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)
- The wild-type (normal) fruit fly, Drosophila melanogaster, has straight wings and long bristles. Mutant strains have been isolated that have either curled wings or short bristles. The genes representing these two mutant traits are located on separate chromosomes. Carefully examine the data from the following five crosses shown below (running across both columns). (a) Identify each mutation as either dominant or recessive. In each case, indicate which crosses support your answer. (b) Assign gene symbols and, for each cross, determine the genotypes of the parents.arrow_forwardYou are studying 5 genes: a, b, c, d, and e. In an attempt to map them, you perform a series of two-point crosses with the results listed below. Select the appropriate gene order based on the results of your experiment. a-b = 20 map units; a-c = 16 map units; a-d = 12 map unit; b-d = 8 map units; b-e = 4 map units; c-d = 28 map units; c-e = 40 map units; d-e = 12 map units. Group of answer choices 1)e-b-a-c-d 2)e-b-d-a-c 3)a-c-e-b-d 4)a-b-c-d-earrow_forwardThe coloration of calico cats is also the result of the inactivation of one of a female cat's sex chromosomes. In females, two X chromosomes are present in all cells but only one is active, the inactive one is called a BARR BODY. This is why the coat color is random, even among cloned calico cats that have identical genomes. Calico is a coat color found in cats, which is caused by a SEX-LINKED, CODOMINANT allele.arrow_forward
- A yeast geneticist irradiates haploid cells of a strain that is an adenine-requiring auxotrophic mutant, caused by mutation of the gene ade1. Millions of the irradiated cells are plated on minimal medium, and a small number of cells divide and produce prototrophic colonies. These colonies are crossed individually with a wildtype strain. Two types of results are obtained:(1) prototroph × wild type : progeny all prototrophic(2) prototroph × wild type : progeny 75% prototrophic, 25% adenine-requiring auxotrophsa. Explain the difference between these two types of results.b. Write the genotypes of the prototrophs in each case.c. What progeny phenotypes and ratios do you predict from crossing a prototroph of type 2 by the original ade1auxotroph?arrow_forwardYou are attempting to genotype a series of cells at gene A through restriction enzyme digestion. You know that the wild type allele “A” has 2 restriction enzyme cutting sites, creating bands that are 100bp, 200bp, and 300bp. A mutation creating allele “a” removes one of these cutting sites, creating bands that are 100bp and 500bp. How many separate bands would you expect to see on a gel for a cell that is heterozygous Aa? please show work. thanksarrow_forwardConsider 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 PC 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 PC 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?arrow_forward
- When doing a lab that involves Extraction of Genomic DNA from adult Drosophila melanogaster. - What are the controls you should use and why use these? What is special about the genomic DNA? What do you expect to find (using these genomic DNA samples)? please help!arrow_forward2. You are making a genetic map and use a cross to measure the distance between genes H and J as 40cMand the distance between J and M as 20cM. When you measure the distance between genes H and M, you get 50 cM. Do the following (A-C). (A) Draw and label a map of genes H, J, and M. (B) What is the genetic distance between genes H and M? How do you know? (Answer in 1 sentence) (C) Complete the table (right). For each pair of genes, write “yes” or “no” to indicate whether the pair of genes are“linked genes” and/orpart of the same“linkage group”.arrow_forwardTwo normal looking fruit flies were crossed and in their progeny, there were 202 females and 98 males. What is unusual about this result? provide a genetic explanation for this anomaly and provide a test of your hypothesisarrow_forward
- Suppose that you are studying the role of Protein B, which you believe plays a role in regulating PCD/Apoptosis in mice. You create two lines of mutant mice. One (bb) is homozygous for a loss-of-function allele of gene B. The other (Bb) is heterozygous, with one wild-type allele and one loss-of function allele. Initially you pay particular attention to two phenotypes of the resulting mice:(i) The morphology of their paws (see picture) (ii) The size of their brains & shape of their skulls. The bb mice have unusually large brains and unusual protrusions from their skulls. Based on these data, does it appear that Protein B, when present and active, favors or inhibits PCD/Apoptosis?Briefly explain your reasoning. The answer should address both the paw and brain/skull data.arrow_forwardConsider 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?arrow_forwardDNA sequencing of your own two β-globin genes (one from each of your two Chromosome 11s) reveals a mutation in one of the genes. given this information alone, should you worry about being a carrier of an inherited disease that could be passed on to your children? What other information would you like to have to assess your risk?arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning