Hardy-Weinberg Equilibrium PDF

HARDY- WEINBERG EQUILIBRI UM 3-2 WHY SEX? GENETIC VARIATION Mutation is the source of all new genetic variation. Sexual reproduction shuffles alleles around in a population Quantifying evolution Evolution = change in change in allele frequencies over time Alleles that e...

HARDY- WEINBERG EQUILIBRI UM 3-2 WHY SEX? GENETIC VARIATION Mutation is the source of all new genetic variation. Sexual reproduction shuffles alleles around in a population Quantifying evolution Evolution = change in change in allele frequencies over time Alleles that enhance survival and reproduction become more common each generation Population genetics - the study of allele frequencies and changes in allele frequencies over time in populations Change in allele frequencies over time in populations causes phenotypic changes How to quantify evolution in natural populations? It is hard to measure changes in gene frequencies in many organisms. Some populations can be hard to sample: Or both: And can live for many years: How to quantify evolution in natural populations? 1. Calculate expected genotype Some populations can be hard to sample: frequencies when NO evolution is occurring (i.e., just random mating) 2. Sample a population of interest and see if the observed genotype frequencies are significantly different when no evolution is occurring. And can live for 3. If different, this is evidence many years: that the population is evolving. How to tell if evolution is occurring in nature? Organisms can be hard to sample: In other words, compare observed genotype frequencies to those expected when no evolution is occurring. And can live for many years: A population of 15 individuals GENOTYPES Homozygous- 2 of the same alleles – Homozygous dominant (AA)- 2 dominant alleles – Homozygous recessive (aa)- 2 recessive alleles Heterozygous (Aa)- different alleles Law of Dominance – The dominant allele will hide the presence of the recessive trait – Heterozygotes will appear the same as homozygous dominant A population of 15 individuals  A Each individual has two alleles, or variants, of each gene = genotype Homozygous The two alleles for A the trait are the AA aa same A Heterozygous The two alleles for Aa a A is dominant and a is the trait are recessive different A population has a frequency of genotypes AA Aa aa AA Aa Aa AA aa AA Aa AA AA aa AA AA Total number = 15, frequency of AA = Aa = aa = AA Aa aa AA Aa Aa AA aa AA Aa AA AA aa AA AA Total number = 15, frequency of AA = 8/15 (0.53) Aa = 4/15 (0.27) aa = 3/15 (0.20) AA Aa aa AA Aa Aa AA aa AA Aa AA AA aa AA AA The gene pool is the collective name for all the alleles in a population: Individuals have 2 alleles for each gene AA Aa aa For a single gene, the total number of alleles in the gene pool of a diploid organism = 2 x # individuals If there are 15 individuals in a population of rabbits how many TOTAL alleles (for a single gene) are in their gene pool? A. 2 B. 15 C. 30 D. Not enough information is given to answer the questions If there are 15 individuals in a population of rabbits how many TOTAL alleles (for a single gene) are in their gene pool? A. 2 B. 15 C. 30 (2 alleles per locus * 15 individuals = 30 alleles) D. Not enough information is given to answer the questions A population has a frequency of alleles in its gene pool AA Aa aa AA Aa Aa AA aa AA Aa AA AA aa AA AA Total number of alleles = 30, frequency of A= a= AA Aa aa AA Aa Aa AA aa AA Aa AA AA aa AA AA Total number of alleles = 30, frequency of A = 20/30 (0.67) a = 10/30 (0.33) AA Aa aa AA Aa Aa AA aa AA Aa AA AA aa AA AA How does evolution act on population gene and genotype frequencies? We will model the genetics of a population of organisms. With this population, we will: 1. Derive an equation that allows us to predict genotype frequencies when there is no evolution occurring – only random mating each generation. 2. Examine the effects of natural Echinus esculentus selection and other evolutionary forces on population gene and phenotype frequencies. For a population of urchins, consider one gene with two possible alleles that determine color, with one allele (R) for red dominant to the other (r) for white Each individual has two alleles for this gene and together they determine its color Homozygous Heterozygous Homozygous RR Rr rr  When sea urchins reproduce, they release eggs and sperm (gametes) into the water. Sperm and eggs randomly pair during fertilization. Each gamete contains a single allele for color. Literally, a gene pool! p = frequency of R allele in the gene pool q = frequency of r allele in the gene pool p+q=1 p = frequency of R allele Gene pool q = frequency of r allele p+q=1 When fertilization occurs, the gametes pair and form genotypes p = frequency of R allele Gene pool q = frequency of r allele p+q=1 RR Rr rr When fertilization occurs, the gametes pair and form genotypes WHAT HAVE YOU USED TO DEAL WITH PROBABILITIES IN BIOLOGY? PUNNETT SQUARES!! We can use Punnett squares to predict the genotype frequencies for a population in Hardy-Weinberg equilibrium.  R r R RR Rr r rR rr  p = frequency of R allele Gene pool q = frequency of r allele p+q=1 frequency of (RR) = probability of (R AND R) RR Rr rr The frequency of each genotype produced by mating is related to the probability of pairing the two alleles.  