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BCHM 2700 Recitation 7 | Nucleic Acid Structures & Methods | Questions and Answers Updated 2026.

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BCHM 2700 Recitation 7 | Nucleic Acid Structures & Methods | Questions and Answers | Updated 2026.

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BCHM 2700 Recitation #7: Nucleic Acid Structures and Methods

1. Bacteriophage T2 has a DNA with a molecular weight of 120 x 10 6 Da.
Assuming an average molecular weight for a nucleotide pair of 650 Da, how
long is the DNA of this bacteriophage, and how does this compare to the
length of the phage head (=210 nm)?

120 x 106 divided by 650 = 1.846 x 105 nucleotide pairs. In B-form DNA, the
double helix moves upwards by 3.4 Å/base, implying that the DNA of the phage
is 1.846 x 105 times 3.4 = 6.277 x 105 Å long. That is equivalent to 62,800 nm,
which is 299 times longer than the length of the head. Packaging required!


2. Do Chargaff’s rules hold true for RNA? Explain why or why not.

No, RNA is typically single-stranded so base-pairing rules don’t apply. RNA may fold
over to form duplex regions but won’t be entirely complementary. Also, some RNAs
have non-standard bases that would complicate things as well.


3. One strand of a double-helical DNA has the sequence:
5′-GCGATGCTAAGCTTGTAG-3′.



a. Write out the sequence of the other strand.

Complementary strand: 5′-CTACAAGCTTAGCATCGC-3′ (the convention is
to write it out 5′ to 3′)


b. Is it possible that this sequence contains a restriction site?

Yes, there is a palindromic site: 5′-AAGCTT-3′ (or 5′-AGCT-3′)

, 4. What is the molecular mass and length of a segment of B-DNA that specifies a
40 kDa protein? How many helical turns does this DNA have and what is its
axial ratio (length to width)? Assume an average amino acid has a mass of
110 Da, and an average deoxynucleotide residue has a mass of 325 Da.

A 40-kDa protein consists of 40,000 Da/110 Da = 364 amino acid residues and
hence is encoded by a B-DNA of (3 bp/ a. .a residue) × 364 residues = 1092 bp. Its
molecular mass is (2 × 325) Da/bp ×1092 bp = 709,800 Da ≈ 710 kDa. Since a B-
DNA helix has a rise of 3.4 Å/bp, the length of the DNA encoding the protein is 3.4
Å/bp × 1092 bp = 3713 Å.

B-DNA has 10.4 bp/turn and hence the DNA encoding the protein ideally has 1092
bp/(10.4 bp/turn) = 105 helical turns. Since B-DNA has a diameter of ~20 Å, its
axial ratio is 3713 Å/20 Å = 186.

End of Chapter problems from Berg et al. 9th ed.:

5. Meselson and Stahl used the replication of 15N-labeled parental DNA in 14N-
media, followed by density-gradient centrifugation, to provide evidence for
the semiconservative model of replication. What would their results have
looked like, after 1 or 2 generations, if DNA replicated conservatively?

Original results:




Hypothetical conservative replication results:




After 1 generation, the 15N:14N ratio should be 1:1. After 2 generations, it should be
3:1. There would be no 15N-14N hybrid band with conservative replication.

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