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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Textbook Question
Chapter 8, Problem 17PDQ
Drosophila may be monosomic for chromosome 4, yet remain fertile. Contrast the F1 and F2 results of the following crosses involving the recessive chromosome 4 trait, bent bristles:
- (a) monosomic IV, bent bristles × diploid, normal bristles;
- (b) monosomic IV, normal bristles × diploid, bent bristles.
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Drosophila may be monosomic for chromosome 4, yet remain fertile.Contrast the F1 and F2 results of the following crosses involvingthe recessive chromosome 4 trait, bent bristles:(a) monosomic IV, bent bristles * normal bristles;(b) monosomic IV, normal bristles * bent bristles.
A mutant sex-linked trait called “notched” (N) is deadly in Drosophila when homozygous in females. Males who have a single N allele will also die. The heterozygous condition (Nn) causes small notches on the wing. The normal condition in both male and females is represented by the allele n.
a) Indicate the phenotypes of the F1 generation from the following cross: XNXn x XnY
b) Explain why dead females are never found in the F1 generation no matter which parents are crossed.
c) Explain why the mating of female XNXn and a male XNy is unlikely.
In Drosophila,, the curled mutation (cu, chromosome 3, position 50.0) results in wings that curl up,
while ebony (e, chromosome 3, position 70.7) results in a dark body. True breeding, wild type females
are mated with true breeding males with curled wings and ebony bodies.
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Chapter 8 Solutions
Concepts of Genetics (11th Edition)
Ch. 8 - A human female with Turner syndrome (47, X) also...Ch. 8 - When two plants belonging to the same genus but...Ch. 8 - What is the effect of a rare double crossover (a)...Ch. 8 - Prob. 1CSCh. 8 - Prob. 2CSCh. 8 - Aquatic vegetation overgrowth, usually controlled...Ch. 8 - HOW DO WE KNOW? In this chapter, we have focused...Ch. 8 - Review the Chapter Concepts list on page 171....Ch. 8 - Define these pairs of terms, and distinguish...Ch. 8 - For a species with a diploid number of 18,...
Ch. 8 - What evidence suggests that Down syndrome is more...Ch. 8 - What evidence indicates that humans with aneuploid...Ch. 8 - Contrast the fertility of an allotetraploid with...Ch. 8 - Describe the origin of cultivated American cotton.Ch. 8 - Predict how the synaptic configurations of...Ch. 8 - Inversions are said to suppress crossing over. Is...Ch. 8 - Contrast the genetic composition of gametes...Ch. 8 - Human adult hemoglobin is a tetramer containing...Ch. 8 - Discuss Ohnos hypothesis on the role of gene...Ch. 8 - What roles have inversions and translocations...Ch. 8 - The primrose, Primula kewensis, has 36 chromosomes...Ch. 8 - Certain varieties of chrysanthemums contain 18,...Ch. 8 - Drosophila may be monosomic for chromosome 4, yet...Ch. 8 - Mendelian ratios are modified in crosses involving...Ch. 8 - Having correctly established the F2 ratio in...Ch. 8 - The mutations called bobbed in Drosophila result...Ch. 8 - The outcome of a single crossover between...Ch. 8 - A couple planning their family are aware that...Ch. 8 - In a cross in Drosophila, a female heterozygous...Ch. 8 - A woman who sought genetic counseling is found to...Ch. 8 - Prob. 25ESPCh. 8 - In a recent cytogenetic study on 1021 cases of...Ch. 8 - A boy with Klinefelter syndrome (47,XXY) is born...Ch. 8 - Prob. 28ESPCh. 8 - Prob. 29ESPCh. 8 - A 3-year-old child exhibited some early indication...Ch. 8 - A normal female is discovered with 45 chromosomes,...Ch. 8 - Prob. 32ESP
