Essentials of Genetics (9th Edition) - Standalone book
9th Edition
ISBN: 9780134047799
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino
Publisher: PEARSON
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Chapter 5, Problem 13PDQ
An attached-X female fly, XXY (see the “Insights and Solutions” box), expresses the recessive X-linked white-eye
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An attached-X female fly, XX ¬Y (see the “Insights and Solutions” box), expresses the recessive X-linked white-eye mutation. It is crossed to a male fly that expresses the X-linked recessive miniature-wing mutation. Determine the outcome of this cross in terms of sex, eye color, and wing size of the offspring.
In Drosophila melanogaster, red eyes are dominant over white and the variation for this characteristic is on the X chromosome. Vestigial wings (v) are recessive to normal (V) for an autosomal gene. Predict the appearance of offspring of the following crosses: XW/XwV/v×Xw/Y v/v, Xw/XwV/v×XW/Y V/v.
In Drosophila melanogaster, red eyes are dominant over white and the variation for this characteristic is on the X chromosome. Vestigial wings (v) are recessive to normal (V) for an autosomal gene. Predict the appearance of offspring of the following crosses: XW/Xw V/v×Xw/Y v/v, Xw/Xw V/v × XW/Y V/v.
Chapter 5 Solutions
Essentials of Genetics (9th Edition) - Standalone book
Ch. 5 -
CASE STUDY | Doggone it!
A dog breeder...Ch. 5 - CASE STUDY| Doggone it! A dog breeder discovers...Ch. 5 - CASE STUDY| Doggone it! A dog breeder discovers...Ch. 5 -
CASE STUDY | Doggone it!
A dog breeder...Ch. 5 - HOW DO WE KNOW? In this chapter, we have focused...Ch. 5 - Review the Chapter Concepts list on p. 84. These...Ch. 5 - As related to sex determination, what is meant by...Ch. 5 -
4. Contrast the life cycle of a plant such as...Ch. 5 - Prob. 5PDQCh. 5 -
6. Describe the major difference between sex...
Ch. 5 - How do mammals, including humans, solve the...Ch. 5 -
8. What specific observations (evidence) support...Ch. 5 - Describe how nondisjunction in human female...Ch. 5 -
10. An insect species is discovered in which the...Ch. 5 -
11. Given your answers to Problem 10, is it...Ch. 5 - When cows have twin calves of unlike sex...Ch. 5 -
13. An attached-X female fly, XXY (see the...Ch. 5 -
14. Assume that on rare occasions the attached X...Ch. 5 - It is believed that any male-determining genes...Ch. 5 -
16. What is a Barr body, and where is it found...Ch. 5 - Indicate the expected number of Barr bodies in...Ch. 5 - Define the Lyon hypothesis.Ch. 5 - Can the Lyon hypothesis be tested in a human...Ch. 5 - Predict the potential effect of the Lyon...Ch. 5 -
21. Cat breeders are aware that kittens...Ch. 5 -
22. What does the apparent need for dosage...Ch. 5 - In mice, the Sry gene (see Section 5.2) is located...Ch. 5 - The genes encoding the red- and...Ch. 5 - In mice, the X-linked dominant mutation Testicular...Ch. 5 -
