PRE-LAB ASSIGNMENT: GMBI, VIBRIO PCR & COMPLEMENTATION Name GSI & Sect # Station

Question

PRE-LAB ASSIGNMENT: GMBI, VIBRIO PCR & COMPLEMENTATION Name GSI & Sect # Station # 7. Sequencing reactions typically use only one type of primer: The primer molecules anneal and provide a free 3 oH group to which nucleotides can add. Nucleotides base pair with the template strand and are linked in a 5' to 3' direction by DNA polymerase. If double stranded DNA is used as a template for DNA sequencing, the double stranded DNA is denatured to yield single stranded DNA. a. How many identical molecules of DNA are present in a 0.5 mg sample? Of course, the number of molecules depends on the length of the DNA molecule and whether it is single- or double-stranded. Assume the identical DNA molecules are single-stranded and 4,000 nucleotides long. The average molecular weight of a nucleotide is 300 grams/mole. Show your work and include all units (include cancellation of units) A sequencing reaction has a ratio of 300 molecules of dATP's to 1 molecule of ddATP. What is the probability that termination will occur for a given position in the growing strand when it is opposite a T in the template strand? In other words, what percentage of the time will a ddATP be incorporated versus a dATP in the growing strand? b. Double stranded DNA will convert to single-stranded DNA if heated high enough; this temperature is known as the "DNA melting point" Which DNA sequence would you expect to have a higher DNA melting point, one with 60% GC or one with 45% GC? Both are the same number of base pairs long. Explain your answer based upon the number of hydrogen bonds a GC base pair can form (versus an AT base pair).

Explanation / Answer

Question 7. a.

given

length of DNA molecule= 4000 nucleotides

sample= 0.5 mg

Average molecular weight of a nucleotide = 300 grams/mole,and

we have to calculate No. of identical molecules of DNA (4000 nucleotide long each) in 0.5 mg of sample

Ans.

300 grams of DNA contains 1 mole of nucleotides (given in question )

which means 300 grams of DNA contains 6022*10^23 nucleotides (1 mole= 6.022*10^23 moleucules)

so, 0.5 mg (or 0.5*10-3 g)of DNA will contain 6.022*1023/300*0.5*10-3 nucleotides, i..e., 1.00366667*1018 nucleotides

as 1 molecule of DNA is having 4000 nucleotides (given in question), No. of molecules of DNA (4000 nucleotide long each) in 0.5 mg of sample will be,

1.00366667*1018 nucleotides/4000 = 2.50916668 * 1014 molecules

so, 2.50916668 * 1014 molecules of DNA (4000 nucleotide long each) are present in 0.5 mg of sample.

Question 7. b.

probabilty of termination (or probabilty of incorportion of ddATP)= no. of ddATPs available/ no. of dATPs + no. of ddATPs

ratio of ddATPs to dATPs = 1/300 (given in question)

probabilty of termination (or incorportion of ddATP)= 1/300+1 =1/301= 0.003322259

or in other words, 0.33 percentage of the time a ddATP will be inserted instead of a dATP (0.003322259*100)

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