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 18PDQ
Summary Introduction
To determine: The mean weight of the progeny if tomato plants have mean fruit weight of 60g, h2 is 0.3, and fruit average mean value of parents is 80g.
Introduction: Heritability is the measure of the extent to which genes are inherited from parents. It is a narrow sense heritability value obtained by selective breeding. By measuring the response of selective breeding in the offspring, one can estimate the value of realized heritability.
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In a population of tomato plants, mean fruit weight is 60 g and h2 is 0.3. Predict the mean weight of the progeny if tomato plants whose fruit averaged 80 g were selected from the original population and interbred.
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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.)
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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- 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% +arrow_forwardThe 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. Interpret the CV results from each population.arrow_forwardThe phenotypic data below are the shoot lengths of 25 F1 and 25 F2 rice plants at seedling 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. Use the formulas below in the computations. From each distribution calculate the mean, the variance, and the standard error of the mean. Round off answers to two decimal places. What is the main difference between F1 and F2 distributions?arrow_forward
- a. 1 dominant allele will contribute 120/10 = 12 cm to the base height of the plant.b. The height of the parent plant 1 Genotype of the parent plant 1 – D1D1D2D2D3D3d4d4d5d5 The height of the parent plant 2 Genotype of the parent plant 2 – d1d1d2d2d3d3D4D4D5D5Contributing alleles – D4D4D5D5. The height of the plant without any contributing alleles would be 80 cm. The plant with genotype d1d1d2d2d3d3D4D4D5D5 has 4 contributing allele each of which contributes 12 cm to the base. Hence, the height of the plant with genotype d1d1d2d2d3d3D4D4D5D5 would be 80 + 12 + 12 + 12 + 12 = 128 cm. c. Parents – D1D1D2D2D3D3d4d4d5d5 × d1d1d2d2d3d3D4D4D5D5 Gametes – D1D2D3d4d5 × d1d2d3D4D5 F1 generation – D1d1D2d2D3d3D4d4D5d5 The height of the plants of F1 generation = 80 + 12 + 12 + 12 + 12 + 12 = 140 cm Hence, Genotype of the F1 = D1d1D2d2D3d3D4d4D5d5 Phenotype of…arrow_forwardComplete the punnett square below for a cross between AA and aa A 21 a QUESTION 4 A Aa Complete the punnett square below for a cross between Aa and Aa. 03 Aa Aaarrow_forwardA horticulturist runs a test cross with an offspring (F1 generation) purple plant from Question 8. The phenotypic frequencies of the resulting offspring are 50% white and 50% purple. What is the true genotype of this offspring (F1 generation) purple plant?arrow_forward
- Give only typing answer with explanation and conclusion to all parts Two true-breeding varieties of maize, one 11 cm high and the other 47 cm high were crossed and the resultant F1 hybrids were then crossed to generate the F2 . In the F2 there were a total of 13,923 plants with a continuous variation in heights between the two extremes and with only 3 plants as large as 47 cm high and 5 plants of 11 cm high. a) How many i) genes and ii) how many alleles are involved in determining height in this plant? b) What is the contribution of each dominant allele to the phenotype in cm?arrow_forwardA researcher crosses two tall plants, however, one quarter (1/4) of the offspring from this crossing have short stems. Complete the punnett square. 6 b EBE Bb Besides Bb, what is the other possible genotype for a tall plant? What is the genotype of a short plant? Bb bbarrow_forwardWhat genotypic ratio would you expect to see.if you self-fertilized a plant that was heterozygous for two traits? 9:3:3:1 3:1 9:6:1 It is impossible to predict this based on the information givenarrow_forward
- Consider a corn plant dihybrid for the traits of purple and smooth kernels. When allowed to self-fertilize, 489 offspring are produced. How many of these offspring are expected to exhibit the traits of yellow and smooth kernels?arrow_forwardUse the following information to answer the next question. A Venn Diagram Showing the Relationship Between Oogenesis and Spermatogeneis. Oogenesis 1 NOTE: Similarities are represented by area 2. Differences are represented by areas 1 or 3. Match the numbered regions with the descriptions below. Spermatogenesis 3 Four viable gametes produced: Occurs in gonads: Daughter cells equal in size Unequal cytoplasmic division ▶arrow_forwardIn a genetic cross between a homozygous tall plant with homozygous purple flowers and a homozygous short plant with heterozygous purple flowers, how many short plants will you expect in the F2 generation if you sample 36 individuals? Both tall and purple are dominant.arrow_forward
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