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- Answer the following questions. 1. Explain the steps associated with DNA profiling and state its advantages and disadvantages, 2. Describe how mutations are linked to DNA polymorphism. GUIDELINES: Each answer should be 250 words in length. Content should not be copied. References and bibliography should be provided for the content and images. please asapCan you please help with 1b 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…
- What are some difficulties with low copy number DNA analysis? What kinds of checks and balances can a laboratory employ to ensure reliability of DNA profiles coming from low amounts of DNA templateWhat are the benefits of using a mixture classification scheme as outlined in DNA Box 14.1? What would be the advantages of using software for deciphering mixture components?See the attachment and answer the following parts of the question: A) If the binturong genome is 2.87 x 109 base pairs, and the "highly repetitive DNA" fraction is composed entirely of copies of sequence 5'ATGGTCC3' and its complement, how many copies of this sequence are present in the binturong genome? B) Briefly explain, in your own words, why the fraction of the binturong DNA fragments that reannealed relatively slowly took so much longer to renature than the other DNA fragments. C) If you took more of the same randomly generated 1000 bp fragments of binturong DNA (the same sample that you used in the equilibrium density gradient centrifugation experiment described in part a and the C0t curve described in part b of this question) and used them as a sample in agarose gel electrophoresis, how many bands would you expect to find in the gel when you turned off the current and stained the gel with ethidium bromide? Briefly explain why you would predict that number of bands.
- In order to determine the purity of a DNA sample. spectrophotometry can be carried out at _______ to measure DNA, and _______ to measure protein. The ratio _______ is then calculated, and a number between _______ indicates higher purity. 280 nm; 260 nm; A280/A260; 0.5-1 260 nm; 280 nm; A260/A280; 0.5-1 260 nm; 280 nm; A260/A280; 1.5-2 260 nm; 280 nm; A280/A260; 1.5-2Is the DNA isolated from Cheek Cells using Saline solution, Detergent, and Ice-cold alcohol pure or not?Consider the four extraction/purification methods . a. Which extraction method would you use for a touch DNA sample? Explain your reasoning.b. Which extraction method would you use for a sample containing a mixture of primarily non-human and some human DNA? INFO: the 4 extraction methods are QIAamp manual extraction, QIAcube semi-automated extraction, DNA IQ extraction, or FTA punch purification
- 1.) What characteristics of VNTR and STR make them useful for DNA fingerprinting? 2.) How does PCR minimize the problems associated with degraded DNA? 3.) What factors can cause DNA to become degraded? 4.) If Ethidium bromide was not added to a gel, what would happen? 5.) How can you tell if an individual is heterozygous for the D1S80 marker? 6.) If a negative control produces a band, what does this indicate? 7.) In an experiment, a student’s sample amplified for D1S80 produced 3 bands. It was the only DNA sample run on the gel. The student knows that there was no problem with the Thermocycler or primers because the other students in the class had the expected results of only one or two bands. What is the most likely explanation for these results?Below are 9 possible primer pairs.● Determine which primer pair is the best choice.● Explain why the other primers are not good choices.● Calculate the Tm for each primer. Underline or highlight the region of DNA for the primer pair you chose as the best.Forward 1: 5’ gaaataattttgtttaactttaag 3’ Tm =Reverse 1: 5’ gtttaagacaaaatagtctgg 3’ Tm =Forward 2: 5’ gtaactcagctttcaggtcg 3’ Tm =Reverse 2: 5’ tctcggaatgttgcaacagc 3’ Tm =Forward 3: 5’ agattagcggatcctacctg 3’ Tm =Reverse 3: 5’ atgtgtaatcccagcagcag 3’ Tm =Forward 4: 5’ cattgattatttgcacggcg 3’ Tm =Reverse 4: 5’ aaaatcttctctcatccgcc 3’ Tm =Forward 5: 5’ tccataagattagcggatcc 3’ Tm =Reverse 5: 5’ tgcaagcttggctgttttgg 3’ Tm =Forward 6: 5’ gatcctacctgacgcttttta 3’ Tm=Reverse 6: 5’ aaataatgaattcgagctcggt 3’ Tm =Forward 7: 5’ataaaaaaatcgagataaccgtt 3’ Tm =Reverse 7: 5’aggtcgactctagaggatc 3’ Tm =Forward 8: 5’ctacctgttccatggccaac 3’ Tm=Reverse 8: 5’ ttcgggcatggcactcttg 3’ Tm=Forward 9: 5’ tccataagattagcggatcc 3’ Tm =Reverse 9: 5’…Why does evenly distributed peak in DNA chromatogram is an indication of a good sequence?