 # Brachydactyly and evolutionary change

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Brachydactyly and evolutionary change
We know the gene for brachydactyl fingers is dominant to normal fingers A man named Yule suggested that short-fingered people should become more common through time Godfrey Hardy showed this inference was wrong Wilhelm Weinberg derived the same solution to the problem independently

The Hardy-Weinberg Law - the most important principle in population genetics
The law is divided into three parts: a set of assumptions and two major results In an infinitely large, randomly mating population, free from mutation, migration and natural selection (note there are five assumptions here) the frequencies of the alleles do not change and as long as the mating is random, the genotypic frequencies will remain in the proportions p2 (frequency of AA), 2pq (frequency of Aa) and q2 (frequency of aa) where p is the frequency of A and q is the frequency of a The sum of the genotypic frequencies should be: p2 + 2pq + q2 = 1

Figure 22.3 DERIVING THE HARDY-WEINBERG PRINCIPLE-A NUMERICAL EXAMPLE
P1 = frequency of allele A1 = 0.7 1. Suppose that the allele frequencies in the parental generation were 0.7 and 0.3. P2 = frequency of allele A2 = 0.3 Gametes from parent generation 2. 70% of the gametes in the gene pool carry allele A1 and 30% carry allele A2 . 3. Pick two gametes at random from the gene pool to form offspring. Three genotypes are possible. A1 A2 A2 A1 A1 A1 A2 A2 .07 x 0.3=0.21 .03 x 0.7=0.21 0.7 x 0.7 = 0.49 0.3 x 0.3 = 0.09 = 0.42 Homozygous Heterozygous Homozygous 4. Calculate the frequencies of these three combinations of alleles. Gametes from offspring generation 5. When the offspring breed, imagine that their gametes go into a gene pool. Figure: 22-03 Caption: Deriving the Hardy-Weinberg Principle—A Numerical Example 6. Calculate the frequencies of the two alleles in this gene pool. 42% of the gametes are from A1A2 parents. Half of these carry A1 and half carry A2 49% of the gametes are from A1A1 parents. All of these carry A1 9% of the gametes are from A2A2 parents. All of these carry A2 BEHOLD! The allele frequencies of A1 and A2 have not changed from parent generation to offspring generation. Evolution has not occurred. P1 = frequency of allele A1 = ( /2(0.42)) = ( ) = 0.7 P2 = frequency of allele A2 = (1/2(0.42) ) = ( ) = 0.3 Genotype frequencies will be given by: p12 : 2p1p2 : p22 as long as all Hardy-Weinberg assumptions are met

Genotypic frequencies under the Hardy-Weinberg Law
The Hardy-Weinberg Law indicates: At equilibrium, genotypic frequencies depend on the frequencies of the alleles The maximum frequency for heterozygotes is 0.5 If allelic frequencies are between 0.33 and 0.66, the heterozygote is the most common genotype

Albinism in Hopi Native Americans
The incidence of albinism is remarkably common ( or 13 in every Hopis) Assuming Hardy-Weinberg equilibrium, we can calculate q as the square root of = 0.066 p is therefore equal to 0.934 The frequency of heterozygotes in the population is 2pq = 0.123 In other words, 1 in 8 Hopis carries the gene for albinism! Take-home Lesson: For a rare allele, heterozygotes can be relatively common

Extensions of the Hardy-Weinberg Law
X-linked traits (two alleles) Females have two copies of the gene but males are hemizygous Female genotypes should be in the familiar p2: 2pq: q2 ratio Male genotypes should be in a p:q ratio This result shows why X-linked recessive traits are much more frequently observed in males

Extensions of the Hardy-Weinberg Law
Multiple alleles Let p, q and r represent the frequencies of the three alleles p + q + r = 1.0 The Hardy-Weinberg model predicts the genotype frequencies will be (p + q + r)2 The result is a ratio of p2: 2pq: q2: 2pr: r2: 2pq

Testing for Hardy-Weinberg distributions
If we know the frequency of genotypes in a population, we can test to see if those frequencies conform to the expectations from the Hardy-Weinberg model Let’s reconsider the data for scarlet tiger moths Number of BB = 452 Number of Bb = 43 Number of bb = 2

Now we need to determine the allele frequency of B and b
Let’s determine the frequency of allele B p = f(B) = [(2 x 452) + 43]/994 = 0.953 We can determine the value of q by difference: q = 1 - p = = 0.047 Next we determine the expected Hardy-Weinberg frequencies of genotypes (p2:2pq: q2) That ratio is BB: Bb: bb

Now we can apportion the 497 individuals to their expected genotypes
Expected BB = x 497 = 451 Expected Bb = x 497 =45 Expected bb = x 497 = 1 Compare the observed genotypes in the field: 453, 42 and 2 These moths clearly conform to the Hardy-Weinberg expectations