The boss in your laboratory has just heard of a proposal by another laboratory that genes for eye color and the length of body bristles may be linked in Drosophila. Your lab has numerous pure-breeding stocks of Drosophila that could be used to verify of refute genetic linkage. In Drosophila, red eyes
a. Give the genotypes of the pure-breeding parental-files, and the genotype
b. In your experimental design, what are the genotype and phenotype of the line you propose to cross to the F1 to obtain the most useful information about genetic linkage between the eye color and bristle-length genes? Explain why you make this choice.
c. Assume the eye color and bristle-length genes are separated by
d. How would the results of the cross differ if the genes are not linked?
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Genetic Analysis: An Integrated Approach (3rd Edition)
- Consider the following three autosomal recessive mutations in Drosophila:vestigial wings (v); wild type is long (v+)black body color (b); wildtype is gray (b+)plum eyes (p); wildtype is red (p+)A vestigal, gray, red female (homozygous for all three genes) is crossed with a long wing, black, plum male (homozygous for all three genes). The F1 female progeny are mated with triple homozygous recessive males. Here is the phenotypic data for the F2 progeny:vestigal; gray; red 580long wings; black; plum 592vestigal; black; red 45long; gray; plum 40vestigal; black; plum 89long; gray; red 94vestigal; gray; plum 3long; black; red 5A total of 1448 progeny were counted.Which one of the following values is the approximate distance between the plum eye color and black body color loci?arrow_forwardIn Drosophila, the wildtype eye color is black. In the laboratory, you screened for mutants than when homozygous results to different eye phenotypes. You found two mutants 1) red eye (re) and 2) white eye (we). You performed a complementation test and the resulting eye phenotype was gray. From the results, what conclusion can you make? A. There was complementation, thus the two mutations are alleles of different genes. B. There was no complementation, thus the two mutations are alleles of different genes. C. The two mutations failed to complement, thus they are alleles of different genes. D. The two mutations failed to complement, thus they are alleles of the same genes. E. The two mutations complemented, hence they are likely controlled by different genes.arrow_forwardDrosophila can have a yellow-brown body (wild-type [+]) or a black body [b] as a result of a mutation in the black gene (marked b). a mutation in the black gene (marked b). There is also a mutation in the purple gene (marked pr) which purple eyes (mutant phenotype [pr]). Flies with normal eyes are noted [+]. a/ Pure strains of male Drosophila [b] are crossed with female Drosophila [pr] (1st cross). The F1 progeny contains only wild-type Drosophila for the two [+, +] traits. both [+, +] traits. The reciprocal cross gives the same result. F1 females from the first cross are crossed with [b, pr] double mutant males. The progeny is composed of : 944 [b, +] 936 [+, pr] 56 [+, +] 64 [b, pr] How do you explain this segregation? How many percentage recombinationarrow_forward
- A maternal effect gene in Drosophila, called torso, is found as a recessive allele that prevents the correct development of anterior- and posterior-most structures. A wild-type (homozygous) male is crossed to a female of unknown genotype. This mating produces 100% larva that are missing their anterior- and posterior-most structures and therefore die during early development. What is the genotype and phenotype of the female fly in this cross? What are the genotypes and phenotypes of the female fly’s parents? Show COMPLETE cross.arrow_forwardIn Drosophila melanogaster, black body (b) is recessive to gray body (b"), purple eyes (pr) are recessive to red eyes (pr"), and vestigial wings (vg) are recessive to normal wings (vg"). The loci encoding these traits are linked, with the following map distances: pr vg 13 The interference among these genes is 0.5. A fly with a black body, purple eyes, and vestigial wings is crossed with a fly homozygous for a gray body, red eyes, and normal wings. The female progeny are then crossed with males that have a black body, purple eyes, and vestigial wings. If 1000 progeny are produced by this testcross, what will be the phenotypes and proportions of the progeny?arrow_forwardRed eyes (r), brown color (b), and curled wings (c) are 3 recessive mutations that occur in house flies. These genes have already been mapped: r-b = 12 mu b-c = 18 mu r-c = 30 mu A fly with red eyes, brown color, and curled wings is crossed with a fly homozygous for the wild-type traits. The F1 males are crossed with females with the three mutant traits and 2000 progeny are produced. Find the number of observed double crossovers if the interference is 0.38. O 43 O 62 O 27 O 16arrow_forward
