Three different chitin synthase genes control chitin synthesis in S. cerevisiae. Discuss what will happen to the budding yeast if a mutation occurs in each of the genes below: CHSI CHSII CHSIII
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Three different chitin synthase genes control chitin synthesis in S. cerevisiae. Discuss what will happen to the budding yeast if a mutation occurs in each of the genes below:
- CHSI
- CHSII
- CHSIII
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- Brenner’s m mutant phages (m1–m6) described inFig. 8.8 were suppressed when grown in suppressor(su−) mutant bacteria; they produced full-length Mproteins that functioned like wild-type M protein.a. What gene do you think was mutant in the su−bacteria?b. When the m− phages were propagated in the su−bacterial strain, not all of the proteins made by themutant m alleles were identical to wild-type Mprotein. How did some of them differ?A hypothetical gene for cephalosporin resistance is found to be carried by a transposon. Explain what a transposon is. Then explain how the cephalosporin resistance could be horizontally transferred between organisms by transformation, conjugation, and transduction. What steps/events would have to occur to allow the transposon to be transferred by each method. Also, explain how it could be transferred vertically between organisms.In the treatment of acquired immunodeficiency syndrome (AIDS), a possible mode of therapy is to inhibit the reverse transcriptase (RT) of the human immunodeficiency virus (HIV), whcih is required for the retrovirus to be propogated by RNA-directed DNA synthesis. In the figure below, one of the substrates for RT is thymidine; and two drugs, AZT and HBY097 are known to inhibit HIV RT> (a) Thymidine; (b) AZT; (c) HBY097 Look at the structures and predict the type of inhibition (i.e. competitive or non-competitive) likely to be shown by each drug. By using knowledge on enzyme, plan an experiment that would enable you to confirm the type of inhibition by investigating enzyme kinetics and explain how you would interpret the results.Remarks: Not more than 250 words.
- Determine the outcome for the lac operon genotype shown below: I*p*o°z*y/I*p*o*z*x* Assume lactose is absent. Select one: O Functional beta-galactosidase and functional permease are produced. O Only functional beta-galactosidase is produced. Functional permease is NOT produced. O Only functional permease is produced. Functional beta-galactosidase is NOT produced. O Functional beta-galactosidase is NOT produced. Functional permease is NOT produced.Mad cow disease, or bovine spongiform encephalopathy, appears to be caused by a novel infectious agent: a protein that replicates by causing related proteins to modify their structure from a harmless shape to a dangerous one. These prions (short for “proteinaceous infectious particles”) also appear to be the cause of several other spongiform encephalopathy diseases, such as scrapie in sheep, and kuru and Creutzfeldt-Jakob disease in humans. Which of the following observations or experiments would not support the hypothesis that a prion causes spongiform encephalopathy? (a) The brains of many sheep with scrapie contain prion proteins, but the brains of most sheep without scrapie do not. (b) There is a high incidence of kuru in populations of people who consume brain tissue from prion-infected animals. (c) Coyotes that feed on cows with mad cow disease do not subsequently develop spongiform encephalopathy. (d) When introduced into sheep brain cells in culture, prions cause the normal…The locations of the TATA box in two species of yeast, Saccharomyces pombe and Saccharomyces cerevisiae, differ dramatically. The TATA box of S. pombe is about 30 nucleotides upstream of the transcription start site, similar to the location in most other eukaryotic cells. However, the TATA box of S. cerevisiae is 40 to 120 nucleotides upstream of the start site. To better understand what sets the start site in these organisms, researchers at Stanford University conducted a series of experiments to determine which components of the transcription apparatus of these two species could be interchanged (Y. Li et al. 1994. Science 263:805–807). In these experiments, different general transcription factors and RNA polymerases were switched in S. pombe and S. cerevisiae, and the effects of each switch on the level of RNA synthesis and on the starting point of transcription were observed. The results from one set of experiments are shown in the table below. Components cTFIIB, cTFIIE, cTFIIF,…
- 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.What are the reagents and materials used/needed in the extraction of invertase from yeast? for a lab experiment regarding enzymes.Use the blanks on the left hand side to put the following statements in order (use numbers or letters to designate the order). All of the statements refer to the diauxic (sequential) growth of a bacterium on the carbon substrates glucose and lactose (see textbook Fig 4.11 on diauxic growth). Remember, the lac operon contains the genes necessary for the lactose catabolism. The glucose substrate is completely consumed, at which time the repressor protein on the lac operon unbinds from the DNA molecule. The lactose substrate is catabolized by the active B-galactosidase enzyme. Glucose, as the preferred carbon substrate, is first consumed by catabolic enzymes that are constitutive (always present). After completion of translation, the B-galactosidase enzyme undergoes folding to form into an active enzyme that can breakdown lactose into its constituent monosaccharides. The messenger RNA code from the lac operon is translated by ribosomal RNAS to form lac-related enzymes, such as…
- Duchenne muscular dystrophy is caused by a mutation in a gene that comprises 2.5 million base pairs and specifies a protein called dystrophin. However, less than 1% of the gene actually encodes the amino acids in the dystrophin protein. On the basis of what you now know about gene structure and RNA processing in eukaryotic cells, provide a possible explanation for the large size of the dystrophin gene.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 + + + > > >For the following genotypes of E. coli strains, state whetherenzyme activity is inducible (I), constitutive (C), or uninducible (U).Key to symbols in table:z = structural gene for -galactosidasey = structural gene for permeasea = structural gene for transacetylasep = promoter (for lac operon)i = structural gene for the lac repressor (regulator gene)o = operator (binding site for repressor)Genotype -gal (z) Permease (y) Transacetylase (a)i+ p+ oC z- y- a+i- p+ o+ z+ y- a- /i+ p- oc z- y+ a-i- p+ o+ z+ y- a+ /i+ p+ oc z- y+ a-i+ p+ o+ z- y- a+i- p- o+ z+ y- a+ /i- p+ o+ z- y+ a-iS p+ o+ z- y- a+i- p+ o+ z+ y- a+ /i+ p- oc z- y+ a-is p+ o+ z- y+ a+ /i+ p+ oc z+ y- a-i- p- o+ z+ y- a+ /i- p+ o+ z- y- a-is p+ oc z+ y- a+i+ p+ oc z+ y- a+