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.
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- Half of the worlds population eats rice at least twice a day. Much of this rice is grown in flooded conditions, and different strains of rice are tolerant (survive) or intolerant (die) under these conditions. Rice breeders used genetic crosses to test whether tolerance to flooding is a dominant trait. Researchers used three true-breeding flood-tolerant strains, FR143, BKNFR, and Kurk, and two true-breeding flood- intolerant strains, IR42 and NB, in the crosses. Results were obtained from three sets of crosses and are reported in the Table below: Results of cross of F1 to tolerant parent: F1 plants were crossed with the tolerant parent of the cross. Number of Plants Progeny Analyzed from Intolerant Tolerant Cross Alive Dead Total 1. F2 results of cross: IR42 FR13A 187 77 264 IR42 BKNFR 192 73 265 NB Kurk 142 52 195 2. Results of cross of F1 to intolerant parent: (F1 of IR42 FR13A) IR42 14 17 31 (F1 of IR42 BKNFR) IR42 15 10 25 (F1 of NB Kurk) NB 21 35 56 3. Results of cross of F1 to tolerant parent: (F1 of IR42 FR13A) FR13A 31 0 31 (F1 of IR42 BKNFR) BKNFR 28 0 28 (F1 of NB Kurk) Kurk 40 0 40 Do the data support the hypothesis that the tolerance trait is dominant? Justify your conclusion by explaining the results from each of the three sets of crosses in terms of genotypes and phenotypic ratios. Source: T. Setter et al. 1997. Physiology and genetics of submergence tolerance in rice. Annals of Botany 79:6777.In normal plants, the probability that an offspring of a heterozygous parent is heterozygous is 0.5. If the survival of heterozygous offspring differs from that of homozygous offspring, the probability that a surviving offspring is heterozygous may not be equal to 0.5. For the following values of the probability, write a discrete-time dynamical system for the fraction of heterozygous offspring over time, find the solution, and compute the fraction that will be heterozygous after ten generations. How does this compare with the fraction for a normal plant? The probability that an offspring is heterozygous is 0.6.A wide-ranging survey of Nicotonia growing in its natural environment recorded a variation in corolla length ranging from 12mm to 47mm with a variance of 36.5. Subsequently, collected seeds were grown in a greenhouse and it was found that the range was now very much lower with most plants having similar corolla lengths and the variance was now only 8.4. After the plants had grown to maturity and formed seed, seeds were collected from plants with either the shortest and or the longest corollas in the population and planted separately in the greenhouse. When flowers were formed it was found that the variance of the plants with the shortest flowers was now 4.2 while that of the flowers from the longest seeds had become 13.7 Calculate the values for heritability in the different groups of plants and explain why this difference may arise.
- A wide-ranging survey of Nicotonia growing in its natural environment recorded a variation in corolla length ranging from 12mm to 47mm with a variance of 36.5. Subsequently, collected seeds were grown in a greenhouse and it was found that the range was now very much lower with most plants having similar corolla lengths and the variance was now only 8.4. After the plants had grown to maturity and formed seed, seeds were collected from plants with either the shortest and or the longest corollas in the population and planted separately in the greenhouse. When flowers were formed it was found that the variance of the plants with the shortest flowers was now 4.2 while that of the flowers from the longest seeds had become 13.7 So,Calculate the new values for heritability in the different groups of plants and explain why this difference may arise.A wide-ranging survey of Nicotonia growing in its natural environment recorded a variation in corolla length ranging from 12mm to 47mm with a variance of 36.5. Subsequently, collected seeds were grown in a greenhouse and it was found that the range was now very much lower with most plants having similar corolla lengths and the variance was now only 8.4. After the plants had grown to maturity and formed seed, seeds were collected from plants with either the shortest and or the longest corollas in the population and planted separately in the greenhouse. When flowers were formed it was found that the variance of the plants with the shortest flowers was now 4.2 while that of the flowers from the longest seeds had become 13.7 Thus, calculate the new values for heritability in the different groups of plants and explain why this difference may arise.Flower color variation in Phlox drummondii living either in sympatry or allopatry with Phlox cuspidata has previously been studied. Phlox drummondii exist in four colors: light blue, dark blue, light red, and dark red. Each flower color is controlled by a specific combination of alleles. In allopatry, butterflies randomly pollinate Phlox drummondii flowers of all color types. In contrast, non-random pollination occurs between differently colored flowers when both Phlox species are living in sympatry. The graph depicts the relative fitness of the alleles that drive flower color in Phlox drummondii under conditions of sympatry or allopatry with Phlox cuspidata. Label each half of the graph with the term sympatry or allopatry based on the observed trend in the relative fitness of the alleles that control all four flower colors in Phlox drummondii.
- A population sample of 300 individuals is studied for the electrophoretic mobility of an enzyme that varies according to the genotype determined by 2 alleles, E and T of a single gene. The results are 7 individuals with genotype EE, 106 with genotype ET, and 187 with genotype TT. What are the allele frequencies of E and T, and what are the expected numbers of the 3 genotypes if random mating is assumed?In a natural population of Drosophila melanogaster, thealcohol dehydrogenase gene has two alleles calledF (fast) and S (slow) with frequencies of Adh-F at 0.75and Adh-S at 0.25. In a sample of 480 flies from thispopulation, how many individuals of each genotypicclass would you expect to observe under Hardy–Weinberg equilibrium?In the plant Lotus corniculatus, cyanogenic glycoside protects againstinsect pests and even grazing by cattle. The presence of this glycoside inan individual plant is due to a simple dominant allele. A population of L.corniculatus consists of 77 plants that possess cyanogenic glycoside and56 that lack the compound. What is the frequency of the dominant allele responsible for the presence of cyanogenic glycoside in this population?
- Two chromosome inversions are commonly found in populations ofDrosophila pseudoobscura: Standard (ST) and Arrowhead (AR). Whenthe flies are treated with the insecticide DDT, the genotypes for theseinversions exhibit overdominance, with the following fitnesses:Genotype FitnessST/ST 0.47ST/AR 1AR/AR 0.62What will the frequencies of ST and AR be after equilibrium has beenreached?The phenotypic data below are the shoot lengths of 25 F1 and 25 F2 rice plants atseedling stage, produced from the cross of IR29 and Hasawi rice varieties. The mean lengths of the shoot from IR29 and Hasawi are 23.1 cm and 46.7 cm, respectively. Tabulate and plot the frequency distributions of the F1 and F2 generations. From each distribution calculate the mean, the variance,and the standard error of the mean. What is the main difference between F1 and F2 distributions? Formulas needed:Range (R) = maximum – minimum (Use the same no. of decimal places as original data.)No. of phenotypic classes (K) = 1 + 3.3logn (Round up answer to an integer. Number may still be increased or decreased as needed.)n = total no. of valuesClass interval (CI) = R/K (Use the same no. of decimal places as original data.)In a certain population in the US, the frequency of CCR5-Δ32 homozygous individuals is 1%. Assuming genetic equilibrium, what is the frequency of heterozygotes in this population?