What is the most likely reason a southern blot may not be able to detect KSS in the clinic? O A nonsynonymous or missense mutation may be present in the mitochondrial DNA O Another biopsy from a distinct site may show a different result Your mitochondria may contain a nonsynonymous or missense mutation in the mitochondrial DNA and another biopsy from a distinct site may show a different result Mitochondrial DNA from different cells do not have the same genome size O A biopsy from your mother is needed to interpret the test
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- I read that vinyl chloride exposure is associated with an increased risk of a rare form of liver cancer (hepatic angiosarcoma), as well as brain and lung cancers, lymphoma, and leukemia. But my question is what gene(s) are being mutated by this type of taxic gas?Some people have a genetic predisposition for developing priondiseases. Examples are described in Table 25.6. In the case ofGerstmann-Straüssler-Scheinker disease, the age of onset istypically 30–50 years, and the duration of the disease (whichleads to death) is about 5 years. Suggest a possible explanationwhy someone can live for a relatively long time withoutsymptoms and then succumb to the disease in a relativelyshort time.The intermediates A, B, C, D, E, and F all occur inthe same biochemical pathway. G is the product of thepathway, and mutants 1 through 7 are all G−, meaningthat they cannot produce substance G. The followingtable shows which intermediates will promote growthin each of the mutants. Arrange the intermediates inorder of their occurrence in the pathway, and indicatethe step in the pathway at which each mutant strain isblocked. A + in the table indicates that the strain willgrow if given that substance, an O means lack of growth.SupplementsMutant A B C D E F G1 + + + + + O +2 O O O O O O +3 O + + O + O +4 O + O O + O +5 + + + O + O +6 + + + + + + +7 O O O O + O +
- . In a certain plant, the flower petals are normally purple.Two recessive mutations arise in separate plants and arefound to be on different chromosomes. Mutation 1 (m1)gives blue petals when homozygous (m1/m1). Mutation2 (m2) gives red petals when homozygous (m2/m2).Biochemists working on the synthesis of flower pigments in this species have already described the following pathway:colorless (white)compoundblue pigmentred pigmentenzyme Aenzyme Ba. Which mutant would you expect to be deficient inenzyme A activity?b. A plant has the genotype M1/m1 ; M2/m2. Whatwould you expect its phenotype to be?c. If the plant in part b is selfed, what colors of progenywould you expect and in what proportions?d. Why are these mutants recessive?Which of the following experimental observations suggest that adisease has a genetic basis?A. The frequency of the disease is less likely in relatives that liveapart compared with relatives that live together.B. The frequency of the disease is unusually high in a small groupof genetically related individuals who live in southern Spain.C. The disease symptoms usually begin around the age of 40.D. It is more likely that both monozygotic twins will be affectedby the disease than will dizygotic twins.C. Anagnostopoulos and I. P. Crawford isolated and studied a series of mutations that affected several steps in the biosynthetic pathway leading to tryptophan in the bacterium Bacillus subtilis . Seven of the strains that they used in their study are listed here, along with the mutation found in each strain.Strain Mutation T3 T− 168 I− 168PT I− TI I− TII I− T8 A− H25 H− To map the genes for tryptophan synthesis, they carried out a series of transformation experiments on strains having different mutations and determined the percentage of recombinants among the transformed bacteria. Their results were as follows:(refer picture) On the basis of these two-point crosses, determine the order of the genes and the distances between them. Where more than one cross was completed for a pair of genes, average the recombination rates from the different crosses. Draw a map of the…
- Six strains of E. coli (mutants 1–6) that had one of thefollowing mutations (i–vi) affecting the lac operonwere isolated.i. deletion of lacYii. ocmutationiii. missense mutation in lacZiv. inversion of the lac operon (but not an inversion ofthe lacI gene)v. superrepressor mutationvi. inversion of lacZ, Y, and A but not lacI, P, oa. Which of these mutations would prevent the strainfrom utilizing lactose?b. The entire lac operon (including the lacI gene andits promoter) from each of the six E. coli strainswas cloned into a plasmid vector containing an ampicillin resistance gene. Each recombinant plasmidwas transformed into each of the six strains to create partial diploids. In analysis of these strains, mutant 1 was found to carry a deletion of lacY, so thisstrain corresponds to mutation i in the list above.Which of the other types of mutations would be expected to complement mutant 1 in these partial diploids so as to allow lactose utilization?c. In part (b), each strain was plated on…Both the isolation buffer and assay buffer that was used in isolation of mitochondria contained 0.3M mannitol. What was the purpose of including mannitol at this concentration?Robert Bost and Richard Cribbs studied a strain of E. coli (araB14)that possessed a nonsense mutation in the structural gene that encodes Lribulokinase,an enzyme that allows the bacteria to metabolize the sugararabinose (R. Bost and R. Cribbs. 1969. Genetics 62:1–8). From thearaB14 strain, they isolated some bacteria that possessed mutations thatcaused them to revert back to the wild type. Genetic analysis of theserevertants showed that they possessed two different suppressormutations. One suppressor mutation (R1) was linked to the originalmutation in L-ribulokinase and probably occurred at the same locus. Byitself, this mutation allowed the production of L-ribulokinase, but theenzyme produced was not as effective in metabolizing arabinose as theenzyme encoded by the wild-type allele. The second suppressormutation (SuB) was not linked to the original mutation. In conjunctionwith the R1 mutation, SuB allowed the production of L-ribulokinase, butSuB by itself was not able to suppress the…
- Recently, scientists discovered that a rare disorder called polkadotism is caused by a bacterial strain, polkadotiae. Mice injected with this strain (P) develop polka dots on their skin. Heat-killed P bacteria and live D bacteria, a nonvirulent strain, do not produce polka dots when injected separately into mice. However, when a mixture of heat-killed P cells and live D cells were injected together, the mice developed polka dots. What process explains this result? Describe what is happening in the mouse to cause this outcome.When 1 million cells of a culture of haploid yeastcarrying a met− auxotrophic mutation were plated onpetri plates lacking methionine (Met), five coloniesgrew. You would expect cells in which the originalmet− mutation was reversed (by a base change back tothe original sequence) would grow on the media lackingmethionine, but some of these apparent reversions couldbe due to a mutation in a different gene that somehowsuppresses the original met− mutations. How wouldyou be able to determine if the mutations in your fivecolonies were due either to a precise reversion of theoriginal met− mutation or to the generation of a suppressor mutation in a gene on another chromosome?nvestigate references dealing with the technical and ethical challenges surrounding mitochondrial replacement therapy (MRT). Consider the following questions: Differentiate between MRT methods: pronuclear transfer (PNT) versus maternal spindle transfer (MST). Much of the controversy surrounding MRT methods has been triggered by the phrase “three-parent babies” in media headlines. Do you think that this phrase is an accurate description of children born following mitochondrial replacement? Summarize the ethical arguments used to support and oppose the use of MRT in humans. In your opinion, which arguments have validity, and why