A3.1
Question 1 Single best answer 1 point
A coordinate file contains the line ATOM 193 CA ASP A1025 12.904 14.115 7.362 1.00 18.52 C. What does this line record?
- The carbonyl carbon atom of the aspartate in chain A
- The alpha-carbon atom of the aspartate in chain A
- A calcium ion on the aspartate in position A
- The alpha-carbon atom of the aspartate in position A
Question 2 Single best answer 1 point
A crystal-structure entry contains SEQRES records. What do they list?
- The residues of the protein chain that the model places
- The residues in the construct that differ from the natural one
- The sequence of the construct that was made and crystallized
- The sequence of the natural protein the entry is labeled with
Question 3 Single best answer 1 point
An enzyme works as a homodimer, but the ATOM lines of its crystal-structure entry list only chain A. Where, if anywhere, does the file record the dimer, and why?
- REMARK 350, because symmetry copies of the listed chain build the dimer
- REMARK 350, because it lists the second chain as residues not modeled
- Nowhere, because the ATOM lines would list both chains of a dimer
- SEQRES, because it lists the sequence again for the second chain
Question 4 Single best answer 1 point
A crystal-structure entry reports a resolution. What does this value measure, and which direction means more detail?
- The number of features resolved per ångström; a larger value means finer detail
- The smallest spacing the data resolve; a larger value means finer detail
- The smallest spacing the data resolve; a smaller value means finer detail
- How closely the model agrees with the data; a smaller value means finer detail
Question 5 Single best answer 1 point
A crystal-structure file lists one position for each atom of a protein. What does one such position represent, and why?
- The position that most molecules share, because the model keeps the majority conformation
- The position of one representative molecule, because the crystal is made of identical copies
- The average over the molecules in the beam during data collection, because each spot sums them
- The average over the molecules in the beam at one instant, because the exposure freezes them
Question 6 Single best answer 1 point
A classmate says the atom coordinates in a deposited crystal structure are read off an image of the protein. What is wrong with this account?
- The coordinates are a model fitted to the reflections, not positions read off an image
- Coarse-resolution coordinates are a model, but fine-resolution ones are an image
- The coordinates are separate measurements of each atom, not positions read off an image
- The coordinates do come from an image, but a blurred one, so each atom is placed roughly
Question 7 Single best answer 1 point
In one crystal-structure model, atom G has a lower B-factor than atom H. Which atom do the data place more precisely, and why?
- Neither, because B tracks the atom’s motion, a property of the molecule, not the data
- Atom G, because a lower B means the atom sits at that spot in more of the copies
- Atom H, because a larger B means stronger support from the data for that atom
- Atom G, because a smaller B means less blur in that atom’s position across molecules
Question 8 Single best answer 1 point
In one model, atom N has a B-factor several times that of atom M. How does the displacement \(u\) of atom N compare with that of atom M, and why?
- Larger by a bigger factor than B, because \(u\) grows as the square of B
- Larger by the same factor as B, because \(u\) measures the same blur as B
- Larger by a smaller factor than B, because \(u\) grows as the square root of B
- Larger by a smaller factor than B, because \(u\) measures the blur along one axis
Question 9 Single best answer 1 point
In one model, atom K has a low B-factor. You measure its distance to atom V, which has a high B-factor, and its distance to atom W, which has a low B-factor. Which distance do the data fix less tightly, and why?
- Neither; the entry’s resolution sets one coordinate error, so both are known equally well
- Neither; atom K anchors both distances, so its blur sets how well each one is known
- The distance to V, because V is the more blurred atom of the pair and limits the distance
- The distance to V, because a high B marks an atom the model has put in the wrong place
Question 11 Single best answer 1 point
Peptides with the sequences of residues 1-4 and 96-99 of HIV-1 protease inhibit the enzyme. How do the peptides inhibit it?
- They bind the dimer away from its active site and distort the site’s shape
- They sit in the active site, so they compete with the substrate for it
- They hold the dimer’s flaps open, so the site cannot grip its substrate
- They stop the two chains from pairing, so the active site is never assembled
Question 12 Single best answer 1 point
A two-chain entry of HIV-1 protease numbers chain A from 1 and chain B from 101. Which residue of chain B is equivalent to Leu 97 of chain A, and why?
- Leu 97, because equivalent residues in the two chains carry the same number
- Leu 197, because it sits at the same sequence position in the partner chain
- None, because chains numbered from different starts do not share one sequence
- Leu 197, because entries number a second chain 100 above the first by convention
Question 13 Single best answer 1 point
An HIV-1 protease entry labeled wild-type carries the substitution Q7K. Why does the crystallized protein carry it, and where does the file flag it?
- It blocks a site where the protease would cut itself; SEQADV lines flag it
- It blocks a site where the protease would cut itself; SEQRES lines flag it
- It was selected by an inhibitor as a resistance change; SEQRES lines flag it
- It was selected by an inhibitor as a resistance change; SEQADV lines flag it
Question 14 Single best answer 1 point
Substrate-bound entries of HIV-1 protease, such as 1F7A, carry the substitution D25N. Why was the enzyme made with this change?
- Asn25 is the residue the virus carries, so the change restores the natural protease
- Asp25 lets a lone chain cut too, so the change keeps lone chains from cleaving the substrate
- Asn25 helps the molecules pack, so the complex forms crystals that diffract better
- Asp25 does the cutting, so an unchanged protease would cleave the substrate it holds
Question 15 Single best answer 1 point
Two unliganded entries of one protein differ widely in flap-tip position, with low tip B-factors in both. Their sequence records match, their symmetry mates sit far from the flaps, and one entry’s R-free sits far above its R-work. Which causes of the disagreement remain open?
- Real state and model error; the contact and sequence checks exclude the other two
- Real state, crystal packing, and model error; the sequence check excludes the construct
- Model error; the large R-free gap shows one model is wrong, excluding the other three
- Real state; a wide shift at low B-factors excludes the other three causes