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
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Textbook Question
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?
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Drosophila females heterozygous for three recessive mutations, a, b, and c , were crossed to males homozygous for all three mutations.The cross yielded the following results:
in the image
Q. Construct a linkage map showing the correct order of these genes and estimate the distances between them.
In Drosophila, a cross was made between females—all expressing the three X-linked recessive traits scute bristles (sc), sable body (s), and vermilion eyes (v)—and wild-type males. In the F1, all females were wild type, while all males expressed all three mutant traits. The cross was carried to the F2 generation, and 1000 offspring were counted, with the results shown in the following table. Phenotype Offspring sc s v 314 + + + 280 + s v 150 sc + + 156 sc + v 46 + s + 30 sc s + 10 + + v 14 No determination of sex was made in the data. (a) Using proper nomenclature, determine the genotypes of the P1 and F1 parents. (b) Determine the sequence of the three genes and the map distances between them. (c) Are there more or fewer double crossovers than expected? (d) Calculate the coefficient of coincidence. Does it represent positive or negative interference?
Females of wild-type Strain A and males of mutant Strain B, as well as females of mutant Strain B and males of wild-type Strain A, make reciprocal crosses. Explain why reciprocal crosses are needed in genetics experiments involving Drosophila fruit flies.
Chapter 5 Solutions
Concepts of Genetics (12th 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 - 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
Additional Science Textbook Solutions
Find more solutions based on key concepts
Figure 1.18 In the example below, the scientific method is used to solve an everyday problem. Which part in the...
Concepts of Biology
What is the difference between histology and radiography?
Human Anatomy (8th Edition)
Why are mutants used as test organisms in the Ames test?
Laboratory Experiments in Microbiology (11th Edition)
How does trandlation differ from transcription?
Microbiology: Principles and Explorations
Some people compare DNA to a blueprint stored in the office of a construction company. Explain how this analogy...
Biology: Concepts and Investigations
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- Female Drosophila heterozygous for three recessive mutations e (ebony body), st (scarlet eyes), and ss (spineless bristles) were testcrossed, and the following progeny were obtained: Phenotype Number wild-type 67 ebony 8 ebony, scarlet 68 ebony, spineless 347 ebony, scarlet, spineless 78 scarlet 368 scarlet, spineless 10 spineless 54 (a) What indicates that the genes are linked? (b) What was the genotype of the original heterozygous females? (c) What is the order of the genes? (d) What is the map distance between e and st? (e) Between e and ss? (f) What is the coefficient of coincidence? (g) Diagram the crosses in this experiment.arrow_forwardThe phenotype of crooked wings (cw) in Drosophila melanogaster is caused by a recessive mutant gene that independently assorts with a recessive mutant gene for hairy (h) body. Assume that a cross is made between a fly with normal wings and a hairy body and a fly with crooked wings and normal body hair. All F1 flies from this cross were wild-type, and these flies were crossed among each other to produce 288 F2 offspring. Which phenotypes would you expect among the offspring in the F2 generation, and how many of each phenotype would you expect?arrow_forwardIn Drosophila, the gene loci for curved wings and purple eyes are 20 centimorgans (map units) apart from each other on an autosomal chromosome. Wild-type flies were mated with double mutant flies with cv cv p p genotypes and curved wing/purple eye phenotypes. AlIl of the F1 flies had wild type phenotypes. The F, flies were crossed with double mutant flies with cv cv p'p genotypes and curved wing/purple eye phenotypes. What are the predicted percentages of flies in the F2 generation with wild type wings and eyes, curved wings and wild type eyes, wild type wings and purple eyes, and curved wings and purple eyes?arrow_forward
- In Drosophila the genes forked bristles (f and f+) and vermilion eyes (v and v+) are ~24 m.u. apart on the X chromosome. From a cross of f v / f+ v+ females with f v / y males, what % of the progeny do you expect to have the f+ v+ phenotypes? (Don’t let the X chromosome throw you off – this question is essentially asking what % of the female’s gametes will be f+ v+).arrow_forwardThe phenotype of vestigial (short) wings (vg) in Drosophila melanogaster is caused by a recessive mutant gene that independently assorts with a recessive gene for hairy (h) body. Assume that a cross is made between a fly that is homozygous for normal wings and has a hairy body and a fly with vestigial wings that is homozygous for normal body. The wild-type F1 flies were crossed among each other to produce 1024 F2 offspring. Which phenotypes would you expect among the F2 offspring, and how many of each phenotype would you expect? Group of answer choices 192 wild type, 256 vestigial, 64 hairy, and 192 vestigial and hairy All vestigial and hairy. 576 wild type, 192 vestigial, 192 hairy, and 64 vestigial and hairy All wild type 256 wild type; 256 vestigial, 256 hairy, and 256 vestigial and hairyarrow_forwardIn a cross in Drosophila, a female heterozygous for the autosomallylinked genes a, b, c, d, and e (abcde/ + + + + +) was testcrossedwith a male homozygous for all recessive alleles (abcde/abcde).Even though the distance between each of the loci was at least3 map units, only four phenotypes were recovered, yielding thefollowing data: Phenotype No. of Flies+ + + + + 440a b c d e 460+ + + + e 48a b c d + 52 Total = 1000 Why are many expected crossover phenotypes missing? Can anyof these loci be mapped from the data given here? If so, determinemap distances.arrow_forward
