In Drosophila, two genes, one for body color and one for eye color, are carried on the same chromosome. The wild-type gray body color is dominant to black body color, and wild-type red eyes are dominant to purple eyes. You make a cross between a fly with gray body and red eyes and a fly with black body and purple eyes. Among the offspring, about half have gray bodies and red eyes and half have black bodies and purple eyes. A small percentage have: (a) black bodies and red eyes; or (b) gray bodies and purple eyes. What alleles are carried together on the chromosomes in each of the flies used in the cross? What alleles are carried together on the chromosomes of the F1 flies with black bodies and red eyes, and those with gray bodies and purple eyes?
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Chapter 13 Solutions
Biology: The Dynamic Science (MindTap Course List)
- . In fruit flies (Drosophila melanogaster), the bn+ allele for normal dull red eyes is dominant to the bn allele that gives brown eyes. Another gene affects wing shape; for this gene, the ct+ allele for normal wings is dominant to the ct allele, which gives “cut” wings, with jagged edges. A fly with dull red eyes and normal wings was crossed with a fly that had dull red eyes and cut wings, and the following progeny were obtained: 16 dull red eyes, normal wings 14 dull red eyes, cut wings 5 brown eyes, normal wings 5 brown eyes, cut wings What were the genotypes of the parents?arrow_forwardIn Drosophila, vermilion eye color is due to a recessive allele (v) located on the X chromosome. Curved wings are due to a recessive allele (cu) located on one autosome, and ebony body is due to a recessive allele (e) located on another autosome. A vermilion male is mated to a curved, ebony female, and the F1 males are phenotypically wild-type. If these males were backcrossed to curved, ebony females, what proportion of the F2 offspring will be wild-type males?arrow_forwardThe genes for mahogany eyes and ebony body are approximately 18 map units apart on chromosome III in Drosophila. Assume that a mahogany-eyed female was mated to an ebony-bodied male, and the resulting F1 phenotypically wild-type females were mated to mahogany-ebony males. Of 942 offspring, what would be the expected phenotypes and in what numbers would they be expected?arrow_forward
- The mutant genes for vestigial wings and singed bristles are approximately 30 map units apart on chromosome II in Drosophila. Assume that a vestigial-winged female was mated to a singed-bristle male, and that the resulting F1 phenotypically wild type females were mated to vestigial singed males. Of 1000 offspring, which phenotype class would represent the product of crossing over between the genes, and how many would you expect? a) vestigial, 300 flies b) vestigial, 150 flies wild type, 300 flies d) singed, 150 flies e) vestigial, singed double mutants, 150 fliesarrow_forwardIn Drosophila, a cross was made between a yellowbodied male with vestigial (not fully developed)wings and a wild-type female (brown body). The F1generation consisted of wild-type males and wild-typefemales. F1 males and females were crossed, and theF2 progeny consisted of 16 yellow-bodied males withvestigial wings, 48 yellow-bodied males with normalwings, 15 males with brown bodies and vestigialwings, 49 wild-type males, 31 brown-bodied femaleswith vestigial wings, and 97 wild-type females.Explain the inheritance of the two genes in questionbased on these results.arrow_forwardWhat are the F1 phenotypes expected from a cross between a heterozygous red-eyed female fruit fly (XWXw) and a hemizygous red-eyed male fruit fly (XWY)? half of the F1 males will have white eyes, and all F1 females will have red eyes half of the F1 females will have white eyes, and half of the F1 males will have red eyes all F1 males will have white eyes, and all F1 females will have red eyes all F1 males will have white eyes, and all F1 females will have white eyes all F1 progeny will have red eyesarrow_forward
- Two pure-breeding strains of flies are mated, and the F1 are intercrossed. The first strain has curled wings and black bodies. The second strain has straight wings and brown bodies. The F2 progeny are 271 straight wings with brown bodies, 31 curled wings with black bodies, 94 curled wings with brown bodies and 90 straight wings with black bodies. If the F1 were backcrossed to the straight, wing brown bodied parent, what phenotypes would be produced among the progeny? What would be the proportion of each phenotype?arrow_forwardA 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 symbolsarrow_forwardTwo pure-breeding strains of flies are mated, and the F1 are intercrossed. The first strain has curled wings and black bodies. The second strain has straight wings and brown bodies. The F2 progeny are 271 straight wings with brown bodies, 31 curled wings with black bodies, 94 curled wings with brown bodies and 90 straight wings with black bodies. If instead of the above, assume the wing shape gene and the body color gene are completely linked. From parents that are curled winged with brown bodies mated to straight winged with black bodies, what would be the outcome of an F1 intercross? (Specify the phenotypes and the frequency of each expected).arrow_forward
- You are doing a cross with Drosophila using the following two traits. Curly wings is dominant over straight wings, and round eyes is dominant over elliptical eyes. You cross a female fly that is known to be heterozygous for both genes with a male that is heterozygous for the wing gene but has elliptical eyes. This cross produces 74 flies with curly wings and round eyes, 61 with curly wings and elliptical eyes, 24 with straight wings and round eyes, and 21 with straight wing and elliptical eyes. Calculate the expected phenotype ratios for this cross, then use the chi-square test to see if the observed data are consistent with the expected numbers.arrow_forwardIn flies, assume black (B) is dominant and green (b) is recessive for abdomen color, and straight (S) is dominant and bent (s) is recessive for antenna shape. Assume also that these genes are found on the same chromosome. If single mutants for each of these genes (green abdomen x bent antenna) are crossed with one another, and you then testcross the resulting offspring, you get the following numbers of each phenotype: black, straight 17 green, bent 12 black, bent 337 green, straight 364 What is the map distance between B and S? You must show and clearly label ALL workarrow_forwardIn the fruit fly Drosophila melanogaster, the trait of black body is due to a gene on chromosome 2 and black body b is recessive to wild type body b + . The trait of purple eyes is controlled by a gene that is also on chromosome 2 and purple eyes p is recessive to wild type eyes p + . A true-breeding wild type strain is crossed with a true breeding strain that has black bodies and purple eyes. The F1 generation is then testcrossed to the black body, purple eye strain and 500 progeny are produced as follows: 224 wild type for both body and eye 236 black body and purple eye 18 wild type body and purple eye 22 black body and wild type eye. What is the recombination frequency and genetic map distance between the two genes?arrow_forward
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