Download Biochemistry I (2010) MOED A

BIOCHEMISTRY I (2009) MOED A
1. A blood test of a patient suspect to have a myocardial infarct indicated high levels of Troponin T however no LDH-2
increase was detected. The reason for this could be one of the following.
a. The changes indicate a Myocardial infarct incidence occurring only few hours prior to test.
b. The results do not indicate a Myocardial infarct.
c. LDH-2 does not increase in a Myocardial infarct, only Troponin-T.
d. Troponin C is the Myocardial infarct marker.
2. Which of the ollowing statements regarding Transthyretin (TTR) is not correct.
a. The long half-life of TTR makes it unsuitable for a malnutrition indicator.
b. Retinol-binding protein half life is increased upon binding to TTR.
c. Mutations in TTR cause amyloid fibril deposits in nerve fibers.
d. TTR inhibits initial plack formation in early stages of Alzheimer's disease.
e. None of the above.
3. Which of the following statements regarding Alpha1 anti-trypsin (A1AT) mutations is not correct
a. Pittsburgh (Met358Arg) mutation reduces Elastase inhibition while increasing thrumbin inhibition.
b. Z (Glu342Lys) mutation increases Elastase inhibition thereby causes emphysema.
c. Z (Glu342Lys) mutation causes polymerization in the ER which can lead to Cirrhosis.
d. Smokers bearing a ZZ mutation in A1AT are not hyper susceptible to A1AT inactivation by Met358 oxidation due to
the initial low enzyme activity.
e. B+D
4. A 40 year old woman arrived at the emergency room (ER) complaining of upper abdominal pain and nausea. The
doctor noticed yellow sclera. Laboratory tests results: Urobilirubineria (Urinary bilirubin); Alkaline phosphatase 12 mg/dL;
Direct bilirubin 6 mg/dL; Indirect bilirubin 1 mg/dL. What could be the diagnosis?
a. Physiological jaundice
b. Haemolysis
c. Viral hepatitis
d. Post-hepatic jaundice due to gallstones
e. Alcoholic Cirrhosis
5. Presence of increased levels of conjugated bilirubin in plasma may result from all EXCEPT.
a. chirrosis
b. hepatitis C
c. biliary obstruction
d. haemolysis
e. Genetic mutation in MRP2
6. Conjugated Bilirubin is secreted across the canalicular membrane by:
a. ATP-dependent transporter multidrug resistance protein 2 (MRP2)
b. ATP-independent organic anionic transporter (OAT)
c. Ligandins
d. UGT1 (UDP-glucuroncsyl transferase 1)
e. Albumin
7. The metabolism of heme involves all the following reactions EXCEPT.
a. Rupture of the a-methylene bridge of the heme molecule by the heme oxygenase enzyme.
b. Release of Fe3+ from heme.
c. Production of CO2.
d. Formation of bilirubin that can be converted into lumirubin upon phototherapy.
e. Formation of the hydrophilic pigment biliverdin IX-31(?).
8. Which of the following conditions will cause the largest increase in the blood citruline concentration?
a. A defect in argininosuccinase enzyme.
b. A defect in the argininosuccinic acid synthetase enzyme.
c. A defect in carbamoyl phosphate synthetase enzyme.
d. A defect in argininosuccinase enzyme in patients treated with arginine.
e. A defect in argininosuccinic acid synthetase enzyme in patients treated with arginine.
9. An experimental drug designed to target the heme metabolism was given to a highly jaundiced baby. After the drug
treatment the baby developed a green discoloration of the skin and the urine appeared to be greenish as well. According
to your knowledge of heme metabolism, which of the following statements is most correct regarding the possible targets
and effects of the drug?
a. The drug inhibits heme oxygenase I enzyme, which will be characterized by decreased elimination of CO by the lungs.
b. The drug induces the expression of heme oxygenase I enzyme, causing hyperbiliverdinemia (increased biliverdin
levels) and decreasing bilirubin levels.
c. The green discoloration of the skin indicates that the direct bilirubin levels in the blood of the baby should be higher
than normal.
d. The drug inhibits biliverdin reductase enzyme causing hyperbiliverdinemia (increased biliverdin levels) and decreasing
bilirubin levels.
e. The drug induces biliverdin reductase enzyme causing hyperbiliverdinemia (increased biliverdin levels) and
decreasing bilirubin levels.
