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Genetic constitution of a population Gene and Genotype Frequencies Lecture 3 Aim • To equip students with more principles of population genetics as basis for quantitative genetics Objectives • By the end of the topic, students should be able to • Define gene and genotype frequencies • Relate gene and genotype frequencies • Describe HW equilibrium • Explain conditions required to fulfill HW equilibrium Genetic constitution of a population Can be described by specifying their genotypes and stating the number of genotypes in the group Specifying the genes carried by the population that influence a particular trait and stating the number or proportions of different alleles at a locus For simplicity sake, we assume a single locus A that has two alleles, designated A1, A2; or A, a Genotype frequency From two alleles A, a , there are three possible combinations (genotypes) i.e. an individual can be A A Aa aa In a population A A a A a a The genetic constitution of the group or population would be described by the proportion or percentage of individuals that belong to each genotype A population of sheep Dorper and Red Masai sheep Genotype frequency These proportions are called Genotype frequencies Definition: Proportion or frequency of a particular genotype among individuals in a population is called Genotype frequency Genotype frequency Assume there are 100 individuals in a population and would like to describe the genetic structure of the population You count how many belong to AA Aa aa In a population Naturally the frequencies of all genotypes must add up to unity or 100 % Calculation of genotype frequency At the SAU, there are 57 chickens of three different phenotypes Yakuda (assume BB) Chiphulusa (assume Bb) Kawangi (assume bb) 30 17 10 If these represent three genotypes, their frequencies are Calculation of genotype frequency N = 57 BB (Yakuda) Bb (Chiphulusa) bb (Kawangi) 30/57 = 0.526 17/57 = 0.298 10/57 = 0.175 Two populations are different if they have different genotype frequencies Gene Frequency This is a proportion of a particular gene in a given pool of genes In terms of a population, it is the proportion in a given population of the loci having a given allelic series occupied by a particular gene It is a fraction of all genes at a given locus that are of specific type Gene Frequency The genetics of a population is concerned with genetic constitution of individuals and also with the transmission of genes from one generation to the next In this transmission, genotypes of parents are broken down and it is genes that are transmitted in the gametes New set of genotypes are therefore formed in the progeny Gene Frequency AA A Aa A AA A aa a Aa a Parents a aa in gametes Progeny Gene Frequency The genes carried in a population have continuity from generation to generation but not the genotypes in which they appear The genetic constitution of a population referring to genes it carries is described by Gene frequencies Gene frequencies can be determined from knowledge of genotype frequencies Calculation of gene frequency From chick genotypes BB Bb bb Total Individuals Genes 30 17 10 57 B 60 17 0 77 b 0 17 20 37 Calculation of gene frequency Each individual contains two genes at a locus In a population of N individuals, there are 2N loci for any allelic series For a locus with only 2 alleles (B and b) the frequency of B N B B b B 2N is the number of individuals in a population Calculation of gene frequency In the chick example There are 2N f(B) f(b) = 114 genes = 77 / 114 = 0.675 = 37 / 114 = 0.325 Gene frequency notation The following designations are used p q f(B) f(b) In a natural, random mating population, p + q = 1 Genotype frequency notation When the two parents mate and produce a zygote, the following genotype is observed p q p p2 pq q pq q2 Genotype frequency will be p2 2pq q2 Gene and Genotype frequencies In a population with natural forces p2 + 2pq + q2 p + q = = 1 or unity 1 or unity If genotype frequencies are known, gene frequencies can be estimated Gene Frequency = (homozygous genotype frequency + ½ 1 heterozygous f BB f 2 AB 2 genotype frequency) f B / Total Total Gene and Genotype frequencies Homozygous are BB (p2) and bb(q2) Heterozygous are Bb (pq) These appear in the ratio of 1:2:1 1 p 2 pq 2 2 p 2 p pq 2 p 2 pq q 2 Similarly, q q pq 2 Gene and Genotype frequencies Genotype frequency in subsequent generations depends on the gene frequency in the gametes of the previous generation Expected or predicted genotype