- In drosophila, a recessive mutation (m-) of a maternal effect gene results in an abnormal phenotype wherein homozygous (m-m-) females produce eggs that cannot support embryonic development. Homozygous (m-m-) males, however, can still produce viable sperm. (A) Using m+ to denote a normal gene, determine the genotypes and phenotypes of the F1s produce by a cross between a heterozygous female and a recessive male. (B) From the offspring, backcross the recessive female with the paternal strain. What are the genotypes and phenotypes of the F2s? (C) If m-m- females produce useless eggs, then how are m-m- produced?arrow_forwardIn wheat, aleurone cells form a thin layer of the seed coat that is critical to early gene expression in plant development. The color of this layer of cells is controlled by two alleles of a gene [colored aleurone (R) is dominant to colorless (r)]. A second gene is known to control the color of leaf tips [green leaf tip (G) is dominant to yellow (g)]. Two plants, each heterozygous for both characteristics, are test crossed to homozygous recessives, and their progeny are combined to produce the following totals: colored green 102 colored yellow 98 colorless green 103 colorless yellow 97 a) Use chi-square analysis to test these data for an independent assortment of the two characteristics (table provided). Please show work, how your expected values are calculated, and explain what your results indicate about the data. b) You decide to be cautious in your analysis, and decide to analyze the progeny from each of the crosses individually (instead of adding them together as shown above).…arrow_forwardBlack body (b) and purple eye (pr) are recessive autosomal mutations in Drosophila. Bridges are crossed b/b females with pr/pr males and in the F2 observed 684 wild type, 371 black-bodied, and 300 purple-eyed flies. Do these results indicate that the b and pr genes are closely linked? Explain. (Remember that there is no crossing-over in male Drosophila)..arrow_forward
- Females of wild-type Strain A and males of mutant Strain B, as well as females of mutant Strain B and males of wild-type Strain A, make reciprocal crosses. Explain why reciprocal crosses are needed in genetics experiments involving Drosophila fruit flies.arrow_forwardAn undergraduate researcher in your lab is studying mutations affecting the wings of Drosophila melanogaster. She has identified two mutant phenotypes of interest: bent wings (be), which are recessive to the wild-type straight wings (be+), and apterous (ap) mutants (which are wingless). The apallele is recessive to the wild-type allele (ap+), which allows wings to develop. If a homozygous bent-winged fly (which possesses the normal allele of apterous) is crossed with a homozygous wingless fly (which possesses the normal allele of bent wings), what phenotypic ratio would you expect to observe in the F2 generation of this cross? a) Please indicate the ratio, including the genotypes and phenotypes of all phenotypic classes. Phenotype: Genotype(s) corresponding to this phenotype Phenotypic ratio: (Be sure to NAME the classes in the ratio). B) Please NAME and DEFINE the type of gene interaction illustrated in this example.arrow_forwardIn Drosophila melanogaster, black body (b) is recessive to gray body (b*), purple eyes (pr) are recessive to red eyes (pr*), and vestigial wings (vg) are recessive to normal wings (vg*). The loci encoding these traits are linked, with the following map distances: b pr vg 6- 13 The interference among these genes is 0.5. A fly with a black body, purple eyes, and vestigial wings is crossed with a fly homozygous for a gray body, red eyes, and normal wings. The female progeny are then crossed with males that have a black body, purple eyes, and vestigial wings. If 1000 progeny are produced by this testcross, what will be the phenotypes and proportions of the progeny?arrow_forward
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