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Genetic Change Hardy Weinberg Within Populations: The HardyWeinberg Rule Genetic Change Within Populations: The Hardy-Weinberg Rule Population genetics is the study of the properties of genes in a population Genetic variation in populations puzzled scientists Dominant alleles were believed to drive recessive alleles out of populations In 1908, G. Hardy and W. Weinberg pointed out that in large populations with random mating, allele frequencies remain constant Dominant alleles do not, in fact, replace recessive ones • A population that is in Hardy-Weinberg equilibrium is not evolving – Hardy and Weinberg came to their conclusion by analyzing allele frequencies in successive generations – Frequency = Number of individuals falling within a category Total number of individuals being considered • If a population of 100 cats has 84 black and 16 white – Then the frequencies of black and white phenotypes are 0.84 and 0.16, respectively • B allele Black color • b allele White color • By convention – The more common allele (B) is designated p – The less common allele (b) is designated q –p+q=1 • The Hardy-Weinberg equilibrium can be written as an equation Individuals homozygous – (p + q)2 = p2 + 2pq + q2 for allele b Individuals homozygous for allele B Individuals heterozygous for alleles B and b • The equation allows calculation of allele frequencies – Frequency of white (bb) cats = 16/100 = 0.16 – => q2 = 0.16 – => q = √ 0.16 = 0.4 – p + q =1 => p = 1 – q = 1 – 0.4 = 0.6 • What about genotype frequencies? – Frequency of the homozygous dominant genotype is 36 out of 100 cats are black (BB) • p2 = (0.6)2 = 0.36 – Frequency of the heterozygous genotype is 48 out of 100 cats are black (Bb) • 2pq = 2(0.6)(0.4) = 0.48 Hardy-Weinberg Assumptions The Hardy-Weinberg equation is true only if the following five assumptions are met 1. Large population size 2. Random mating 3. No mutation 4. No migration 5. No natural selection Why Allele Frequencies Change • Five evolutionary forces can significantly alter the allele frequencies of a population – 1. Mutation – 2. Migration – 3. Genetic drift – 4. Nonrandom mating – 5. Selection Mutation • Errors in DNA replication • The ultimate source of new variation • Mutation rates are too low to significantly alter allele frequencies on their own Migration • Movement of individuals from one population to another – Immigration: movement into a population – Emigration: movement out of a population • A very potent agent of change Genetic Drift • Random loss of alleles – More likely to occur in smaller population • Founder effect – Small group of individuals establishes a population in a new location • Bottleneck effect – A sudden decrease in population size to natural forces Nonrandom Mating • Mating that occurs more or less frequently than expected by chance • Inbreeding – Mating with relatives – Increases homozygosity • Outbreeding – Mating with non-relatives – Increases heterozygosity Selection • Some individuals leave behind more offspring than others • Artificial selection – Breeder selects for desired characteristics • Natural selection – Environment selects for adapted characteristics Conditions for Hardy-Weinberg Equilibrium Describes a hypothetical population in which these five conditions are met Extremely large population size No gene flow No mutations Random mating No natural selection Hardy-Weinberg Equation Hardy-Weinberg Equation Can be used to estimate the percentage of a population carrying an allele for a specific trait If p = A Then: And q=a p2 + 2pq + q2 = 1 Hardy-Weinberg Equilibrium If p = W, wings And q = w, no wings In a population of 100 buffalo 70 % have wings (WW and Ww) 30% have no wings (ww) Use the equation to figure out the allele frequencies for W and w p2 + 2pq + q2 = 1 p+q = 1 Applications Sample problem 01 Consider a population of 100 jaguars, with 84 spotted jaguars and 16 black jaguars. The frequencies are 0.84 and 0.16. Based on these phenotypic frequencies, can we deduce the underlying frequencies of genotypes ? If the black jaguars are homozygous recessive for b (i.e. are bb) and spotted jaguars are either homozygous dominant BB or heterozygous Bb, we can calculate allele frequencies of the 2 alleles. Let p = frequency of B allele and q = frequency of b allele. (p+q)2 = p2 + 2pq + q2 where p2 = individuals homozygous for B pq = heterozygotes with Bb q2 = bb homozygotes If q2 = 0.16 (frequency of black jaguars), then q = 0.4 (because0.16 = 0.4) Therefore, p, the frequency of allele B, would be 0.6 (because 1.0 – 0.4 = 0.6). The genotype frequencies can be calculated: There are p2 = (0.6)2 X 100 (number of jaguars in population) = 36 homozygous dominant (BB) individuals The heterozygous individuals (Bb) = 2pq = (2 * 0.6 * 0.4) * 100 = 48 heterozygous Bb individuals Applications Sample problem 02 In 1986, Henry Horn counted 133 Gray Squirrels and 25 Black Squirrels (16% Black) at Princeton University. In 1994, he counted 43 Gray Squirrels and 9 Black Squirrels (17% Black). This indicates that the ratio of Gray:Black squirrels may be in Hardy-Weinberg equilibrium Assuming that the gene for the black morph is autosomal dominant (not sex-linked) what are the gene frequencies in the population ? Black squirrels, which could be Black homozygotes (BB) or black heterozygotes (Bb) = 0.17 of the population in 1994 Thus, the proportion of gray recessives (bb) in the population is q2 = 1 – 0.17 = 0.83 (or 83%) q = 0.83 = 0.91 (or 91%) of the genes in this gene pool Therefore, the frequency of the dominant allele (for Black) = 1 – 0.91 = 0.09 Thus, black individuals that are homozygous dominants (BB) = p2 = 0.09 * 0.09 = 0.0081 (=0.08% of the population). The black individuals that are heterozygotes (Bb) are 2pq = 2 * 0.09 * 0.91 = 0.162 (= 16.2% of the population) The population therefore consists of: 83% homozygous (recessive) gray 16.2% heterozygous black 0.08% homozygous (dominant) black This example shows that dominants can be less common than recessives and that there is no evidence that the dominant character will eliminate the recessive Sample problem 03 The allele for cystic fibrosis is present in Caucasians at a frequency q of 22 per 1000 individuals (= 0.022) What proportion of Caucasians is expected to express this trait ? The frequency of double recessives (q2) is 0.022 * 0.022 = 0.000484, or approx. 1 in every 2000 individuals. What proportion is expected to be heterozygous carriers ? If the frequency of the recessive allele q is 0.022, then the frequency of the dominant allele p is 1 – 0.022 = 0.978 The frequency of heterozygous individuals (2pq) is 2 * 0.978 * 0.022 = 0.043, or 43 people in every 1000