Concepts of Genetics (11th Edition)
11th Edition
ISBN: 9780321948915
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino
Publisher: PEARSON
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Chapter 5, Problem 17PDQ
Drosophila females homozygous for the third chromosomal genes pink and ebony (the same genes from Problem 16) were crossed with males homozygous for the second chromosomal gene dumpy. Because these genes are recessive, all offspring were wild type (normal). F1 females were testcrossed to triply recessive males. If we assume that the two linked genes, pink and ebony, are 20 mu apart, predict the results of this cross. If the reciprocal cross were made (F1 males—where no crossing over occurs—with triply recessive females), how would the results vary, if at all?
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Chapter 5 Solutions
Concepts of Genetics (11th Edition)
Ch. 5 - In a family with one autistic child the risk for...Ch. 5 - Given that the prenatal test can provide only a...Ch. 5 - Prob. 3CSCh. 5 - Prob. 4CSCh. 5 - Consider two hypothetical recessive autosomal...Ch. 5 - With two pairs of genes involved (P/p and Z/z), a...Ch. 5 - In Drosophila, a heterozygous female for the...Ch. 5 - HOW DO WE KNOW? In this chapter, we focused on...Ch. 5 - Review the Chapter Concepts list on page 94. Most...Ch. 5 - Describe the cytological observation that suggests...
Ch. 5 - Why does more crossing over occur between two...Ch. 5 - Explain why a 50 percent recovery of...Ch. 5 - Why are double-crossover events expected less...Ch. 5 - What is the proposed basis for positive...Ch. 5 - What two essential criteria must be met in order...Ch. 5 - The genes dumpy (dp), clot (cl), and apterous (ap)...Ch. 5 - Colored aleurone in the kernels of com is due to...Ch. 5 - In the cross shown here, involving two linked...Ch. 5 - In a series of two-point mapping crosses involving...Ch. 5 - Two different female Drosophila were isolated,...Ch. 5 - In Drosophila, a cross was made between femalesall...Ch. 5 - Another cross in Drosophila involved the...Ch. 5 - In Drosophila, Dichaete (D) is a mutation on...Ch. 5 - Drosophila females homozygous for the third...Ch. 5 - In Drosophila, two mutations, Stubble (Sb) and...Ch. 5 - If the cross described in Problem 18 were made,...Ch. 5 - Are mitotic recombinations and sister chromatid...Ch. 5 - What possible conclusions can be drawn from the...Ch. 5 - An organism of the genotype AaBbCc was testcrossed...Ch. 5 - Based on our discussion of the potential...Ch. 5 - Traditional gene mapping has been applied...Ch. 5 - DNA markers have greatly enhanced the mapping of...Ch. 5 - In a certain plant, fruit is either red or yellow,...Ch. 5 - Two plants in a cross were each heterozygous for...Ch. 5 - A number of humanmouse somatic cell hybrid clones...Ch. 5 - A female of genotype produces 100 meiotic tetrads....Ch. 5 - In laboratory class, a genetics student was...Ch. 5 - Drosophila melanogaster has one pair of sex...Ch. 5 - In Drosophila, a female fly is heterozygous for...Ch. 5 - The gene controlling the Xg blood group alleles...Ch. 5 - Prob. 34ESP
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- In Drosophila, an X-linked recessive mutation, Xm causes miniature wings. List the F2 phenotypic ratios if: a miniature-winged female is crossed with a normal male and a miniature-winged male is crossed with a normal female. What would the phenotypic ratio from (a) be if the miniature-winged gene were autosomal? Assume in all cases that the P1 individuals are true-breeding.arrow_forwardThe recessive allele s causes Drosophila to have small wings, and the s+ allele causes normal wings. This gene is known to be X linked. If a small-winged male is crossed with a homozygous wild-type female, what ratio of normal to small-winged flies can be expected in each sex in the F1? If F1 flies are intercrossed, what F2 progeny ratios are expected? What progeny ratios are predicted if F1 females are backcrossed with their father?arrow_forwardThe genes for mahogany eyes and ebony body are approximately 18 map units apart on chromosome III in Drosophila. Assume that a mahogany-eyed female was mated to an ebony-bodied male, and the resulting F1 phenotypically wild-type females were mated to mahogany-ebony males. Of 942 offspring, what would be the expected phenotypes and in what numbers would they be expected?arrow_forward
