Concepts of Genetics (12th Edition)
12th Edition
ISBN: 9780134604718
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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Chapter 25 Solutions
Concepts of Genetics (12th Edition)
Ch. 25 - A homozygous plant with 20-cm-diameter flowers is...Ch. 25 - The following table shows measurements for fiber...Ch. 25 - The following cable gives the percentage of twin...Ch. 25 - At an interview with a genetic counselor, a couple...Ch. 25 - Prob. 2CSCh. 25 - At an interview with a genetic counselor, a couple...Ch. 25 - HOW DO WE KNOW? In this chapter, we focused on a...Ch. 25 - CONCEPT QUESTION Review the Chapter Concepts list...Ch. 25 - Define the following: (a) polygenic, (b) additive...Ch. 25 - A dark-red strain and a white strain of wheat are...
Ch. 25 - Height in humans depends on the additive action of...Ch. 25 - An inbred strain of plants has a mean height of 24...Ch. 25 - Erma and Harvey were a compatible barnyard pair,...Ch. 25 - In the following table, average differences of...Ch. 25 - What kind of heritability estimates (broad sense...Ch. 25 - List as many human traits as you can that are...Ch. 25 - Corn plants from a test plot are measured, and the...Ch. 25 - The following variances were calculated for two...Ch. 25 - The mean and variance of plant height of two...Ch. 25 - Prob. 14PDQCh. 25 - Prob. 15PDQCh. 25 - In an assessment of learning in Drosophila, flies...Ch. 25 - Prob. 17PDQCh. 25 - Prob. 18PDQCh. 25 - In a population of 100 inbred, genotypically...Ch. 25 - Many traits of economic or medical significance...Ch. 25 - A 3-inch plant was crossed with a 15-inch plant,...Ch. 25 - In a cross between a strain of large guinea pigs...Ch. 25 - Type A1B brachydactyly (short middle phalanges) is...Ch. 25 - In a series of crosses between two true-breeding...Ch. 25 - Students in a genetics laboratory began an...Ch. 25 - Prob. 26ESPCh. 25 - Canine hip dysplasia is a quantitative trait that...Ch. 25 - Floral traits in plants often play key roles in...Ch. 25 - In 1988, Horst Wilkens investigated blind...
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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_forwardExample 3.5.7 Fish Vertebrae Consider the distribution of vertebrae given in Table 3.5.1. In Exam- ple 3.5.5 we found that the mean of Y is µy = 21.49. The variance of Y is VAR(Y) = of = (20 – 21.49) x Pr{Y = 20} 21) %3D + (21 – 21.49)? x Pr{Y + (22 – 21.49)? x Pr{Y = 22} + (23 – 21.49) x Pr{Y = 23} = (-1.49) x 0.03 + (-.49) x 0.51 + (0.51) x 0.40 + (1.51)² × 0.06 = 2.2201 x 0.03 + .2401 x 0.51 + .2601 x 0.40 + 2.2801 x 0.06 = 0.066603 + 0.122451 + 0.10404 + 0.136806 = 0.4299. The standard deviation of Y is ay = V0.4299 - 0.6557. %3Darrow_forwardIn the F2 generation, 306 rabbits with red eyes and 71 with a white eye phenotype suppose the calculated x2 value is 0.35. Find the x2 range using the distribution chart. What is the p-value range? using these information do you accept or reject the null hypothesis? The distribution chart is attached below.arrow_forward
- 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.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 So，Calculate the new values for heritability in the different groups of plants and explain why this difference may arise.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 Thus, calculate the new values for heritability in the different groups of plants and explain why this difference may arise.arrow_forward
- true 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_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_forwardin a randomly mating laboratory population of Drosophila, 4% of the flies have black bodies, and 96% have brown bodies. Flies with brown bodies always have at least one parent with a brown body. What is the frequency of heterozygotes in this population? 0.04 0.16 0.08 0.32 The period gene of Drosophila melanogaster encodes for a stretch of Thr-Gly repeated in tandem. In natural populations, the three most common alleles encode for 17, 20 and 23 Thr-Gly repeats. The amplification by PCR of the allele encoding for 20 Thr-Gly repeats produces a fragment of 320 bp. Using the same set of primers, what is the size expected when amplifying the 17 Thr-Gly allele? 317 303 314 302 in a certain species of plant loci A, B and C have an additive effect on the colour of the flower. Alleles A, B, and C are dominant and alleles a, b and c are recessive. Knowing that a plant with genotype AAbbCc has a…arrow_forward
- The following data for the genotypes at the alcohol dehydrogenase locus were observed from a sample of Drosophila melanogaster. Sample size 1000 FF 550 FF=0.55; FS=0.34; SS=0.11 O F-0.72; S=0.28 FF=0.078; FS=0.40; SS=0.52 FS What are the Hardy-Weinberg equilibrium genotype frequencies for each genotype based upon this sample? OFF=0.52; FS=0.40; SS=0.078 340 SS 110arrow_forwardFlower 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.arrow_forwardAssume Hardy-Weinberg conditions. In the American Caucasian population approximately 70% of people can taste the chemical phenylthiocarbamide (PTC) (the dominant phenotype), while 30% are non-tasters (the recessive phenotype). Determine the expected frequency of: Round up your answer to the nearest 100th. If your answer is 0.4, put 0.4. Both 0.40 and .4 will be graded as incorrect answers. homozygous recessive phenotype (q²) a) b) the recessive allele (q) c) the dominant allele (p) d) homozygous tasters (p²) e) heterozygous tasters (2pq)arrow_forward
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