In a study, an undergraduate student discovered a new enzyme involved in the metabolism of serotonin. This enzyme is made up of 3 subunits of the same protein. To characterize this enzyme, the student used genetic approaches to induce mutations in the coding region of the gene that codes for this protein and performed crosses to measure the effects of genotypes on enzyme activity.
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- The synthesis of flower pigments is known to be dependent on enzymatically controlled biosynthetic pathways. For the crosses shown here, postulate the role of mutant genes and their products in producing the observed phenotypes: (a) P1: white strain A * white strain B F1: all purple F2: 9/16 purple: 7/16 white (b) P1: white * pink F1: all purple F2: 9/16 purple: 3/16 pink: 4/16 whiteIn the fungus Neurospora, a strain that is auxotrophic for thiamine (mutant allele t) was crossed with a strain that isauxotrophic for methionine (mutant allele m). Linear asci were isolated and classified into the following groups: a. Determine the linkage relations of these two genes to their centromere(s) and to each other. Specify distances in map units. b. Draw a diagram to show the origin of the ascus type with only one single representative (second from right).For the following genotype, indicate if beta-galactosidase will be made in the presence of lactose and absence of lactose. P₁+1+PL+OcZ/ P₁1 P₁O+Z+ O O O beta-galactosidase will be made in both the presence and absence of lactose (+; +) beta-galactosidase will be made in the presence but not in the absence of lactose (+;-) beta-galactosidase will not be made in the presence but will be made in the absence of lactose (-; +) beta-galactosidase will not be made in either the presence or absence of lactose (-;-)
- A yeast geneticist irradiates haploid cells of a strain thatis an adenine-requiring auxotrophic mutant, caused bymutation of the gene ade1. Millions of the irradiatedcells 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 ofresults.b. Write the genotypes of the prototrophs in each case.c. What progeny phenotypes and ratios do you predictfrom crossing a prototroph of type 2 by the original ade1auxotroph?Schizosaccharomyces pombe, also known as "fission yeast," is a powerful model organism in molecular and cell biology. While performing a genetic screen, you discover an auxotrophic S. pombe strain that is unable to synthesize one or more vitamins. The following table represents the key experiments you performed during your genetic screen. Fill in the table with the outcome of each experiment for your mutant strain (using + for growth and - for no growth). Medium Rich media Minimal media Minimal media + all vitamins Minimal media + all amino acids Growth Wild-type + + + + Mutant + + + > > >Parents who both have "sickle-cell trait", i.e, are heterozygous for HbS have a child who is tested at birth, and is found to be homozygous for HbS (both alleles affected). What is the molecular reason why the child presents with no symptoms until 6 months of age? a.) the mutation affects the beta chain, which is not dominant at birth b.) the mutation affects the alpha chain, which is not dominant at birth c.) babies cannot be exposed to low oxygen that triggers symptoms d.) babies cannot be exposed to high oxygen that triggers symptoms
- A 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.Starting with an AA, 2n = 14 genome of Triticum urartu (wild einkorn wheat) and a BB, 2n = 14 genome of Triticum speltoides (wild grass), show a pathway to obtain AABBDD, 2n = 42 Triticum spelta (spelt wheat) that is commonly used in cereals, pasta, and bread.Leber Congenital Amaurosis (LCA) causes progressive vision loss due to defects in the gene that encodes RPE65 isomerase. Affected individuals are homozygous recessive for mutant alleles of the RPE65 gene. You are trying to determine the molecular nature of the mutations in three individuals with LCA. For ease of analysis, you may assume that each individual is homozygous for the same mutant allele (though the three individuals have different mutations than each other). You use the polymerase chain reaction to amplify DNA from each patient and you determine the sequence of the DNA and compare it to unaffected individuals. You identify the following differences. Note that the non-template strand of DNA is given and the changes are highlighted using red boldface. You can assume that the sequences are in the first reading frame (eg. the first three nucleotides of each sequence is a codon). The coding region of the gene is 1602 bp and the position of the sequences shown below is…
- In E. coli, four Hfr strains donate the following markers,shown in the order donated:Strain 1: M Z X W CStrain 2: L A N C WStrain 3: A L B R UStrain 4: Z M U R BAll these Hfr strains are derived from the same F+ strain.What is the order of these markers on the circularchromosome of the original F+?What is the simplest explanation for why patients have been identified with only one copy of the phosphofructokinase-1 gene (heterozygous), but no patients have been identified that lack both copies of the phosphofructokinase-1 gene (homozygous)? Patients lacking both copies of the phosphofructokinase-1 genes will be found once DNA sequencing technology can sequence whole genomes. Phosphofructokinase-1 is needed for nitrogen metabolism and there are no enzymes to replace this function, so cells die from ammonia toxicity. Phosphofructokinase-1 is a required enzyme for carbohydrate metabolism in all living cells, complete loss of this enzyme would be lethal. There are 6 phosphofructokinase-1 paralogous genes in humans and it is impossible to lack both type 1 copies when there are also types 4, 5, and 6.A mutation creates a dominant negative allele of a particular gene. The gene encodes a protein that forms a trimer within the cell. If one or more of the subunits has the mutant structure, the entire trimeric protein is inactive. In a heterozygous cell, if the proteins of both alleles are present at the same levels, what percent of the trimers present in the cell will be active? A) 100% B) 5% C) 50% D) 33% E) 5%