In a complementation cross between Drosophila wiltype(brown eyes). you get a 9:3:3:1 ratio with white eyes being the lowest ratio. what would account for the viabilty of this white mutant strain compared to the brown eyes wild type. additionally why would we see a more robust female population than male population.
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- The following table summarises the results of the 2022-2023 Drosophila three-point cross involving the loci white eyes (w), miniature wings (m) and singed bristles (sn). The data is also available in the Excel file 'Drosophila Counts 2022-2023'. We strongly recommend working in Excel during this exercise.Phenotype Count+ + + 584w m sn 324w + + 227+ m sn 150+ m + 134w + sn 196 + + sn 134 w m + 92The aim of today's tutorial is to use the above data to establish the genetic map for these three loci. We have been told that all three loci are on the X chromosome; however, as scientists, we shouldn't simply take someone else's word for it - we need to test whether our data does, in fact, indicate linkage.As a first step, we will therefore conduct a chi-square test in order to test our data against the null hypothesis of independent assortment (i.e., "no…Sunflowers with flowers 10 cm in diameter are crossed with a plant that has 20-cm flowers. The F1 plants have flowers 15 cm in diameter. In the F2 generation, 4 flowers are 10 cm in diameter and 4 are 20 cm in diameter. Between these are 5 phenotypic classes with diameters intermediate to those at the extremes. a. Assuming that the alleles that contribute to flower diameter act additively, how many genes control flower size in this strain of sunflowers? b. How much does each additive allele contribute to flower diameter? c. What size flower makes up the largest phenotypic class?In fruit flies, you are mapping three genes in a three point cross. The mutants are hairy body (h), sepia colored eyes (se) and female sterility (g). You cross a heterozygous parent with a homozygous recessive parent and obtain the following results: Type Number h se g. 5 + se + 450 + se g 27 ++g_ h se + + + + h + g. h + + TOTAL is the gene in the middle and the distance in map units between se and g is Oh; 16.4 se; 7.1 Oh; 7.1 70 82 7 327 32 1000 se; 16.4
- A Drosophila strain with two mutated phenotypes was crossed with a wild-type Drosophila strain. The wild-type phenotypes were present in all of the F1 flies. The F1 flies were bred with one another. The F2 generation, on the other hand, did not have the predicted 9:3:3:1 phenotypic ratio. Explain why these outcomes occurred.The data set attached summarizes F2 numbers from an F1 cross arising from two, true-breeding Drosophila strains (P generation), which differ with respect to two mutant traits. Here are the hypothesis: Leg length - The wild-type and mutant alleles for leg length are incomplete dominant relative to each other. Justification: The data set includes three phenotypic categories for leg length: wild type (long leg), medium leg, and truncated wings. The presence of three distinct phenotypes suggests an incomplete dominance pattern, where the heterozygous individuals exhibit an intermediate leg length phenotype (medium leg). The absence of purebred short-legged individuals supports the idea that the long leg allele is dominant over the short leg allele. This shows that mode of inheritance is incomplete dominance of the alleles relative to each other. Since the data does not mention any specific differences between males and females, we can assume that the mode of inheritance for the trait is…Give only typing answer with explanation and conclusion to all parts Two true-breeding varieties of maize, one 11 cm high and the other 47 cm high were crossed and the resultant F1 hybrids were then crossed to generate the F2 . In the F2 there were a total of 13,923 plants with a continuous variation in heights between the two extremes and with only 3 plants as large as 47 cm high and 5 plants of 11 cm high. a) How many i) genes and ii) how many alleles are involved in determining height in this plant? b) What is the contribution of each dominant allele to the phenotype in cm?
- Please explain how you do this question step by step I am very confused! thank you:) You have three independent mutant alleles in the Drosophila gene no-antenna: nan1, nan2 and nan3. You determine the phenotype of Drosophila that are heterozygous for the three alleles (heterozygous for a wild-type allele and a mutant allele), and that are homozygous for the three mutant alleles. The antenna is composed of three segments that are followed at the distal end by a feathery arista (that is the antenna is composed of three segments and an arista). Allele nan1 nan2 nan3 heterozygous Wild-type No arista Wild-type homozygous No arista No antenna No antenna nan1 is a __x__ allele, nan2 is a __y__ allele, and nan3 is a __z__ allele. X Y Z A Dominant negative Null Hypomorphic B Null Dominant negative Hypomorphic C Null Hypomorphic Dominant negative D Hypomorphic Dominant negative Null E Dominant negative Hypomorphic Null Referencing the table above, select the…A cross between a wild-type fruit fly strain and various true-breeding mutant strains (all at a single locus) for bristle mutations resulted in the following phenotypes: Mutant Bristle Phenotype 1 0.20 shorter 2 0.30 longer 3 0.75 shorter 4 bent and damaged, non-functional 5 no bristlesIn Drosophila (fruit flies), jammed wings (J), daughterless (da), curly wings (Cy), star eyes (S), and a black body (b) are determined by genes located on the same chromosome. Gene Combination Recombination Frequency J and Cy 34.9 J and da 1.7 S and Cy 4.8 Cy and b 42.4 S and da 38 b and S 47.2 What is the map unit distance between b and da?Answer map unitsRecord your answer as a value rounded to one decimal place.
- What is the genetic distance between the eye colour locus (w) and the bristle locus (sn)? answer in Mu (Map units) and to 2 decimal places? The following table summarises the results of the 2022-2023 Drosophila three-point cross involving the loci white eyes (w), miniature wings (m) and singed bristles (sn). The data is also available in the Excel file 'Drosophila Counts 2022-2023'. We strongly recommend working in Excel during this exercise.Phenotype Count+ + + 584w m sn 324w + + 227+ m sn 150+ m + 134w + sn 196 + + sn 134 w m + 92Cross Cross A Cross A Cross B Cross B Phenotype F1 generation F2 generation F1 generation F2 generation Male red eyes 132 150 0 99 Female red eyes 135 295 110 101 Male white eyes 0 147 105 93 Female white eyes 0 0 0 95 Using “+” to indicate the wildtype red-eyed allele and “w” to indicate the mutant white-eyed allele, state the genotypes of the following: Wildtype red-eyed and white-eyed parental flies from cross A and cross B. Males and females from the F1 generation flies from cross A and cross B Males and females, F2 generation flies from cross A and cross B.Kernel color in wheat Red kernel color in wheat results from the interaction between two dominant alleles. With only one dominant allele, the phenotype is brown while in the absence of any dominant allele, the phenotype is white. Suppose that plants of a variety that is true breeding for red kernels are crossed with plants true breeding for white kernels. What specific type of interaction can be observed? A. duplicate dominant genes B. duplicate genes with cumulative effects C. complementary genes D. dominant epistasis (case 2)