p = frequency of R allele Gene pool q = frequency of r allele p+q=1 freq (RR) = RR Rr rr The frequency of each genotype is related to the probability of pairing the two alleles.  p = frequency of R allele Gene pool q = frequency of r allele p+q=1 freq (RR) = prob (R) x prob (R) RR Rr rr The frequency of each genotype is related to the probability of pairing the two alleles.  p = frequency of R allele Gene pool q = frequency of r allele p+q=1 freq (RR) = prob (R) x prob (R) prob (R) = p RR Rr freq (RR) = p x p = p2 rr The frequency of each genotype is related to the probability of pairing the two alleles.  p = frequency of R allele Gene pool q = frequency of r allele p+q=1 freq (rr) = RR Rr rr The frequency of each genotype is related to the probability of pairing the two alleles.  p = frequency of R allele Gene pool q = frequency of r allele p+q=1 freq (rr) = prob (r) x prob (r) RR Rr rr The frequency of each genotype is related to the probability of pairing the two alleles.  p = frequency of R allele Gene pool q = frequency of r allele p+q=1 freq (rr) = prob (r) x prob (r) prob (r) = q RR Rr freq (rr) = q x q = q2 rr The frequency of each genotype is related to the probability of pairing the two alleles.  p = frequency of R allele Gene pool q = frequency of r allele p+q=1 freq (Rr) = RR Rr rr The frequency of each genotype is related to the probability of pairing the two alleles.  p = frequency of R allele Gene pool q = frequency of r allele p+q=1 freq (Rr) = prob (R) x prob (r) + prob (r) x prob (R) RR Rr rR rr The frequency of each genotype is related to the probability of pairing the two alleles.  p = frequency of R allele Gene pool q = frequency of r allele p+q=1 freq (Rr) = prob (R) x prob (r) + prob (r) x prob (R) RR Rr rR freq (Rr) = pq + qp = 2pq rr The frequency of each genotype is related to the probability of pairing the two alleles.  p = frequency of R allele Gene pool q = frequency of r allele p+q=1 freq(RR) + freq (Rr) + freq (rr) = 1 p2 + 2pq + q2 = 1 RR Rr rr The genotype frequencies all have to add to one. For a population of urchins, consider one gene with two possible alleles that determine color, with one allele (R) for red dominant to the other (r) for white Each individual has two alleles for this gene and together they determine its color Homozygous Heterozygous Homozygous RR Rr rr Based on this gene pool, what are the predicted genotype frequencies after fertilization? p= 0.5 q= 0.5 p2 = 2pq = q2 = Based on this gene pool, what are the predicted genotype frequencies after fertilization? p= 0.5 q= 0.5 p2 = 0.52 = 2pq = 2*0.5*0.5 = q2 = 0.52 = Based on this gene pool, what are the predicted genotype frequencies after fertilization? p= 0.5 q= 0.5 p2 = 0.52 = 0.25 2pq = 2*0.5*0.5 = 0.50 If each genotype (RR vs. Rr vs. rr) is equally likely to survive until reproduction, what do you think will happen to the allele and genotype frequencies in the next 2 generation? p = ??? 2pq = ??? q2 = ??? p= 0.5 p q= = ??? 0.5 q p2 = 0.52 = 0.25 = ??? 2pq = 2*0.5*0.5 = 0.50 If each genotype (RR vs. Rr vs. rr) is equally likely to survive until reproduction, then the allele frequencies and genotype frequencies will remain the 2 same each generation. p = 0.25 2pq = 0.50 q2 = 0.25 p= 0.5 p= q= 0.5 0.5 q= p2 = 0.52 = 0.25 0.5 2pq = 2*0.5*0.5 = 0.50  p = frequency of R allele Gene pool q = frequency of r allele p+q=1 freq(RR) + freq (Rr) + freq (rr) = 1 p2 + 2pq + q2 = 1 RR Rr rr Hardy-Weinberg equilibrium: Under random mating only, a population’s genotype frequencies will stay at these values – this is our expectation if our population is not evolving. How does evolution act on population gene and genotype frequencies? We will model the population genetics of a population of sea urchins. With this population, we will: 1. Derive an equation that allows us to predict genotype frequencies when there is no evolution occurring – only random mating each generation. Echinus esculentus 2. Examine the effects of natural selection and other evolutionary forces on population gene and Based on this gene pool, what are the predicted genotype frequencies after fertilization? p= 0.6 q= 0.4 p2 = 0.62 = 2pq = 2*0.6*0.4 = q2 = 0.42 = Based on this gene pool, what are the predicted genotype frequencies after fertilization? p= 0.6 q= 0.4 p2 = 0.62 = 0.36 2pq = 2*0.6*0.4 = 0.48 What happens to the gene pool if the white genotype (rr) has lower survival than the other two genotypes? p= 0.6 p q= = ??? 0.4 q p2 = 0.62 = 0.36 = ??? 2pq = 2*0.6*0.4 = 0.48 What happens to the gene pool if the white genotype (rr) has lower survival than the other two genotypes? p2 = ??? 2pq = ??? q2 = ??? p= 0.6 p= q= 0.8 0.4 q= p2 = 0.62 = 0.36 0.2 2pq = 2*0.6*0.4 = 0.48 What happens to the gene pool if the white genotype (rr) has lower survival than the other two genotypes? p2 = 0.64 2pq = 0.32 q2 = 0.04 p= 0.6 p= q= 0.8 0.4 q= p2 = 0.62 = 0.36 0.2 2pq = 2*0.6*0.4 = 0.48 PRACTICE PROBLEM In wild pigs, one gene controls ear posture. Having erect ears (E) is dominant to having droopy ears (e). You sample a population of pigs and find that the frequency of the e allele is 0.3. 1. What is the expected frequency of each genotype (EE, Ee, and ee) if the population is in Hardy-Weinberg equilibrium? PRACTICE PROBLEM In a population of 100 butterflies, brown wings (B) are dominant over white wings (b). You count 40 white butterflies. 1. What are the expected allele and genotype frequencies if the population is in Hardy-Weinberg equilibrium?