1. Use the observed genotype frequencies from Day 7 data to calculate the frequencies of the C allele (p) and the C' allele (a). (Remember that the frequency of an allele in a gene pool is the number of copies of that allele divided by the total number of copies of all alleles at that locus.)
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- You are studying an isolated population of irises on an island off the coast of Maine. Plants in this population exhibit three different phenotypes for stem hair. ShSh plants are hairy, SsSs plants are smooth, and ShSs plants have intermediate amounts of stem hair. After characterizing stem phenotypes for a sample of the population, you classify 450 plants as having hairy stems, 701 as having intermediate hairy stems, and 112 as having smooth stems. Is the population in Hardy-Weinberg Equilibrium with respect to stem hair? ___YES______NO # Justify your answer. Edit View Insert Format Tools Table Paragraph✓ BI U 12pt v 80 F3 $ 000 000 F4 % F5 <In a population of the annual, self-compatible, Ipomoea purpurea, allele frequencies at a neutral genetic marker with two alleles are p=0.7; q=0.3 in generation 1. Assume this population in generation 1 was initially in Hardy Weinburg equilibrium. In this year, pollinators are absent and all plants self-fertilize, thus producing only self-fertilized seeds. Is the population in this second generation (i.e the offspring) still in Hardy Weinburg equilibrium? Show your work or explain your answer.A botanist studying water lilies in an isolated pond observedthree leaf shapes in the population: round, arrowhead, and scalloped.Marker analysis of DNA from 125 individuals showed theround-leaf plants to be homozygous for allele r1, while the plantswith arrowhead leaves were homozygous for a different allele atthe same locus, r2. Plants with scalloped leaves showed DNA profileswith both the r1 and r2 alleles. Frequency of the r1 allele wasestimated at 0.81. If the botanist counted 20 plants with scallopedleaves in the pond, what is the inbreeding coefficient F forthis population?Entomologists at the New York State Department of Agriculture are interested in determiningthe connection between pest insects infesting crop plants with populations of the same insect-infesting native plants in natural habitats. Long-term trapping and monitoring studies have estimated that on average 3% of the populations move between habitats (farm to natural and vis versa) each generation. A new insecticide resistance allele (∆K) has begun to increase in frequency in agricultural populations. A genotyping survey at this locus of 50 individuals in each population has revealed the following genotype counts:4a. Based on the effects of migration alone, what will the frequency of ∆K be in the forestpopulation in the next generation? 4b. If migration was acting in here without selection, what would the frequency of ΔK be in the agricultural population in the next generation? 4c. If the natural forest population was substantially larger than the agricultural population, how might…A graduate student is studying a population of bluebonnets along aroadside. The plants in this population are genetically variable. Shecounts the seeds produced by each of 100 plants and measures the meanand variance of seed number. The variance is 20. Selecting one plant,the student takes cuttings from it and cultivates them in a greenhouse,eventually producing many genetically identical clones of the sameplant. She then transplants these clones into the roadside population,allows them to grow for one year, and then counts the seeds produced byeach of the cloned plants. The student finds that the variance in seednumber among these cloned plants is 5. From the phenotypic variancesof the genetically variable and the genetically identical plants, shecalculates the broad-sense heritability.Small population size causes genetic drift because ofchance sampling of different alleles from one generation to the next. We can predict how much geneticdrift occurs for a given population size using binomialsampling statistics. With a population of size N, wecan estimate that 95% of the time the allele frequency(p) in the next generation will be withinthe confidence interval of p ± 1.96 (√p(1 − p)2N ),where p(1 − p)2Nis an estimate of the statistical variancein allele frequencies from one generation to the nextwith random sampling of 2N alleles each generation. a. What is the confidence interval for p = 0.5 whenN = 100,000?b. What is the confidence interval for p = 0.5 whenN = 10?c. How are the results in parts (a) and (b) related tothe consequences of a population bottleneck?Let’s suppose that weight in a species of mammal is polygenic, andeach gene exists as a heavy and light allele. If the allele frequenciesin the population are equal for both types of alleles (i.e., 50%heavy alleles and 50% light alleles), what percentage of individualswill be homozygous for the light alleles in all of the genesaffecting this trait, if the trait was determined by the followingnumber of genes?A. TwoB. ThreeC. FourA large, genetically heterogeneous group of tomato plants was usedas the original breeding stock by two different breeders, namedMary and Hector. Each breeder was given 50 seeds and began aselective breeding strategy, much like that described in Figure24.11. The seeds were planted, and the breeders selected the 10plants with the highest mean tomato weights as the breeding stockfor the next generation. This process was repeated over the courseof 12 growing seasons, and the following data were obtained: A. Explain these results.B. Another tomato breeder, named Martin, got some seeds fromMary’s and Hector’s tomato strains (after 12 generations),grew the plants, and then crossed them to each other. The mean weight of the tomatoes in these hybrids was about 1.7 pounds.For a period of 5 years, Martin subjected these hybrids to thesame selective breeding strategy that Mary and Hector had followed,and he obtained the following results: Explain Martin’s data. Why was Martin able to obtain…For a locus with two alleles (B and b) in a population at riskfrom an infectious neurodegenerative disease, 16 individualshave the genotype BB, 92 individuals have Bb and 12individuals have bb. Use the Hardy-Weinberg equation todetermine whether this population appears to be evolving.Calculate the frequencies of the AA, Aa and aa genotypes after one generation if the initialpopulation consists of 0.2 AA, 0.6 Aa and 0.2 aa genotypes and meets the requirements ofthe Hardy-Weinberg relationship. What genotype frequencies will occur after a secondgeneration?The sd gene causes a lethal disease of infancy in humanswhen homozygous. One in 100,000 newborns die eachyear of this disease. The mutation rate from Sd to sd is2 × 10−4. What must the fitness of the heterozygote be toexplain the observed gene frequency in view of the mutation rate? Assign a relative fitness of 1.0 to Sd /Sd homozygotes. Assume that the population is at equilibriumwith respect to the frequency of sd.Please refer to the Chi squared step by step document for an example to follow. 4a. A large random sample taken from a natural population of flowering plants found 300 plants with red flowers (RR genotype), 150 with pink flowers (Rr genotype) and 38 white flowers (rr genotype). Is the population in HWE? Step 1 Calculate the frequencies of alleles R and r. Include the formula for calculating allele frequencies in a population. Ans: Step 2 Based on the allele frequencies how many individuals are expected to a certain genotype? Calculate by multiplying the number of individuals in population (n) x frequencies of alleles for that genotype: Ans: T Step 3 Fill in the Table Below Genotype Ans: RR Rr rr Observed Expected Obs - Exp Step 3 Use the following formula to calculate the Chi squared value., Note calculate for each genotype and get the sum. (Obs - Exp)² Exp ³x²³ = [ (O - E)² E Step 4 Determine the Chi Square value for 1 Degree of Freedom and for the probability 0.05 from the table…SEE MORE QUESTIONS