26. Shown here are graphs that plot the...Ch. 5 -
27. In chickens, a key gene involved in sex...
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- In Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn+ and ct+ is crossed to a sn ct male. The F1 flies are interbred. The F2 males are distributed as follows sn ct 36 sn ct+ 13 sn+ ct 12 sn+ ct+ 39 What is the map distance between sn and ct?arrow_forwardProduce a Punnett square to illustrate the dihybrid cross described below: There are two common alleles for the TAS2R38 gene on Chromosome 7. This gene encodes a seven-transmembrane G-protein coupled receptor. This receptor controls the ability to taste glucosinolates. Phenylthiocarbamide (PTC) is a synthetic glucosinolate. The recessive TAS2R38 allele produces a non-functional receptor. The father in this dihybrid cross is heterozygous for these alleles, meaning that he can taste PTC. The mother is homozygous recessive, meaning that she cannot taste PTC The father has X-Linked Protoporphyria which means that he is very sensitive to sunlight exposure, he is hemizygous for the dominant causative mutation. The mother is homozygous wild type at the same locus. Add a file here showing your diagram.arrow_forwardIn Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn and ct+ is crossed to a sn+ct male. The F1 flies are interbred. The F2 males are distributed as follows: genotype number sn ct 15 sn ct+ 34 sn+ ct 33 sn+ct+ 18 What is the map distance between sn and ct?arrow_forward
- Barr bodies are formed to adjust X chromosome dosage in species other than humans, such as Drosophila (fruit flies) and C. elegans (nematodes). True or False ?arrow_forwardIn fruit flies, the following X-linked traits are found: white eyes are recessive to red eyes, ebony body is recessive to gray body, and short wings is recessive to long wings. A cross was made between wild-type males with red eyes, long wings, and gray bodies and females with white eyes, short wings, and ebony bodies. Female heterozygote resulting from this cross, which had red eyes, long wings, and gray bodies, were then crossed with males with white eyes, short wings, and ebony bodies. The F2 generation data is obtained below: 1299 white eyes, short wings, ebony bodies 1367 red eyes, long wings, gray bodies 99 white eyes, short wings, gray body 89 red eyes, long wings, ebony bodies 49 white eyes, long wings, ebony bodies 49 red eyes, short wings, gray bodies 1 red eyes, short wings, ebony bodies 1 white eyes, long wings, gray bodies A) Calculate the map distance separating the three genes B) Which gene is in the middle?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, 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_forwardLet’s suppose that two different X-linked genes exist in mice,designated with the letters N and L. Gene N exists in a dominant,normal allele and in a recessive allele, n, that is lethal. Similarly,gene L exists in a dominant, normal allele and in a recessive allele,l, that is lethal. Heterozygous females are normal, but males thatcarry either recessive allele are born dead. Explain whether or notit would be possible to map the distance between these two genesby making crosses and analyzing the number of living and deadoffspring. You may assume that you have strains of mice in whichfemales are heterozygous for one or both genes.arrow_forward: In Drosophila, yellow body is due to an X-linked gene that is recessive to the gene forgray body.(a) A homozygous gray female is crossed with a yellow male. The F1 are intercrossed toproduce F2. Give the genotypes and phenotypes, along with the expected proportions, of theF1 and F2 progeny.(b) A yellow female is crossed with a gray male. The F1 are intercrossed to produce the F2.Give the genotypes and phenotypes, along with the expected proportions, of the F1 and F2progeny.(c) A yellow female is crossed with a gray male. The F1 females are backcrossed with graymales. Give the genotypes and phenotypes, along with the expected proportions, of the F2progeny.(d) If the F2 flies in part b mate randomly, what are the expected phenotypic proportions offlies in the F3??arrow_forward
- In Drosophila, the X-linked recessive mutation vermilion (v) causes bright red eyes, in contrast to the brick-red eyes of wild type. A separate autosomal recessive mutation, suppressor of vermilion (su-v), causes flies homozygous or hemizygous for v to have wildtype eyes. In the absence of vermilion alleles, su-v has no effect on eye color. Determine the F1 and F2 phenotypic ratios from a cross between a female with wild-type alleles at the vermilion locus, but who is homozygous for su-v, with a vermilion male who has wildtype alleles at the su-v locusarrow_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_forwardThe following X-linked recessive traits are found in fruit flies:vermilion eyes are recessive to red eyes, miniature wings are recessiveto long wings, and sable body is recessive to gray body. A cross wasmade between wild-type males with red eyes, long wings, and graybodies and females with vermilion eyes, miniature wings, and sablebodies. The heterozygous female offspring from this cross, whichhad red eyes, long wings, and gray bodies, were then crossed tomales with vermilion eyes, miniature wings, and sable bodies. Thefollowing data were obtained for the F2 generation (including bothmales and females):1320 vermilion eyes, miniature wings, sable body1346 red eyes, long wings, gray body102 vermilion eyes, miniature wings, gray body90 red eyes, long wings, sable body42 vermilion eyes, long wings, gray body48 red eyes, miniature wings, sable body2 vermilion eyes, long wings, sable body1 red eyes, miniature wings, gray bodyWhat information do you know based on the question and your…arrow_forward
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