Biology: The Dynamic Science (MindTap Course List)
4th Edition
ISBN: 9781305389892
Author: Peter J. Russell, Paul E. Hertz, Beverly McMillan
Publisher: Cengage Learning
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Please answer only question D i.e 4d alone.
Other were answered in my previous post.
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- 4. You find that all four flower color genes map to the second chromosome, and perform complementation tests with deletions for each gene. You obtain the following results: (mutant a = blue, mutant b = white, mutant c = pink, mutant d = red) wolod Results of Complementation tests suld Jostum Mutant a b с Del (2.2 -2.6) blue white pink purple Del (2.3-2.8) blue white pink red Del (2.1 -2.5) blue purple pink purple Del (2.4-2.7) purple white pink red C d Indicate where each gene maps: a b ori ai indW (anioq 2) .8arrow_forwardPlease find the attachmentarrow_forward2. Drosophila females heterozygous for three recessive x-linked markers y (yellow bodies), ct (cut wings), and m (miniature wings) were crossed to y ct m males and the following progeny were obtained. F1 Phenotype Yellow, cut, miniature Wildtype Yellow Cut, miniature miniature Number cut 60 63 20 13 Yellow, cut 15 Yellow, miniature 5 22 2 a) F1 Genotype b) Recombinant or Parental? c) SCO, DCO, or NONE? a) Fill in the table with the genotypes of each progeny type, use + to indicate wildtype/dominant traits, show separate homologues. b) In the table, identify whether each progeny type is Recombinant (R) or Parental (P). c) In the table, identify whether each progeny type results from a single crossover (SCO), a double crossover (DCO), or no crossing over (NONE). d) Diagram the parental cross, clearly indicating allelic contributions from each homologue of parental flies. Heterozygous wild-type Female Fly X Hemizygous mutant Male fly // // e) Draw the genetic map showing correct order and…arrow_forward
- 3. Short-tailed pup distribution. In mice, the dominant T allele results in a short tail. Homozygous T/T genotype is lethal, which means mouse embryos with this genotype die before they are born. Homozygous t/t is normal, with normal tail. A cross between two short-tailed mice produces a litter of 5 pups. a). What are the genotype(s) of the two short-tailed mice used in the cross? b). What are the possible genotypes and phenotypes of the 5 pups? What are their perspective ratio?arrow_forwardI. 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…arrow_forward2. The following is a pedigree of a family in which a rare form of X-linked colorblindness is found (filled-in symbols). II 2 II 3 4 a. Label the genotypes of as many family members as the available information permits. Remember that this gene is located on the X-chromosome when writing out genotypes. b. Is the colorblindness trait recessive or dominant? Explain how you know.arrow_forward
- Miniature wings, X in Drosophila melanogaster result from an X-linked allele that is recessive to the allele for long wings, 9 + X. Match the genotypes for each parent in the crosses. Male parent phenotye long miniature miniature 111 long long Female parent phenotype long long long miniature long m m X X Male offspring phenotypes 231 long, 250 miniature 610 long 410 long, 417 miniature 753 miniature 625 long m X Y Female offspring phenotypes 560 long 632 long 412 long, 415 miniature 761 long 630 long Answer Bank ++ X X Male parent genotype + X Y Female parent genotype + X X m 00arrow_forward1. A mechanism for equalizing the expression of X-linked genes in males and females of a species. [ Choose ] Conjugation Reciprocal Crosses Transduction Binary Replication Haplotype Interference Transformation Gain of Function Mutation Allelic Series Blending Theory Dosage Compensation Rolling Circle Replication Pleiotropy Coefficient of Coincidence Reciprocal Cross Loss of Function Mutation Synaptonemal Complex Allelic Phase Allele Chiasma Law of Independent Assortment 2. A group of alleles of a gene that show a hierarchy of dominance relationship among them. [ Choose ] Conjugation Reciprocal Crosses Transduction Binary Replication…arrow_forward2. The homozygous mutant a, an allele of the albinism gene in the mouse, has a white coat and dark eyes. The homozygous mutant p has pink eyes. Mice of genotype a+/+p are testcrossed with mice homozygous for the two mutations. The progeny obtained are shown below. Calculate the recombination frequency and map distance between the two genes. Phenotype of Progeny White coat, dark eyes White coat, pink eyes Dark coat, white eyes Dark coat, pink eyes Number 240 31 34 474arrow_forward
- (X^X^) and one male (X"). 12. The following are the gene order on the chromosomes of an individual who is heterozygous for this translocation. translocations (a.unbalanced breciprocal C. Robertsonian MN OP QR N067 89 34 5P QR 3 4 5 6 7 8 9 a. 0 LMNOP QR LMN067 89 345P QR 3456789 hmm Draw the synapsis (Prophase/Metaphase I) configuration dark spots b. What type of translocation is depicted by these chromosomes and what are the consequences of this chromosome rearrangement to the individual? this type of translocation is unbalanced and the major consequence here is that the chromosome number is not reciprocated leading to all sorts of problems with non-disjunction.arrow_forwardGenetic Crosses that Involve 2 Traits In rabbits, black hair is dominant to brown hair. Also in rabbits, long straight ears are dominant to floppy ears. These letters represent the genotypes and phenotypes of the rabbits: BB = black hair EE = long ears Ee = long ears ee = floppy ears %3D %3D Bb = black hair bb = brown 1. A male rabbit with the genotype BBee is crossed with a female rabbit with the genotype bbEe The square Is set up below. Fill it out and determine the phenotypes and proportions in the offspring. How many out of 16 have black hair Ве Ве Ве Ве and long ears? bE How many out of 16 have black hair and floppy ears? be How many out of 16 have brown hair and long ears? bE How many out of 16 have brown hair be and floppy ears?arrow_forward2. In Drosophila sp. Red eyes are dominant over white eyes. Normal wings are dominant over vestigial wings. Homozygous red eyes normal wings females were crossed with homozygous white eyes and vestigial wings males. The F2 generation after a test cross are as below Phenotype Red eyes, normal wings White eyes, vestigial wings Red eyes, vestigial wings White eyes, normal wings Amount 1670 1590 425 435 a) Are the wings and eyes colour genes linked? b) Show all the crosses c) Calculate the recombination frequency value. d) What is the distance between the wings and eyes colour genes?arrow_forward
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