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7-1 Biology 1001 Lab 7: POPULATION GENETICS PREPARTION Read this exercise before you come to the laboratory. Review the lecture notes from October 15 (Hardy-Weinberg Equilibrium) OBECTIVES At the end of this lab you should be able to: 1. Explain and define the terms population genetics, genetics, diploid, gene, genotype, phenotype, allele, homozygote, and heterozygote. 2. Explain the difference between dominant and recessive alleles and how these affect the phenotype (or survival) of an organism. 3. Apply the Hardy-Weinberg principle to populations and properly use the Hardy-Weinberg equation. MATERIALS (per laboratory team) Calculator 1 500-ml beaker 100 light-colored beads 100 dark-colored beads 10 toothpicks INTRODUCTION Population genetics is the study of allele frequency distribution and change under the influence of four main evolutionary processes: 1) natural selection; 2) genetic drift; 3) mutation and 4) gene flow. In other words, population genetics focuses on the genetic composition of a population and how it changes with time. Genetics is the science of genes, heredity and variation in living organisms. Inheritance in organisms occurs by means of discrete traits called genes. In a diploid organism (an organism with paired chromosomes) two homologous chromosomes (i.e. two chromosomes with genes for the same characteristic at the same location or loci) are inherited from the parents. One chromosome is from the mother and one from the father. This allows for the possibility of two alleles (alternate versions of a gene or two different forms of the DNA sequence) of each gene in an organism. Different DNA sequences (alleles) can result in different traits, such as blue, brown or hazel eye color. An organism’s genotype is its set of alleles (i.e. its genetic makeup). Observable traits in an organism are its phenotype. If an organism receives two copies of the same allele then this organism is homozygous at this gene locus. If an organism receives two different alleles of a given gene then it is heterozygous at this gene locus. When organisms are heterozygous at a gene locus, often one allele is called dominant as its qualities dominate the phenotype of the organism, 7-2 while the other allele is called recessive as its qualities recede and are not observed. For example, in humans the allele for brown eyes (indicated by B) is dominant over the blue allele (indicated by b). Therefore, BB homozygotes and Bb heterozygotes express a brown-eyed phenotype while bb homozygotes express a blue-eyed phenotype. Co-dominance also occurs where the contributions of both alleles (in heterozygous individuals) are visible in the phenotype of an organism. The Hardy-Weinberg principle states that both allele and genotype frequencies in a population remain constant from generation to generation assuming certain conditions are met. Requirements for Hardy-Weinberg Equilibrium: 1. Very large population size 2. No gene flow (no movement of genetic material between populations) 3. No mutations 4. Random mating 5. No natural selection The Hardy-Weinberg equation can be found below. RESEARCH QUESTION How do population size and natural selection influence gene frequencies? EXERCISE 1: Modeling Genetic Drift 1. Select 10 toothpicks to represent a population of 10 animals. 2. Place 50 dark-colored beads and 50 light-colored beads into a beaker. These represent genes in a gamete pool with allelic frequencies 0.5 A (p) and 0.5 a (q). Stir up the beads to ensure random genetic recombination. 3. Calculate expected frequencies of AA, Aa, and aa individuals, based on Hardy-Weinberg expectations. Multiply these expected frequencies by 10 and round off to the nearest whole numbers to calculate how many AA, Aa and aa individuals to expect in your population. Record these numbers in DATA TABLE 1 (pg 7-5) p+q=1 p2+2pq+q2 =1 Where p=the frequency of allele A q=the frequency of allele a Therefore, the frequency of AA individuals is p2, the frequency of aa individuals is q2 and the frequency of Aa individuals is 2pq. 4. Without looking at the beaker, draw out two beads and slip them onto a toothpick. If you draw two dark beads, this animal has genotype AA. Two light beads means aa. One of each means Aa. Repeat this process, placing two beads on each of the 10 toothpicks. How do your actual genotypic frequencies compare to the expected frequencies calculated in step 3? Can you explain the reason for any discrepancies? 7-3 5. Calculate gene frequencies in your population by counting “genes” in your population of 10 animals: p = f(A) = (# dark beads)/20, and q = f(a) = (# light beads)/20. Record your results for p and q in DATA TABLE 2 (pg 7-5). 6. Take all of the beads off your 10 toothpicks, and generate a new gamete pool by changing the ratio of light and dark beads in the beaker to match the p and q values you calculate in step 5. Keep the size of the gamete pool constant at 100 beads, but take out or put in beads to create the correct proportions. For example, if your population has values p = 0.55 and q – 0.45, you would fill the beaker with 55 dark-colored beads and 45 light-colored beads. Since you started with 50 of each, return all beads to the beaker, then take out five light-colored beads and replace them with five dark-colored ones to make a new p = 55/100 and q = 45/100. 