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- Choose the correct gel electrophoretic pattern that would be seen in dideoxy sequence analysis of the DNADNA molecule shown below. pGGCGACCGATTAGTCCCATCGATGGG−OHHow is the color of the spot coverted into useful data if data is collected from a Vicrochip DNA microarray that's from the colors of spots that illuminate when the spots have a laser shine on them? From the following which one is the best option The color of the spot is converted to a number that represents the intensity of green or red, so that the numerical intensity values can be compared between spots by a computer program The color of the spot is bright so that it can be interpreted visually by trained scientists The color of the spot is present on the chip, counted and a ratio of red-yellow and green-yellow is calculated which can be done by hand The color of the spot can't be converted None of the aboveCan you please help with 1d 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…
- Match the PCR sample with the predicted result on the agarose gel. 1) Sample from individual homozygous for their PV92 locus with the Alu insert. Five bands of 100, 200 500, 700 and 1000bp Single band of 300 bp Two bands one of 300 bp and one of 641bp Tow bands one of 641 bp and one of 941 bp No Bends Single band of 941 bp Single band of 641bp 2) Sample from individual homozygous for their PV92 locus without the Alu insert. Five bands of 100, 200 500, 700 and 1000bp Single band of 300 bp Two bands one of 300 bp and one of 641bp Tow bands one of 641 bp and one of 941 bp No Bends Single band of 941 bp Single band of 641bp 3) Sample from individual heterozygous for their PV92 locus having the Alu insert. Five bands of 100, 200 500, 700 and 1000bp Single band of 300 bp Two bands one of 300 bp and one of 641bp Tow bands one of 641 bp and one of 941 bp No Bends Single band of 941 bp Single band of 641bp 4) Negative control sample. Five bands of 100, 200 500, 700 and 1000bp…For the given restriction enzyme: BslI (CCNNNNN^NNGG) i) Approximately how many fragments would result from digestion of the human genome (3 × 109 bases) with the enzyme? ii) Estimate the average size of the pieces of the human genome produced by digestion with the enzyme. iii) State whether the fragments of human DNA produced by digestion with the given enzyme would have sticky ends with a 5′ overhang, sticky ends with a 3′ overhang, or blunt ends. iv) If the enzyme produces sticky ends, would all the overhangs on all the ends produced by that enzyme on all fragments of the human genome be identical, or not? The recognition sequence on one strand for the enzyme is given in parentheses, with the 5′ end written at the left. N means any of the four nucleotides; R is any purine—that is, A or G; and Y is any pyrimidine—that is, C or T. ^ marks the site of cleavage. Note that the recognition sites containing Ys and Rs are not always rotationally symmetrical.ARE THEY TRUE OR FALSE? a) In a caesium chloride density gradient centrifugation method performed in the presence of ethidium bromide, supercoiled plasmid DNA binds more ethidium bromide than linear DNA. B)Tth DNA polymerase shows DNA-dependent DNA polymerase activity as well as RNA-dependent DNA polymerase activity. C)Magnesium ions stimulate polymerase activity. Therefore, it is better to use the highest concentration of magnesium in PCR amplification. D)When lambda bacteriophage infects E. coli cell lysis never occurs, and the infected bacterium can continue to grow and divide. E)The copy number refers to the number of molecules of an individual plasmid that are normally found in a single bacterial cell. F)RNase H selectively hydrolyzes phosphodiester bonds of RNA molecules in RNA:DNA duplexes.
- 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…DNA from 100 unrelated individuals from one population of Chinook salmon were amplified at a single microsatellite locus, and run on an agarose gel. The results from gel electrophoresis shows three different fragment lengths (i.e. alleles/band positions) corresponding to 3 alleles; Allele F (250bp), Allele R (180bp), and Allele Y (100bp). The numbers at the top of each lane is the number of fish observed with that particular genotype or banding pattern in the population. Note that a single band means a homozygote for that allele (band size). a) Calculate the allele frequency for Allele Y. (b) Calculate the genotype frequencies for the following genotypes: FF, FR, and RY. (c) What is the expected number of Chinook salmon with homozygous genotype for allele Y in the study population? (d)What is the name of the statistical test that you could conduct to test whether this population of Chinook salmon is in Hardy Weinberg EquilibriumA method for detecting methylated CpGs involvesthe use of a chemical called bisulfite, which convertscytosine to uracil but leaves methylated cytosine untouched. You want to know whether a particularCpG dinucleotide at one location in the genome ismethylated on one or both strands in a tissue sample.The genomic sequence containing this CpG is:5’...TCCATCGCTGCA…3’. You take genomicDNA from the sample tissue, treat it exhaustivelywith bisulfite, and then use flanking primers toPCR-amplify the region including this CpGdinucleotide. You then want to Sanger sequence(see Fig. 9.7) the amplified PCR product. a. After you treat genomic DNA with bisulfite, the twoDNA strands will melt into single strands. Why?b. Your answer to part (a) introduces a potential complication, because if you do not account for this result of bisulfite treatment, the PCR primers willnot amplify the DNA. What special considerationswould be necessary when you design your PCRprimers for this experiment? Could one pair…
- High doses of caffeine interfere with the DNAdamage response in mammalian cells. Why then do yousuppose the Surgeon General has not yet issued an appro-priate warning to heavy coffee and cola drinkers? A typicalcup of coffee (150 mL) contains 100 mg of caffeine (196 g/mole). How many cups of coffee would you have to drinkto reach the dose (10 mM) required to interfere with theDNA damage response? (A typical adult contains about 40liters of water.)In eukaryotes, Cot DNA reassociation curves are not smooth. They have bumps. Why? What are the bumps?Suppose you are constructing a human genomic library in BAC vectors where the human DNA fragments are on average 100,000 bp. a. What is the minimum number of different recombinant BACs you need to construct in order to havea greater than zero chance of having a completelibrary—meaning one in which the entire genomeis represented?The simple statistical equation that follows allows youto determine the size that a genomic library needs tobe (that is, the number of independent recombinantclones you need to make) for a given likelihood thatthe entire genome is represented in the library.N = ln (1 − P)ln (1 − f )In the equation, N is the number of independent recombinant clones; P is the probability that any particular partof the genome is represented at least one time; f is thefraction of the genome in a single recombinant clone.(Note: ln is the natural log, sometimes written as loge.)b. Calculate f for the genomic library described in part (a).c. How many different recombinant BAC…