Consider the experiment conducted by Meselson and Stahl in which they used 14N and 15N in cultures of E. coli and equilibrium density gradient centrifugation. Draw pictures to represent the bands produced by bacterial DNA in the centrifuge tube before the switch to medium containing 14N and after one, two, and three rounds of replication in that medium. Use separate sets of drawings to show the bands that would appear if replication were (a) semiconservative; (b) conservative; (c) dispersive.

Can you please help with 1c please  picture with 1 graph is for question 1a) picture with 4 graphs is for question 1b) 1a) E. coli DNA and binturong DNA are both 50% G-C. If you randomly shear E. coli DNA into 1000 bp fragments and put it through density gradient equilibrium centrifugation, you will find that all the DNA bands at the same place in the gradient, and if you graph the distribution of DNA fragments in the gradient you will get a single peak (see below). If you perform the same experiment with binturong DNA, you will find that a small fraction of the DNA fragments band separately in the gradient (at a different density) and give rise to a small "satellite" peak on a graph of the distribution of DNA fragments in the gradient (see below). Why do these two DNA samples give different results, when they're both 50% G-C? 1b) If you denatured the random 1000 bp fragments of binturong DNA that you produced in question 1a by heating them to 95ºC, and then cooled them down to 60ºC…

You are analyzing the intracellular DNA intermediates formed during the replication of the single-stranded (ss) DNA phage M13. In a replication mutant, you find accumulation of DNA molecules with the following properties:Results of a 0.7% agarose gel electrophoresis run at pH 8.0 – gel is running top tobottom, with the origin at the top of the diagram. The smudges indicate several bandswith slightly different migration rates.   open circle M13closed circle M13 Left lane — markers (wild-type open circle M13 and closed circle M13)2nd lane — DNA extracted from mutant M13 phage cells3rd lane — DNA extracted from mutant M13 phage cells denatured4th lane — DNA extracted from normal M13 phage cellsRight lane — DNA extracted from normal M13 phage cells denaturedProfiles from a 5 – 20% sucrose gradient centrifuging of the mutant M13 phage DNA   The arrow on the neutral (pH 7.4) gradient shows the position of the covalently closed circular DNA. On the alkaline (pH 12.5) gradient, note two peaks…

What result would Meselson and Stahl have obtained if the replication of DNA were conservative (i.e., the parental double helix stayed together)? Give the expected distribution of DNA molecules after 1.0 and 2.0 generations for conservative replication.

What is the role of GelRed® in Agarose gel electrophoresis of DNA fragments? Please select the single answer that is most correct GelRed® moves down the agarose gel in response to the electric current and enables visualisation of the position of A the nucleic acids within in the agarose gel. GelRed® intercalates with the Nucleic acid and, under UV light, fluoresces to enable visualisation of the position of the nucleic acids in the agarose gel. GelRed® intercalates with the Nucleic acid and enables visualisation of the position of the nucleic acids in C the agarose gel. GelRed® intercalates with the amino acids in the agarose gel and enables visualisation of the position of their in D the agarose gel.

Can you please help with 1f please picture with 1 graph is for question 1a) picture with 4 graphs is for question 1b) 1a) E. coli DNA and binturong DNA are both 50% G-C. If you randomly shear E. coli DNA into 1000 bp fragments and put it through density gradient equilibrium centrifugation, you will find that all the DNA bands at the same place in the gradient, and if you graph the distribution of DNA fragments in the gradient you will get a single peak (see below). If you perform the same experiment with binturong DNA, you will find that a small fraction of the DNA fragments band separately in the gradient (at a different density) and give rise to a small "satellite" peak on a graph of the distribution of DNA fragments in the gradient (see below). Why do these two DNA samples give different results, when they're both 50% G-C? 1b) If you denatured the random 1000 bp fragments of binturong DNA that you produced in question 1a by heating them to 95ºC, and then cooled them down to 60ºC and…

The transformation results below were obtained with 10 ul of intact plasmid DNA at nine concentrations. The following numbers of colonies are obtained when 100 ul of transformed cells are plated on selective medium: Fill in the following table: Concentration # colonies DNA mass of Fraction of Mass Transformation PGREEN (Concentration x volume OR X spread = x 10ul plasmid solution) PGREEN in cell Cell efficiency Y÷ A suspension suspension spread = 100 ul - total vol cell susp. (Colonies - Mass spread) C x Z = A See (510 ul) HINT: this calculation is constant Given= X Given=Y С. Z. 0.00001 ug/ul | 4 0.00005 ug/ul 12 0.0001 ug/ul 0.0005 ug/ul 32 125 0.001 ug/ul 442 0.005 µg/ul 0.01 ug/ul 0.05 ug/ul 0.1 ug/ul 542 507 475 516 0.5 ug/ul 505

Can you please help with 1c please  picture with 1 graph is for question 1a) picture with 4 graphs is for question 1b) 1a) E. coli DNA and binturong DNA are both 50% G-C. If you randomly shear E. coli DNA into 1000 bp fragments and put it through density gradient equilibrium centrifugation, you will find that all the DNA bands at the same place in the gradient, and if you graph the distribution of DNA fragments in the gradient you will get a single peak (see below). If you perform the same experiment with binturong DNA, you will find that a small fraction of the DNA fragments band separately in the gradient (at a different density) and give rise to a small "satellite" peak on a graph of the distribution of DNA fragments in the gradient (see below). Why do these two DNA samples give different results, when they're both 50% G-C? 1b) If you denatured the random 1000 bp fragments of binturong DNA that you produced in question 1a by heating them to 95ºC, and then cooled them down to 60ºC…

Can you please help with 1c please  picture with 1 graph is for question 1a) picture with 4 graphs is for question 1b) 1a) E. coli DNA and binturong DNA are both 50% G-C. If you randomly shear E. coli DNA into 1000 bp fragments and put it through density gradient equilibrium centrifugation, you will find that all the DNA bands at the same place in the gradient, and if you graph the distribution of DNA fragments in the gradient you will get a single peak (see below). If you perform the same experiment with binturong DNA, you will find that a small fraction of the DNA fragments band separately in the gradient (at a different density) and give rise to a small "satellite" peak on a graph of the distribution of DNA fragments in the gradient (see below). Why do these two DNA samples give different results, when they're both 50% G-C? 1b) If you denatured the random 1000 bp fragments of binturong DNA that you produced in question 1a by heating them to 95ºC, and then cooled them down to 60ºC…

A neutrophil detects a pathogen, which step would most likely happen next?     Increase in intracellular levels of Arp2/3     Decrease in intracellular levels of Arp2/3     Neutrophils do not respond to pathogens.     Decreases in capZ