Please refer to the table below. If only data on the variances for F1, F2 and BC1 populations were available, how will the broad sense heritability estimate change? Show solutions. 1. A soybean breeder is aiming to breed for early maturation. He was able to cross variety A which has desirableagronomic characteristics but late maturation with variety B which has less-than-ideal agronomic characteristicsbut early maturation. In addition to the F₁ population, F2 and BC1 population were also generated. The variancesfor each population are presented below.P₁3.3 days²PopulationVarianceP24.7 days²F₁6.6 days²F219.2 days²BC₁12.1 days²

Answered: Image /qna-images/answer/9b899379-a4ae-47d1-8cb1-91cda8db83fb.jpg

Answered: If only data on the variances for F1,…

Science

Biology

If only data on the variances for F1, F2 and BC1 populations were available

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter12: Mendel, Genes, And Inheritance

Section: Chapter Questions

Problem 2ITD

See similar textbooks

Similar questions

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 corn breeder wishes to improve yield. As part of their work, the narrow-sense heritability for corn yield is calculated to be 0.3. If the average yield in the starting population is 400 and the breeder selects for breeding plants with an average yield of 500, what will be the expected average yield among the offspring of the selected plants? A. 530 B. 330 C. 130 D. 430 E. 230
A 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…
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.
Population A consists of 100 hens that are fully isogenicand that are reared in a uniform environment. Theaverage weight of the eggs they lay is 52 g, and the variance is 3.5 g2. Population B consists of 100 geneticallyvariable hens that produce eggs with a mean weight of52 g and a variance of 21.0 g2. Population B is raised in anenvironment that is equivalent to that of Population A.What is the environmental variance (Ve) for egg weight?What is the genetic variance in Population B? What is thebroad-sense heritability in Population B?
A large, genetically heterogeneous group of tomato plants was usedas the original breeding stock by two different breeders, named Maryand Hector. Each breeder was given 50 seeds and began an artificialselection strategy, much like the one described in the figure. Theseeds were planted, and the breeders selected the 10 plants with thehighest mean tomato weights as the breeding stock for the next generation. This process was repeated over the course of 12 growing seasons, and the following data were obtained: Mean Weight of Tomatoes (pounds)Year Mary’s Tomatoes Hector’s Tomatoes1 0.7 0.82 0.9 0.93 1.1 1.24 1.2 1.35 1.3…
A genetically diverse population (Population 1) of monkeyflowers is growing on an alpine slope. The seeds produced by 100 of these plants are counted and the mean and variance are calculated. The variance is 25. One plant is selected and cuttings are taken from it. These cuttings are cultivated in the greenhouse, producing many genetically identical clones of the same plant. These clones (Population 2) are then transplanted into the alpine population, allowed to grow for a year, and then the numbers of seeds produced by each of the cloned plants is counted. The variance in seed number among these cloned plants is 10. Assume that the alpine slope environment does not vary across the years of the experiment, and that there is no genotype-by-environmental variation for seed number. What is the broad-sense heritability for seed number for Population 1?
1- If additive genes contribute to all of the observed phenotypic variation for a quantitative characteristic, then what must be true? a-Narrow-sense heritability for this characteristic will be equal to 0.5 b-The mean phenotype of the parental generation will be equal to the phenotype of the individuals in F1 c-Broad-sense-heritability for this characteristic will be greater than narrow-sense heritability. 2- How can a geneticist find and map quantitative trait loci (QTLs)? a-By dividing the covariance by the product of the standard deviations of traits x and y b-By determining whether different genotypes for a known DNA marker also show variation in a quantitative characteristic c-By determining the selection response "R" and the selection differential "s" d-By calculating the genetic variance for a quantitative from the phenotypic and environmental variance asap please.
A graduate student is studying a population of bluebonnets along a roadside. The plants in this population are genetically variable. She counts the seeds produced by each of 100 plants and measures the mean and 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 same plant. She then transplants these clones into the roadside population, allows them to grow for one year, and then counts the seeds produced by each of the cloned plants. The student finds that the variance in seed number among these cloned plants is 5. From the phenotypic variances of the genetically variable and the genetically identical plants, she calculates the broad-sense heritability. a. What is the broad-sense heritability of seed number for the roadside population of bluebonnets?
John, noted acarologist (he studies mites) and fancy humming bird breeder hobbyist, is attempting to breed giant puffing humming bird that he can present at the fancy balls given in his glittering social circle. After years of experimenting with humming bird, John has found that the slope of the least squares regression of Midoffspring on Midparent values is 0.75. All the humming birds were allowed to breed, and a first batch of offspring was produced. The table below presents data on the puffing ability (in cm3) of 10 breeding families of humming bird in his dovecote: Family midparent mid offspring (first batch) 1 55 56 2 68 72 3 77 81 4 37 40 5 25 30 6 34 32 7 44 44 8 23 49 9 60 39 10 67 68 All the humming birds were allowed to breed, and a first batch of offspring was produced. These offspring were used to calculate the midoffspring values above, and when John calculated the slope of the least squares regression of Midoffspring on Midparent values, he…
John, noted acarologist (he studies mites) and fancy humming bird breeder hobbyist, is attempting to breed giant puffing humming bird that he can present at the fancy balls given in his glittering social circle. After years of experimenting with humming bird, John has found that the slope of the least squares regression of Midoffspring on Midparent values is 0.75. All the humming birds were allowed to breed, and a first batch of offspring was produced. The table below presents data on the puffing ability (in cm3) of 10 breeding families of humming bird in his dovecote: Family midparent mid offspring (first batch) 1 55 56 2 68 72 3 77 81 4 37 40 5 25 30 6 34 32 7 44 44 8 23 49 9 60 39 10 67 68 John was disappointed in the puffiness of the first batch of offspring, so he shipped all these offspring to market. John desperately wanted to breed an even bigger, puffier humming bird, so for a second batch of offspring, he let only certain humming bird breed.…
The mean and standard deviation 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. Complete the table by calculating the variances and coefficient of variation for each population and answer the questionswhich follow (See image) 1. Explain the possible reasons for the observed differences in the sample means. Account forthe differences in the sample means of P1 and P2. Similarly, account for the differences in thesample means of the F1 and F2. Compare the difference in the parental generations with thatin the filial generations.2. Interpret the CV values from each population.3. Compare the sample variances of P1 and P2. Account for any differences. Similarly, comparethe sample variances of the F1 and F2 generations, and account for any differences. Give thepossible causes of variation in each generation.4. Calculate the broad-sense heritability of plant height in this species. Interpret your results.