Study Guide for Campbell Biology
11th Edition
ISBN: 9780134443775
Author: Lisa A. Urry, Michael L. Cain, Steven A. Wasserman, Peter V. Minorsky, Jane B. Reece, Martha R. Taylor, Michael A. Pollock
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 15, Problem 9TYK
Summary Introduction
Introduction: Locus is the specific location of a gene in a chromosome. The inheritance of two different characters is affected by the gene linkage. The production of the offspring with traits that are combined is studied in recombination. Map units measure genetic linkage.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
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 =1
In Drosophila, a cross was made between a yellow-bodied male with vestigial wings and a wild-type (WT) female(brown body and normal wings). The F1 generation consisted of WT males and WT females. The F1 males and females were crossed, and the F2 progeny consisted of 16 yellow males with vestigial wings, 48 yellow males with WT wings, 15 brown males with vestigial wings, 49 WT males, 31 brown females with vestigial wings, and 97 WT females. Based on these results, explain the inheritance of the two genes (i.e. autosomal or sex-linked, dominant or recessive).
The 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?
Chapter 15 Solutions
Study Guide for Campbell Biology
Ch. 15 - Complete the following summary of Morgans crosses...Ch. 15 - Two normal color-sighted individuals have two...Ch. 15 - In a testcross between a heterozygote tall,...Ch. 15 - With unlinked genes, an equal number of parental...Ch. 15 - The following recombination frequencies have been...Ch. 15 - a. What is the difference between an organism with...Ch. 15 - Prob. 7IQCh. 15 - Prob. 8IQCh. 15 - Mendels law of independent assortment applies to...Ch. 15 - You have found a new mutant phenotype in fruit...
Ch. 15 - Prob. 3SYKCh. 15 - Prob. 4SYKCh. 15 - Thomas Hunt Morgan firmly established the location...Ch. 15 - Prob. 2TYKCh. 15 - Sex-linked traits a. are coded for by genes...Ch. 15 - Prob. 4TYKCh. 15 - Prob. 5TYKCh. 15 - Prob. 6TYKCh. 15 - Prob. 7TYKCh. 15 - Prob. 8TYKCh. 15 - Prob. 9TYKCh. 15 - Prob. 10TYKCh. 15 - Consider three genes on the X chromosome: A, B,...Ch. 15 - Prob. 12TYKCh. 15 - Genomic imprinting a. explains cases in which the...Ch. 15 - Prob. 14TYKCh. 15 - Prob. 15TYKCh. 15 - Suppose that alleles for an X-linked character for...Ch. 15 - Some girls who fail to undergo puberty are found...Ch. 15 - Prob. 18TYKCh. 15 - The genetic event that results in Turner syndrome...Ch. 15 - Prob. 20TYKCh. 15 - Prob. 1GPCh. 15 - Prob. 2GPCh. 15 - Prob. 3GPCh. 15 - Prob. 4GPCh. 15 - Prob. 5GPCh. 15 - Red-green color blindness is caused by a...
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 with cinnabar eye (cn) and vestigal wings (vg) were mated to males with roof wings (fr). The F1 were all wild type. When the F1 females were tested and crossed with male homozygous for all thrre traits, the following result were obtained. For this problem, rf, cn, vg is for roof wing / cinnabar eye / vestugal wing rf + cn + vg all wild typr(no definite phenotypr described.arrow_forwardIn 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?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
- In Drosophila, the dominant Bar mutation (B, chromosome X, position 57) results in thin bar- shaped eyes, while the recessive singed (sn, chromosome X, position 21) results burnt looking bristles. True breeding, wild type females are mated with true breeding males with Bar eyes and singed bristles. 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, the sepia mutation (se, chromosome 3, position 26) results in dark brown eyes, while cinnabar (cn, chromosome 2, position 57.5) results in bright orange-red eyes. True breeding, wild type females are mated with true breeding males homozygous recessive for both traits. 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 =1arrow_forwardIn 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_forward
- In Drosophila, the allele for red eyes (pt) is wild-type and the allele for purple eyes (p¯) is mutant. The allele for grey body (b+) is wild-type and the allele for black body (b¯) is mutant. Flies with p*p* b*b* genotypes are mated with flies that have p p- b¯b¯ genotypes. A testcross was then performed in which the F1 offspring with p*p¯ b*b¯ genotypes were mated with flies with p p¯ b¯b genotypes. Ten-thousand flies were produced from this test cross. The following results were observed: 4,300 red eye, grey body flies 550 red eye, black body flies 4,500 purple eye, black body flies 650 purple eye, grey body flies Which F2 phenotypes are parental types? Which F2 phenotypes are recombinant types? What is the distance between the gene loci for eye color and body color? Use the equation for Hardy-Weinberg equilibrium for the following questions. p+q = 1 p2 + 2pq + q2 = 1arrow_forwardA cross was performed using Drosophila melanogaster involving a female known to be heterozygous for both ebony body and sepia eyes and a male known to be homozygous wild type male. The resulting progeny were allowed to mate with one another to produce the data set. Three repetitions of the experiment were conducted. The following data were produced from the crosses. Test these data to determine if they are significantly different from the expected phenotypic ratio. Use the 5% level of significance. Your answer should include the hypothesized cross in genotypes, the Chi-squared value, the critical value and whether you reject or do not reject for each experiment. Wild eye Wild body – 112, Wild eye Ebony body – 40, Sepia eye Wild body – 35, Sepia eye Ebony body – 11arrow_forwardFemale Drosophila with cinnabar eye (cn) and vestigial wings (vg) were mated to males with roof wings (rf). The F1 were all wild-type. When the F1 females were test crossed with males homozygous for all three traits the following result were obtained. For this problem, rf cn vg is for roof wing / cinnabar eye / vestigial wing rf + cn + vg + all wild (no definite phenotype described) Give the genotype of the offspring (this time simply follow the given sequence above and separate the genes with single space) F2: Phenotype Frequency Genotype cinnabar, vestigial 382 roof 401 cinnabar 3 roof, vestigial 4 roof, cinnabar, vestigial 59 Wild 67 roof, cinnabar 44 vestigial 40 2. Give the phenotype and genotype of the female parent: phenotype: ____ genotype: _______ 3. Give the phenotype and genotype of the male parent: phenotype: _______ genotype: _______ 4. What…arrow_forward
- The parental genotypes for a series of crosses are wild-type male fruit flies mated to females with white eyes (wh) and miniature (min) wings. The phenotypes of the F1 generation were wild-type females, and males with white eyes, and miniature wings. These flies were allowed to mate with each other and produced the following offspring: Red eyes, long wings White eyes, miniature wings Red eyes, miniature wings White eyes, long wings 770 716 401 318 Total 2205 A. Are these genes linked? Why or why not?arrow_forwardIn 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_forwardYou are doing a cross with Drosophila using the following two traits. Curly wings is dominant over straight wings, and round eyes is dominant over elliptical eyes. You cross a female fly that is known to be heterozygous for both genes with a male that is heterozygous for the wing gene but has elliptical eyes. This cross produces 74 flies with curly wings and round eyes, 61 with curly wings and elliptical eyes, 24 with straight wings and round eyes, and 21 with straight wing and elliptical eyes. Calculate the expected phenotype ratios for this cross, then use the chi-square test to see if the observed data are consistent with the expected numbers.arrow_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