10. The following best represents the backbone arrangement of two peptide bonds?
a. C-N-C-C-C-N-C-C
b. C-N-C-C-N-N
c. C-N-C-C-C-N
d. C-C-N-C-C-N
e. C-C-C-N-C-C-C
11. The following peptide was isolated.
Gly-Ser-Cys-Glu-Lys-Arg-Cys-Arg
|
|
S---------------S
The charge of the peptide is about.
a. -2 at pH > 13.5
b. -1.5 at pH 12.52
c. +1 at pH 7.1
d. +2 at pH 3.1
12. Which of the following is the characteristic of globular proteins?
a. Hydrophilic amino acids tend to be on the inside.
b. Hydrophobic amino acids tend to be on the outside.
c. Tertiary structure is formed by hydrophobic and electrostatic interactions between amino acids, and hydrogen bonds
between amino acids and between amino acids and water.
d. Secondary structure is formed principally by hydrophobic interactions between amino acids.
e. Covalent disulfide bonds are necessary to hold a protein in a rigid conformation.
13. Methanol (CH3OH) is converted by alcohol dehydrogenase to formaldehyde (CHO), a compound that is highly toxic in
humans. Patients who have ingested toxic levels of methanol are sometimes treated with ethanol (CH3CH2OH) to inhibit
methanol oxidation by alcohol dehydrogenase. Which of the following statements provides the best rational for this
treatment?
a. Ethanol is a structural analog of methanol and might therefore be an effective noncompetitive inhibitor.
b. Ethanol would be expected to alter the Vmax of alochol dehydrogenase for oxidation of methanol to formaldehyde.
c. Ethanol is a structural analog of methanol that would be expected to compete with methanol for its binding site on
the enzyme.
d. Ethanol would be expected to inhibit the enzyme by binding to the formaldehyde and preventing its binding to the
enzyme.
14. Which of the following statements about allosteric control o enzymatic activity is false?
a. Allosteric effectors give rise to segmoidal V0 vs [S] kinetic plot.
b. Allosteric proteins are generally composed of several subunits.
c. An effector may either inhibit or activate the enzyme.
d. Binding of the effector changes the conformation of the enzyme.
e. Heteroallosteric effectors compete with the substrate for binding sites.
15. 5x10-4 umoles of an enzyme working at Vmax catalyze the conversion of 60umol of substrate to product in 3 min.
What is the enzyme turnover number (Kcat)?
a. Cannot be calculated with the presented data.
b. 656 S-1
c. 5x10-4 umol
d. 4x10-4 S-1
e. 0.003 min-1
16. A 7 year old African American male is admitted to hospital with indicated anemia that is suspected as sickle cell
anemia. The molecular event triggering this disease is which of the following?
a. A loss of quaternary structure of hemoglobin molecule
b. An increase in oxygen binding to hemoglobin.
c. A gain of ionic interactions, stabilizing the "T" form of hemoglobin.
d. An increase in hydrophobic interactions between deoxyhemoglobin molecules.
e. An alteration in hemoglobin secondary structure leading to loss of a-helix.
17. A biochemist purifies the enzyme aspartate transcabamoilase over affinity chromatography column. He/she tries two
approaches, in the first PALA is covalently linked to the column and aspartate is used in solution to elute the enzyme from
the beads. In the second Aspartate is covalently linked to the column and PALA is used in solution to elute the enzyme.
Which of the above approaches is likely better?
a. The First since the affinity of the enzyme to PALA is higher and will allow better purification.
b. The second since the affinity of the enzyme to aspartate is higher and will allow better purification.
c. There is no difference between the approaches in both cases purification will be significant.
d. The first, since aspartate can efficiently compete out PALA and elute the enzyme.
e. The second, since PALA can efficiently compete out aspartate and elute the enzyme.
