Find the mean height of a population of palm trees with the following information: Trees with heights of 140 feet are bred, and the average height of the progeny is 128 feet. The selection response, R, is 70, and the selection differential, S, is 100
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- Plant height is controlled by 5 gene pairs. The homozygous dominant is 200 cm tall while the homozygous recessive exhibits 80 cm height. Assume that the alleles have equal contribution and have cumulative effects. The F1 and F2 of the two parents below were studied. Parent 1 Parent 2 DID,D2D2D3D3dąd, dsds dıdıd2d2d3d3DąD4 D5D5 a. Compute for the contribution of the dominant allele. b. Compute for the height of parent 1 and parent 2. Provide the genotype and phenotype of the F1. d. Compute for the frequencies of the different individuals in the F2 using the Pascal's triangle. e. Provide the phenotypes of the F2 and their frequencies. С.The mean of plant height from two rice plants (P1 and P2) and their progeny (F1 and F2) and a backcross generation (P1 x F1) are shown below. Population Mean (in) P1 34.1 P2 65.2 F1 44.1 F2 46.2 BC1 39.6 Explain the possible reasons for the observed differences in the sample means. Account for the differences in the sample means of P1 and P2. Similarly, account for the differences in the sample means of the F1 and F2. Compare the difference in the parental generations with that in the filial generations.A gardener grows two different colors of apples under exactly the same environmental conditions. Red apples have a heritability of 0.14 for the value of apple weight, and green apples have a heritability of 0.41 for the value of apple weight. The gardener plotted the number of apples present at each weight. Which graph would you hypothesize represents the red apples? Number of apples Weight GRAPH A Number of apples Weight GRAPH B O Graph A O Graph B There is not enough information to make a hypothesis.
- To analyze: Imagine the length of the maize ears which has narrow sense heritability (h²) of 0.70 A population yields ears that have an average length of 28 cm, and a breeder selects a plant harvesting 0.70 cm ears from this population and cross by self-fertilization. Find the expected selection differential (S) and the response to selection (R) for this cross.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.Figure 19-16 shows the results of a QTL fine-mapping experiment. Which gene would be implicated as controlling fruit weight if the mean fruit weight for each linewas as follows?Line Fruit weight (g)1 181.42 169.33 170.74 171.25 171.46 182.27 180.68 180.79 181.810 169.3
- Give the genotype of the parents and determine the linkage map of the three genes bm (brown midrib), v (virescent seedling), and pr (purple aleurone) in maize from the results of the cross below. Genotypes of offspring Total and Frequency percentage V bm 230 467 pr + 237 42.1% + + bm 82 161 pr V 79 14.5% V 200 395 pr bm 195 35.6% pr bm 44 86 V + 42 7.8% +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.A new kind of tulip is produced that develops only purple or pink flowers. Assume that flower color is controlled by a single-gene locus and that the purple allele (C) is dominant to the pink allele (c). A random sample of 1000 tulips from a large cultivated field yields 847 purple flowers (out of which 476 are heterozygous) and 153 pink flowers. If p = frequency of C allele and q = frequency of the c allele. Calculate p and q. A. p = 0.4 q = 0.6 B. p = 0.61 q = 0.39 C. p = 0.847 q = 0.153 D. p = 0.82 q = 0.18
- Which of the following statements does NOT apply to the Hardy-Weinberg expression: p2 + 2pq + q2? Group of answer choices p2 is the frequency of individuals with the homozygous recessive genotype. 2pq is the frequency of individuals with the heterozygous genotype. It can be used to determine the genotype and allele frequencies of the previous and the next generations. Knowing either p2 or q2, you can calculate all the other frequenciesA 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?A species of butterfly shows variation in the length of the antennae. You measure antennae in a butterfly population and find the mean to be 15 mm. You calculate a heritability of antennae length to be 1 (h2 = 1). %3D That summer, there is a severe drought. Many of the butterflies die and only a few survive to reproduce. The surviving butterflies have a mean antennae length of 17 mm. a) What is the predicted mean limb-length of the offspring of the surviving butterflies? [ Select ] b) What type of selection was acting on antennae length in these butterflies? [ Select ] c) Five years later, you go back and measure antennae length in this butterfly population again. Every butterfly in the population has antennae measuring 16 mm. What is the heritability of antennae length in this population now? [Select ]