Identity gene 1 (bl, pr, vg) Identity gene 2 ( bl, pr, vg) Identity gene 3 ( bl,pr, vg) How many map units separate genes 1 and 2 ( Distance 1)? How many map units separate genes 2 and 3 ( distances 2) ?
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- 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.Let’s suppose that a species of mosquito has two different types ofsimple transposons that we will call X elements and Z elements.The X elements appear quite stable. In a population of 100 mosquitoes,it is found that every mosquito has 6 X elements, and theyare always located in the same chromosomal locations amongdifferent individuals. In contrast, the Z elements seem to movearound quite a bit. Within the same 100 mosquitoes, the number of Z elements ranges from 2 to 14, and the locations of the Z elementstend to vary considerably among different individuals. Explainhow one simple transposon can be stable and another simple transposoncan be mobile, within the same group of individuals.We used/chose a human Rab protein with no direct yeast equivalent as an “outgroup” for this study (image). Question: Why have we chosen a human Rab protein with no direct yeast equivalent as an “outgroup” for this study (image given)?
- 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..Consider the first category of test-cross offspring shown in figure 8.2 (+b, LS). Consider also that the parents of the heterozygous female flies in the test cross had the following genotypes: bb, SS, and +, LL. A. What would be the physical phenotype of these flies? B. If PCR was conducted with the DNA of one of these flies using the primers for the molecular marker, what would be the appearance of the bands on an electrophoresis gel with the PCR products? C. If the gene for black body and the locus for the molecular marker (L long or S short) were unlinked, what proportion of the test-cross progeny would be black flies that are heterozygous for the molecular marker? What proportion would be flies with normal body color, which are homozygous for one form of the molecular marker? D. If the gene for black body and the locus for the molecular marker were linked, how would the proportion of flies be different?Because of the relatively high frequency of meiotic errors that lead to developmental abnormalities in humans, many research efforts have focused on identifying correlations between error frequency and chromosome morphology and behavior. Tease et al. (2002) studied human fetal oocytes of chromosomes 21, 18, and 13 using an immunocytological approach that allowed a direct estimate of the frequency and position of meiotic recombination. Below is a summary of information [modified from Tease et al. (2002)] that compares recombination frequency with the frequency of trisomy for chromosomes 21, 18, and 13. (Note: You may want to read appropriate portions of Chapter 8 for descriptions of these trisomic conditions.) Trisomic Mean Recombination Frequency Live-born Frequency Chromosome 21 1.23 1/700 Chromosome 18 2.36 1/3000–1/8000 Chromosome 13 2.50 1/5000–1/19,000 (a) What conclusions can be drawn from these data in terms of recombination and nondisjunction frequencies? How might…
- 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) 6EC4. In a test-cross of trihybrid drosophila the following phenotypes and number of offspring were found. Without doing any calculations, answer the following questions: +++ 669 abc 653 ++c 121 ab+ 139 +b+ 2280 a+c 2215 +bc 3 a++ 2 a) What were the genotypes of the parental strains? (fill in) b). What is the gene order?In an in situ hybridization experiment, a certain clonebound to only the X chromosome in a boy with no diseasesymptoms. However, in a boy with Duchenne musculardystrophy (X-linked recessive disease), it bound to theX chromosome and to an autosome. Explain. Could thisclone be useful in isolating the gene for Duchenne muscular dystrophy?
- in 3-5 sentences max. Describe how it is that when a short tandem repeat (STR) genomic region is PCR-amplified you can get 1 or 2 different-sized products using the same primer pair with only 1 person’s genomic DNA as the template?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 + 92Answer the following MCQs: 1:This database is most likely to have data about gene expression. GenBank BLAST dbEST RefSeq PDB 2: Dot matris noise reduction: window size is 7 by 7 and mismatch is 2. What is the minimal number of matches in this window that allows us to place the dot? 2 3 1 5 4 3: Find the optimal global alignment of two DNA sequences _ACAAC and AC. Match = 2, mismatch =0, gap opening =-3, gap extention=-1. What is the total score of alignment? 2 -1 0 -3 1