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A wild-type fruit fly (heterozygous for gray body color and red eyes) is mated with a black fruit fly with purple eyes. The offspring are wild-type, 721; black-purple, 751; gray purple, 49; black-red, 45. What is the recombination frequency between these genes for body color and eye color? Using information from Figure 12.9, what fruit flies (genotypes and
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Campbell Biology in Focus; Modified Mastering Biology with Pearson eText -- ValuePack Access Card -- for Campbell Biology in Focus (2nd Edition)
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- You cross a true-breeding yellow-bodied, smooth-winged female fly with a true-breeding red-bodied, crinkle-winged male. The red body phenotype is dominant to the yellow body phenotype and smooth wings are dominant to crinkled wings. Use B or b for body color alleles, and W or w for wing surface alleles.(4 points) a) What are the genotypes of the P generation flies? b) What will be the genotype(s) and phenotype(s) of the F1 offspring? c) You discover that the genes for body color and wing surface are linked. You perform a dihybrid test cross between the F1 flies from part (b) with a true-breeding yellow-bodied, crinkle-winged fly. Use the following results of this cross to determine the recombination frequency (%) between the body color and wing surface genes. (Remember that the recombinants are the ones that do not resemble the parental types from the P generation.) Body Color Wing Surface # of Individuals red smooth 102 yellow smooth 404 red crinkled 396 yellow crinkled…arrow_forwardA homozygous red pigmented female beetle (bb) is bred with a black pigmented male of unknown genotype. It’s unknown if the male homozygous dominant (BB) or heterozygous (Bb) for the pigment trait. The color of their offspring will help identify the genotype of the male. What do you call this method used to determine the genotype of the male beetle? Explain your answer by using Punnett square to show possible offspring genotypes.arrow_forwardThe recessive allele s causes Drosophila to have small wings, and the s+ allele causes normal wings. This gene is known to be X linked. If a small-winged male is crossed with a homozygous wild-type female, what ratio of normal to small-winged flies can be expected in each sex in the F1? If F1 flies are intercrossed, what F2 progeny ratios are expected? What progeny ratios are predicted if F1 females are backcrossed with their father?arrow_forward
- In 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.arrow_forwardA wild-type fruit fly (heterozygous for gray body color andred eyes) is mated with a black fruit fly with purple eyes. Theoffspring are wild-type, 721; black purple, 751; gray purple, 49;black red, 45. What is the recombination frequency betweenthese genes for body color and eye color? Using informationfrom problem 3, what fruit flies (genotypes and phenotypes)would you mate to determine the order of the body color, wingsize, and eye color genes on the chromosome?arrow_forwardIf a fruit fly that has long wings and a gray body (LIGG) mated with another fruit fly that has long wings and a gray body (LIGG), what would be the phenotypic ratio of their offspring? Also, list the different phenotypes. Do the cross! L=long wings, l=short wings, G=gray body, g=black body. 6) Phenotypic ratio:arrow_forward
- The genetic map was based on crosses in Drosophila involving the three sex-linked genes a, b and c. “a” gives red eyes, “b” gives normal wings and “c” gives black body. The recombination frequencies between these genes are as follows; a and b is 23.8, b and c is 2.6 and a and c is 28.1, respectively. Could you draw a basic genetic map based on distance between these genes using dots to show distance(s)? Could you make one fundamental comment using these distances based on genetic linkage?arrow_forwardIn an insect species, the genes for white eyes (w) and smooth legs (s) are linked and are located 16 map units apart. A white-eyed, smooth-legged female was mated to a true-breeding wild-type male; the resulting F1 phenotypically wild-type females were mated to white-eyed, smooth-legged males. In an F2 of 100 individuals, what would be the expected number of F2 individuals for each phenotype? wild-type: white-eyed: smooth-legged: white-eyed and smooth-legged:arrow_forwardA female from true breeding line of Drosophila with white eyes is crossed with a male from a true breeding line with brown eyes. All of the offspring have wild type brick red eyes. Which of the following explanations is most likely? A) There are many alleles for the single gene for eye color. Wild type brick red eyes result only when the fly is heterozygous. B) The alleles for brown, white, and brick red eyes are alleles for a single locus. The allele for brown eye color is dominant to the allele for brick red eye color and to the allele for white eyes. C) There is more than one gene for eye color. The brown mutation and the white mutation occur in separate genes and are both recessive to the wild type alleles. The offspring are heterozygous for both genes, so they are phenotypically wild type. D) None of the above. It is not possible for a cross between a white-eyed and a brown-eyed fly to produce wild type offspring.arrow_forward
- 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 =1arrow_forwardBees have a haplodiploid sex determination. The recessive allele ivory causes bees to have white eyes. A white-eyed female is mated to a wild-type male. An F1 female is mated to a white-eyed male and some of her eggs are fertilized. She lays eggs after the mating. What are possible phenotypes of the offspring? i) wildtype males ii) wildtype females iii) white-eyed males iv) white-eyed females Group of answer choices A. only i and iii are possible oucomes B. only i and ii are possible oucomes C. all four (i, ii, iii, and iv) are possible phenotypes D. only ii and iii are possible oucomes E. only i,ii, and iii are possible oucomesarrow_forwardIn sesame plants, the one-pod condition (P) is dominant to the three-pod condition (p), and normal leaf (L) is dominant to wrinkled leaf (l). Pod type and leaf type are inherited independently. Determine the genotypes for the two parents for all possible matings producing the following offspring: 318 one-pod, normal leaf and 98 one-pod, wrinkled leaf. show your solution using punnet squarearrow_forward
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