A female fly is heterozygous for three X-linked genes. The dominant alleles are ABC and the recessive alleles are abc. She is crossed with an abc/Y male. The following progeny phenotypes are produced. These phenotypes are equally divided between male and female flies. Abc 420 aBC 435 abc 55 ABC 60 abC 14 ABC 16 What is the distance between a and c in mu?
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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, 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?Miniature 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 /xm
- In Drosophila, two genes w and s are X-linked and are 25 map units apart. A heterozygous female fly with a genotype w* s*/ws (w* and s* are wild-type alleles) is crossed to a male wild-type fly. What percent of progeny will have only the wild- type alleles of the w and s genes? O 50% 75% O 12.5% 37.5% 25%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 P1 cross? X X ++ e + + + O+ X + X + ■ + X + + + 3+ X X X X + + Y Y cu cu cu + cu cu J e e e e e (D e + cu cu (DIN 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.
- A mutant sex-linked trait called “notched” (N) is deadly in Drosophila when homozygous in females. Males who have a single N allele will also die. The heterozygous condition (Nn) causes small notches on the wing. The normal condition in both male and females is represented by the allele n. a) Indicate the phenotypes of the F1 generation from the following cross: XNXn x XnY b) Explain why dead females are never found in the F1 generation no matter which parents are crossed. c) Explain why the mating of female XNXn and a male XNy is unlikely.In Drosophila, the brown mutation (bw, chromosome 2, position 104.5) results in brown eyes, while miniature (min, chromosome X, position 36.1) results in wings that are 2/3 the length of wild type. True breeding, wild type females are mated with true breeding males with brown eyes and miniature wings. 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 =1In 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 =1
- A wild-type male and a wild-type female Drosophila with red eyes and full wings are crossed. Their progeny are shown below: Males Females 3/8 Full wing, Red eye ¾ Full wing, Red eye 3/8 Miniature wing, Red eye ¼ Full wing, Purple eye 1/8 Full wing, Purple eye 1/8 Miniature wing, Purple eye 1. What is/are the genotype(s) of females with purple eye? Of males with purple eye and miniature wing? Draw both out with appropriate symbolsIn 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.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 +