Experiment Mouse injected with type S Mouse injected with type R Lived or Died Die Live Mouse injected with dead type Live S Mouse injected with a mix of Die type R and dead type S 1. Fill in the table above saying whether or not the indicated mice lived or died for each experiment. 2. What was concluded from this experiment? Note: nothing was concluded about DNA from this experiment. Later experiments showed that the relevant factor is DNA.
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- Why is the company Qiagen has more refined DNA extraction steps than a normal Strawberry DNA extraction practical? Summary of Qiagen DNA extraction steps Add ATL buffer and grind with sample. Add 20 microliters of enzyme Proteinase K to degrade protein into a 1.5-2ml microcentrifuge tube. Add 200 microlitres AL lysis buffer, and mix by vortexing for 5–10 seconds, which breaks cell membrane allowing DNA to be released. Incubate the sample at 56 degrees for 10 minutes. Mix the cell lysate with 200 microlitres ethanol by pipetting it at the side of the microcentrifuge wall so DNA precipitates. The DNA forms a white layer and the remaining liquid is discarded. Pipet the mixture into DNeasy Mini spin column placed in a 2 ml collection tube. Centrifuge for a minute at 8000 rpm. Place the mini spin column into a 2 ml collection tube, add 500 µl Buffer AW1, and centrifuge for 1 min at 8000 rpm. Then add it to a new 2 ml collection tube (provided), add 500 µl Buffer AW1, and centrifuge for 1…An AGE run was set at 100V for 30 min. The 3 ul of the ladder was loaded into the gel, while 10 ul of the DNA samples plus an appropriate amount of 6x loading buffer were added into the gel. Find the amount of the 6x loading buffer added to 10 ul DNA samples in order to make the samples sink in the gel? Tip. From 6x final conc of the loading buffer should be 1x and answer shoud be in ulCan you explain thie each of the statement given i dont really understand the dna recombinant is used as a molecular cloning and application for recombinant dna
- Part A Why are most recombinant human proteins produced in animal or plant hosts instead of bacterial host cells? Bacteria do not process any eukaryotic proteins. Bacterial cells might be harmfull for the researchers. Bacteria might not process all the eukaryotic proteins that might be usefull to process the protein molecule. O Only transgenic eukaryotes are able to process eukaryotic proteins. Submit Request AnswerCopy and paste the link below and watch the video on Youtube and Answer the Questionshttps://www.youtube.com/watch?v=g-dNJdOvBM4 Polymerase Chain Reaction Questions: 1. What are the materials used for the polymerase chain reaction? 2. Draw a schematic diagram of the procedure in PCR. 3. Why is it important to design the primers at the start of the laboratory procedure? 4. What are the components of the PCR buffer and what is its pH range? What is the purpose of the buffer? 5. What is the use for magnesium chloride? 6. How much template DNA is added? What is the concentration of the primers? 7. At what temperatures does denaturation, annealing and extension occur? Why are the processes placed in that temperature? 8. In this particular PCR experiment, how many cycles was used? 9. Can this PCR be used on its own to find out if a person has Covid or not on its own? Why or why not?3 different pcr tubes were obtained each tube containing a variation of hemoglobin E. tube A is a normal hemoglobin E,tube B is a diseased hemoglobin E, and tube C is a hemoglobin E trait. conduct a graph showing which one would move further electrophoresis gel well if an experiment was performed. explain your answer.
- Here is a DNA agarose gel showing PCR products from a mouse genotyping experiment. Genotyping tells us whether each mouse is a wild type mouse (i.e. not genetically modified) or a mutant mouse. Interpret the results for each mouse 1-3.The results of gel electrophoresis of 4 different DNA samples given in the figure. 16 ul was loaded into each well (6 μL diluted DNA* + 8 μL water + 2 μL sample buffer). The ladder is in lane 5, with the size and amount (ng) in each band indicated. *1:5 dilutions of the original DNA sample were made, and the diluted samples were used for AGE. Q. These are the conclusions made about sizes and concentrations of the original (undiluted) DNA samples from lanes 1 and 2. Are these conclusions reasonable? Explain specifically, and if any of the conclusions are not reasonable, explain why not and what should we conclude instead. Sample in lane 1:size is 2.32 kb concentration of original DNA sample is 9.2 ng/ul Sample in lane 2:size is 4.36 kb concentration of original DNA samples is 100 ng/ul Q. The sample in lane 3 was expected to be about 2 kb in size. What is a possible explanation for the results observed?Give only typing answer with explanation and conclusion Recombinant human insulin produced by bacteria carrying a cloned insulin gene, is now the major form of insulin used to treat diabetes. The human insulin gene encodes an mRNA only 333 nucleotides long, but the entire gene spans more than 4000 nucleotides. There are three exons and two introns 1. Every cell in the human body has the same DNA, so very cell has an insulin gene. However in order to use the technique, you described in b, you would have to start with cells from the pancreas the only body cells that actually produce the insulin protein. Why are these the only cells that would work.
- ABO blood group system is defined by the presence of agglutinogens (A and B molecules) at the surface of red blood cells. Enzyme A which leas to the production of the molecule A is coded by all ele A, while enzyme B which leads to the production of the molecule B is coded by allele B, and enzyme O which cannot lead to the production of any molecule is coded by allele O. A part of the coding DNA strand for enzyme A: GAC GTG CGC GCC A part of the coding DNA strand for enzyme B: GAG GTG GcC GCC 5. Compare the non-transcribed strand coding for enzyme A to that coding for enzyme B. 6. Identify the type of mutation involved in this case. Justify. 7. Write the amino acid sequence for both enzymes. 8. "Mutations can lead to diseases or to genetic diversity"Justify by refering to parts A and B. Second letter A G UCU UCC UCA UUG Leu ucG UUU T Phe UUC UAU1- Tyr UACJ Ser UAA Stop UGA Stop UGU), UGCJ UUA UAG Stop UGG Trp CAUTHIS CCU CC CCA CCG CUU CÚC CÁCJ Pro CAA CGU] CGC Arg FLeu CGA CGG CỦA Gln…1/V (uM x min-1)-1 1. To the right is a Lineweaver-Burk plot for an enzyme that can cleave both DNA (D) and RNA (R). 2200 2000 1800 a. What is the KM when cleaving DNA? I 1/V R 1600 • 1/V D b. What is the KM when cleaving RNA? 1400 1200 C. What is the Vmax when cleaving DNA? Z 1000 800 d. What is the Vmax when cleaving RNA? 600 e. On the graph, draw an appropriate line for a 400 noncompetitive inhibitor for cleaving DNA. Label clearly. f. On the graph, draw an appropriate line for an uncompetitive inhibitor for cleaving RNA. Label clearly. 200 8 10 11 1/S (µM)-1O Off target effects are not really a concern. Question 20 What happens after a double stranded break is induced in the DNA? Select the statement that is FALSE. O HR which will lead to a small indel if template DNA is absent O Microhomology-Mediated End Joining O Non-Homologous End Joining O HR if template DNA is present Question 21 See below for four STR profiles from four different boys, as depicted in an electropherogram. The peak local