2-Female Drosophila heterozygous for three recessivemutations e (ebony body), st (scarlet eyes), and ss(spineless bristles) were testcrossed, and the followingprogeny were obtained:Phenotypewild-typeebonyebony, scarletebony, spinelessebony, scarlet, spinelessscarletscarlet, spinelessspineless(a)(b)Number13416(c)(d)(f)(h)13669415673620108What indicates that the genes are linked? 1 markWhat was the genotype of the originalheterozygous females? 1 markWhat is the order of the genes? 1 markWhat is the map distance between e and st? 1 markBetween e and ss? 1 mark? 1 markWhat is the coefficient of coincidence? 1 markCalculate the interference? 1 markDiagram the crosses in this experiment. 1 mark

a) When all the progeny are not shown at equality then this indicates that the genes are linked…

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 (g) 136 694 156 108 What indicates that the genes are linked? 1 mark What was the genotype of the original heterozygous females? 1 mark 736 20 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 mark

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 (g) 136 694 156 108 What indicates that the genes are linked? 1 mark What was the genotype of the original heterozygous females? 1 mark 736 20 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 mark

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter13: Genes, Chromosomes, And Human Genetics

Section: Chapter Questions

Problem 5TYK

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Similar questions

The gene for flower position in pea plants exists as axial or terminal alleles. Given that axial is dominant to terminal, list all of the possible F1 and F2 genotypes and phenotypes from a cross involving parents that are homozygous for each trait. Express genotypes with conventional genetic abbreviations.
Does the phenotype indicated by the red circles and squares in this pedigree show an inheritance pattern that is autosomal dominant, autosomal recessive, or X-linked?
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?
7-5. In corn a pair of genes determines leaf shape and another pair determines pollen shape. A ragged-leafed plant with round pollen was crossed to a ragged-leafed plant with angular-pollen, and the resultant progeny were classified as follows: Class Phenotypes 186 ragged-leaf round-pollen 11 IV 174 ragged-leaf angular-pollen 57 smooth-leaf round-pollen Total 63/480 smooth-leaf angular-pollen (a) Using alphabetical letters of your choice, designate the genes for the different leaf and pollen characters. (b) On the basis of the symbols given in (a) 1 provide genotypes two parents. (c) (forth)/(nbers) According to your hypothesis what would you have expected for each of the four classes of progeny? (d) After re viewing Chapter 8, use the chi-square method (p. 132) to test your hypothesis and indicate whether you accept or reject it.
13 Black coat color (B) in cocker spaniels is dominant to white coat color (b). Solid coat pattern (S) is dominant to spotted pattern (s). The pattern arrangement is located on a different chromosome than the one for color, and its gene segregates independently of the color gene. A male that is black with a solid pattern mates with three females. Indicate the genotypes of the parents. a) The mating with female A, which is white, solid, produces four pups: two black, solid, and two white, solid. b) The mating with female B, which is black, solid, produces a single pup, which is white, spotted. c) The mating with female C, which is white, spotted, produces four pups: one white, solid; one white spotted; one black, solid; one black, spotted.
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…
13 - Assume that you have a garden and some pea plants have solid leaves and others have striped leaves. You conduct a series of crosses [{a) through (e)] and obtain the results given in the table. Define gene symbols and give the possible genotypes of the parents of cross e. Cross Progeny solid striped solid x striped solid x solid 55 60 (a) (b) (c) (d) (e) 36 striped x striped solid x solid 65 92 30 solid × striped 44 a) O solid is dominant to striped Ss x Ss b) O striped is dominant to solid Ss x ss C) O solid is dominant to striped SS x ss d) O striped is dominant to solid SS x ss e) solid is dominant to striped Ss x ss
18 The genes dumpy(dp) and clot(cl) are linked on chromosome II of Drosophila. A heterozygous female was testcrossed with a male. The progeny were • 11% dumpy. • 996 clot, • 4196 wildtype. 3996 dumpy-clot. What was the genotype of the heterozygous female? a) O dp dp"/cl cl D)O dp'dp/cl" cl DO dp'dp cl"cl 0 dp cl/ dp cl* e) O dpt cl/dp cl Boş bırak
8-9. White eyes (w) is sex-linked and recessive in Drosophila. Cruved wings(cu) and black body (b) are autosomal recessive alleles on the same chromosome, separated by 20 map units. If one breeds a homozygous white, curved winged female fly to a homozygous black male fly to produce an F1, and inbreeds the F1 males and females, at what frequency is a phenotypically wild-type male expected? (note: male flies do not cross over) A. 0.25 es B.0.05 C.0.10 D. 0.20 E.0.125
7 Cinnabar eyes is a X-linked recessive characteristic in fruit flies. If a female having cinnabar eyes is crossed with a normal wild-type male, what percentage of the F1 males will have cinnabar eyes?
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.
23- In pea plants, axial flower position is dominant over terminal flower position. A student crosses a heterozygous axial flower with a plant with terminal flowers. What percentage of the plants will exhibit a homozygous genotype? a) 100% b) 0% c) 50% d) 75% 24- In pea plants, round seed is completely dominant over wrinkled seed. What genotype will result if a homozygous round seed plant is cross with a wrinkled seed plant? a. RR b. Rr c. Rr d. All of the above can be obtained