7. Repeat steps 4 through 6 a total of 10 times to simulate 10 generations of genetic drift. Record p and q values on the calculation page each time you recalculate them. Since departures from original gene frequencies are random, you cannot predict which way genetic drift will take your small population. If p becomes 1.00, we say that the A gene has gone to fixation and the a gene to extinction. Fixation of the a gene could also happen. ASSIGNMENT: Based on this exercise answer questions 1 and 2 on page 7-7. EXERCISE 2: Modeling Selection In this simulation, we will assume that a is a recessive lethal gene. It causes no harm in heterozygotes (Aa), but is fatal to aa individuals. Rather than a random change in gene frequency, here we will see how natural selection against one allele can make a directional change in gene frequencies. 1. Select 10 toothpicks to represent a population of 10 animals as you did in the first simulation. 2. Begin the simulation, as you did in Exercise 1, with 50 dark-colored beads and 50 lightcolored beads in the beaker. These represent genes in a gamete pool with allelic frequencies 0.5 A and 0.5 a. Stir up the beads to ensure random genetic recombination. 3. Draw two beads to place on each toothpick as you did before. Look at the resulting genotypes. In this simulation, assume that aa genotypes produce a fatal genetic illness. Since A is dominant, Aa individuals do not have this disease. Record frequencies for all three genotypes in DATA TABLE 3 (pg 7-6), then remove the aa toothpicks to represent selection against the aa genotype. 4. Calculate p and q for the surviving population, record the p and q values in DATA TABLE 4 (pg 7-6). This time, p = f(A) = (# dark beads)/(all beads in surviving animals), and q = f(a) = (# light beads)/(all beads in surviving animals). 7-4 5. Remove beads from toothpicks and refill the beaker with dark and light beads to generate a gamete pool of 100 total beads, in the proportion of p and q that you just calculated. Mix the beads thoroughly. 6. Repeat steps 3 through 5 for 10 generations, record the p and q values each time. ASSIGNMENT: Based on this exercise answer questions 3-5 on page 7-8. 7-5 DATA TABLE 1: Exercise 1- Expected and observed genotypic frequencies Generation Frequency of AA Expected Observed Frequency of Aa Expected Observed 1 (initial generation) 2 3 4 5 6 7 8 9 10 DATA TABLE 2: Exercise 1- Gene frequencies Generation 1 (initial generation) 2 3 4 5 6 7 8 9 10 (f)A =p 0.5 (f)a =q 0.5 Frequency of aa Expected Observed 7-6 DATA TABLE 3: Exercise 2- Genotypic frequencies Generation 1 (initial generation) 2 3 4 5 6 7 8 9 10 Frequency of AA Frequency of Aa DATA TABLE 4: Exercise 2- Gene frequencies Generation 1 (initial generation) 2 3 4 5 6 7 8 9 10 Frequency of A (p) 0.5 Frequency of a (q) 0.5 Frequency of aa 7-7 ASSIGNMENT QUESTIONS (Handed in for marking) 1. In horses, the genes for white coat color and red coat color are codominant. Heterozygotes have a light red coloration, called roan. If you located a population of wild mustangs in a valley that had 476 red horses (AA), 323 roan horses (Aa), and 51 white horses (aa), could you say the population is in Hardy-Weinberg equilibrium? First calculate p and q, then use the HardyWeinberg formula to calculate expected genotypic frequencies. Show all your calculations! 2. Based on this exercise, you can explain why zoos go to so much trouble to exchange rare animals such as tigers or gorillas rather than maintaining small separate breeding populations? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ 7-8 3. Most serious genetic diseases are caused by recessive alleles rather than dominant alleles. Based on this exercise, can you explain why a recessive lethal gene could persist in the population, while a dominant lethal gene could not? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ 4. Suppose your model of selection represents the change in frequency of plain white coquina clams because predatory birds see and remove them more quickly from a beach with dark-colored sand. If larval offspring from this population drift to a different beach made of light-colored sand, could selection go the other way? What does this say about fitness of a particular gene? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ 5. In this simulation, you sampled the gene pool without replacing beads in the beaker after you drew each one. Thus, f(A) and f(a) in the gene pool changed slightly after each bead was drawn. For example, if you begin with 50 light and 50 dark beads, the probability of drawing a dark bead the first time is 50/100 = 0.500. The beaker would then contain 49 dark beads and 50 light beads, so the probability of drawing a second dark bead becomes 49/99 = 0.495. Does this make your simulation slightly less realistic? In small natural populations, does one mating change the gene pool available for the next mating, or not? What biological factors must be considered in answering this question? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________