- In Drosophila, males from a true-breeding stock with raspberry-colored eyes were mated to females from a true-breeding stock with sable-colored bodies. In the F1 generation, all the females had wild-type eye and body color, while all the males had wild-type eye color but sable-colored bodies. When F1 males and females were mated to each other, the F2 was composed of: 216 females with wild-type eyes and wild-type bodies 223 females with wild-type eyes and sable bodies 191 males with wild-type eyes and sable bodies 188 males with raspberry eyes and wild-type bodies 23 males with wild-type eyes and bodies 27 males with raspberry eyes and sable bodies Which statements are consistent with the above data? (Select all correct answers.) The alleles causing the raspberry-colored eye and sable-colored body phenotypes are dominant to the corresponding wild-type alleles The genes controlling raspberry-colored eyes and sable-colored bodies map…arrow_forwardIn Drosophila, ebony body colour is produced by a recessive gene a and wild-type (gray) body colour by its dominant allele a+. Vestigial wings are governed by a recessive gene vg, and normal wing size (wild type) by its dominant allele vg+. If wild-type dihybrid flies are crossed and produce 256 progeny, how many of these progeny flies are expected in each phenotypic class?arrow_forwardThe genes for the recessive traits of mahogany eyes and ebony body are approximately 30 map units apart on chromosome III in Drosophila. Assume that a mahogany-eyed female was mated to an ebony-bodied male and that the resulting F1 females (all phenotypically wild-type) were then mated to mahogany, ebony males. Of 1000 offspring, what would be the expected phenotypes, and in what numbers would they be expected? Group of answer choices None of these is correct wild-type = 200; mahogany and ebony = 200; mahogany = 300; ebony = 300 mahogany = 200; ebony = 200; wild-type = 300; mahogany and ebony = 300 wild-type = 350; mahogany and ebony = 350; mahogany = 150; ebony = 150 mahogany = 350; ebony = 350; wild-type = 150; mahogany and ebony = 150arrow_forward
- You have been given a virgin Drosophila female. You notice that the bristles on her thorax are much shorter than normal. You mate her with a normal male (with long bristles) and obtain the following F1 progeny1 3 short-bristled females, 1 3 long-bristled females, and 1 3 long-bristled males. A cross of the F1 long-bristled females with their brothers gives only long-bristled F2. A cross of short-bristled females with their brothers gives 1 3 short-bristled females, 1 3 long-bristled females, and 1 3 long-bristled males. Provide a genetic hypothesis to account for all these results, showing genotypes in every cross.arrow_forwardIn Drosophila, males from a true-breeding stock with raspberry-colored eyes (instead of normal brownish-red eyes) were mated to females from a true-breeding stock with sable- colored bodies (instead of normal brown bodies). In the F, generation, all the females had wild-type eye and body color, while all the males had wild-type eye color but sable-colored bodies. When F, males and females were mated, the F generation was composed of: 216 females with wild-type eyes and bodies 223 females with wild-type eyes and sable bodies 191 males with wild-type eyes and sable bodies 188 males with raspberry eyes and wild-type bodies 23 males with wild-type eyes and bodies 27 males with raspberry eyes and sable bodies. Explain these results by diagramming the crosses and calculating any relevant map distances.arrow_forwardIn Drosophila, a cross was made between a yellowbodied male with vestigial (not fully developed)wings and a wild-type female (brown body). The F1generation consisted of wild-type males and wild-typefemales. F1 males and females were crossed, and theF2 progeny consisted of 16 yellow-bodied males withvestigial wings, 48 yellow-bodied males with normalwings, 15 males with brown bodies and vestigialwings, 49 wild-type males, 31 brown-bodied femaleswith vestigial wings, and 97 wild-type females.Explain the inheritance of the two genes in questionbased on these results.arrow_forward
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