PROBLEM SET
Chapter 9 -- Chemistry of the gene

P. 232, #6 Following is a section of a single strand of DNA. Supply a strand, by the rules of complementarity, that would turn this into a double helix. What RNA bases would primase use if this segment initiated an Okazaki fragment? In which direction would replication proceed?
5'-A T T C T T G G C A T T C G C-3'   original DNA

3'-T A A G A A C C G T A A G C G-5'   complementary strand of DNA

3'-U A A G A A C C G U A A G C G-5'   complementary strand of RNA

DNA replication always proceeds 5' ---> 3'
P. 232, #11. What are the differences between continuous and discontinuous DNA replication? Why do both exist?
Continuous DNA replication occurs in the 5' to 3' direction on the leading strand of DNA, which is oriented 3' to 5' with the 5'-end leading towards the Y-junction.
Discontinuous DNA replication occurs in the 5' to 3' direction as well, but on the lagging strand where the 3' to 5' direction is oriented away from the Y-junction. Thus DNA replication starts and then proceeds away from the Y-junction, until it reaches a previous section of newly synthesized DNA. Replication must then restart further up the lagging strand towards the Y-junction. Both exist because DNA synthesis can not proceed in the 3' to 5' direction on the lagging strand.

P.232, #12. Describe the synthesis of an Okazaki fragment.
a) Synthesis of a short section of RNA that is complementary to the DNA strand. This is done by RNA polymerase (primase).
b) Binding of DNA polymerase and synthesis of a DNA strand that is complementary to the lagging strand. DNA synthesis continues until it reaches the previous RNA primer.

P. 232, #16. For each of the following nucleic acid molecules, deduce whether it is DNA or RNA and singe stranded or double stranded.
Molecule     %A        %G        %T        %C       %U 
a.           33        17        33        17        0
b.           33        33        17        17        0
c.           26        24         0        24       26
d.           21        40        21        18        0
e.           15        40         0        30       15
f.           30        20        15        20       15
Strand a is double stranded DNA. We know that it is DNA because it contains thymine and no uracil and we know that it is double stranded because [A] = [T] and [G] = [C].
Strand b is single stranded DNA. We know that it is DNA because it contains thymine and no uracil and we know that it is single stranded because [A] does not equal [T].
Strand c is double stranded RNA. We know that it is RNA because it contains uracil and no thymine and we know that it is double stranded because [A] = [U] and [G] = [C].
Strand d is single stranded DNA. We know that it is DNA because it contains thymine and no uracil and we know that it is single stranded because [G] does not equal [C].
Strand e is single stranded RNA. We know that it is RNA because it contains uracil and no thymine and we know that it is single stranded because [G]does not equal [C].
Strand f is a hybrid strand of RNA and DNA. We know this because it contains both thymine and uracil. The RNA strand and the DNA strand form a duplex strand; we know this because [G] = [C] and [A] = [T + U].

P. 232, #17. A double stranded DNA molecule is 28% guanosine (G).
a. What is the complete base composition of this molecule?

In double stranded DNA the concentration of guanosine equals the concentration of cytosine, thus the molecule is also 28% cytosine. This leaves 44% for adenosine and thymine combined (1-28%-28% = 44%). As the concentration of adenosine equals the concentrations of thymine, the concentration of adenosine must be 22% and the concentration of thymine must also be 22%.
b. Answer the same question, but assume the molecule is double-stranded RNA.
For double stranded RNA the reasoning is the same except that uracil replaces thymine. Thus for double stranded RNA the base composition is 28% guanosine, 28% cytosine, 22% adenosine, and 22% uracil.

P. 232, #18. The following are melting temperatures for five DNA molecules. Arrange these DNAs in increasing amount of G-C pairs: 73 C, 69 C, 84 C, 78 C, 82 C
Because there are 3 hydrogen bonds between G and C but only 2 between A and T, the greater the G-C content of a double stranded DNA causes the DNA to have a higher melting point. Thus to arrange these DNAs in increasing amount of G-C pairs, we simply arrange them in increasing melting temperature.
69
73
78
82
84

P. 232, # 23 In a DNA molecule, if the amount of G is twice the amount of A, the amount of T is three times the amount of C, and the ratio of pyrimidines to purines is 1.5, what is the base composition of the DNA?
The trick is, do not assume that the DNA is double stranded. If you do, the problem can not be worked. The problem reduces to three equations with four unknowns.
(1) G = 2 * A
(2) T = 3 * C
(3) T + C = 1.5 * (G + A)
substitute 3C for T and 2A for G in equation 3
(4) 3C + C = 1.5 *(2A + A)
(5) 4C = 4.5A
(6) C = 1.125A
substitute 1.125A for C in equation 2
(7) T = 3 * 1.125A
(8) T = 3.375A
now state each equation so that each base is a function of the amount of A
(9) G = 2.000A
(10) C = 1.250A
(11) T = 3.375A
(12) A = 1.000A
now add the coefficient of A in the four equations 2.000 + 1.125 + 3.375 + 1.000 = 7.5
the percent of G equals (2.000/7.5)100 = 26.66%
the percent of C equals (1.125/7.5)100 = 15.00%
the percent of T equals (3.375/7.5)100 = 45.00%
the percent of A equals (1.000/7.5)100 = 13.33%

P. 232, #24 A double-stranded DNA measures 6.5 um in length. Approximately how many base pairs does it contain?
One base pair is about 0.34 nm long. To determine the number of base pairs, simply divide the length of the strand by the length of a base pair. The result is the number of base pairs in the piece of DNA.
     6.5 um          6.5 x 10^3 nm
---------------- =  ---------------- = 1.91 x 10^4 base pairs
0.34 nm/base pair    0.34 nm/bp

Last updated on 22 August 1996.
Provide comments to Dwight Moore at mooredwi@esumail.emporia.edu.
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