250+ TOP MCQs on Structural Study – X-ray Crystallography – 2 and Answers

Protein Engineering MCQs on “Structural Study – X-ray Crystallography – 2”.

1. Which of the following is not a component needed for the X – ray crystallography of proteins?
A. Protein crystal
B. Source of X – rays
C. Detector
D. Monochromator
Answer: D
Clarification: Monochromator is not a component needed for the X – ray crystallography of proteins. Protein crystal, source of X – rays, and detector are the components needed for the X – ray crystallography of proteins.

2. What is the first step in the X – ray crystallography of proteins?
A. Creating an electron density map
B. X – ray diffraction
C. Precipitation of protein
D. Purification of protein
Answer: D
Clarification: Purification of protein is the first step in the X – ray crystallography of proteins. Then the pure protein solution is obtained, which is then used for the precipitation of proteins. Crystals obtained in this step are then irradiated with X – rays and detected by the detector. The last step is to create an electron density map.

3. X – ray crystallography requires the formation of pure crystals to acquire accurate results.
A. False
B. True
Answer: B
Clarification: The above statement is true. X – ray diffraction requires the formation of pure crystals to acquire accurate results. Recrystallization is an incredibly important technique that is used for the purification of a substance.

4. Which among the following is not a method to generate X -rays?
A. Synchrotron radiation source
B. Radioactive decay process
C. Bombarding a metal source with a beam of high energy electrons
D. Bombarding a metal source with a beam of high energy electromagnetic radiation
Answer: D
Clarification: Bombarding a metal source with a beam of high energy electromagnetic radiation does not generate X – rays. Hence, it is not a method to generate X – rays. Synchrotron radiation source, radioactive decay process, and bombarding a metal source with a beam of high energy electrons generate X – ray.

5. For which of the following path difference, the superimposing waves will not interfere constructively?
A. 3λ
B. 4λ
C. 11λ
D. 11.5λ
Answer: D
Clarification: A path difference of 11.5λ between the two waves will not result in constructive interference. For the waves to interfere constructively, the differences in the travel path must be equal to integer multiples of the wavelength.

6. What is produced by the analysis of X – ray diffraction data sets?
A. Molecular density maps
B. Nuclei density maps
C. Atom density maps
D. Electron density maps
Answer: D
Clarification: Electron density maps are produced by the analysis of X – ray diffraction data sets. This is because the X – rays are diffracted by electrons. These electron density maps are then used to construct the structure of the sample molecule.

7. What is the device on which the crystal (grown in a laB. is mounted known as?
A. X – ray source
B. Collimator
C. Refractometer
D. Goniometer
Answer: D
Clarification: A crystal grown in the lab, is mounted on a goniometer and exposed to X – rays produced by rotating anode generators or a synchrotron facility.

8. Which of the following detector is used for proteins in the X – ray crystallography?
A. Scintillation counters
B. Proportional counters
C. Optical cameras
D. CCD cameras
Answer: D
Clarification: CCD cameras are used for proteins in the X – ray crystallography. A diffraction pattern of regularly spaced dots known as reflections is recorded on a detector. CCD cameras/image plates are most frequently used as detectors for protein X – ray crystallography.

9. The wavelength of X – rays used in X – ray crystallography is 8 x 10^-10 m. It undergoes the first-order reflection at a glancing angle of 4.59°. What is the spacing between the atomic planes?
A. 4 x 10^-9 m
B. 15 x 10^-9 m
C. 10 x 10^-9 m
D. 5 x 10^-9 m
Answer: D
Clarification: The spacing between the atomic planes is 5 x 10^-9m.
Formula: 2dsinθ=nλ (Bragg’s law)
Where, d=spacing of the atomic planes in the crystal,
θ=glancing angle,
n=order of reflection,
λ=wavelength.
Therefore, d=(frac{nlambda}{2sintheta})
d=(frac{1 times 8 times 10^{-10}}{2 times sin (4.59)})
d≈5 x 10^-9 m.

10. An X – ray having a wavelength of 1.67 x 10^-10 undergoes the first-order reflection. What is the glancing angle (θ) if the spacing between the two planes is 6 x 10^-10?
A. 6.25°
B. 8.99°
C. 7°
D. 8°
Answer: D
Clarification: The glancing angle (θ) is 8°.
Formula: 2dsinθ=nλ (Bragg’s law)
Where, d=spacing of the atomic planes in the crystal,
θ=glancing angle,
n=order of reflection,
λ=wavelength.
Therefore, sinθ=(frac{nlambda}{2d})
sinθ=(frac{1.67 times 10^{-10}}{2 times 6 times 10^{-10}})
sinθ=0.139167
θ≈8°.

11. If a beam of X – ray has a wavelength of 1 A°, then what is the energy of this wave?
A. 39.78 x 10^-16erg
B. 19.89 x 10^-16erg
C. 39.78 x 10^-16J
D. 19.89 x 10^-16J
Answer: D
Clarification: The energy of the wave is 19.86 x 10^-16J.
Formula: E=hf=(frac{hc}{lambda})
Where, E=Energy of the wave,
h=plank’s constant=6.63 x 10^-34m²/kg.s,
c=speed of light=3 x 10⁸m/s,
λ=wavelength.
Therefore, E=(frac{hc}{lambda})=(frac{6.63 times 10^{-34} times 3 times 10^8}{10^{-10}})(m²/kg.s)x(m/s)÷m
E=19.89 x 10^-16 kg.m²/s².
E=19.89 x 10^-16J.

12. Which of the following phenomenon is depicted in the figure shown below?

A. Diffraction
B. Non-interference
C. Constructive interference
D. Destructive interference