1. Thomas Morgan's choice of Drosophila melanogaster has been proven useful even today. Which of the following has/have continued to make it a most useful species? I. its four pairs of chromosomes II. a very large number of visible as well as biochemically mutant phenotypes III. casy and inexpensive maintenance IV. short generation time and large number of offspring (a) I and IV only (b) II and III only (c) I, II, and IIl only (d) II, III, and IV only (e) I, II, III, IV, and V

1. Thomas Morgan's choice of Drosophila melanogaster has been proven useful even today. Which of the following has/have continued to make it a most useful species? I. its four pairs of chromosomes II. a very large number of visible as well as biochemically mutant phenotypes III. casy and inexpensive maintenance IV. short generation time and large number of offspring (a) I and IV only (b) II and III only (c) I, II, and IIl only (d) II, III, and IV only (e) I, II, III, IV, and V

Biology (MindTap Course List)

11th Edition

ISBN:9781337392938

Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Publisher:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Chapter17: Developmental Genetics

Section: Chapter Questions

Problem 10TYU

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6. A diploid strain of yeast was made by mating a haploidstrain with a genotype w−, x−, y−, and z− with a haploidstrain of opposite mating type that is wild type for thesefour genes. The diploid strain was phenotypically wildtype. Four different X-ray-induced diploid mutantswith the following phenotypes were produced fromthis diploid yeast strain. Assume a single new mutation is present in each strain.Strain 1 w− x+ y− z+Strain 2 w+ x− y− z−Strain 3 w− x+ y− z−Strain 4 w− x+ y+ z+When these mutant diploid strains of yeast go throughmeiosis, each ascus is found to contain only two viablehaploid spores.a. What kind of mutations were induced by X-rays tomake the listed diploid strains?b. Why did two spores in each ascus die?
4. a. Give at least three examples of types of mutationsthat would disrupt the process of mitotic chromosomesegregation. That is, explain in what DNA structuresor in genes encoding what kinds of proteins youwould find these segregation-disrupting mutations.b. How could you use yeast artificial chromosomes(YACs) to find such mutations in S. cerevisiae?
. A diploid strain of yeast was made by mating a haploidstrain with a genotype w−, x−, y−, and z− with a haploidstrain of opposite mating type that is wild type for thesefour genes. The diploid strain was phenotypically wildtype. Four different X-ray-induced diploid mutantswith the following phenotypes were produced fromthis diploid yeast strain. Assume a single new mutation is present in each strain.Strain 1 w− x+ y− z+Strain 2 w+ x− y− z−Strain 3 w− x+ y− z−Strain 4 w− x+ y+ z+When these mutant diploid strains of yeast go throughmeiosis, each ascus is found to contain only two viablehaploid spores.a. What kind of mutations were induced by X-rays tomake the listed diploid strains?b. Why did two spores in each ascus die?c. Are any of the genes w, x, y, or z located on thesame chromosome?d. Give the order of the genes that are found on thesame chromosome
I. Male Drosophila from a true-breeding wild-typestock were irradiated with X-rays and then mated withfemales from a true-breeding stock carrying the following recessive mutations on the X chromosome:yellow body (y), crossveinless wings (cv), cut wings(ct), singed bristles (sn), and miniature wings (m).These markers are known to map in the order:y - cv - ct - sn - mMost of the female progeny of this cross were phenotypically wild type, but one female exhibited ct and snphenotypes. When this exceptional ct sn female wasmated with a male from the true-breeding wild-typestock, twice as many females as males appearedamong the progeny.a. What is the nature of the X-ray-induced mutationpresent in the exceptional female?b. Draw the X chromosomes present in the exceptional ct sn female as they would appear duringpairing in meiosis.c. What phenotypic classes would you expect to seeamong the progeny produced by mating the exceptional ct sn female with a normal male from a truebreeding wild-type…
