n 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?
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In 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?
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- In 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?In Drosophila melanogaster, red eyes are dominant over white and the variation for this characteristic is on the X chromosome. Vestigial wings (v) are recessive to normal (V) for an autosomal gene. Predict the appearance of offspring of the following crosses: XW/Xw V/v×Xw/Y v/v, Xw/Xw V/v × XW/Y V/v.In Drosophila melanogaster, red eyes are dominant over white and the variation for this characteristic is on the X chromosome. Vestigial wings (v) are recessive to normal (V) for an autosomal gene. Predict the appearance of offspring of the following crosses: XW/XwV/v×Xw/Y v/v, Xw/XwV/v×XW/Y V/v.
- In Drosophila, the sepia mutation (se, chromosome 3, position 26) results in dark brown eyes, while cinnabar (cn, chromosome 2, position 57.5) results in bright orange-red eyes. True breeding, wild type females are mated with true breeding males homozygous recessive for both traits. Using Drosophila notation, diagram the P1 and F1 crosses. P1 F1 Fill in the chart with phenotypic ratios that would be expected in the F2 generation. Use the space provided to show your work. Phenotype Females Males Overall (♀and ♂) =1 =1In silkmoths (Bombyx mori), red eyes (re) and white-banded wings (wb) are encoded by two mutant alleles that are recessive to those that produce wild-type traits (re+ and wb+); these two genes are on the same chromosome. A moth homozygous for red eyes and white-banded wings is crossed with a moth homozygous for the wild-type traits. The F1 have wild-type eyes and wild-type wings. The F1 are crossed with moths that have red eyes and white-banded wings in a testcross. The progeny of this testcross are wild-type eyes, wild-type wings red eyes, wild-type wings wild-type eyes, white-banded wings red eyes, white-banded wings a. What phenotypic proportions would be expected if the genes for red eyes and for white-banded wings were located on different chromosomes? b. What is the rate of recombination between the gene for red eyes and the gene for white-banded wings?In Drosophila,, the curled mutation (cu, chromosome 3, position 50.0) results in wings that curl up, while ebony (e, chromosome 3, position 70.7) results in a dark body. True breeding, wild type females are mated with true breeding males with curled wings and ebony bodies. Considering Drosophila notation, which of the following correctly diagrams the F1 cross? X X 3+ cu e + X X e + + + + + cu e + O + ■ 3+ X X X X Y Y + + ■ cu cu cu ' + ■ cu ■ ' + e + e e e e e + cu +
- In Drosophila, the dominant Bar mutation (B, chromosome X, position 57) results in thin bar- shaped eyes, while the recessive singed (sn, chromosome X, position 21) results burnt looking bristles. True breeding, wild type females are mated with true breeding males with Bar eyes and singed bristles. Using Drosophila notation, diagram the P1 and F1 crosses. P1 F1 Fill in the chart with phenotypic ratios that would be expected in the F2 generation. Use the space provided to show your work. Phenotype Females Males Overall (♀and ♂) =1 =1 =1Miniature wings (Xm) in Drosophila result from an X-linked allele that is recessive to the allele for long wings (X*). Give the genotypes of the parents in the following cross: Male parent Female parent Male offspring Female offspring Long Miniature 750 miniature 761 long O male: X* / X* and female X™ /x+ O male: X*/Xt and female Xm /xm O male: X*/ Y and female Xm /xm O male: Xm/ Y and female Xm /xmIn Drosophila, a fully heterozygous female with the X-linked recessive genes a, b, and c (not necessarily in that order on the chromosome) was mated to a male that was genetically a, b, c (not necessarily in that order on the chromosome). The offspring occurred in the following phenotypic ratios: Phenotypes: Numbers: What is the cis/trans arrangement in the heterozygous parent? Wild 426 а, с, b 428 Which gene is in the middle? a 23 c, b 22 If you added 23, 22, 3, and 2, it would give you the map distance between genes C 49 b, a 46 What calculation would you make to determine if interference was occurring? (you don't have to complete the calculation) b. C, a Total 1000 3.
- In Drosophila fruit flies, the genes for warped wings (dwp), rumpled bristles (rmp), and pallid wings (pld) are linked. A trihybrid female for all three allleles is crossed with homozygous recessive male for all three alleles and the offspring obtained showed the following phenotypes: 3 pld rmp dwp 428 pld rmp + 427 + + dwp 48 + rmp + 47 pld + dwp 23 pld + + 2 + + + 22 + rmp dwp What is the order and map distance between these three alleles?Bar (B) is a dominant sex-linked mutant of D. melanogaster. Held-out wings and ebony body are recessive, autosomal mutants on chromosome III, mapping 12 map units apart. A Bar-eyed, ebony-bodied male was crossed to a held-out female, and the resulting F1 progeny WERE INBRED to produce the F2. At what frequency do you expect wild-type flies from this cross? (Reminder: Crossing over does not occur in the male). Based on the information in Problem 3, how often would you get a bar ebony but not held-out progeny?In Drosophila, an X-linked recessive mutation, Xm causes miniature wings. List the F2 phenotypic ratios if: a miniature-winged female is crossed with a normal male and a miniature-winged male is crossed with a normal female. What would the phenotypic ratio from (a) be if the miniature-winged gene were autosomal? Assume in all cases that the P1 individuals are true-breeding.