A plot showing the % of denautration as a function of temperature for a melting point of a DNA sample under 0.12 M NaCl gives an equation of a line of Y= 0.0074 X - 0.044 (Where Y is the % of denautration and X is the temperature in degree C) This is all the information that was given. 1. Calcuate the melting point of this DNA in degree C
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A plot showing the % of denautration as a function of temperature for a melting point of a DNA sample under 0.12 M NaCl gives an equation of a line of
Y= 0.0074 X - 0.044
(Where Y is the % of denautration and X is the temperature in degree C)
This is all the information that was given.
1. Calcuate the melting point of this DNA in degree C
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Solved in 3 steps
- A plot showing the % of denaturation as a function of temperature for a melting point of a DNA sample under 0.12 M NaCl gives an equation of a line of Y= 0.0074 X - 0.044. (where Y is the % of denaturation and X is the temperature in °C) Question: Calculate the melting point of this DNA in °C.You extract DNA from 200, You pipette 200 microlitres of this extraction sample into a 3-ml cuvette and make up to 3.0 ml using buffer. The absorbance at 260 nm (A260) of the solution in the cuvette is 0.2 (Remember: an A 260 of 1.0 corresponds to a DNA concentration in the cuvette solution of 50 micrograms per ml). Calculate the total amount of DNA in 100g wheat germ, assuming that the extraction efficiency of DNA from wheat germ during your lab protocol is 40%. Please fill out each of the boxes in the Table below What to Calculate Calculation / Result Marks (x / 100) 12/100 DNA concentration in cuvette solution (in microgram per ml) Dilution factor of DNA sample in cuvette 12/100 12/100 Total amount (microgram) of DNA in the entire extraction sample Total amount (milligram) of DNA in 100g wheat germ 24/100 assuming an extraction efficiency of 40%if this DNA has a molecular weight of 1.20 ×108 Dalton which contains a head in a about 200 nm long. Calculate the length of the DNA assuming the molecular weight of a nucleotide pair is 600 Dalton and assume that the DNA is a B-form and that there are 10 base pair per turn which makes is 34 Å per turn. ( 1nm = 10Å)
- Show the calculations for the preparation of 0.01%, 0.001%, 0.0001%, 0.00001% standard DNA solutions via serial dilution from a 0.1% stock solution. Prepare 10.0 mL of 0.10% standard DNA solution using acetate buffer pH 4.6 as solvent. Prepare four different concentrations of the 0.10% standard DNA solution from step 1 via serial dilution to obtain 0.01%, 0.001%, 0.0001%, 0.00001% standard DNA solutions. Obtain seven clean and dry test tubes. Place 1.50 mL solutions in each tube as follows: Table 2.1. Set-up for the diphenylamine assay. Test tube # Content 1 (blank) acetate buffer pH 4.6 2 0.10% standard DNA 3 0.01% standard DNA 4 0.001% standard DNA 5 0.0001% standard DNA 6 0.00001% standard DNA 7 DNA extract from Part I Add 3.50 mL diphenylamine reagent to each tube, swirl each tube to thoroughly mix the contents and heat for 10 mins in a boiling water bath. Cool immediately under tap water. Read absorbances at 595 nm.…Assume that the molar percentage of thymine in a double stranded DNA is 20. What are the percentages of the four bases (G, C, T, A)? b. The base content of a sample of DNA is as follows: A=31% G=31% T=19% C=19%. What conclusion can be drawn from this information?A 10 μL-aliquot of a resuspended genomic DNA stock solution was obtained and further diluted by adding 990 μL TE buffer. The A260 of the resulting solution was 0.316. Determine its concentration in µg/µL.
- Please answer You have received a dehydrated sample of DNA primer at a concentration of 19.9 microM. what volume (in microlitres) of buffer would you add to achieve a solution of 100nM of this primer? Show workings.The figure shows that the average distance between base pairs measured parallel to the axis of a DNA molecule is 3.4 Å. The average molar mass of a pair of nucleotides is 650 g•mol-1. Estimate the length in cm of a DNA molecule of molar mass 5.1x109 4.0 .26 g•mol-1. cm Roughly how many base pairs are contained in this molecule? [4.0 784615 x base pairs CG G C OH TA Thymine Adenine H. H. OH CH2 N-H.......O CH3 H H N……·H–N HLH TA OH 0-P H2C CG G C H. CH2 Guanine H OH Cytosine ... H-N TA AT H' N.....H-N HLH N-H.......O H2C OH H -o. -P3D0 TA 0- OH H. -CH2 (b) The most common structure of DNA, which is a right-handed double helix. The two strands are held together by hydrogen bonds (a) Base-pair formation between adenine (A) and thymine (T) and between cytosine (C) and guanine (G). and other intermolecular forces.The complementary strands of DNA in the double helixare held together by hydrogen bonds: G ≡ C or A = T.These bonds can be broken (denatured) in aqueous solutions by heating to yield two single strands of DNA(see Figure 1-13a). How would you expect the relativeamounts of GC versus AT base pairs in a DNA doublehelix to affect the amount of heat required to denatureit? How would you expect the length of a DNA doublehelix in base pairs to affect the amount of heat requiredto denature it?
- Assume the energy of hydrogen bonds per base pair to be 5.86 kJ•mol-1. Given two complementary strands of DNA containing 145 base pairs each, calculate the ratio of two separate strands to hydrogen-bonded double helix in solution at 319 K. ratio = Supnorting MatorialsThe melting temperature Tm of DNA can be predicted by calculation without actually measuring it. Calculate the Tm of the DNA double strand shown in (1) to (3), and discuss the results. The numbers in parentheses indicate the degree of polymerization of nucleotides.(1) A(10) + T(10), (2) A(15) + T(15), (3) G(10) + C(10)The figure shows that the average distance between base pairs measured parallel to the axis of a DNA molecule is 3.4 Å. The average molar mass of a pair of nucleotides is 650 g•mol-1. Estimate the length in cm of a DNA molecule of molar mass 5.1x109 g•mol-1. 4.0 cm Roughly how many base pairs are contained in this molecule? 4.0 7.1e6 X base pairs AT CG OH O-P=D0 G C TA Adenine Thymine OH CH2 N-H.......O CH3 H -N N..... G C H A Он C G G C TA AT OH CH2 Cytosine Guanine H. O.......H-N H H' H N…**H-N HLH O N-H..... H2C OH TA OH CH2 (b) The most common structure of DNA, which is a right-handed double helix. The two strands are held together by hydrogen bonds and other intermolecular forces. (a) Base-pair formation between adenine (A) and thymine (T) and between cytosine (C) and guanine (G).