GENETIC ANALYSIS: AN INTEG. APP. W/MAS
2nd Edition
ISBN: 9781323142790
Author: Sanders
Publisher: Pearson Custom Publishing
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Chapter 5, Problem 22P
T. H. Morgan’s data on eye color and wing form, shown in Figures
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A large ear of corn has a total of 433 grains, including 271 Purple & starchy, 73 Purple & sweet, 63 Yellow & starchy, and 26 Yellow & sweet. Calculate the chi-squared value for the null hypothesis that the ear of corn was produced by a dihybrid cross (PpSs x PpSs) involving the null hypothesis that the two genes are not linked for color and sweetness. Explain the chi-square test results.
When a chi square analysis is applied to solve a linkage problem,explain why an independent assortment hypothesis is proposed?
Seed weight in a particular plant species is determined by pairs of alleles attwo loci (a+ a− and b+ b−) that are additive and equal in their effects. Plantswith genotype a− a− b− b− have seeds that average 1 g in weight, whereasplants with genotype a+ a+ b+ b+ have seeds that average 3.4 g in weight. Aplant with genotype a− a− b− b− is crossed with a plant of genotype a+ a+ b+b+.a. What is the predicted weight of seeds from the F1 progeny of this cross?b. If the F1 plants are intercrossed, what are the expected seed weights andproportions of the F2 plants?
Chapter 5 Solutions
GENETIC ANALYSIS: AN INTEG. APP. W/MAS
Ch. 5 - For parts a, b, and c, draw a diagram illustrating...Ch. 5 - 5.2 In a diploid species of plant, the genes for...Ch. 5 - A pure-breeding tall plant producing oval fruit as...Ch. 5 - 5.4 Genes E and H are syntenic in an experimental...Ch. 5 - In tomato plants, purple leaf color is controlled...Ch. 5 - 5.6 In Drosophila, the map positions of genes are...Ch. 5 - 5.7 Genes A, B, and C are linked on a chromosome...Ch. 5 - Gene G recombines with gene T at a frequency of...Ch. 5 - Genes A, B, C, D, and E are linked on a...Ch. 5 - Syntenic genes can assort independently. Explain...
Ch. 5 - The recombination frequency between linked genes...Ch. 5 - On the DrosophilaX chromosome, the dominant allele...Ch. 5 - Researchers cross a corn plant that is pure -...Ch. 5 - 5.14 syndrome is an autosomal disorder affecting...Ch. 5 - 5.15 Three dominant traits of corn seedlings,...Ch. 5 - 5.16 In a diploid plant species, an with the...Ch. 5 - Prob. 17PCh. 5 - The Rh blood group in humans is determined by a...Ch. 5 - 5.19 Genetic linkage mapping for a large number of...Ch. 5 - 20. Neurospora with the genotype form tetrads in...Ch. 5 -
Gene and gene are genetically linked. Answer...Ch. 5 -
T. H. Morgan’s data on eye color and wing form,...Ch. 5 - Prob. 23PCh. 5 -
The boss in your laboratory has just heard of a...Ch. 5 - In rabbits, chocolate-colored fur (w+) is dominant...Ch. 5 - Prob. 26PCh. 5 - 5.27 In tomatoes, the allele for tall plant height...Ch. 5 - 5.28 Neurofibromatosis is an autosomal dominant...Ch. 5 - A 2006 genetic study of a large American family...Ch. 5 - 5.30 A experiment examining potential genetic...Ch. 5 - 31. As part of their analysis of intragenic...Ch. 5 - In experiments published in 1918 that sought to...
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- 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. Compare the sample variances of P1 and P2. Account for any differences. Similarly, compare the sample variances of the F1 and F2 generations, and account for any differences. Give the possible causes of variation in each generation.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_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
- E. W. Lindstrom crossed two corn plants with green seedlings and obtained the following progeny: 3583 green seedlings, 853 virescentwhite seedlings, and 260 yellow seedlings . Q. Explain how color is determined in these seedlings.arrow_forwardWhen a chi square test is applied to solve a linkage problem, explain why an independent assortment hypothesis is proposed.arrow_forwardConsidering the Mendelian traits round versus wrinkledand yellow versus green, consider the crosses belowand determine the genotypes of the parental plants byanalyzing the phenotypes of their offspring.Parental Plants Offspring(a) round, yellow * round, yellow 3/4 round, yellow1/4 wrinkled, yellow(b) wrinkled, yellow * round, yellow 6/16 wrinkled, yellow2/16 wrinkled, green6/16 round, yellow2/16 round, green(c) round, yellow * round, yellow 9/16 round, yellow3/16 round, green3/16 wrinkled, yellow1/16 wrinkled, green(d) round, yellow * wrinkled, green 1/4 round, yellow1/4 round, green1/4 wrinkled, yellow1/4 wrinkled, greenarrow_forward
- In pea plants, a dihybrid for the recessive a and b is testerossed. The distribution of the phenotypes is as follows: Phenotype AB Number 138 a b 132 a B 69 A b 61 (i) Are the genes assorting independently? Test the hypothesis with a chi-square test. State your hypothesis, estimate the p value using the x table below and clearly state your conclusion. Table 1: Chi-square table Chi Square Values and Probability Degrees of Freedom P = 0.99 0.95 0.80 0.50 0.20 0.05 0.01 0.000157 0.00393 0.0642 0.455 1.642 3.841 6.635 2 0.020 0.103 0.446 1.386 3.219 5.991 9.210 3. 0.115 0.352 1.005 2.366 4.642 7.815 11.345 0.297 0.711 1.649 3.357 5.989 9.488 13.277 (ii) If the genes are linked, calculate the map distance between the genes in cM. '.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: Phenotype Number Kn + + 451 Kn f + 134 + + + 97 + f bz 436 Kn + bz 18 + + bz 119 + f + 24 Kn f bz 86 Total: 1,365 Determine the sequence (order) of the three genes and the calculate the distance between them.arrow_forwardWe think that dwarfism in river birch might be inherited as a single-locus, simple Mendelain trait. We have taken a pure breeding normal plant and crossed it with a a pure breeding dwarf plant. The resulting plants were self-pollinated with the following results: Normal River Birch 805 Dwarf River Birch 246 Perform a chi square analysis on this data using the following hypothesis: The mode of inheritance for plant height in the river birch is simple Mendelian. Calculate the chi square value without rounding and then round your final answer to three decimal digits and put it in the space provided.arrow_forward
- 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.)arrow_forwardA test cross between a plant of genotype PpSs and the tester white plant with wrinkled seed coat (ppss) gives the following numbers of progeny in four phenotypic types. 14:87:83:16 (purple flower + smooth seed coat: purple flower + wrinkled seed coat: white flower + smooth seed coat: white flower + wrinkled seed coat). a. What is the expected ratio of progeny phenotypes assuming independent assortment of alleles? b. Explain how ratios of progeny show that the two genes are linked. c. How many map units separate the purple and smooth genes? Show your calculations. d. What is the “parental” genotype of the heterozygous parent? (i.e. Which alleles of the P and S loci are present on each of the two chromosomes of the doubly heterozygous parent of this test cross?)arrow_forwardE. W. Lindstrom crossed two corn plants with green seedlings and obtained the following progeny: 3583 green seedlings, 853 virescentwhite seedlings, and 260 yellow seedlings . Q. Give the genotypes for the green, virescent-white, and yellow progeny.arrow_forward
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