GENETIC ANALYSIS: AN INTEG. APP. W/MAS
2nd Edition
ISBN: 9781323142790
Author: Sanders
Publisher: Pearson Custom Publishing
expand_more
expand_more
format_list_bulleted
Concept explainers
Textbook Question
Chapter 20, Problem 15P
What
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
You are interested in studying position effect variegation in Drosophila using the chromosome depicted below:
Deactivation of the w+ gene gives a white eye phenotype and deactivation of the rst+ gene gives a rough eye
phenotype; the normal phenotypes are red and smooth. Because the rst+ and w+ genes have now been placed
close to a heterochromatic domain, some sections (or sectors) of the eye display mutant phenotypes due to gene
deactivation while others display the normal, wild type phenotype. Which phenotype would you not expect to see
rst w
Inverted X chromosome
white smooth eye sectors
white rough eye sectors
red smooth eye sectors
red rough eye sectors
You have identified a Drosophila gene that is expressed exclusively in the odd-numbered "stripes" in the cellular blastoderm. Assuming that this gene is not redundant, what
would be the most likely phenotype cause by a loss-of-function mutation in this gene?
an embryo missing odd numbered segments
an adult fly with a second pair of wings instead of halteres
an embryo with two anterior ends
an embryo missing even-numbered segments
an embryo missing larval segments 3-10
Here are schematic diagrams of mutant Drosophila larvae.
The left side of each pair shows a wild-type larva, with gray boxes showing the sections that are missing in the mutant larva. Which type of gene is defective in each larva: a gap gene, a pair-rule gene, or a segment-polarity gene?
Chapter 20 Solutions
GENETIC ANALYSIS: AN INTEG. APP. W/MAS
Ch. 20 - 18.1 Explain why many developmental genes encode...Ch. 20 - Bird beaks develop from an embryonic group of...Ch. 20 - 18.3 How is positional information provided along...Ch. 20 - Early development in Drosophila is atypical in...Ch. 20 - 18.5 Consider the evenskipped regulatory sequences...Ch. 20 - What is the difference between a parasegment and...Ch. 20 - Why do loss-of-function mutations in Hox genes...Ch. 20 - 18.8 Compare and contrast the specification of...Ch. 20 - Prob. 9PCh. 20 - Ablation of the anchor cell in wild type C....
Ch. 20 - 18.11 In gain-of-function and. elegans mutants,...Ch. 20 - Prob. 12PCh. 20 - Prob. 13PCh. 20 - 18.14 Given that maternal Bicoid activates the...Ch. 20 - What phenotypes do you expect in flies homozygous...Ch. 20 - The pair rule gene fushitarazu is expressed in...Ch. 20 - 18.17 In contrast to Drosophila, some insects...Ch. 20 - Prob. 18PCh. 20 - 18.19 You are traveling in the Netherlands and...Ch. 20 - 19.20 A powerful approach to identifying genes of...Ch. 20 - Prob. 21PCh. 20 - The Hoxd 913 genes are thought to specify digit...Ch. 20 - Three-spined stickleback fish live in lakes formed...Ch. 20 - In C. elegans there are two sexes: hermaphrodite...Ch. 20 - The flowering jungle plant Lacandoniaschismatica,...Ch. 20 - 18.24 Homeotic genes are thought to regulate each...Ch. 20 - Prob. 27PCh. 20 - Basidiomycota is a monophyletic group of fungi...Ch. 20 - In Drosophila, recessive mutations in the...
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
- Part 1. Given the offspring ratios and genotypes, what must the genotype be of each of the parent flies? Be sure to indicate the HAPLOTYPE of the parents (ie, indicate each chromosome separately) Explain briefly how you got this answer.arrow_forward) Explain how and why dorsal/ventral polarity will be affected in fly Question 3 (1. embryos carrying the following mutations; also in each case darken in the area of the cells in the cross-sectional view of the fly embryo which are expected to express the paulie gene. D = dorsal; V = ventral. a) a mutation which results in the deletion of the cytoplasmic domain of the Cookie protein. b) A mutation which results in a constitutively active Bombe protein, i.e. the Bombe protein is always in an activated state. c) A mutation which causes the Pickle protein to be retained in the cytoplasm of the embryo.arrow_forwardIn drosophila melonegaster, nuclei of somatic cells contain 8 chromosomes while nuclei of sperm cells contain only 4 chromosomes. What does "n" equal for this species? O 2 4 8. O 16arrow_forward
- Researchers have exploited Minute mutations in orderto study the phenotypes associated with recessive lethal mutations (l−) that decrease the rate of cell divisionand thus make only very tiny homozygous mutant clones that are difficult to analyze. Many differentstrains of Drosophila carry dominant loss-of-functionMinute (M) mutations in a variety of genes encodingribosomal protein subunits. The M genes are haploinsufficient; flies with only one wild-type M+ gene copyhave a slower pace of cell division, and thus prolongeddevelopment and subtle morphological abnormalities.To circumvent the tiny clone problem, researchersgenerate GFP-marked homozygous l−/ l− clones thatare also M+/ M+, in flies that are l−/ l+ and M−/ M+.The loss of the Minute mutation only in cells withinthe clone gives the l−/ l− cells a growth advantageover their neighbors, enabling the mutant clone togrow large enough to study. Diagram chromosomesthat could be used to generate such clonesarrow_forwardIn a haploid yeast strain, eight recessive mutationswere found that resulted in a requirement for theamino acid lysine. All the mutations were found to revert at a frequency of about 1 × 10−6 except mutations5 and 6, which did not revert. Matings were madebetween a and α cells carrying these mutations. Theability of the resultant diploid strains to grow onminimal medium in the absence of lysine is shown inthe following chart (+ means growth and − means nogrowth.)1 2 3 4 5 6 7 81 − + + + + − + −2 + − + + + + + +3 + + − − − − − +4 + + − − − − − +5 + + − − − − − +6 − + − − − − − −7 + + − − − − − +8 − + + + + − + −a. How many complementation groups were revealedby these data? Which