The MOLECULESof LIFEPhysical and Chemical Principles

Selected Solutions for Students

Prepared by James Fraser and Samuel Leachman

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

Nucleic Acid Structure

Chapter 2

Problems and Solutions

True/False and Multiple Choice

2. Which of the following is not a stabilizing force for the structure and stability of double-stranded nucleic acids?

a. base stacking b. hydrogen bonding c. disulfide bondsd. electrostatic forces

4. Genomic DNA can become deformed from its normal B-form by DNA binding proteins, such as the histone proteins and the TATA-box binding protein.

True/False

6. Classify the following RNA structural elements as secondary or tertiary structure:

a. coaxial helices b. pseudoknot c. hairpin d. junction e. adenosine platform f. ribose zipper g. bulge

Answer: Secondary: c, d, g; tertiary a, b, e, f

Fill in the Blank

8. The modified RNA base in which two methyl groups are added to guanine is __________.Answer: N,N-dimethylguanine. The corresponding nucleoside is N,N-dimethylguanosine.

10. Hoogsteen base pairs, where the hydrogen-bonding interactions utilize the Watson-Crick base-pairing edge on one base and the major groove edge in the other base, can be utilized to form an RNA _______ helix.

Answer: triple

Quantitative/Essay

12. What physical factors force RNA to adopt only the C3ʹ endo sugar configuration, but allow DNA to adopt either the C2ʹ endo or the C3ʹ endo sugar configurations?

Answer:The C2ʹ endo configuration results in close contact (1.9 Å) between one of the oxygen atoms of the

3ʹ phosphate group and the hydrogen atom at the 2ʹ position of the ribose ring in DNA. Compared with DNA, RNA has the 2ʹ hydrogen replaced by hydroxyl group (OH) and the two oxygen atoms (2ʹ ribose and 3ʹ phosphate) repel each other strongly. Rather than distorting the structure away from the Watson-Crick model, it is energetically favorable for RNA to switch to the C3ʹ endo conformation.

14. Label the atoms, the bases, the hydrogen bonds across bases, the major and minor grooves, and the interactions along the grooves for the following base pair (oxygen and nitrogen atoms are not identified explicitly):

Is the base pair a standard Watson-Crick base pair?

Answer:

Problem set (quantitative question 2) (na_56_v1)

2Q14

major groove

H

H

N

N

N

N

N

N

N

H

H

H

H H

O

O

OU

Gminor groove

2 ChAPTEr 2: Nucleic Acid Structure

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

Interactions on minor groove: H-bond acceptor, H-bond donor, H-bond acceptor.

Interactions on major groove: hydrogen atom, H-bond acceptor, H-bond acceptor, H-bond acceptor.

The G-U base pair is not a standard Watson-Crick base pair.

16. In water, why is base stacking relatively more important than hydrogen bonding to forming a DNA double helix?

Answer:

Water can make H bonds with the DNA base H-bond donor and acceptors, so the energetic gain in interactions when those H bonds are satisfied by other nucleotides is not great. However, the many small polar interactions along the nucleotide rings that are formed upon base stacking are energetically more favorable than the interactions with water.

18. Why is the R (purine) in GNRA tetraloops required rather than a pyrimidine?

Answer:

The R position, adjacent to the A in sequence and space, is functionally significant because it provides a large surface area (the base has the two rings of a purine, as opposed to one ring for a pyrimidine) against which the A can base stack.

20. Which of the following DNA sequences is most likely to adopt the Z-form? Why?

a. GCGCGCGCATATGCGCGCGCCb. AGAGAGCTCTCTCTCTAAAAT

Answer:

Sequence a is more likely to adopt Z-form because it alternates purine and pyrimidines in a GC-rich sequence. This alternating pattern of 2ʹ endo and 3ʹ endo puckers yields the zig-zag pattern, where the smallest repeating unit is two base pairs, characteristic of Z-form DNA.

22. The structure of the large ribosomal subunit from Haloarcula marismortui (PDB code: 1FFK) has been solved by x-ray crystallography. The 23S RNA contains 2922 nucleotides (758 A, 889 G, 739 C, and 536 U).

a. Assuming a random distribution of nucleotides, how many four-mer sequences with the sequence G-N (any base)-R (purine)-A are possible?

b. There are actually 21 GNRA tetraloops in the structure. What percentage of possible GNRA sequences actually formed tetraloops in the structure?

Answer:

a. Probability of G = total G/total nucleotides = 889/2922

Probability of N = 1.

Probability of R = (total A + total G)/total nucleotides = (758 + 889)/2922.

Probability of A = total A/total nucleotides = 758/2922.

Each four-mer has a GRNA probability = (probability of G) × (probability of N) × (probability of R) × (probability of A) = 0.0445.

Total four-mers = 2922 – 3 = 2919.

Probable GRNA four-mers = probability × total four-mers = 0.0445 × 2919 = 130.

b. If only 21 GNRA tetraloops formed out of a possible 130, then16% of the sequences form a stable tertiary GNRA tetraloop motif.