18. The first stage of the catalytic reaction of chymotrypsin, the enzyme serine residue (195) forms a covalent bond with a
carbon of the polypeptide substrate. Which of the following statements is true?
a. The serine becomes electrophilic upon binding of the substrate.
b. This stage is known as a tetrahedral transition state which is very stable.
c. This stage is possible due to the basic neighboring histidine residue.
d. This stage occurs since histidine 57 transfers its hydrogen to the amino group of the substrate.
e. This stage is stabilized by aspartate 102 that accept the hydrogren from the active serine residue.
19. The affinity of an enzyme to its substrate is 20nM and the turnover number of the enzyme is 40,000 S-1. 10nmoles of
competitive inhibitor that its affinity to the enzyme is 5nM is added. The addition of the inhibitor affects.
a. Km by decreasing its value to 10nM
b. Vmax by decreasing its value to 20.000 nM/sec
c. Km by increasing its value to 40nM
d. Vmax by increasing its value to 30,000 nM/sec
e. Km by increasing its value to 60nm
f. Kcat by increasing its value to 40nM and decreasing Vmax to an unknown value.
20. Reseachers investigated the genes controlling size difference in dogs, comparing Huskies and Chihuahuas. Gene A
encodes a transcription factor, protein A that regulates growth in dogs. Protein A regulates expression of the gene
encoding the Growth Hormone protein. The DNA sequence of Gene is identical in Huskies and Chihuahuas. However,
Gene A is highly methylated in all Huskie tissue, but completely unmethylated in Chihuahua tissue. Pregnant Huskie
mothers were fed a diet including the DNA methylation inhinitor 5-AZA-C. Huskie pups weighed 3X less at birth. In
comparison, researches made transgenic strain of Chihuahuas knocked out for Gene A. The Chihuahua pups weighed 4X
more at birth.
a. DNA methylation inhibits gene expression.
b. DNA methylation activates gene expression.
c. Gene A encodes a protein that is a transcriptional repressor.
d. Gene A encodes a protein that is a transcriptional activator.
e. Answers A and C are possible.
f. Answers A and D are possible.
21. A new anitbiotic was developed that inhibits DNA replication in bacteria. DNA was isolated from bacteria after antibiotic
treatment and compared to untreated control bacteria. The treated strain had two striking anomalies.
1. Only one strand of DNA was replicated efficiently, always the same strand.
2. The remnant of the second strand synthesis contains many short molecules (of differing sequence) that were RNA, not
containing DNA.
a. The antibiotic prevents DNA priming by inhibiting the enzyme DNA primase.
b. The antibiotic prevents DNA polymerase recognition of RNA on the leading strand.
c. The antibiotic prevents DNA polymerase recognition of DNA on the lagging strand.
d. Leading strand DNA polymerase synthesis activity is inhibited by the antibiotic.
e. DNA ligase activity is inhibited by the enzyme.
f. Answers B and D are possible.
22. A strain of archebacteria was recently found in soil that highly resembled fossils of bacteria dating back 100 million
years ago. Remarkably, scientists succeeded to sequence and compare the genome of the "new" soil bacteria to a DNA
fossil sample that was highly preserved. To the scientists' surprise, after sequencing the genome, they found absolutely
no sequence differences between the "old" and "newly" isolated bacteria.
a. This strain of bacteria is highly evolved, because it does not need mutations.
b. This strain of bacteria has lost proofreading activity in its DNA polymerase enzyme.
c. This strain of bacteria has very strong proofreading activity in its DNA polymerase enzyme.
d. This strain has strong proofreading activities in its amino acyl tRNA synthethase enzymes.
e. This strain of bacteria has strong DNA polymerase 5' to 3' exonuclease enzymatic activity.
f. This strain is very sensitive to DNA damage agents in the soil.
23. A new brain diesease was found by the TEAMs Technion medical students. There was severe disruption in brain
activity and a few lucky students donated a small portion of their brain for biopsies. Investigators examined the total
amino acid content of proteins in the diseased versus normal brains. Suprisingly, they foudn that while brain proteins of
helthy people contain 5% alanine and 8% serine the Technion medical students brain proteins contained 7% alanine and
5% serine.
a. A mutation occurred in a gene transcribing serine tRNA, the mutated tRNA is now also recognized by the alanine
amino acyl tRNA synthethase enzyme.
b. A mutation occurred in a gene transcribing alanine tRNA, the mutated tRNA is now also now recognized recognized by
the serine amino acy tRNA synthethase enzyme.
c. A mutation occurred in the gene encoding serine amino acyl tRNA synthethase enzyme it now recognizes alanine
tRNA.
d. A mutation occurred in the gene encoding alanine amino acyl tRNA synthethase enzyme it now recognizes serine.
e. Serine tRNA has undergone a deletion mutation.
f. Answers B, C or D are possible.
