To review:
The results of the cross between two Drosophila melanogaster with long wings and red eyes—the wild type
(a) Determine the curved-wing allele autosomal recessive, autosomal dominant, sex-linked recessive, or sex-linked dominant.
(b) Determine the lozenge-eyed allele autosomal recessive, autosomal dominant, sex-linked recessive, or sex-linked dominant.
(c) The genotype of the female parent.
(d) The genotype of the male parent.
Introduction:
Drosophila melanogaster are the fly species that are also known as vinegar fly. Further, due to the accessible features of the Drosophila, it is used as a model system extensively used in laboratories for the research purpose. The results of the cross between two Drosophila melanogaster with long wings and red eyes—the wild type phenotype is given in table below.
Females | Males |
600 long wings, red eyes | 300 long wings, red eyes |
200 curved wings, red eyes | 100 curved wings, red eyes |
300 long wings, lozenge eyes | |
100 curved wings, lozenge eyes |
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Biological Science (6th Edition)
- You are mapping three linked loci in Drosophila melanogaster (the common laboratory fruit fly). You cross flies that are triply mutant for apricot (pale eyes), bristle (extra bristles) and clipped (notched wings) to wild-type flies. The F+ flies are wild-type in appearance. You then backcross the F+ females to pure-breeding (apricot, bristle, clipped) males and score the phenotypes of 1000 F progeny for all three loci. Here are the results: 359 wild-type 361 apricot, bristle, clipped 89 bristle, clipped 91 apricot 42 apricot, bristle 38 clipped 9 apricot, clipped 11 bristle Using these data, first determine what gametes from the F; trihybrid produced each of the eight F2 categories. Note that apricot = aa (recessive to wild-type A); bristle = bb (recessive to wild-type B); and clipped = cc (recessive to wild-type C). Then determine if each gamete is recombinant (R) or nonrecombinant (R) for each pair of alleles (that is, for each genetic interval). Complete the table by dragging the…arrow_forwardIn Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn and ct+ is crossed to a sn+ct male. The F1 flies are interbred. The F2 males are distributed as follows: genotype number sn ct 15 sn ct+ 34 sn+ ct 33 sn+ct+ 18 What is the map distance between sn and ct?arrow_forwardImagine that you set up a three-point mapping cross to determine the order of three X-linked genes. You have recessive mutations available for all three genes: yellow, white, and echinus. A heterozygous F1 Drosophila female is crossed to hemizygous mutant male. The observed numbers and phenotypes of F2 are as follows. 4685 yellow, white, echinus 4759 wild type 80 yellow 70 white, echinus 193 yellow, white 207 Echinus 3yellow, echinus 3white Based on this data which of the above phenotypic classes represent the double crossing over? Each answer yields a clue useful in following question. Choose all that applyarrow_forward
- Imagine that you set up a three-point mapping cross to determine the order of three X-linked genes. You have recessive mutations available for all three genes: yellow, white, and echinus. A heterozygous F1 Drosophila female is crossed to hemizygous mutant male. The observed numbers and phenotypes of F2 are as follows. 4685 yellow, white, echinus 4759 wild type 80 yellow 70 white, echinus 193 yellow, white 207 Echinus 3yellow, echinus 3white Based on this data which of the above phenotypic classes represent the PARENTAL types? Each answer yields a clue useful in following question. Choose all that applyarrow_forwardPhenotypically wild-type F1 female Drosophila, whose mothers had light eyes (lt) and fathers had straw (stw) bristles, produced the following offspring when crossed to homozygous light eyed, straw bristled males: Phenotype Number light-straw 140 wild-type 160 light 360 straw 340 Total 1000 Compute the map distance between the light and straw loci. Group of answer choices 70 map units 3 map units 7 map units 0.03 map units 30 map unitsarrow_forwardFemale 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.arrow_forward
- In fruit flies, long wings (M) is dominant to miniature wings (m) and red eyes (B) is dominant to brown eyes (b). You testcross a long-winged, red-eyed fly (genotype Mm Bb) female to a mini-winged, brown-eyed male (genotype mm bb) and get the following progeny: 256 long wings, red eyes 261 mini wings, red eyes 240 long wings, brown eyes 243 mini wings, brown eyes Are these genes for wing length and eye color linked or unlinked? How do you know?arrow_forwardThe data set attached summarizes F2 numbers from an F1 cross arising from two, true-breeding Drosophila strains (P generation), which differ with respect to two mutant traits. Here are the hypothesis: Leg length - The wild-type and mutant alleles for leg length are incomplete dominant relative to each other. Justification: The data set includes three phenotypic categories for leg length: wild type (long leg), medium leg, and truncated wings. The presence of three distinct phenotypes suggests an incomplete dominance pattern, where the heterozygous individuals exhibit an intermediate leg length phenotype (medium leg). The absence of purebred short-legged individuals supports the idea that the long leg allele is dominant over the short leg allele. This shows that mode of inheritance is incomplete dominance of the alleles relative to each other. Since the data does not mention any specific differences between males and females, we can assume that the mode of inheritance for the trait is…arrow_forwardThe production of eye-color pigment in Drosophila requires the dominant allele A. The dominant allele P of a second independent gene turns the pigment to purple, but its recessive allele leaves it red. A fly producing no pigment has white eyes. Two pure lines were crossed with the following results:P red-eyed female white-eyed maleF1 purple-eyed femalesred-eyed malesF1 F1purple eyed 38red eyed 38white eyed 28F both males and females: 2Explain this mode of inheritance, and show the genotypes of the parents, the F1, and the F2.arrow_forward
- Two Drosophila flies that had normal (transparent, long) wings were mated. In the progeny, two new phenotypes appeared, dusky wings (having a semi-opaque appearance) and clipped wings (with squared ends). The progeny were as follows: Females: 179 transparent, long 58 transparent, clipped Males: 92 transparent, long 89 dusky, long 28 transparent, clipped 31 dusky, clipped a) Provide a genetic explanation for these results, showing genotypes of parents and of all progeny classes under your model. b) Design a test for your model.arrow_forwardIn Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn+ and ct+ is crossed to a sn ct male. The F1 flies are interbred. The F2 males are distributed as follows sn ct 36 sn ct+ 13 sn+ ct 12 sn+ ct+ 39 What is the map distance between sn and ct?arrow_forwardF1 female Drosophila heterozygous for cinnabar eye (cn), vestigial wings (vg) and roof wings (rf) genes were test crossed with males homozygous for all three traits. The following result were obtained. 382 cinnabar, vestigial 401 roof 3 cinnabar 4 roof, vestigial 59 cinnabar, roof, vestigial 67 wild 44 cinnabar, roof 40 vestigial The chromosomal interference is equal to? 0.24 0.32 0.42 0.58 0.49arrow_forward
- Biology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage Learning