Note that the two parental gamete types (+ + and bl pu) are the most abundant, as expected. Use the data to calculate the recombination frequency and the genetic map distance between the two genes. Record the map distance in your notes.
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- QUESTION:- In Drosophila, assume that the gene for scute bristles (s) is located at map position 0.0 and that the gene for ruby eyes (r) is at position 25.0. Both genes are located on the X chromosome and are recessive to their wild-type alleles. A cross is made between scute-bristled females and ruby-eyed males. Phenotypically wild-type F1 females were then mated to homozygous double mutant males, and 1000 offspring were produced. Give the phenotypes and frequencies expected..Karyogram and Identification of Chromosomal Aberrations List down five human chromosomal aberrations. Give the corresponding karyotype for each type then briefly describe the traits of affected individuals.Analyzing Karyotypes 1. Originally, karyotypic analysis relied only on size and centromere placement to identify chromosomes. Because many chromosomes are similar in size and centromere placement, the identification of individual chromosomes was difficult, and chromosomes were placed into eight groups, identified by the letters A to G. Today, each human chromosome can be readily identified. a. What technical advances led to this improvement in chromosome identification? b. List two ways this improvement can be implemented. c. What clinical information does a karyotype provide?
- mapping gene The genes for ruby eyes (rb), tan body (t) and cut wings (ct) are all found on the X-chromosome of Drosophila melanogaster. All of these are recessive traits. They map in the order rb, ct, t with 12.5 map units between rb and ct and 7.5 map units between ct and t. Suppose you cross a cut wing male with a homozygous female that is both tan and has ruby eyes. What will the F1 females look like? Draw map of the section of the X chromosomes that has these 3 genes for the F1 females Assume you testcross your F1 females. What progeny classes would you expect? ii. Give approximate numbers for each class based on a total of 2000 progeny. Assuming the i=1 and there are no double crossovers. Assuming the i=0 and there are the expected number of double crossovers.Inquiry In a cross between a wild-type femalefruit fly and a mutant white-eyed male, what coloreyes will the F1 and F2 offspring have?D) Genetics Problem show the table and word In certain breeds of chickens, the allele "B" is responsible for black feathers whereas the contrasting allele "b" produces feathers that are white. Another r pair of alleles influences the shape of the feathers. "F" produces straight feathers whereas the contrasting gene for "f results in the frizzled condition. Give the genotypic and phenotypic ratios to be expected from the following cross: white, frizzled hen with a heterozygous black, homozygous straight rooster. Genotypic Ratio: Phenotypic Ratio:
- Give typed full explanation You are studying three linked genes in snapdragons. The flower color locus is in the center. There are 13.8 cM between the flower color locus and the plant height locus. There are 14.5 cM between the flower color locus and the leaf type locus. The coefficient of coincidence is 0.8. Pure-breeding tall, red-flowered plants with fuzzy leaves were crossed to pure-breeding dwarf, blue-flowered plants with smooth leaves. The F1 were testcrossed. Calculate the proportion of the testcross progeny that are expected to be dwarf with red flowers. Round properly to 4 decimal digits.Need help. Knowing that the Curly leaf (Cy) is a dominant gene on chromosome 6, as it is useful in tracking other genes using trisomics for chromosome 6. Assume a Cy Cy cy plant used as a pollen parent where disomic pollen does not function crossed to a Cy cy cy female where disomic eggs do function. A) What ratio would be predicted in the progeny of this cross? B) What if the reciprocal cross was made?Cinnabar eyes (cn) and reduced bristles (rd) are linked autosomal recesive characters in Drosophila fruit flies. A homozygous wild-type female was crossed to a reduced, cinnabar male, and the F1 females were then crossed with cinnabar reduced males to obtain the F2. Of the 200 F2 offspring obtained, 78 were wild type, 18 were cinnabar, 22 were reduced and 82 were reduced and cinnabar.. What is the map distance between the cn and rd alleles? 18 mu 7.8 mu 40 mu 20 mu 22 mu
- Two plants in a cross were each heterozygous for two gene pairs (AB/ab) whose loci are linked and 10 map units (mu) apart. (Recall that 1 mu is equal to 1% recombination between two genes.) Assuming that crossing over occurs during the formation of both male and female gametes and that the A and B alleles are dominant, determine the phenotypic ratio of their offspring. Part D If the two genes are 15 mu apart and the plant is (Ab/aB), what proportion of gametes from a signal plant will be ab? Part E What proportion of the offspring of two plants ( both (Ab/aB)) will be A_B_ if the genes are 15 mu apart? Part F What proportion of the offspring of two plants ( both (Ab/aB)) will be A_bb if the genes are 15 mu apart? Part G What proportion of the offspring of two plants ( both (Ab/aB)) will be aaB_ if the genes are 15 mu apart? Part H What proportion of the offspring of two plants ( both (Ab/aB)) will be aabb if the genes are 15 mu apart? How would I solve these?This problem leads you through the derivation of acorrected equation for RF in yeast tetrad analysis thattakes into account double crossover (DCO) meioses. A yeast strain that cannot grow in the absence ofthe amino acid histidine (his−) is mated with a yeaststrain that cannot grow in the absence of the aminoacid lysine (lys−). Among the 400 unordered tetrads resulting from this mating, 233 were PD, 11 wereNPD, and 156 were T.a. What types of spores are in the PD, NPD, andT tetrads?b. Are the his and lys genes linked? How do you know? c. Using the simple equation RF = 100 × [NPD +(1/2)T]/total tetrads, calculate the distance in mapunits between the his and lys genes.d. If you think about all the kinds of meiotic eventsthat could occur (refer to Fig. 5.24), you can seethat the calculation you did in part (c) may substantially underestimate RF. What kinds of meioses(NCO, SCO, or DCO) generated each of the tetradtypes in this cross? e. What incorrect assumptions does the simple RFequation…Complementation tests of distinct recessive mutants, 1 through 8, produce the data in the matrix below. A plus (+) indicates complementation, meaning the phenotype of the combined alleles is wild type, and a minus (-) indicates a failure to complement meaning that a mutant phenotype results. Assume that the missing mutant combinations would yield data consistent with the entries that are shown. How many complementation groups are formed by these eight mutants? (Picture attached) A) 2 B) 3 C) 4 D) 5 E) 6