Explain why the experimental procedures performed by Lenski (and his lab) caused the fitness of the bacteria to increase over time (especially early on) in all 12 of his populations.
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Bacterial Genomics
The study of the morphological, physiological, and evolutionary aspects of the bacterial genome is referred to as bacterial genomics. This subdisciplinary field aids in understanding how genes are assembled into genomes. Further, bacterial or microbial genomics has helped researchers in understanding the pathogenicity of bacteria and other microbes.
Transformation Experiment in Bacteria
In the discovery of genetic material, the experiment conducted by Frederick Griffith on Streptococcus pneumonia proved to be a stepping stone.
Plasmids and Vectors
The DNA molecule that exists in a circular shape and is smaller in size which is capable of its replication is called Plasmids. In other words, it is called extra-chromosomal plasmid DNA. Vectors are the molecule which is capable of carrying genetic material which can be transferred into another cell and further carry out replication and expression. Plasmids can act as vectors.
Explain why the experimental procedures performed by Lenski (and his lab) caused the fitness of the bacteria to increase over time (especially early on) in all 12 of his populations.
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- What is the 1% extinction value of the protein used to calibrate this experiment? I obtained a gradient of 0.0021, into which units should this be presented?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…Please help with all of these! thank you so much! Neutral regions of a species’ genome that are distant from any selected sites have apairwise diversity (?) of 0.01. Other regions of the genome have reduced genetic variation,and selective sweeps could be responsible for the reduction. Assume that selective sweepshappen at a rate of 10^-10 per bp, and they take 500 generations to reach fixation. Therecombination rate in regions where selective sweeps might occur is 10^-8 per bp. Theeffective population size of the species is 10^5. A. Find the expected genetic diversity (? per bp) in regions where selective sweepsmight be occurring.B. Find the deleterious mutation rate (per bp) that would be required to produce anequivalent reduction in genetic diversity as a result of background selection (BGS)?Assume the same recombination rate as above.
- 54. 35delG mutation in the connexin 26 gene is an autosomal recessive condition that has been identified to cause congenital deafness. In a population of 367 Greek children, 75 of the children carried two copies of the 35delG mutation.In order to calculate the percentage of individuals who are heterozygous for the 35delG mutation, the calculation that must be performed is Select one: a. 1 – (p2 – 2pq) b. 1 – (p +q) c. 1 – (p2 + q2) d. 1 – q2The electrophoresis gel shown below had a 4th lane (cut off and not shown) which contained 7 proteins of known molecular weight. A plot of these proteins was made with log MW on the y-axis and migration distance (cm) on the x-axis.The equation for the linear regression line is: Y = -0.1 X + 4.79If the distance between the well and the 3rd band of the Red sample is 4 cm, what is the approximate MW, expressed in daltons, of the protein at that location?Cold sensitive mutation (Cs) results in a mutant phenotype below a particular temperature. Bacteria with mutation ess-2 (Ts) can form colonies at 32oC but not at 37oC and 42oC, while bacteria with mutation ess-5 (Cs) can form colonies at 42oC but not at 32oC and 37oC. What would be the phenotype of an ess-2 (Ts) and ess-5 (Cs) double mutant?
- When Meselson and Stahl grew E. coli in 15N mediumfor many generations and then transferred the cells to 14Nmedium for one generation, they found that the bacterial DNA banded at a density intermediate between that ofpure 15N DNA and pure 14N DNA following equilibrium density centrifugation. When they allowed thebacteria to replicate one additional time in 14N medium, they observed that half of the DNA remained atthe intermediate density, while the other half banded atthe density of pure 14N DNA. What would they haveseen after an additional generation of growth in 14Nmedium? After two additional generations?In pine tree it was found three genotypes (R2R2, R2R3, R3R3) at a locus encoding the enzyme peroxidase. The observed numbers of trees of these genotypes were genotypes number observed R2R2 135 R2R3 44 R3R3 11 Are the pine trees population in Hardy-weinberg equillibrium at the peroxidase locus? use chi square testRecombinant protein production by a genetically modified Escherichia coli strain is proportional to cell growth. Ammonia is used as a nitrogen source for aerobic glucose respiration. The recombinant protein has the general formula CH1,55O0,31N0,25, while that of the cellular biomass is CH1,77O0,49N0,24. The biomass yield from glucose equals 0.50 g/g, while the recombinant protein yield from glucose corresponds to 20% of the cell yield from substrate.a) How much ammonia is required? What is the oxygen demand? (b) If the biomass yield remains the same, what are the ammonia and oxygen requirements for a wild-type strain of E. coli, with cell biomass of the same elemental composition, but unable to synthesize the recombinant protein? (c) On an industrial scale, cultivation takes place in a continuous fermenter at 28°C and the desired recombinant protein production rate is 7 g/h. Since the viscosity of the culture broth is considerable, the energy input due to agitation cannot be neglected.…
- Production of a recombinant protein by E. coli is proportional to cell growth. Ammonia is used as a source of N and glucose as a source of C, under aerobic conditions. The recombinant protein has a general formula CH1.55N0.31O0.25 and the cell CH1.77N0.4900.24. The yield of biomass from glucose is 0.48 gcel/gglic, and the yield of recombinant protein from glucose is about 20% of that for biomass.a) How much ammonia is needed to produce 50 g of cells producing the recombinant protein?b) What is the oxygen demand in this process?c) For the cultivation of a wild E. coli not producing the recombinant protein, how different would the ammonia and oxygen demand be if the biomass yield remained at 0.48 gcel/gglyc ?Price et al. [(1999). J. Bacteriol. 181:2358–2362] conducteda genetic study of the toxin transport protein (PA) of Bacillusanthracis, the bacterium that causes anthrax in humans. Withinthe 2294-nucleotide gene in 26 strains they identified five pointmutations—two missense and three synonyms—among differentisolates. Necropsy samples from an anthrax outbreak in 1979revealed a novel missense mutation and five unique nucleotidechanges among ten victims. The authors concluded that thesedata indicate little or no horizontal transfer between differentB. anthracis strains. Question: Which types of nucleotide changes (missense or synonyms)cause amino acid changes?Price et al. [(1999). J. Bacteriol. 181:2358–2362] conducteda genetic study of the toxin transport protein (PA) of Bacillusanthracis, the bacterium that causes anthrax in humans. Withinthe 2294-nucleotide gene in 26 strains they identified five pointmutations—two missense and three synonyms—among differentisolates. Necropsy samples from an anthrax outbreak in 1979revealed a novel missense mutation and five unique nucleotidechanges among ten victims. The authors concluded that thesedata indicate little or no horizontal transfer between differentB. anthracis strains. Question: On what basis did the authors conclude that evidence ofhorizontal transfer is absent from their data?