To analyze:
Dr. Sophila observed that fruit flies with normal behavior kept at
Introduction:
The fruit flies in the given question show different phenotypes at different temperatures. The probable cause for this might be the inactivity of the gene at a high temperature which is important for the normal behavior or phenotype. Therefore, an organism showing temperature-sensitive mutation is called a temperature sensitive mutant. Temperature sensitive mutant grown in the permissive condition means the mutated gene product behaves normally at the low temperature and shows altered behavior at high temperature. The temperature at which they show normal behavior is called permissive temperature. A temperature sensitive mutant returns to its normal behavior if placed in permissive temperature.
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Genetic Analysis: An Integrated Approach (3rd Edition)
- Cap, EA1, and Sap are all genes/proteins of interest in this study. For each gene, what gene product is encoded and where is the gene (the literal DNA sequence) located physically in the cell? I need help fimiding this in the artticle and answer as short as possible https://www.ncbi.nlm.nih.gov/pmc/articles/PMC106848/arrow_forwardC. elegans can be viewed under a dissecting scope; their movement normally is sinusoidal, actually not unlike a snake slithering. Some C.elegans mutants have a mutations that cause them to roll instead of quickly move sinusoidally. You've collected two roller mutants, whose gene identity is currently unknown, but you temporarily give the names "A" and "B." You examine their phenotype in two different conditions. 15 degrees Celsius 26 degrees Celsius Wild type rol-? (A) sinusoidal movement rolls sinusoidal movement rolls Which C. elegans strain is temperature-sensitive? A Which temperature is the permissive condition? 26 degrees rol-? (B) sinusoidal movement rollsarrow_forwardA panel of cell lines was created by human–mouse somatic-cell hybridization. Each cell line was examined for the presence of human chromosomes and for the production of three enzymes. The following results were obtained:On the basis of these results, give the chromosomal locations of the genes encoding enzyme 1, enzyme 2, and enzyme 3.arrow_forward
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- E. coli cells are simultaneously infected with two strains of phage λ. One strain has a mutant host range, is temperature sensitive, and produces clear plaques (genotype h st c); another strain carries the wildtype alleles (genotype h+ st+ c+). Progeny phages are collected from the lysed cells and are plated on bacteria. The following numbers of different progeny phages are obtained: Progeny phage genotype Number of plaques h+ c+ st+ 321 h c st 338 h+ c st 26 h c+ st+ 30 h+ c st+ 106 h c+ st 110 h+ c+ st 5 h c st+ 6 a. Determine the order of the three genes on the phage chromosome. b. Determine the map distances between the genes. c. Determine the coefficient of coincidence and the interferencearrow_forwardA panel of cell lines was created by human–mouse somatic-cell hybridization. Each line was examined for the presence of human chromosomes and for the production of an enzyme. The following results were obtained:On the basis of these results, which chromosome has the gene that encodes the enzyme?arrow_forward8) You perform flow cytometry on a population of cells treated with a mutagen to identify new factors required for cell cycle progression. You isolate two new mutants that display a change in the rate of the cell cycle. You measure DNA content with flow cytometry from a control cell population as well as from these two mutants. The results from this experiment are shown below. Control Mutant 1 Mutant 2 1 Relative amount of DNA per cell Use these data to answer the following questions. 1) If you measured levels of M-cyclin protein from these different populations, which would have the HIGHEST. a. Control b. Mutant 1 c. Mutant 2 2) After genome sequencing, you identify a loss-of-function mutation in Cdc20 in one of these cell populations. Based on these data, which population would you expect to have this mutation? a. Mutant 1 b. Mutant 2 3) If DNA damage is sensed during replication and cell cycle progression is stopped, which of the graphs would you expect to observe? a. Control b.…arrow_forward
- If you wanted to make a mouse model for any of the following human genetic conditions (a–d), indicate which of thefollowing types of mice (i–vi) would be useful to your studies. If more than one answer applies, state which type ofmouse would most successfully mimic the human disease:(i) transgenic mouse overexpressing a normal mouse protein; (ii) transgenic mouse expressing normal amounts of amutant human protein; (iii) transgenic mouse expressing adominant negative form of a protein; (iv) a knockout mouse;(v) a conditional knockout mouse; and (vi) a knockin mousein which the normal allele is replaced with a mutant allelethat is at least partially functional. In all cases, the transgeneor the gene that is knocked out or knocked in is a form of thegene responsible for the disease in question.a. Marfan syndrome (a dominant disease caused byhaploinsufficiency for the FBN1 gene);b. A dominantly inherited autoinflammatory diseasecaused by a hypermorphic missense mutation in thegene PLCG2;c.…arrow_forwardSam wants to alter his interrupted mating experiment by placing a filter in between the donor and the recipient cells in order to separate the two. Is Sam’s suggestion for this modification to the procedure utilized in interrupted mating experiments valid? Explain.arrow_forwardSuppose a researcher has three different Drosophila strains that have mutations in the bicoid gene called bicoid-A, bicoid-B, and bicoid-C; the wild type is designated bicoid +. To study these mutations, phenotypically normal female flies that are homozygous for the given bicoid mutation were obtained, and their oocytes were analyzed using a Northern blot to determine the size and/or amount of the bicoid mRNA and in situ hybridization to determine the bicoid mRNA location within the oocyte. A wild-type strain was also analyzed as a control. In both cases, the probe was complementary to the bicoid mRNA and the results are shown below. (Anterior is on the left; posterior is on the right.) Northern blot 1 2 - 3 4 In situ hybridization Wild type Lane 1. Wild type (bicoid*) Lane 2. bicoid-A Lane 3. bicoid-B Lane 4. bicoid-C bicoid-B bicoid-A bicoid-C Which mutation is likely to cause the embryo to develop two "anterior" ends? bicoid-B Obicoid-A bicoid-Carrow_forward
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