In Drosophila, assume that the gene for scute bristles (s) is located at map position 0.0 and that the gene for ruby eyes (r) is at position 25.0. Both genes are located on the X chromosome and are recessive to their wild-type alleles. A cross is made between scute-bristled females and ruby-eyed males. Phenotypically wild-type F1 females were then mated to homozygous double mutant males, and 1000 offspring were produced. Give the phenotypes and frequencies expected..
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QUESTION:-
In Drosophila, assume that the gene for scute bristles (s) is located at map position 0.0 and that the gene for ruby eyes (r) is at position 25.0. Both genes are located on the X chromosome and are recessive to their wild-type alleles. A cross is made between scute-bristled females and ruby-eyed males. Phenotypically wild-type F1 females were then mated to homozygous double mutant males, and 1000 offspring were produced. Give the
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- Problem Solving: In Drosophila, the gene that controls red eye color (dominant) versus white eye color is on the X chromosomes. What are the expected phenotypic results if a heterozygous female is crossed with a white-eyed male? In cats, S = short hair; s = long hair; XC = black coat; Xc = yellow coat; XCXc = tortoiseshell (calico) coat. If a long-haired yellow male is crossed with a tortoiseshell female homozygous for short hair, what are the expected phenotypic results?In Drosophila the genes forked bristles (f and f+) and vermilion eyes (v and v+) are ~24 m.u. apart on the X chromosome. From a cross of f v / f+ v+ females with f v / y males, what % of the progeny do you expect to have the f+ v+ phenotypes? (Don’t let the X chromosome throw you off – this question is essentially asking what % of the female’s gametes will be f+ v+).Homozyogous wild type male Drosophila PPQQRRSSTTUUVV (all linked) are irradiated to induce the formation of chromosomal deletions and then mated with homozygous recessive females. In several of these matings a unique pattern of pseudodominance could be correlated with the loss of specific polytene chromosome bands, as shown in the table... Cross Chromosomal Bands Deleted Pattern of Pseudodominance PqRs TuV 1 2 3 4 5 5-6 1-2 3-5 6-8 1-3 PQRSTUV PqrSTUV PQRSTUV pQrSTUV In which chromosomal band is gene R located? A. 3 OB. 4 OC. 2 D.8 O E. none of these answers are correct
- Drosophila (fruit-flies) mutants A and B have the same phenotype: the absence of red pigment in the eyes. Homozygous mutants were crossed, and all offspring possessed red eyes. Does the appearance of red eyes in the progeny indicate complementation or a failure to complement? [Select] Do the mutants likely have mutations in the same or different genes? [ Select]Another gene in Drosophila determines wing length. The dominant wild-type allele of this gene produces long wings; a recessive allele produces vestigial (short) wings. A female that is true- breeding for red eyes and long wings is mated with a male that has purple eyes and vestigial wings. F1 females are then crossed with purple-eyed, vestigial-winged males. From this second cross, a total of 600 offspring are obtained with the following combinations of traits: 252 with red eyes and long wings 276 with purple eyes and vestigial wings 42 with red eyes and vestigial wings 30 with purple eyes and long wings Are the genes linked, unlinked, or sex-linked? If they are linked, how many map units separate them on the chromosome?Female Drosophila heterozygous for three recessive mutations e (ebony body), st (scarlet eyes), and ss (spineless bristles) were testcrossed, and the following progeny were obtained: Phenotype Number wild-type 67 ebony 8 ebony, scarlet 68 ebony, spineless 347 ebony, scarlet, spineless 78 scarlet 368 scarlet, spineless 10 spineless 54 (a) What indicates that the genes are linked? (b) What was the genotype of the original heterozygous females? (c) What is the order of the genes? (d) What is the map distance between e and st? (e) Between e and ss? (f) What is the coefficient of coincidence? (g) Diagram the crosses in this experiment.
