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Here are schematic diagrams of mutant Drosophila larvae.
The left side of each pair shows a wild-type larva, with gray boxes showing the sections that are missing in the mutant larva. Which type of gene is defective in each larva: a gap gene, a pair-rule gene, or a segment-polarity gene?
![(a)
(b)
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- Drosophila 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. What information can you draw from this?A contig map for a chromosome in Drosophila is shown below. The chromosome has 10 segments, numbered 1- 10. Drosophila Chromosome 1 2 4 7 9 10 A Genomic Clones C D E If an EST hybridizes with genomic clones C, D and E, but not with the other clones, in which segment of the chromosome is the EST found? Enter a number. If a clone of gene GLO hybridizes only with genomic clones F and G, in which chromosome segment is gene GLO located? Enter a number. If a restriction fragment hybridizes with only one of the genomic clones, in which chromosome segment(s) could the fragment be found? Enter numbers for each segment ordered from smallest to largest separated by commas and no spaces. (Le. 4.8,10)Researchers have exploited Minute mutations in orderto study the phenotypes associated with recessive lethal mutations (l−) that decrease the rate of cell divisionand thus make only very tiny homozygous mutant clones that are difficult to analyze. Many differentstrains of Drosophila carry dominant loss-of-functionMinute (M) mutations in a variety of genes encodingribosomal protein subunits. The M genes are haploinsufficient; flies with only one wild-type M+ gene copyhave a slower pace of cell division, and thus prolongeddevelopment and subtle morphological abnormalities.To circumvent the tiny clone problem, researchersgenerate GFP-marked homozygous l−/ l− clones thatare also M+/ M+, in flies that are l−/ l+ and M−/ M+.The loss of the Minute mutation only in cells withinthe clone gives the l−/ l− cells a growth advantageover their neighbors, enabling the mutant clone togrow large enough to study. Diagram chromosomesthat could be used to generate such clones
- 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?Red 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 16In c. elegans, genetics model organism, movement problems (unc) and small body size (sma) are encoded by two mutant alleles that are recessive to those that produce wild-type traits (unc+ and sma+). A worm homozygous for movement problems and small body is crossed with a worm homozygous for the wild-type traits. The F1 have normal movement and normal body size. The F1 are then crossed with worms that have movement problems and small body size in a testcross. The progeny of this testcross is: Normal movement, normal body size 210 Movement problems, normal body size 9 Normal movement, small body size 11 Movement problems, small body size 193 a)From the test cross results, can you tell if the two genes are on the same chromosome or not? Explain your reasoning. b)What phenotypic proportions would be expected if the genes for round eyes and white body were located on different chromosomes? (please explain hot to get to these conclusions)
- Consider the following variations in Drosophila melanogaster, relative to the wild-type: White eyes are a recessive trait—the gene of which is found in Chromosome I (X). Vestigial wings are a recessive trait—the gene of which is found in Chromosome II. Aristapedia is a dominant trait—the gene of which is found in Chromosome III. Being homozygous for this condition is lethal. Cross the following mutant females with a wild-type (homozygous) male. Show the Punnett square and obtain the genotypic and phenotypic ratios of the first filial generation (F1). Female with white eyes Q4: Show the Punnett squares and obtain the genotypic and phenotypic ratios of the first filial generation (F1) and second filial generation (F2) of the following crosses. Note: The F2 generation can be obtained by crossing one male and one female from the F1 generation. Female with white eyes and vestigial wings and wild-type maleIn Drosophila, the X-linked recessive mutation vermilion (v) causes bright red eyes, in contrast to the brick-red eyes of wild type. A separate autosomal recessive mutation, suppressor of vermilion (su-v), causes flies homozygous or hemizygous for v to have wildtype eyes. In the absence of vermilion alleles, su-v has no effect on eye color. Determine the F1 and F2 phenotypic ratios from a cross between a female with wild-type alleles at the vermilion locus, but who is homozygous for su-v, with a vermilion male who has wildtype alleles at the su-v locusWhat would you predict to be the phenotype of a Drosophila larva whose mother was homozygous for a loss-of-function allele in the nanos gene?
- In Drosophila, the genes st (scarlet eyes), ss (spineless bris- tles), and e (ebony body) are located on chromosome 3, with map positions as indicated: st SS e 44 58 70 Each of these mutations is recessive to its wild-type allele (st*, dark red eyes; ss*, smooth bristles; e*, gray body). Phenotypically wild-type females with the genotype st ss e*/st* st* ss+ e were crossed with triply recessive males. Predict the phenotypes of the progeny and the frequen- cies with which they will occur assuming (a) no interfer- ence and (b) complete interference.Two Drosophila flies that had normal (transparent, long) wings were mated. In the progeny, two new phenotypes appeared, dusky wings (having a semi-opaque appearance) and clipped wings (with squared ends). The progeny were as follows: Females: 179 transparent, long 58 transparent, clipped Males: 92 transparent, long 89 dusky, long 28 transparent, clipped 31 dusky, clipped a) Provide a genetic explanation for these results, showing genotypes of parents and of all progeny classes under your model. b) Design a test for your model.The data set attached presents the results of a testcross using female flies heterozygous for three traits and male flies, which are homozygous recessive. For simplicity, mutant alleles are shown with letters a, b, and c and wildtype alleles are indicated by a “+” symbol. For this part of the report do the following in order: a) Determine the gene order (which gene is in the middle?)d) Construct a genetic map for the three genes, including the map distances between them. Clearly indicate the logic you followed and show all your calculations. Include the full distance calculations for the two most distanced genes (do not just add the other 2 distances). Ensure the work is neat and clear and does not contain spelling or grammatical errors so that it is understandable. Make sure to double check the solution provided.
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