Group 2: Snakes were placed on an island where there is only once size of burrow available for the snakes to live in. Number of Snake at Each Length 2cm 4cm бст 8cm 10сm Year 1980 4 12 16 10 3 1984 10 18 3 1988 6 24 2. 1. Graph the data and paste your graph here. Make sure and include title, x and y labels, units and a key to indicate which color represents each year.
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- The table below shows four human populations (CLM, MXL, PEL, PUR). A single nucleotide polymorphism with alleles A and G has been genotyped in each of these populations. For each genotype, the frequency is shown, and the number of individuals with that genotype in the sample is in brackets. What is the frequency of the A allele in the CLM population? 0.80 0.76 0.88 0.84Which of the following populations is not in Hardy-Weinberg equilibrium? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a. A population with 23 homozygous recessive individuals (yy), 7 homozygous dominant individuals (YY), and 4 heterozygous individuals (Yy) b. A population that receives new individuals from a normally distant population. c. q + p = 1 d. A population in which the allele frequencies do not change over time1. Did both populations (with and without natural selection) conform to Hardy-Weinberg equilibrium? What happened to observed allele frequencies in each population? Only answer question 1 below is an information about the question . Data result of Testing Hardy-Weinberg Equilibrium with natural selection Chi-square of results from bean model for F1: a. Total of (obs-exp)2/exp = Chi- square value for F1 = 3.1 The resulting chi-squared value is 3.1. b. For a p-value of 0.05 and 2 degrees of freedom, the critical value is 5.99. Our value of 3.1 is smaller than the critical value (5.99), Thus, we cannot reject the null hypothesis. c. The population is in Hardy-Weinberg Equilibrium, and is not evolving or It does conform to null hypothesis. Chi-square of results from bean model for F2: a. Total of (obs-exp)2/exp = Chi- square value for F2 = 6.5 The resulting chi-squared value is 6.5. b. For a p-value of 0.05 and 2 degrees of freedom, the critical value is 5.99. Our value of…
- Which parameter from the software must you adjust in order to give a "selection advantage" to organisms with particular phenotypes? Select one. 1. Starting frequency of allele A1 2. Fitness of genotype A1A1 3. Mutation rate from A1 to A2 4. Fraction migrants each generation 5. Population size 6. Inbreeding coefficient (F)1. Which parameter from the software you are using in lab this week must you adjust in order to decrease the population size of the organisms in the simulation? 2. Which parameter from the software you are using in lab this week must you adjust in order to change the frequency of the A2 allele in the simulation? 3. If you wanted to simulate a population that experiences no natural selection pressures in AlleleA1, you would:Write true if the statement is true or false if the statement is false. 1. As recently as 200 years ago, many people believed that Earth was only 6,000 years old. 2. Artificial selection occurs when nature selects for beneficial traits. 3. The individual Galápagos Islands are all similar to each other. 4. Malthus argued that human populations grow faster than their resources. 5. Lamarck was one of the first scientists to propose that species evolve by natural selection. 6. Lyell was one of the first to say that Earth must be far older than most people believed. 7. Lamarck's inheritance of acquired characteristics is has become a widely accepted scientific theory. 8. Fossils proved to Darwin that species can evolve. 9. The term fitness to refer to an organism's ability to outrun its hunters. 10. Darwin published his findings soon after returning to England from the voyage of the Beagle. 11. According to Darwin, natural selection is what occurs, and evolution is how it happens. 12.…
- According to the Hardy-Weinberg Equilibrium (p2 + 2pq + q2 = 1, within 5% variation of 1) which of the genes in the following three populations is actively undergoing evolution? population 1: AA = 80%; Aa = 10%; aa = 10% population 2: AA = 55%; Aa = 35%; aa = 10% population 3: AA = 25%; Aa = 50%; aa = 25% the gene in populations 1 and 3 are undergoing evolution the gene in populations 1, 2, and 3 are undergoing evolution the gene only in population 1 is undergoing evolution the gene in none of the populations are undergoing evolutionIn smurfs, blue tails and red tails are codominant to white tails (this is similar to blood type in humans). In a population of 200 smurfs, 6 have white tails, 93 have blue tails, 51 have red tails, and 50 have purple tails. Papa smurf has learned that the frequency of the blue tail allele is 0.53. Is this population in Hardy-Weinberg Equilibrium? Be sure to do a Chi-square test and show your work.Consider a set of genotypes with fitnesses: AA = 1.12 Aa = 1.04 aa = 1.0 Where the frequency of the "A" allele is 0.4. A. What is delta p, the change in the allele frequency to the next generation? B. What is the new frequency of the "A" allele after one generation of selection? (Note: provide both values to the nearest 0.0001)
- 1. Would it have made a difference if natural selection had acted on the dominant allele instead of the recessive one? Why or why not Only answer question 1 below is an information about the question. Data result of Testing Hardy-Weinberg Equilibrium with natural selection Chi-square of results from bean model for F1: a. Total of (obs-exp)2/exp = Chi- square value for F1 = 3.1 The resulting chi-squared value is 3.1. b. For a p-value of 0.05 and 2 degrees of freedom, the critical value is 5.99. Our value of 3.1 is smaller than the critical value (5.99), Thus, we cannot reject the null hypothesis. c. The population is in Hardy-Weinberg Equilibrium, and is not evolving or It does conform to null hypothesis. Chi-square of results from bean model for F2: a. Total of (obs-exp)2/exp = Chi- square value for F2 = 6.5 The resulting chi-squared value is 6.5. b. For a p-value of 0.05 and 2 degrees of freedom, the critical value is 5.99. Our value of 6.5 is greater than the critical…3. a. Why are most populations not in Hardy Weinberg Equilibrium? b. There is an ancient village population of humans. We know very little about this population. How can we use genetics to determine if the societal system of the village was matrilocal or patrilocal? c. There was variation in a rat phenotype (coat color.) The coat colors ranged from dark color coats to light color coats. As the rats migrated into the basements of campus, the light color rats were more likely to be caught by the campus cat. Over time, the basement rat population shifted to entirely dark color coats. This is an example of what kind of selection? d. Explain how a genetic bottleneck could lead to higher susceptibility of a disease, such cancer, in that population.Five conditions are required to maintain the Hardy–Weinberg equilibrium in a population. Closed population Large population Random mating No net mutations No natural selection If any of these conditions is not being met, the allele frequencies in the population will change, leading to microevolution in the population.Match each of the following scenarios to the Hardy–Weinberg condition that is NOT being met:An increase in antibiotic resistance among bacteria exposed to antibiotics occurs. AnswerCaribou from one herd move to a new area and breed with caribou of a completely different herd. Answer Among eastern bluebirds, more brightly coloured males breed with more brightly coloured females. AnswerDue to overhunting, there is little genetic diversity in the current population of bearded vultures, which have all descended from a population of only 36 birds. Answer