Drift.
5. Individuals select mates at random. › Individuals do not prefer to mate with individuals of a certain genotype. If this assumption is violated allele frequencies will not change, but genotype frequencies might.
Assume two alleles A 1 and A 2 with known frequencies (e.g. A 1 = 0.6, A 2 = 0.4.) Only two alleles in population so their allele frequencies add up to 1.
Can predict frequencies of genotypes in next generation using allele frequencies. Possible genotypes: A 1 A 1, A 1 A 2 and A 2 A 2
Assume alleles A 1 and A 2 enter eggs and sperm in proportion to their frequency in population (i.e. 0.6 and 0.4) Assume sperm and eggs meet at random (one big gene pool).
Then we can calculate genotype frequencies. A 1 A 1 : To produce an A 1 A 1 individual, egg and sperm must each contain an A 1 allele. This probability is 0.6 x 0.6 or 0.36 (probability sperm contains A 1 times probability egg contains A 1 ).
Similarly, we can calculate frequency of A 2 A 2. 0.4 x 04 = 0.16.
Probability of A 1 A 2 is given by probability sperm contains A 1 (0.6) times probability egg contains A 2 (0.4). 0.6 x 04 = 0.24.
But, there’s a second way to produce an A 1 A 2 individual (egg contains A 1 and sperm contains A 2 ). Same probability as before: 0.6 x 0.4= 0.24. Overall probability of A 1 A 2 = 0.24 + 0.24 = 0.48.
Genotypes in next generation: A 1 A 1 = 0.36 A 1 A 2 = 0.48 A 2 A 2 = 0.16 Adds up to one.
General formula for Hardy-Weinberg. Let p= frequency of allele A 1 and q = frequency of allele A 2. p 2 + 2pq + q 2 = 1.
If three alleles with frequencies P 1, P 2 and P 3 such that P 1 + P 2 + P 3 = 1 Then genotype frequencies given by: P 1 2 + P 2 2 + P 3 2 + 2P 1 P 2 + 2P 1 P 3 + 2P 2 P 3
Allele frequencies in a population will not change from one generation to the next just as a result of assortment of alleles and zygote formation. If the allele frequencies in a gene pool with two alleles are given by p and q, the genotype frequencies will be given by p 2, 2pq, and q 2.
The frequencies of the different genotypes are a function of the frequencies of the underlying alleles. The closer the allele frequencies are to 0.5 the greater the frequency of heterozygotes.
You need to be able to work with the Hardy-Weinberg equation. For example, if 9 of 100 individuals in a population suffer from a homozygous recessive disorder can you calculate the frequency of the disease causing allele? Can you calculate how many heterozygotes are in the population?
p 2 + 2pq + q 2 = 1. The terms in the equation represent the frequencies of individual genotypes. P and q are allele frequencies. It is vital that you understand this difference.
9 of 100 (frequency = 0.09) of individuals are homozygotes. What term in the H-W equation is that equal to?
It’s q 2. If q 2 = 0.09, what’s q? Get square root of q 2, which is 0.3. If q=0.3 then p=0.7. Now plug p and q into equation to calculate frequencies of other genotypes.
p 2 = (0.7)(0.7) = 0.49 2pq = 2 (0.3)(0.7) = 0.42 Number of heterozygotes = 0.42 times population size = (0.42)(100) = 42.
There are three alleles in a population A 1, A 2 and A 3 whose frequencies respectively are 0.2, 0.2 and 0.6 and there are 100 individuals in the population. How many A 1 A 2 heterozygotes will there be in the population?
Just use the formulae P 1 + P 2 + P 3 = 1 and P 1 2 + P 2 2 + P 3 2 + 2P 1 P 2 + 2P 1 P 3 + 2P 2 P 3 = 1 Then substitute in the appropriate values for the appropriate term 2P 1 P 2 = 2(0.2)(0.2) = 0.08 or 8 people out of 100.
Hardy Weinberg equilibrium principle identifies the forces that can cause evolution. If a population is not in H-W equilibrium then one or more of the five assumptions is being violated.