24. We have designed the following antisense RNA molecule to block translation of mRNA encoding the Gene X protein.
Antisense RNA forms a double stranded RNA complex with the mRNA to clock its translation in the ribosome. Gene X is a
gene causing leukemia and we have delivered this antisense RNA to the lymphocytes of cancer patients. The target
sequence is gene X that will be black is 5' CATGCCGGATC 3'. The correct antisense sequence is.
a. '5 GTACGGCCTAG 3'
b. 5' CUAGGCCGUAC 3'
c. 5' GATCCGGCAG 3'
d. '5 GUACGGCCUAG 3'
e. 5' GAUCCGGCAUG 3'
f. 5' CTAGGCCGTAC 3'
25. A mutation has occurred in a strain of mice. The mutation in Gene T causes the mutant mice to have a long tail which
is over 35 cm long. Gene T encodes the protein that is expressed in the tail. The normal T protein is 60,000 kd in size but
in mutant mice it is 75,000 kd. The length of the primary transcript of T mRNA in the nucleus is identical in both normal
and mutant mice. In both normal and mutant mice, researchers sequenced the first 300 nucleotides of the 5' end and the
last 300 nucleotides of the 3' end of the cytoplasmic mRNA and found that the sequences were identical between the two
strains.
a. The mutant T mRNA has undergone alternative polyadenylation thus the transcript and protein in the cytoplasm are
longer/larger.
b. The mutant T mRNA is an alternatively spliced version of the normal gene.
c. The mutant primary gene T mRNA has failed to remove one of its introns thus the transcript and protein in the
cytoplasm are longer/larger.
d. The mutant T mRNA contains extra exon, thus the transcript and protein in the cytoplasm are longer/larger.
e. The mutant and normal gene T mRNAs are the same length.
f. Answers B and/or D are possible.
26. In bacteria a mutation inhibiting RNA polymerase activity could inhibit its...
a. Recognition of the origin of replication.
b. Recognition of TTP.
c. DNA Helicase activity.
d. Proofreading activity.
e. 5' to 3' DNA nuclease activity.
f. None of the answers are correct.
27. Researchers have discovered a new antibiotic that inhibits protein synthesis in bacteria. When examining mRNA
isolated from strains of antibiotic treated bacteria they found that the mRNA was coated with many ribosomes covering the
complete 5' to 3' end of the transcript. Peptides of varying sizes were bound to the ribosomes that were stalled along the
entire length of the mRNA.
a. This antibiotic inhibits EF-G mediated translocation of the ribosome along the mRNA.
b. This antibiotic inhibits the peptide bond formation activity of the peptidyl transferase enzyme of the ribosome.
c. This antibiotic inhibits activity of RF (release factor) induced hydrolystic activity of the peptidyl transferase enzyme of
ribosome.
d. This enzyme inhibits EF-TU's ability to recognize tRNA
e. Answer A or D is possible.
f. Answer B or C is possible.
28. A single point mutation has arisen in gene Y which causes a shorter protein Y to form in mice. This protein is 25,000
kd in size in mutant mice versus to the normal protein of 35,000 kd in both mutant and normal cells. (1) the cytoplasmic
and nuclear size of the Y mRNA transcripts is identical, (2) the Y mRNA amounts are identical and (3) besides the one
nucleotide change, the sequence of the Y mRNAs are identical. What is the nature of the mutation?
a. It is in the promoter and enhancer region of Gene Y.
b. It is a splicing site mutation.
c. It is a polyadenylation site mutation.
d. It is a mutation turning a normal codon into a stop codon.
e. It is a mutation in the 5' UTR Shine-Dalgarno region of gene Y.
f. Answer B or C is correct.
