Concepts of Genetics Plus Mastering Genetics with Pearson eText -- Access Card Package (12th Edition) (What's New in Genetics)
12th Edition
ISBN: 9780134811390
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino, Darrell Killian
Publisher: PEARSON
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Chapter 25, Problem 16PDQ
In an assessment of learning in Drosophila, flies were trained to avoid certain olfactory cues. In one population, a mean of 8.5 trials was required. A subgroup of this parental population that was trained most quickly (mean = 6.0) was interbred, and their progeny were examined. These flies demonstrated a mean training value of 7.5. Calculate realized heritability for olfactory learning in Drosophila.
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The narrow-sense heritability of wing length in a population of Drosophila melanogaster is 0.8. The narrow-sense heritability of head width in the same population is 0.9. The genetic correlation between wing length and head width is −0.86. If a geneticist selects for increased wing length in these flies, what will happen to head width?
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Chapter 25 Solutions
Concepts of Genetics Plus Mastering Genetics with Pearson eText -- Access Card Package (12th Edition) (What's New in Genetics)
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- In a population of Drosophila melanogaster reared in thelaboratory, the mean wing length is 0.55 mm and therange is 0.35 to 0.65. A geneticist selects a female withwings that are 0.42 mm in length and mates her with amale that has wings that are 0.56 mm in length.a. What is the expected wing length of their offspringif wing length has a narrow-sense heritability of 1.0?b. What is the expected wing length of their offspring ifwing length has a narrow-sense heritability of 0.0?arrow_forwardThe following table summarises the results of the 2022-2023 Drosophila three-point cross involving the loci white eyes (w), miniature wings (m) and singed bristles (sn). The data is also available in the Excel file 'Drosophila Counts 2022-2023'. We strongly recommend working in Excel during this exercise.Phenotype Count+ + + 584w m sn 324w + + 227+ m sn 150+ m + 134w + sn 196 + + sn 134 w m + 92The aim of today's tutorial is to use the above data to establish the genetic map for these three loci. We have been told that all three loci are on the X chromosome; however, as scientists, we shouldn't simply take someone else's word for it - we need to test whether our data does, in fact, indicate linkage.As a first step, we will therefore conduct a chi-square test in order to test our data against the null hypothesis of independent assortment (i.e., "no…arrow_forwardIn the nematode C. elegans, homozygosity for the e mutant allele causes an extreme "uncoordinated" phenotype, where the worm completely loses its ability to move. Examination of 100 individuals with genotype e/e reveals that 60 mutant worm can't move at all, 35 show a very reduced ability to move, and the remaining 5 seem to have a completely wild-type phenotype with respect to movement ability. These observations suggest that e has O incomplete penetrance and variable expressivity O low expressivity and variable penetrance Ohigh expressivity and variable penetrance O high penetrance and incomplete expressivityarrow_forward
- What is the genetic distance between the eye colour locus (w) and the bristle locus (sn)? answer in Mu (Map units) and to 2 decimal places? The following table summarises the results of the 2022-2023 Drosophila three-point cross involving the loci white eyes (w), miniature wings (m) and singed bristles (sn). The data is also available in the Excel file 'Drosophila Counts 2022-2023'. We strongly recommend working in Excel during this exercise.Phenotype Count+ + + 584w m sn 324w + + 227+ m sn 150+ m + 134w + sn 196 + + sn 134 w m + 92arrow_forwardWhat is the heritability of corolla length?arrow_forwardA 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. What is the heritability of corolla length?arrow_forward
- Do 9, 10 and 11arrow_forwardtrue for false? 1a) In Drosophila flies, there are wing shapes such as curly, apterous, miniature and others. These would be considered continuous traits (as opposed to discontinuous traits). 1b) The reason we assess the narrow sense heritability is because there are hidden genetic components that are simply too difficult to include in the calculation of VG 1c) You can figure out the heritability of a trait if you compare the variation of the trait in a clonal population of organisms to a wild type (heterogeneous) population.arrow_forwardGiven the following testcross data for corn in which the genes for fine stripe (f), bronze aleurone (bz) and knotted leaf (Kn) are involved: PhenotypeNumber Kn + +451 Kn f +134 + + + 97 + f bz436 Kn + bz18 + + bz119 + f +24 Kn f bz86 Total:1,365 Maize geneticists tend to be like Drosophila geneticists utilizing a + to indicate the wild type allele (which also is dominant) and a lower case letter for the mutant allele (which is recessive). The first thing I suggest in attacking this problem is to use the typical convention of upper case letters for dominant alleles and lower case letters for recessive alleles as follows: F = wild type, f = fine stripe; B = wild type, b = bronze aleurone K = wild type, k = knotted leaf Then we can rewrite the phenotypes in a “language” that is more easily understood: PhenotypeNumber k F B451 k f B134 K F B97 K f b436 k F b18 K F b119 K f B24 k f b86 Total:1,365 a) Determine the sequence (order) of the…arrow_forward
- In c. elegans, genetics model organism, movement problems (unc) and small body size (sma) are encoded by two mutant alleles that are recessive to those that produce wild-type traits (unc+ and sma+). A worm homozygous for movement problems and small body is crossed with a worm homozygous for the wild-type traits. The F1 have normal movement and normal body size. The F1 are then crossed with worms that have movement problems and small body size in a testcross. The progeny of this testcross is: Normal movement, normal body size 210 Movement problems, normal body size 9 Normal movement, small body size 11 Movement problems, small body size 193 a)From the test cross results, can you tell if the two genes are on the same chromosome or not? Explain your reasoning. b)What phenotypic proportions would be expected if the genes for round eyes and white body were located on different chromosomes? (please explain hot to get to these conclusions)arrow_forwardA 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 new values for heritability in the different groups of plants and explain why this difference may arise.arrow_forwardIn corn plants, a dominant allele (1) inhibits the expression of kernel color, while the recessive allele (1) promotes the expression of kernel color when homozygous. Another dominant allele, (P), causes purple kernel color, while the homozygous recessive genotype (pp) causes red kernels. Assuming that these alleles assort independently, predict the phenotypic ratio of the offspring if two plants dihybrid for these characteristics are crossed.arrow_forward
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