point mutations are foundwithin which complementation groups?The same diploid strains are now induced to undergosporulation. The vast majority of resultant spores areauxotrophic; that is, they cannot form colonies whenplated on minimal medium (without lysine). However,particular diploids can produce rare spores…arrow_forwardTo understand the genetic basis of locomotion in the diploid nematode Caenorhabditis elegans, recessive mutations were obtained, all making the worm “wiggle” ineffectually instead of moving with its usual smooth gliding motion. These mutations presumably affect the nervous or muscle systems. Twelve homozygous mutants were intercrossed, and the F1 hybrids were examined to see if they wiggled. The results were as follows, where a plus sign means that the F1 hybrid was wild type (gliding) and “w” means that the hybrid wiggled.a. Explain what this experiment was designed to test. b. Use this reasoning to assign genotypes to all 12 mutants. c. Explain why the phenotype of the F1 hybrids between mutants 1 and 2 differed from that of the hybrids between mutants 1 and 5arrow_forward
- In a wild-type fungus, protein E (encoded by the haplosufficient gene E) normally dimerizes to catalyzes a biochemical reaction necessary for the production of a dark pigment. Ed represents a mutant, dominant negative allele of gene E. What is the predicted phenotype of a fungus cell of genotype E*/Ed, and why? O wild type (normal production of the dark pigment), as E is haplosufficient mutant (no pigment production), as no dimers will form in the heterozygous mutant (no pigment production), as the mutant allele Eg is dominant O wild type (normal production of the dark pigment), as dimers of wild-type and mutant protein E will be formed in the heterozygousarrow_forwardThe normal sequence of markers on a certain Drosophila chromosome is ABCDE*FGHIJK, where the asterisk represents the centromere. Some flies were isolated with a chromosome aberration that has the following structure: ABCDE*FIJK . This represents a O a) deletion of GH segment O b) inversion of GH segment O c) deletion O d) deletion of centromerearrow_forwardConsider the following variations in Drosophila melanogaster, relative to the wild-type: White eyes are a recessive trait—the gene of which is found in Chromosome I (X). Vestigial wings are a recessive trait—the gene of which is found in Chromosome II. Aristapedia is a dominant trait—the gene of which is found in Chromosome III. Being homozygous for this condition is lethal. Cross the following mutant females with a wild-type (homozygous) male. Show the Punnett square and obtain the genotypic and phenotypic ratios of the first filial generation (F1). Female with white eyes Q4: Show the Punnett squares and obtain the genotypic and phenotypic ratios of the first filial generation (F1) and second filial generation (F2) of the following crosses. Note: The F2 generation can be obtained by crossing one male and one female from the F1 generation. Female with white eyes and vestigial wings and wild-type malearrow_forward
- For a haploid fungus, the starting point in the biosynthesis of the amino acid arginine is Compound X, which is always present in and absorbed from the environment. The arginine biosynthetic pathway is: Enzyme A Enzyme B Enzyme Ç Compound X It is know that genes encoding enzymes A and C are on two different chromosomes. Compound Y Compound Z- Arginine A mutant strain of genotype a (lacking only enzyme A) is crossed to a mutant strain of genotype c (lacking only enzyme C) to generate a diploid strain. Sporulation (i.e. meiosis) is subsequently induced in the resulting diploid strain. What proportion of the spores (haploids formed by sporulation) is expected to grow on medium without arginine but supplemented with Compound Y? O 100% 50% 0% 25%arrow_forwardFollowing a mutagenesis experiment to identify novel genes affecting the circadian clock in Drosophila melanogaster you discover several mutants. You start considering two of those mutants that you call c and d. The two homozygous c/c and d/d are arrhythmic (arrhythmic is the definition of their phenotype), whereas the two heterozygous C/c and D/d are rhythmic (rhythmic is the definition of their phenotype) with a 24h period. You make two true-breeding stocks: stock 3 homozygous for c and stock 4 homozygous for d. You cross them in both directions and in both cases you observe complementation with no difference between males and females. Then you take the progeny of one cross, for instance the F1 of Females 3 x Males 4, and you perform a Testcross. Out of 1000 flies resulting from the lestcross only 125 are rhythmic. Using the number of flies expected for the Parental and the Recombinant genotypes and the formula that defines recombination frequency, what is the distance between loci C…arrow_forwardFollowing a mutagenesis experiment to identify novel genes affecting the circadian clock in Drosophila melanogaster you discover several mutants. You start considering two of those mutants that you call c and d. The two homozygous c/c and d/d are arrhythmic (arrhythmic is the definition of their phenotype), whereas the two heterozygous C/c and D/d are rhythmic (rhythmic is the definition of their phenotype) with a 24h period. You make two true-breeding stocks: stock 3 homozygous for c and stock 4 homozygous for d. You cross them in both directions and in both cases you observe complementation with no difference between males and females. Then you take the progeny of one cross, for instance the F1 of Females 3 x Males 4, and you perform a Testcross. Out of 1000 flies resulting from the Testcross only 125 are rhythmic. Considering that out of 1000 flies resulting from the Testcross only 125 are rhythmic how many flies in total are approximatively expected for the Parental genotypes?…arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Biology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage Learning
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