Part This research exploited the principles of Mendelian genetics combined with the MCR technique to genetically manipulate . , you do similar experintent You perform two crosses one that for Mendelan patterns of inheritance and another that uses with CRISPR-Cas to genetically the of fruit fliesIn flies the gene eye color is loceled on the X chromosome (wild type w , Let's see how you can predict genotype and phenotype ratios resulting from the two Genetic cross 1 without MCR ): a whitemale and a homozygous red-eyed Genetic cross 2 with : between MOR-element ( carries the CRISPR-Cas9 ) white- ayed and a homozygous -eyed fortalein mind, this technology porwerts heterozygous genclype homozygous recessive genotype Drag the labels to the table to fill in the genotypic and phenotypic ratios . Not all labels will be used. Hint: Figure in the online paper illustrates the process of the technology and the genetic cross outcomes observed in the research
19. The diagram below represents a process that occurs during sexual reproduction. A B C + - (Not drawn to scale) The process represented by the arrow will ensure that *REQUIR structures A and B contain a complete set of genetic information structure C contains half of the genetic information structures A and B contain half of the genetic information structure C contains a complete set of genetic information *REQUIRED
Drosophila researchers have collected many strainsthat carry a single recombinant P element containing awild-type white gene (a P[w+] transgene) inserted intoa known genomic location. These strains can be used tomap the location of any mutant gene in the fly genome.Investigators performed a testcross to map arecessive mutation rough (ro), which causes rougheyes, relative to a P[w+] element on chromosome 3.Females heterozygous for the P[w+] on one chromosome 3 and a ro− mutation on the other, homologouschromosome 3 were crossed to ro−/ro− males, and theprogeny in the following list were obtained. In boththe parents and the progeny, the endogenous whitegene is nonfunctional—the flies have red eyes onlyif they contain the P[w+] transgene.145 red, smooth (wild-type) eyes152 white, rough eyes2 white, smooth eyes1 red, rough eyesa. Are ro and the P[w+] linked? If so, how many mapunits separate them?b. The data in part (a) do not indicate on which sideof the P[w+] (toward the centromere or…
In 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…
3. The allele b gives Drosophila flies a black body and b+ gives brown, the wild-type phenotype. The allele wx of a separate gene gives waxy wings and wx+ gives nonwaxy, the wild-type phenotype. The allele cn of a third gene gives cinnabar eyes and cn+ gives red, the wild-type phenotype. A female heterozygous for these three genes is testcrossed, and 1000 progeny are classified as follows: 5 wild type; 6 black, waxy, cinnabar; 69 waxy, cinnabar; 67 black; 382 cinnabar; 379 black, waxy; 48 waxy; and 44 black, cinnabar. Are these three genes linked? If so, draw a map of the chromosomes in the heterozygous parent, showing the order, arrangement, and map distance between the linked genes.
A mutation occurs in the bicoid gene in a female fly. What will result if her eggs are fertilized? 1.Her embryos will be normal if they are fertilized by a male fly that contributed a wild type allele for bicoid 2.Her embryos will be normal if the male that fertilizes her eggs also has that same bicoid mutation 3.All of her embryos will be disrupted irrespective of the genotype of the father 4.Her embryos will develop normally but the adults will be sterile
1. Every time a human cell divides, over 50 new mutations can be found in the daughter cells 2. Mutations can accumulate over the course of successive cell divisions 3. In every new kid, over 50 new mutations can be found when comparing them to their parents 4. Most of the new mutations in gametes are probably silent*, but there's a good chance that one of them might occur in a gene, and alter that gene in such a way as to potentially create a mutation *Note: by 'silent' we mean that the mutation doesn't have the potential to alter phenotype, either because it is not in a gene, or is in a gene but in the middle of an intron, or is in the coding sequence but is a silent substitution, etc., etc. 5. Most new gene mutations (i.e. mutations that occur in genes that alter gene expression, splicing or protein sequence) are recessive
5. Design a mating between a female with these three traits: purple (eye color), black (body color), and curved (wing shape), and a wild type male. a. What are the phenotypes of their offspring? + (wildtype) b. What are their genotypes? XW+ XW+ XW+ Y 9 Female PRC Source Design View QFy Flies to be Mated Phenotype Female: + Male: + Total Male Source Design View Results for Cross # 3 Observed Proportion 522 0.5113 499 1021 0.4887 1.0000 Female: PR, C, BL; N=1 8 Parents of Cross #3 Select to Mate Q Select to Mate Offspring of Cross #3 Male N-499 Female N 522 Select to Male 10 Total Male N=1 Phenotype Select to Mate Results for Cross #3 Observed Proportion 1021 1.0000 1021 1.0000