Solutions to Selected End of Chapter 8 Problems Plus One More
Class 1
1. This is not a trivial question! Check out Fig 8-11 which shows the H-bonding between A-T and
G-C base pairs. Those “Watson-Crick” H-bonds that are part of holding the two DNA strands
together which along with hydrophobic stacking allows DNA to form stable double stranded structure.
But consider the protein that will bind to specific sequences in double stranded DNA…these proteins
do not see the Watson-Crick H-bonds at all…they see the sides of DNA through the major and minor
grooves
From the figure can you see the H-bonding possibilities for the R=groups of the proteins amino acids?
The patterns for each base pair are different….therein lies the specificity of proteins binding to only
certain specific sequences of DNA. . And, think about it, those sides are stacked right next to each
other providing structure and H-bonding (donor and acceptors) for a protein’s amino acid R-groups to
bind to ds-DNA.
The figure on the left is DNA that has been punched or squished down so it falsely looks like you
could see the ring structures (of ribose and the bases) from the side. But this is totally false, see
Figure 8-13: all the yellow balls are the atoms at the sides of the DNA strand (more visible in the
major groove, which is the top side of Fig 8-11 right side)…generally a pattern of 4 atoms.
For the AT pair, the H-bond possibilities are: acceptor-donor acceptor-neutral (methyl).
For the GC pair, the H-bond possibilities are: acceptor-acceptor-donor-neutral (cytosine ring carbon).
2. This is super easy: to get the complementary sequence, just remember to make sure it is
antiparallel, just the way DNA is:
(5’)GCGCAATATTTCTCAAAATATTGCGC(3’)
(3’)CGCGTTATAAAGAGTTTTATAACGCG(5’) the complementary strand.
See how the underlined sequences could form a hairpin structure for each single strand, or a
cruciform structure for the double stranded structure.
3. Lets stretch out some DNA out between the earth and the moon. What fun! A hard thing to do in
biochem lab, but a fun mind-problem. Data: the moon is about 320,000 km away from Earth and
double stranded DNA weighs 1 x10-18 grams per 103 base pairs, and each base pair in DNA is 3.4Å
of ds-DNA’s length.
First convert the distance to the moon into Å, then into base pairs (bp):
(3.2 x 105 km)(1012 nm/km)(10 Å/ nm) =3.2 x 1018 Å….from Earth to the moon.
(3.2 x 1018 Å) / (3.4 Å / bp) = 9.4 x 1017 bp …from Earth to the moon.
Now, the weight of this DNA is:
(9.4 x 1017 bp) ( 1 x 10-18 g/103 bp) = 9.4 x 10-4 g
This is rather tiny considering the total DNA in your body weighs 0.5 g. So your DNA could go
to the moon and back 265 times (why make it just one way trips?).
5. DNA and RNA structure in hairpin turns: RNA forms A-helices whereas DNA forms B-helices. So
in the hairpin helix, RNA will be thicker and shorter, DNA longer and a bit thinner (See Fig 8-17, page
291 in the text).
8. Spontaneous DNA damage: Apurinic, or AP sites where the glycosidic link from the ribose to the
purine is broken releasing the purine. This happens most often with G’s. When it happens it makes
ds-DNA unstable at this point (Fig 8-30b) because ribose carbon 1 is now free to open the furanose
ring making the noncyclic aldol of ribose (remember in the ring form it is a hemiacetal). Fortunately
all living organisms have DNA-repair enzyme systems to correct this and other chemical changes to
our ds-DNA.
10. The hyperchromic shift of ds-DNA when it is heated, forming two ss-DNA strands: the UV
absorbency of the DNA solution INCREASES. This makes measuring the melting of DNA easy to
observe in thermally controlled spectrophotometer cuvettes. So from General Chem, remember the
Beer-Lambert law:
Abs = Ex x concentration x light path length
where Ex is the extinction coefficient (absorbency of a 1M solution). Double stranded DNA has the
bases stacked, ss-DNA they are non stacked, so the Extinction coefficient for each is different and
the concentration of DNA doubles (one DNA strand =>2 single strands of DNA, even though the
amount is the same one mole of ds-DNA => 2 moles of ssDNA, the concentration increases).