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- In Drosophila,, the curled mutation (cu, chromosome 3, position 50.0) results in wings that curl up, while ebony (e, chromosome 3, position 70.7) results in a dark body. True breeding, wild type females are mated with true breeding males with curled wings and ebony bodies. Considering Drosophila notation, which of the following correctly diagrams the F1 cross? X X 3+ cu e + X X e + + + + + cu e + O + ■ 3+ X X X X Y Y + + ■ cu cu cu ' + ■ cu ■ ' + e + e e e e e + cu +arrow_forwardAssume that the trihybrid cross AABBrr x aabbRR is made in a plant species. Assume that A and B are dominant alleles, but there is no dominance effect of alleles at the R locus. a) How many different gametes are possible in the F1generation? What are the genotypes of these gametes? b) What is the probability of the parental aabbRR genotype in the F2 progeny? c) What proportion of the F2 progeny would be expected to be homozygous for all three genes?arrow_forwardIN 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_forward
- In corn, the genes v (virescent seedlings), pr (red aleurone), and bm (brown midrib) are all on chromosome 5, but not necessarily in the order given. The cross: v+ pr bm/ v pr+ bm+ with v pr bm/ v pr bm produces 1000 progeny with the following phenotypes: v+ pr bm 226 v pr+ bm+ 229 v+ pr bm+ 153 v pr+ bm 185 v+ pr+ bm 59 v pr bm+ 71 v+ pr+ bm+ 36 v pr bm 41 What is the gene order, and the (b) genetic map of these three genes?arrow_forwardIn corn, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of corn plants (i.e. the tassel) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male-fertile corn plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male-sterile lines. Using the following color-coded circles, simulate the crosses indicated below. Put the illustrations of crosses in the spaces provided. Be sure to include in the labels the genotypes and phenotypes of the offspring in each cross. Big light green circle - male-sterile cytoplasm Big orange circle - male-fertile cytoplasm Small orange circle - FF nucleus Small half-light green-half-orange circle - Ff nucleus Small light-green circle - ff nucleusarrow_forwardFemales heterozygous for the recessive second chromosome mutations px, sp, and cn are mated to a male homozygous for all three mutations. The offspring are as follows: Must show all work 1.) What gene is in the middle? and what are the map distances between the genes in centimorgans? 2.) What is the interference?arrow_forward
- In Drosophila, the brown mutation (bw, chromosome 2, position 104.5) results in brown eyes, while miniature (min, chromosome X, position 36.1) results in wings that are 2/3 the length of wild type. True breeding, wild type females are mated with true breeding males with brown eyes and miniature wings. Using Drosophila notation, diagram the P1 and F1 crosses. P1 F1 Fill in the chart with phenotypic ratios that would be expected in the F2 generation. Use the space provided to show your work. Phenotype Females Males Overall (♀and ♂) =1 =1 =1arrow_forwardThe mutant genes for vestigial wings and singed bristles are approximately 30 map units apart on chromosome II in Drosophila. Assume that a vestigial-winged female was mated to a singed-bristle male, and that the resulting F1 phenotypically wild type females were mated to vestigial singed males. Of 1000 offspring, which phenotype class would represent the product of crossing over between the genes, and how many would you expect? a) vestigial, 300 flies b) vestigial, 150 flies wild type, 300 flies d) singed, 150 flies e) vestigial, singed double mutants, 150 fliesarrow_forwardIn a cross between a white-eyed female (ww) and a red-eyed male (w+Y), nearly all the progeny were either red-eyed females (w+w) or white-eyed males (wY). However, about 1 in every 2000 F1 flies had an "exceptional phenotype" and was either a white-eyed female or red-eyed male. How did Bridges explain this unexpected result? A) Crossing over B) Incomplete cytokinesis C) Incorrect synapsis D) Nondisjunction E) Pseudoautosomal regionarrow_forward
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Mitochondrial mutations; Author: Useful Genetics;https://www.youtube.com/watch?v=GvgXe-3RJeU;License: CC-BY