Biology: The Dynamic Science (MindTap Course List)
4th Edition
ISBN: 9781305389892
Author: Peter J. Russell, Paul E. Hertz, Beverly McMillan
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Textbook Question
Chapter 13, Problem 10TYK
You conduct a cross in Drosophila that produces only half as many male as female offspring. What might you suspect as a cause?
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
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).
In 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 region
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 13 Solutions
Biology: The Dynamic Science (MindTap Course List)
Ch. 13.1 - You want to determine whether genes a and b are...Ch. 13.2 - You have a true-breeding strain of...Ch. 13.3 - What mechanisms are responsible for: (a)...Ch. 13.4 - A man has Simpson syndrome, an addiction to a...Ch. 13.4 - Prob. 2SBCh. 13.5 - Prob. 1SBCh. 13 - In humans, redgreen color blindness is an X-linked...Ch. 13 - The following pedigree shows the pattern of...Ch. 13 - Individuals affected by a condition known as...Ch. 13 - A number of genes carried on the same chromosome...
Ch. 13 - Prob. 5TYKCh. 13 - Discuss Concepts Can a linkage map be made for a...Ch. 13 - In Drosophila, two genes, one for body color and...Ch. 13 - Another gene in Drosophila determines wing length....Ch. 13 - Prob. 9TYKCh. 13 - You conduct a cross in Drosophila that produces...Ch. 13 - Discuss Concepts Crossing-over does not occur...Ch. 13 - Prob. 12TYKCh. 13 - Prob. 13TYKCh. 13 - Prob. 14TYKCh. 13 - Prob. 1ITDCh. 13 - Prob. 2ITDCh. 13 - Prob. 3ITDCh. 13 - Prob. 4ITD
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
- 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_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 P1 cross? X X ++ e + + + O+ X + X + ■ + X + + + 3+ X X X X + + Y Y cu cu cu + cu cu J e e e e e (D e + cu cu (Darrow_forwardImagine Drosophila genes C, D, and E are autosomal genes located close to each other on the same chromosome (same assumptions as the first problem). You cross a C D E homozygote with a c d e homozygote, then cross the F1 females with a c d e homozygous male. Of 400 progeny, you observe the following phenotypes: 135 CDE 139 cde 22 cDE 18 Cde 42 CdE 38 cDe 3 cdE 3 CDe a. What is the order of the genes? Calculate the distance between genes and draw a map to get the final answer. Question 2 options: 1.CED 2. DEC 3. ECD 4. DCEarrow_forward
- The 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 the following cross, imagine that you have a female fly that has two Xs and one Y due to a nondisjunction event in her mother's germ cells. Draw out what the possible gametes are for both the female and the male and also a Punnett square showing the genotypes, phenotypes, and sex of the possible flies as a result of this cross. You do not need to provide the probabilities of each of these. Red-eyed wi C Ở Red-eyed wt XX Y X Y Meiosisarrow_forwardIn the haploid yeast Saccharomyces cerevisiae, the twomating types are known as MATa and MATα. You cross apurple (ad-) strain of mating type a and a white (ad+)strain of mating type α. If ad- and ad+ are alleles of onegene, and a and α are alleles of an independently inherited gene on a separate chromosome pair, what progenydo you expect to obtain? In what proportions?arrow_forward
- In mice, there is a yellow strain that when crossed yields 2 yellow:1 black. How could you explain this observation?arrow_forwardYou 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, 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, 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_forwardPURPLE VESTIGIAL DIHYBRID CROSS In the parental generation, you mate a pure-breeding wild-type female (put/pu+;vg+/vg+) with a pure-breeding purple, vestigial (pu/pu;vg/vg) to produce an F1 generation that is all wild-type (pu*/pu;vg+/vg). Note that the F1 flies are all dihybrid. Next, you mate several F1 dihybrid females (pu*/pu;vg+/vg) with tester males, which are purple, vestigial (pu/pu;vg/vg). The offspring of this dihybrid testcross are: Phenotype Genotype Tester Gamete Dihybrid Gamete Number Wild-type 437 417 77 59 Purple, vestigial Vestigial Purple Copy the table into your notes and derive the dihybrid gametes following the example in the first section. The columns in blue (phenotypes and numbers of offspring) are what you can see and count. The genotypes of the testcross offspring (orange) must be deduced from the phenotypes and knowing that the tester contributed pu vg gametes. Finally, you can deduce the dihybrid gametes (green) by subtracting the tester gamete contribution…arrow_forwardYou are conducting independent research for your Honors Thesis in a Drosophila research lab. One of the graduate student researchers has provided you with two mutant strains of Drosophila. One has scarlet eyes, and the other has brown eyes. You cross homozygous scarlet-eyed male Drosophila with female flies homozygous for the brown-eye color allele, and all of the male and female F1 flies have wild-type eyes. Crosses involving F1 males x F1 females yield the results shown below. You know the traits involved are autosomal, and that reciprocal parental crosses produce similar results. a) What is the phenotypic ration observed among the offspring in the F2 generation? b) How many pairs of genes are involved in determining these traits? Please explain. c) Please EXPLAIN the genetic basis for the pattern of inheritance illustrated in this cross. Be sure to mention the alleles involved.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning
Human Biology (MindTap Course List)BiologyISBN:9781305112100Author:Cecie Starr, Beverly McMillanPublisher:Cengage Learning
Biology (MindTap Course List)BiologyISBN:9781337392938Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. BergPublisher:Cengage Learning
Concepts of BiologyBiologyISBN:9781938168116Author:Samantha Fowler, Rebecca Roush, James WisePublisher:OpenStax College
Biology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage Learning
Human Heredity: Principles and Issues (MindTap Co...
Biology
ISBN:9781305251052
Author:Michael Cummings
Publisher:Cengage Learning
Human Biology (MindTap Course List)
Biology
ISBN:9781305112100
Author:Cecie Starr, Beverly McMillan
Publisher:Cengage Learning
Biology (MindTap Course List)
Biology
ISBN:9781337392938
Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg
Publisher:Cengage Learning
Concepts of Biology
Biology
ISBN:9781938168116
Author:Samantha Fowler, Rebecca Roush, James Wise
Publisher:OpenStax College
Biology: The Dynamic Science (MindTap Course List)
Biology
ISBN:9781305389892
Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan
Publisher:Cengage Learning
Mitochondrial mutations; Author: Useful Genetics;https://www.youtube.com/watch?v=GvgXe-3RJeU;License: CC-BY