- mapping gene The genes for ruby eyes (rb), tan body (t) and cut wings (ct) are all found on the X-chromosome of Drosophila melanogaster. All of these are recessive traits. They map in the order rb, ct, t with 12.5 map units between rb and ct and 7.5 map units between ct and t. Suppose you cross a cut wing male with a homozygous female that is both tan and has ruby eyes. What will the F1 females look like? Draw map of the section of the X chromosomes that has these 3 genes for the F1 females Assume you testcross your F1 females. What progeny classes would you expect? ii. Give approximate numbers for each class based on a total of 2000 progeny. Assuming the i=1 and there are no double crossovers. Assuming the i=0 and there are the expected number of double crossovers.16 - 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 the is homozygous for normal wings with a hairy body and a fly with vestigial wings that is homozygous for normal body hair. The wild-type F1 flies were crossed among each other to produce 1024 offspring. Which phenotypes would you expect among the 1024 offspring, and how many of each phenotype would you expect? a) Phenotypes: wild, vestigial, hairy, and vestigial hairy; Numbers expected: wild (256), vestigial (256), hairy (256), and vestigial hairy (256). b) O Phenotypes: wild, vestigial, hairy, and vestigial hairy; Numbers expected: wild (576), vestigial (192), hairy (192), and vestigial hairy (64). C) O Phenotypes: wild, vestigial, hairy, and vestigial hairy; Numbers expected: wild (192), vestigial (256), hairy (64), and vestigial hairy (192). d) All vestigial…Male Drosophila expressing the autosomal recessivemutations sc (scute), ec (echinus), cv (crossveinless),and b (black) were crossed to phenotypically wildtype females, and the 3288 progeny listed wereobtained. (Only mutant traits are noted.)653 black, scute, echinus, crossveinless670 scute, echinus, crossveinless675 wild type655 black71 black, scute73 scute73 black, echinus, crossveinless74 echinus, crossveinless87 black, scute, echinus84 scute, echinus86 black, crossveinless83 crossveinless1 black, scute, crossveinless1 scute, crossveinless1 black, echinus1 echinusa. Diagram the genotype of the female parent.b. Map these loci.c. Do the data provide evidence of interference?Justify your answer with numbers.
- 2-Female Drosophila heterozygous for three recessive mutations e (ebony body), st (scarlet eyes), and ss (spineless bristles) were testcrossed, and the following progeny were obtained: Phenotype wild-type ebony ebony, scarlet ebony, spineless ebony, scarlet, spineless scarlet scarlet, spineless spineless (a) (b) Number 134 16 (c) (d) (f) (h) 136 694 156 736 20 108 What indicates that the genes are linked? 1 mark What was the genotype of the original heterozygous females? 1 mark What is the order of the genes? 1 mark What is the map distance between e and st? 1 mark Between e and ss? 1 mark? 1 mark What is the coefficient of coincidence? 1 mark Calculate the interference? 1 mark Diagram the crosses in this experiment. 1 markIn drosophila, a recessive mutation (m-) of a maternal effect gene results in an abnormal phenotype wherein homozygous (m-m-) females produce eggs that cannot support embryonic development. Homozygous (m-m-) males, however, can still produce viable sperm. (A) Using m+ to denote a normal gene, determine the genotypes and phenotypes of the F1s produce by a cross between a heterozygous female and a recessive male. (B) From the offspring, backcross the recessive female with the paternal strain. What are the genotypes and phenotypes of the F2s? (C) If m-m- females produce useless eggs, then how are m-m- produced?You are given a Drosophila female that looks wild-type but is heterozygous for mutations intan body (t), miniature wings (m), and white eyes (w). You test cross this female with a tanbodied, miniature winged, and white-eyed homozygous mutant male, and you obtain thefollowing 1400 progeny: Phenotype : number+ + + : 608t m w : 516+ m w : 2t + + : 6+ m + : 39t + w : 46+ + w : 81t m + : 102 Calculate the distance between each pair t-m, m-w, and t-w only using the number ofrecombinants between them (i.e. ignoring the gene in the middle). Draw a linear map with thedistances between genes.