Back in the 1950s, the techniques for isolating DNA from cells all yielded molecules of about 10,000 to 20,000 base pairs. We now know that the DNA molecules in all cells are much longer. Why do you suppose such short pieces were originally isolated?

The Sanger products in the sequencing of a template DNA is shown below: +ddTTP +ddATP +ddCTP +ddGTP Pentanucleotide trinucleotide mononucleotide Dinucleotide Hexanucleotide Octanucleotide tetranucleotide Heptanucleotide nonanucleotide What is the sequence of the template DNA? A. 3'-GAGTTCAGC-5' B. 5'-GAGTTCAGC-3' С. 3-СТСААGТCG-5' D. 5-СТCААGТCG-3' E. 5'-CUCAAGUCG-3'

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…

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…

The region of the normal hemoglobin gene used for genetic testing for sickle cell anemia contains a restriction site such that homozygous normal individuals show two DNA fragments. If a single nucleotide change in hemoglobin destroys that restriction site, then how many DNA fragments will be visible on a gel from individuals that are homozygous mutant?  What about heterozygotes?

If the GAATTC palindrome repeats are randomly found along the DNA strand, then what can you say about the sizes of the fragments that will be produced when the DNA is digested with a restriction enzyme that recognizes that sequence?  How does the total length of the fragments relate to the size of the original DNA fragment?

Nucleosomes can be assembled onto defined DNA segments. When a particular 225-bp segment of human DNA was used to assemble nucleosomes and then incubated with micrococcal nuclease, which digests DNA that is not located within the nucleosome, uniform fragments 147  bp in length were generated. Subsequent digestion of these fragments with a restriction enzyme that cuts once within the original 225-bp sequence produced two well-defined bands at 37 bp and 110 bp. Why do you suppose two well-defined fragments were generated by restriction digestion, rather than a range of fragments of different sizes? How would you interpret this result?

How are DNA molecules visualized in a gel after electrophoresis? Why do DNA molecules migrate toward the + electrode? What determines the rate of their migration? What is the effect of PEG on DNA fragments of different sizes? How is this influenced by the concentration of PEG?

You are working in a biotechnology lab arid are analysing DNA. You obtain a sample of a short dodecamer of DNA that contain 12 base pairs. What must the ratio of adenine to thymine be in your sample? What must be ratio of cytosine, to guanine be in your sample? Assume the counter ions present in your DNA solution are sodium ions. How many sodium ions must there be per dodecamer? Assume the 5’ end phosphate each bear a-1 charge.