Exam 1 Practice Test Questions

Question 1

What are the differences of the 3 domains? (Bacteria,Eukarya, and Archaea)

Answer 1

• Eukarya include all eukaryotes – organisms with nuclei. All animals and plants big enough to see are eukaryotes, but not all eukaryotes are multicellular. Amoebas, yeast, and parameciums are examples of single celled eukaryotes.
• Bacteria and Archaea have no nuclei and used to be lumped together as prokaryotes.

B. Archaean

Question 2

Answer 2

C. T(Change)S

Question 3

Answer 3

B. Nitrogen is Blue

Other Colors

Carbon is colored black

Oxygen is red

Nitrogen is blue

Hydrogen is white

Chlorine is green

Sulfur is yellow

Question 4

What are the primary, secondary, tertiary, and quanternary structures held together by?

Answer 4

• Primary structure = covalent bonds. Amino acids are held together by peptide bonds.
• Secondary structure = repetitive 3D structure, held together by hydrogen bonds between peptide bonds.
• Tertiary structure is non-repetitive larger scale 3D structure. Contributing factors, van der Waals, H bond, ionic forces – but mostly the hydrophobic effect.
• Quaternary structure = more than one chain. All weak forces contribute, but salt bridges are often important.
• E. Covalent Peptide bonds

Question 5

Answer 5

C. Six

Question 6

Answer 6

D. Dialysis

Used dialysis treatment to refold the protein – back to its biologically active native conformation, by removing urea and beta-mercaptoethanol. This process involved placing the unfolded protein with the small molecules, urea and beta-mercaptoethanol into a porous dialysis tubing which allowed the movement of these small molecules out via diffusion.

Question 7

Answer 7

B. #1 is L and #2 is D

Question 8

Answer 8

C. The beta pleated sheet

The right handed alpha helix is in the lower left quadrant, and the beta sheet (and collagen helix) are in the upper left quadrant.

Dark green is where there are likely to be angles plotted from real life proteins. White is “forbidden” but you do see Glycine residues there

Answer 9

E. K and R

Question 10

Answer 10

A. Mass Spectrometry

Matrix Assisted Laser Deorbtion/Ionization Time of Flight

• How long it takes the protein to fly through the field

Question 11

Answer 11

E. Isoelectric focusing

The advantage is that proteins are run in their native state and they are separated by their native charges.

The gel is treated with “amphoteres” so that it forms a pH gradient when the current is turned on.

Proteins migrate toward their isoelectric points (zero charge) and then “focus” at that spot.

Question 12

Answer 12

D. Ph(NCS)

This is Edman Degradation. Formerly, entire proteins were sequenced with this method. Now it is mainly a first step toward sequencing by DNA.

Know the structure of Edman Reagent (Phenylisothiocyanate) F-NCS. Know that the structure is a PTH = phenylthiohydantoin.

An important drug, Dilantin, is also a hydantoin.

Question 13

Answer 13

F. Promoter region

Question 14

Answer 14

A. AGGTTC

Question 15

Answer 15

B. Introns

Question 16

Answer 16

F. Meselson and Stahl

Question 17

Answer 17

This is extremely important, a BLOSUM62 matrix. Relying on IDENTICAL amino acids is kind of silly when we understand which amino acids are similar to which other amino acids. I am not asking you to memorize this whole table but I am asking you to learn certain aspects. Which amino acids are BLUE? (FILMYVW). Which amino acids are RED? (HKRED).

F. F

Question 18

Answer 18

B. BLASTP

Question 19

Answer 19

A. BY simple logic If A~B and B~C then A~C

Question 20

Answer 20

The oxygen is at half of its capacity and it would have 2 out of the 4 oxygens present

D. The solution would be 1/2 Hb(O2)4 and 1/2Hb

Question 21

Answer 21

1. CO2 +, H+ , 2,3 BPG (Negative allosteric Modulator)
2. Glucose (not negative)

B. Glucose

Question 22

Answer 22

The loading curve of MYO-globin is HYPERBOLIC. As soon as there is a tiny bit of oxygen, the protein gets loaded up, so the half-occupied state occurs at a P50 of only 2 torr. A “torr” is a millimeter of mercury, as in atmospheric pressure.

The hyperbolic curve is typical of binary binding states – oxygen is either OFF or ON, there are no other possibilities, and so we expect to see a hyperbolic graph.

D. Hyperbolic

Question 23

Answer 23

Don’t need to know where methyls are. A heme group is a tetrapyrrole – see the pyrrole ring in red. Without the iron it is called Protoporphyrin IX. When the iron is inserted it becomes a heme group and it is capable of binding oxygen molecules.

D. Tetrapyrrole

Question 24

Answer 24

Question 25

Answer 25

Question 26

Answer 26

1. Vanderwalls
2. H-band
3. Hydrophobic

Question 27

(Need to know)

Answer 27

CHARLES

Question 28

Answer 28

Question 29

Answer 29

Question 30

Answer 30

Question 31

Answer 31

Question 32

Answer 32

Question 33

Answer 33

Some viruses have RNA genomes. (This includes HIV AIDS, a retrovirus). Reverse Transcriptase copies the structure of an RNA strand making a new DNA strand. Then Reverse Transcriptase hydrolyzes the RNA and the DNA strand is copied into a second DNA strand.

Question 34

Answer 34

Same job in different species

Question 35

Answer 35

Question 36

Answer 36

Question 37

Answer 37

Question 38

Answer 38

Histidine (HIS)

Question 39

Answer 39

Question 40

Answer 40

D. An Apple

Question 41

Answer 41

2. B Vand der waals

Question 42

Answer 42

B. L

Question 43

Answer 43

C. 6

A peptide bond is planar because the C and the N are sp2. So a plane of six atoms is defined, C C N C plus O and H. Thus a peptide chain could be thought of as a series of playing cards, a series of planar rectangles connecting only at the corners. The corners that connect are the alpha carbons, and that is also where the R groups are attached – perpendicular to the planar portions.

Question 44

Answer 44

E. Peptide H bonds

Question 45

Answer 45

B. Large Hydrophobic