Class 2
12. Solubility of DNA parts: the sugar, a base, and phosphate:
Most to least soluble: phosphate > deoxyribose > guanine. Hope you can see that this (and the
other bases are not very soluble compared to sugars and phosphate. This is important having the
more non-polar, aromatic bases inside the helix and the more hydrophilic components on the outside
of the helix.
13. The logic of Sanger sequencing. Sanger’s big contribution was the development of dideoxynucleotides (ddATP, ddGTP, ddCTP, ddTTP) which terminated DNA synthesis after it is incorporated
into the newly synthesized strand in the 3’ position having no 3’ –OH to participate in further strand
elongation (DNA strand synthesis).
14. Sanger Sequencing….where things don’t always go right (only lanes 3 and 4). Lane 1 bands are
given in the text and serve as molecular length markers: the bottom are the smallest with just one
nucleotide added. The actual sequences for
Lane 1 is
5’-primer-TAATGCGTTCCTGTAATCTG
5’-primer-TAATGCGTTCCTGTAATCT
5’-primer-TAATGCGTTCCTGTAAT
5’-primer-TAATGCGTTCCTGT
5’-primer-TAATGCGTTCCT
5’-primer-TAATGCGTT
5’-primer-TAATGCGT
5’-primer-TAAT
5’-primer-T
Lane 2 is:
5-primer-TAATGCGTTCCTGTAATCTG
5-primer-TAATGCGTTCCTG
5-primer-TAATGCG
5-primer-TAATG
Lane 3 is: a mix that left out dTTP, so DNA polymerase could not go beyond added one nucleotide
which could ONLY be ddTTP. A single band at the bottom.
Lane 4 is: a mix the left out dd-Nucleotide, and could only synthesize the full length, largest product.
A single band at the top.
The results should look like this:
Notice that the bands in Lanes 3 and 4 look bigger, they are! Why?
15. Snake venom phopphsdiesterase hydrolyzes off nucleotides from the 3’ end of any nucleic acid.
Partial digestion means all lengths are possible. So partial digestion of (5’)GCGCCAUUGC(3’)–OH
will produce:
(5’)P–GCGCCAUUGC(3’)–OH
(5’)P–GCGCCAUUG(3’)–OH
(5’)P–GCGCCAUU(3’)–OH
(5’)P–GCGCCAU(3’)–OH
(5’)P–GCGCCA(3’)–OH
(5’)P–GCGCC(3’)–OH
(5’)P–GCGC(3’)–OH
(5’)P–GCG(3’)–OH
(5’)P–GC(3’)–OH
And the released nucleotside-5’ phosphates: GMP, UMP, AMP and CMP.
Why did they not get some products such as: AUUGC, or GCAUU…or others all of which are
sequences in the original nucleic acid?
What is the philosophy of a partial digest?
Narrated Power Point for Chapter 8 has a small section about restriction enzymes. Lets do
solve some problems using the classic plasmid, pBR322:
Things to Note: each restriction enzyme is an
endonuclease that ONLY cuts ds-DNA at specific
sequences that are not found anywhere else. Check
out PvuII, it only has one site in pBR322.
Question a. So if pBR322 was cut with PvuII, then
how many pieces would there be? Answers later on
down the pages, but try to do it before going to the
answers.
Question b. If you cut pBR322 with EcoR1 and
PvuII what are the products and if you
electrophoresed the products on an agarose gel,
what would be the banding pattern.
Question c. If you cut pBR322 with PvuII, BamH1
and Pst1, what would the electrophoresis look like?
Answers
a: only one, it would convert the circular ds-DNA into a linear ds-DNA. It would be the same size,
only the geometry changes.
b. two products of about the same size! One large band on electrophoresis that is about half the size
of the original plasmid..and being two different nucleic acid products making the same band.
c. three products: one from PvuII to EcoR1 site, the largest, the second from PvuII to BamH1, the
next smallest, and three the smallest from BamH1 to EcoR1.