frequency in the offspring is obtained by multiplying the gene frequency in the males by the gene frequency in the females An example A group of cocks produced 1000 sperms comprising 200 dwarf (dw) types, the rest being normal type (N) Then the frequency of dw 200 0.2 q 1000 And the frequency of N 800 0.8 p 1000 An example A group of hens produced 1000 eggs comprising 200 dwarf (dw) types, the rest being normal type (N) Then the frequency of dw 200 0.2 q 1000 And the frequency of N 800 0.8 p 1000 An example We can use these frequencies to predict genotype of possible offspring with respect to this trait 0.8N 0.2dw 0.8N 0.64NN 0.16Ndw 0.2dw 0.16Ndw 0.04dwdw So we would expect 0.64 NN, 0.32 Ndw and 0.04dwdw Reading for More examples • Please go through examples in Falconer, first Chapter dealing with genotype and gene frequencies Assumptions behind examples This assumes no mutations, migration and selection was taking place in the population Under such conditions, the population is said to be in Hardy-Weinberg equilibrium An exercise • For a herd of 100 Shorthorn cattle there are: Red Rhone White Genotype RR Rr rr Number 30 50 20 Phenotype frequencies the f(R) = p = and f(r) = q = Total 100 The Hardy – Weinberg Law In 1908, two scientists working independently G.H. Hardy (English Mathematician) W. Weinberg (German Physician and Geneticist) Developed a relationship between gene and genotype frequencies These developed a law that states that The Hardy – Weinberg Law In a large random – mating population with no forces disturbing the environment ie no selection No mutation No migration Random genetic drift Gene frequencies and genotype frequencies remain unchanging from generation to generation A population with constant gene and genotype frequencies is therefore said to be in Hardy – Weinberg equilibrium The Hardy – Weinberg Law In Hardy – Weinberg equilibrium, there is a simple relationship between gene and genotype frequencies If gene frequencies of two alleles among the parents are p and q, then the genotype frequencies in the progeny will be p2, 2pq, and q2 The Hardy – Weinberg Law Genes in parents A1 A2 p Frequencies q Male Female A1 p A2 q A1 p A1 A1 p2 A1 A2 pq A2 q A1 A2 pq A2 A2 q2 Progeny The Hardy – Weinberg Law Genotypes in parents A1 A1 A1 A2 A2 A2 Frequencies p2 2pq q2 Which is basically squaring the gamete array (p + q)2 p2 is the frequency of A1A1 genotype 2pq is the frequency of A1A2 genotype q2 is the frequency of A2A2 genotype The Hardy – Weinberg Law In a population in HW equilibrium, we may predict Gene frequency Genotype frequency Phenotype frequency Frequency of various mating Carriers of a certain recessive allele We use HW Law to predict mating of sires and dams Conditions to fulfil HW equilibrium The population is infinitely large There is no or at least very infrequent mutation There is no selection Mating is at random There is normal gene segregation Conditions to fulfil HW equilibrium There is equal fertility of parents and equal fertilising ability of gametes and reproduction is sexual Equal viability and equal gene frequencies in male and female parents Gene frequency is same in parents and offspring Generations are not overlapping Gene and Genotype frequencies See details in Falconer ~ check excel Gene frequency of b Genotype frequency 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 Relationship between genotype and gene frequencies for two alleles in a population in HW equilibrium BB Bb bb Testing a population in HW Equilibrium MM MN NN Total No observed 233 385 129 747 No expected 242.4 366.3 138.3 Frequency p2 2pq q2 •First calculate gene frequencies •p = 0.57 •q = 0.43 •To calculate expected frequencies, find frequency of each genotype, then multiply by total Testing a population in HW Equilibrium MM p 2 * Total 0.57 * 747 242.4 2 MN 2 pq * Total 2 * 0.57 * 0.43 * 747 366.3 NN q 2 * Total 0.43 * 747 138.3 2 • Chi-square goodness of fit method used to test whether the population we work with is in HW equilibrium or not Chi-square goodness of fit 2 Observed Expected 2 Expected • Under 1 d.f., X2 test statistic = 1.96 (from tables) • Larger test statistic than 1.96 means less likelihood that deviations occur simply due to chance deviations alone • That is, observed numbers are not in satisfactory agreement with expected numbers, hence population is not in HW Equilibrium Chi-square goodness of fit ~ check excel MM MN NN Total No observed 233 385 129 No expected 242.4 366.3 138.3 Frequency p 2 2pq q 747 2 Sum Chi-Square 0.3645 0.9546 0.6253 1.94