- Black body (b) and purple eye (pr) are recessive autosomal mutations in Drosophila. Bridges are crossed b/b females with pr/pr males and in the F2 observed 684 wild type, 371 black-bodied, and 300 purple-eyed flies. Do these results indicate that the b and pr genes are closely linked? Explain. (Remember that there is no crossing-over in male Drosophila)..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_forwardThe recessive alleles k (kidney-shaped eyes instead ofwild-type round), c (cardinal-colored eyes instead of wildtype red), and e (ebony body instead of wild-type gray)identify three genes on chromosome 3 of Drosophila.Females with kidney-shaped, cardinal-colored eyes weremated with ebony males. The F1 was wild type. When F1females were testcrossed with kk cc ee males, the following progeny phenotypes were obtained:k c e 3k c + 876k + e 67k + + 49+ c e 44+ c + 58+ + e 899+ + + 4Total 2000a. Determine the order of the genes and the mapdistances between them.b. Draw the chromosomes of the parents and the F1.c. Calculate interference and say what you think of itssignificancearrow_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_forwardIn Drosophila, a cross was made between females expressing thethree X-linked recessive traits, scute bristles (sc), sable body (s),and vermilion eyes (v), and wild-type males. All females were wildtype in the F1, while all males expressed all three mutant traits.The cross was carried to the F2 generation and 1000 offspringwere counted, with the results shown in the following table. Nodetermination of sex was made in the F2 data. Question:Calculate the coefficient of coincidence; does this represent positive or negative interference? Phenotype Offspringsc s v 314+ + + 280+ s v 150sc + + 156sc + v 46+ s + 30sc s + 10+ + v 14arrow_forwardIn Drosophila, a female with three recessive traits, miniature wings, (m), ebony body (e) and rough eyes (r) were crossed with true-breeding wild-type males. In the F1, all offspring had wild-type phenotypes. F1 offspring were testcrossed to miniature, ebony, rough individuals and 1000 offspring were counted, shown in the table below. Determine the genotypes of the P1 and F1 parents. Determine the sequence of the three genes and the map distance between them. Are there more or fewer double crossovers than expected? Calculate the interference. Phenotype Offspring m e r 310 + + + 284 + e r 150 m + + 156 m + r 45 + e + 32 m e + 8 + + r 15 Total 1000arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Human Anatomy & Physiology (11th Edition)BiologyISBN:9780134580999Author:Elaine N. Marieb, Katja N. HoehnPublisher:PEARSON
Biology 2eBiologyISBN:9781947172517Author:Matthew Douglas, Jung Choi, Mary Ann ClarkPublisher:OpenStax
Anatomy & PhysiologyBiologyISBN:9781259398629Author:McKinley, Michael P., O'loughlin, Valerie Dean, Bidle, Theresa StouterPublisher:Mcgraw Hill Education,
Molecular Biology of the Cell (Sixth Edition)BiologyISBN:9780815344322Author:Bruce Alberts, Alexander D. Johnson, Julian Lewis, David Morgan, Martin Raff, Keith Roberts, Peter WalterPublisher:W. W. Norton & Company
Laboratory Manual For Human Anatomy & PhysiologyBiologyISBN:9781260159363Author:Martin, Terry R., Prentice-craver, CynthiaPublisher:McGraw-Hill Publishing Co.
Inquiry Into Life (16th Edition)BiologyISBN:9781260231700Author:Sylvia S. Mader, Michael WindelspechtPublisher:McGraw Hill Education
Human Anatomy & Physiology (11th Edition)
Biology
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:PEARSON
Biology 2e
Biology
ISBN:9781947172517
Author:Matthew Douglas, Jung Choi, Mary Ann Clark
Publisher:OpenStax
Anatomy & Physiology
Biology
ISBN:9781259398629
Author:McKinley, Michael P., O'loughlin, Valerie Dean, Bidle, Theresa Stouter
Publisher:Mcgraw Hill Education,
Molecular Biology of the Cell (Sixth Edition)
Biology
ISBN:9780815344322
Author:Bruce Alberts, Alexander D. Johnson, Julian Lewis, David Morgan, Martin Raff, Keith Roberts, Peter Walter
Publisher:W. W. Norton & Company
Laboratory Manual For Human Anatomy & Physiology
Biology
ISBN:9781260159363
Author:Martin, Terry R., Prentice-craver, Cynthia
Publisher:McGraw-Hill Publishing Co.
Inquiry Into Life (16th Edition)
Biology
ISBN:9781260231700
Author:Sylvia S. Mader, Michael Windelspecht
Publisher:McGraw Hill Education
How to solve genetics probability problems; Author: Shomu's Biology;https://www.youtube.com/watch?v=R0yjfb1ooUs;License: Standard YouTube License, CC-BY
Beyond Mendelian Genetics: Complex Patterns of Inheritance; Author: Professor Dave Explains;https://www.youtube.com/watch?v=-EmvmBuK-B8;License: Standard YouTube License, CC-BY