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"the supplement that restored growth would be the molecule that the mutant strain could not synthesize" Explain this statement ?
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- In your laboratory, you have an F− strain of E. coli that is resistantto streptomycin and is unable to metabolize lactose, but it can metabolizeglucose. Therefore, this strain can grow on a medium thatcontains glucose and streptomycin, but it cannot grow on a mediumcontaining lactose. A researcher has sent you two E. colistrains in two separate tubes. One strain, let’s call it strain A, hasan F′ factor (an F prime factor) that carries the genes that are requiredfor lactose metabolism. On its chromosome, it also has thegenes that are required for glucose metabolism. However, it is sensitiveto streptomycin. This strain can grow on a medium containinglactose or glucose, but it cannot grow if streptomycin is addedto the medium. The second strain, let’s call it strain B, is an F−strain. On its chromosome, it has the genes that are required forlactose and glucose metabolism. Strain B is also sensitive to streptomycin.Unfortunately, when strains A and B were sent to you, thelabels had fallen…In a process of production of a recombinant protein by E. coli cells, it was observed accumulation of acetate in the culture medium. In this situation, it can be said that: (a) certainly the process in question was being conducted in anaerobiosis (B).Acetate accumulation is advantageous for the process as the acetate formation reaction generates 1 molecule of ATP (c)Knowing that decreasing the temperature of the process causes a reduction in the rate of glycolysis, this could be a strategy to reduce the accumulation of acetate (d).the acetate formed can be re-assimilated by the cell if the glyoxylate pathway is activated at some point in the cultureA bloactlve hexapeptide was cleaved by trypsln from a larger proteln. The peptlde was sequenced by treatment with varlOus reagents and enzymes. The results are as follows: a) Treatment with cyan0gen bromlde gave a tetrapeptlde of compositlon asp, trp, cys and met. b) Treatment with chymotrypsln gave a tripeptlde of compositlon asp, trp and cys, a dipeptlde wlth the same C-term and an amlno acld. c) Treatment with streptococcal protease gave a dipeptlde and a tetrapeptlde. The composltion of the peptlde Is lys, trp, asp, cys and met 1. What Is the sequence of the hexapeptlde? 2. What is the pl of thls peptlde?
- A bloactlve hexapeptide was cleaved by trypsln from a larger proteln. The peptlde was sequenced by treatment with varlous reagents and enzymes. The results are as follows: a) Treatment wlth cyanOgen bromlde gave a tetrapeptlde of compositlon asp, trp, cys and met. b) Treatment with chymotrypsln gave a tripeptlde of compositlon asp, trp and cys, a dipeptlde wlth the same C-term and an amlno acld. c) Treatment with streptococcal protease gave a dipeptlde and a tetrapeptlde. The composltion of the peptlde Is lys, trp, asp, cys and met 1. What Is the sequence of the hexapeptlde? 2. What is the pl of thls peptlde?Mabelle used the pET vector system to express her prokaryotic amylase enzyme. She added IPTG into her culture broth of DH5a Escherichia coli strain. At the end of the experiment, she discovered that her protein was not expressed. She repeated three more times but her protein of interest was still not produced. (i) (ii) Explain the reason why Mabelle failed to obtain her protein of interest and suggest a solution to troubleshoot this problem. Mabelle plans to express her protein fused to a polyhistidine-tag (His-tag). Explain the importance of His-tag in protein work.In the procedure shown, why was it necessary to link thecoding sequence for the A or B chains to the sequence forβ-galactosidase? How were the A or B chains separated fromβ-galactosidase after the fusion protein was synthesized in E. coli?
- An E. coli colony grew on minimal medium supplemented with arginine and leucine. However, bacteria from this colony are unable to grow and form colonies on minimal medium supplemented with arginine and methionine. What is the genotype of the bacteria in this E. coli colony?The synthesis of arginine by Nuerospora was determined by examining a number of mutant strains that were unable to synthesize the compound. Use the table of bacterial growth below to 1) determine the correct sequence of the synthesis pathway and 2) where in the synthesis pathway each mutation interrupts the synthesis. A “+" indicates growth. Nothing added to Succinate Ornithine added Strain Cirtulline Arginine Added added added growth medium Wild Mutant 1 Mutant 2 Mutant 3 Mutant 4Direct repair of pyrimidine dimer formation in E. Coli can be accomplished by nucleotide excision. true or false?
- Some bacteria may have higher mutation rates than others following exposure to UV light. Discuss a reason why this might be the case. What experiments could you do to determine whether this is a likely possibility?"what is the reason for forming mucoid colonies in some types of bacteria?"When Griffith incubated heat-killed virulent S strain bacteria with live avirulent R strains, he found that R cells were transformed into lethal, disease-causing bacteria. What is the molecule responsible for the transformation of R cells into the S type? Why did the transformation occur? A. Proteins in the heat-killed S cell extract; they were able to synthesize the capsular polysaccharide in the R cells and make them virulent. B. The capsular polysaccharide in the heat-killed S cell extract; the polysaccharide was able to attach to R cells, thus making them virulent. C. DNA in the heat-killed S cell extract; the DNA altered the genetic makeup of R cells, allowing them to synthesize the polysaccharide capsule, thus making them virulent. D. Specialized lipids in the cell membrane of the S cells were able to integrate with the R cells, conferring the ability to evade host immunity.