Molecule Test 4
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- 1.The isoelectric point of Arg is
- A.
5.40
- B.
7.15
- C.
10.73
- D.
12.48
- E.
1.82
- 2.The isoelectric point of Tyr-Arg is
- A.
9.83
- B.
11.46
- C.
5.51
- D.
10.46
- E.
12.48
- 3.The isoelectric point of Ala-His-Asp-Arg-Val-Gln is
- A.
7.95
- B.
11.17
- C.
4.97
- D.
7.0
- E.
3.03
- 4.Which of the following statements does not apply to collagen?
- A.
It contains hydroxylated amino acids.
- B.
One single polypeptide forms a left-handed helix.
- C.
Cross-links between neighbouring hydroxylated Lys residues stabilize the helix.
- D.
Glycine at every third position is essential for the stability of the triple helical structure.
- E.
The triple helical structure twists in the right-handed direction.
- 5.Why is the decreased affinity of fetal hemoglobin for BPG advantageous to the fetus?
- A.
With fewer BPG molecules bound there are more heme residues available for O2 binding
- B.
Decreased BPG binding biases the fetal hemoglobin toward the T state
- C.
More free BPG is available to bind to adult hemoglobin, resulting in a shift of adult hemoglobin to the R state
- D.
BPG is available to bind to fetal myoglobin, helping to release O2 in fetal muscle tissue
- E.
It reduces fetal hemoglobin's P50
- 6.According to Figure 4, which of the following is not likely to be a serine protease?
- A.
Thrombin
- B.
Factor XIa
- C.
Factor IXa
- D.
Tissue factor
- E.
Factor VIIa
- 7.DIPF is a ________ inhibitor of serene proteases
- A.
Competitive
- B.
Noncompetitive
- C.
Uncompetitive
- D.
Reversible
- E.
Irreversible
- 8.Which of the following best describes the kinetics of DIPF on serine protease catalyzed reaction
- A.
Figure 10 with alpha = alpha'
- B.
Figure 9
- C.
Figure 8 with alpha = 2
- D.
Figure 8
- E.
Figure 9 with alpha' = 2
- 9.Transketolase catalyzed reaction follows a
- A.
Ping-pong mechanism
- B.
Second order reaction mechanism
- C.
Hills reaction mechanism
- D.
Allosteric activation mechanism
- E.
Random mechanism
- 10.Which of the following is the best competitive inhibitor to adenosine deaminase catalyzed reaction
- A.
Adenosine
- B.
Inosine
- C.
1,6 Dihydroinosine
- D.
Malonate
- E.
PCNA
- 11.Which of the following has the smallest KI towards adenosine deaminase
- A.
Adenosine
- B.
Inosine
- C.
1,6 Dihydroinosine
- D.
Malonate
- E.
PCNA
- 12.Which of the following corresponds to an enzyme catalyzed reaction with a competitive inhibitor
- A.
Figure 6
- B.
Figure 7
- C.
None of the rest
- D.
Figure 9
- E.
Figure 10
- 13.Which of the following corresponds to an enzyme catalyzed reaction with a uncompetitive inhibitor
- A.
Figure 5
- B.
Figure 7
- C.
Figure 9
- D.
Figure 10
- E.
None of the rest
- 14.Which of the following corresponds to an enzyme catalyzed reaction with a noncompetitive inhibitor
- A.
Figure 10
- B.
Figure 5
- C.
Figure 6
- D.
Figure 8
- E.
None of the rest
- 15.Which of the following accurately describes the effects of a competitive inhibitor on km
- A.
Km decreases because the inhibitor binds to ES to form ESI which decreases ES
- B.
Km unchanged; inhibitor binding does not affect S binding
- C.
Km increases because the inhibitor is decreasing the probability of the substrate's binding to the active site
- D.
Km increases because the inhibitor binds to ES to form ESI which decreases ES
- E.
None of the rest
- 16.Which of the following statements about a plot of V0 vs [S] for and enzyme that follows Michaelis-Menten kinetics is FALSE
- A.
Km is the [S] at which V0 = 0.5 Vmax
- B.
The shape of the curve is a hyperbola.
- C.
The y-axis is initial reaction velocity
- D.
As [S] increases, the initial velocity of reaction, V0, also increases.
- E.
At very high [S], the velocity curve becomes a horizontal line that intersects the y-axis at Km.
- 17.The active compound in Aspirin is acetylsalicylic acid (ASA), which inhibits the cyclooxygenase (COX) enzyme involved in synthesis of prostaglandins. The structure of ASA and COX's natural substrate, arachidonic acid, are given next to Figure 4. Based on this information, you would be right to say ASA is a ___ inhibitor of COX.
- A.
Competitive
- B.
Allosteric
- C.
Dephosphorylating
- D.
Co-structural
- E.
Dissociative
- 18.
- A.
Its linking number is not changed
- B.
The change in its writhing number is 55
- C.
Its original linking number is 400
- D.
The change in its twist is -55
- E.
All answers are correct
- 19.If 20% of the DNA in a relaxed circular-covalently-closed plasmid is induced to change from B-DNA to Z-DNA and then treated by an enzyme that has type IA topoisomerase activity, what happens to the value of the linking number?
- A.
Decreases
- B.
Increases
- C.
Remains the same
- D.
Cannot be calculated without knowing the size of the plasmid
- E.
None of the answers
- 20.Which of the following statements is correct?
- A.
The replication machinery is stationary and the template DNA is reeled through it.
- B.
The replication machinery move along the template DNA like a train on tracks.
- C.
Positive supercoiling facilitates initiation of DNA replication.
- D.
Introduction of negative supercoiling to DNA by helicase is dependent on ATP hydrolysis.
- E.
Introduction of negative supercoiling to DNA by helicase is independent on ATP hydrolysis.
- 21.Which of the following increases the processivity of a DNA polymerase
- A.
Adenosine
- B.
Inosine
- C.
1,6 Dihydroinosine
- D.
Malonate
- E.
PCNA
- 22.According to Figure 12, one of the Mg2+ ions interacts with ____ to _____ its nucleophilicity as it attack(s) the incoming nucleotide
- A.
The 3'O atoms, enhance
- B.
The 2'O atoms, enhance
- C.
The 3'O atoms, reduce
- D.
The 2'O atoms, reduce
- E.
The phosphates, enhance
- 23.Which of the following is true
- A.
Telomerase adds DNA sequence repeats to the 5’ end of the DNA strands in the telomere regions
- B.
Telomerase joins Okazaki fragments on the lagging strand
- C.
In the circular chromosomes of bacteria, DNA replication begins at multiple origins
- D.
In prokaryotes DNA ligase is activated by ATP
- E.
In eukaryotes DNA ligase is activated by ATP
- 24.An individual is diagnosed with a deficiency of DNA repair enzymes that resolve deamination to hypoxanthine. What changes will be seen in this individual's DNA if these deaminations are left unrepaired?
- A.
A G:C base pair will become an A:T base pair
- B.
An A:T base pair will become a G:C base pair
- C.
A G:C base pair will become a U:T base pair
- D.
An A:T base pair will become a U:T base pair
- E.
There will be no change seen in the individual's DNA
- 25.Histones are usually rich in _______ and ________ residues as they are positively charged at physiological pH and can form ion pairs with the negatively charged phosphate groups on the backbone of the DNA molecule.
- A.
Lysine, Arginine
- B.
Arginine, Leucine
- C.
Uracil, Thymine
- D.
Glycine, Proline
- E.
Glutamate, Aspartate
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