29. The PEPCK gene is expressed in both the liver and kidney of mice after birth. In kidney cells, there are 3,000
mRNAs/cell of PEPCK, but in liver cells there are 30,000 mRNAs/cell of PEPCK. A mutation has occurred in mice that causes
PEPCK to be equally expressed in both cells, at a level of 30,000 mRNAs/cell.
a. A mutation has occurred in the enhancer of the PEPCK gene, which increases binding of transcriptional activators in
kidney, but not liver cells.
b. A mutation has occurred in the promoter of a gene encoding a liver specific transcription factor, this mutant gene is
now expressed in both liver and kidney cells.
c. A mutation has occurred in a kidney specific transcriptional repressor protein, which is not expressed in the mutant
kidney cells.
d. The mutation has caused a kidney specific histone acetylase enzyme to be activated that recognizes the PEPCK gene
regulatory region.
e. All the answers are possible.
f. None of the answers are possible.
30. As a side effect, a (hypothetical) drug inhibited the formation of carbamate group at the N termini of the B globin
chains in HbA. Administration of this drug will have the following effects EXCEPT:
a. Carbonic acid will become the exclusive origin of every exhaled CO2 molecule.
b. Whereas The N-termini of Beta globin chains in deoxyhemoglobin A are negatively charged in an untreated individual,
they will be positively charged in individuals that are exposed to the drug.
c. The equilibrium between the R and T conformations of hemoglobin will be shifted toward the R conformation.
d. This is a misleading question. Carbamylation takes place at the C-termini of Beta globin chains in deoxyhemoglobin A
whereas the N-termini of these chains are never carbamylated.
31. As a result of a (hypothetical) mutation in a Beta globin gene the distal histidine residue (His E7) was substituted by a
glycine residue which cannot form a bond with oxygen O2. The following alterations are expected in the behavior of HbA
that contains such mutated chain(s) EXCEPT:
a. The proportion of methemoglobin in the erythrocytes will be elevated.
b. Binding of oxygen to the iron ion in the heme group of the mutated Beta globin chain will be tighter than normal.
c. Rather than being positioned at an angle of 8० off plain perpendicular to the level of the heme group, the proximal
histidine residue will be fully perpendicular to it (90० angle) in both oxy- and deoxy- forms of the mutated hemoglobin.
d. The oxygen binding curve of the mutated hemoglobin will be shifted left-wise relative to normal hemoglobin.
32. An alpha-class globin chain that is absent (undetectable at both the mRNA and protein levels) during both the
embryonic fetal stages and adult life despite its intact encoding, gene is:
a. theta globin
b. epsilon globin
c. zeta globin
d. delta globin
e. Every known alpha-class globin gene is expressed during some point in the in the embryonic fetal or adult phases of
life.
33. Binding and release of oxygen to and from hemoglobin and the resulting sigmoid oxygen curve are the outcome of.
a. The hydrophobic nature of the surface of the hemoglobin molecule.
b. The hydrophobicity of the heme pocket.
c. The direct bonding between the proximal histidine residue (His F8) and the iron ion in the heme group.
d. The lack of direct bonding between the distal histidine residue (His E7) and the iron ion in the heme group.
e. The tetrameric structure of hemoglobin.
f. The absence of beta sheets as elements of the secondary structure of the globin chains.
34. a० thalassemia (homozygous a thalassemia 1; Hb Bart Hydrops fetalis) is characterized by:
a. A significant elevation in the synthesis of γ globin.
b. A significant but clinically insufficient elevation in the synthesis of both γ and δ globins.
c. Continuous synthesis of ζ globin after birth and until the premature death of the affected individuals at 2 to 10 years
of age.
d. Accumulation of Hb Bart and HbH in erythrocytes of an affected fetus during the third trimester of gestation.
e. Accumulation of Hb Bart but not of HbH in erythrocytes of an affected fetus during the third trimester of gestation.
f. Accumulation of HbH but not of Hb Bart in erythrocytes of an affected fetus during the third trimester of gestation.
35. The following statement is correct:
a. Full inactivation of of methemoglobin reductase is likely to increase the overall oxygen carrying capacity of hemoglobin.
b. Increased rate of glycolysis and a resulting accumulation of lactic acid in peripheral tissues is likely to cause left-ward
shift in the oxygen binding/dissociation curve of hemoglobin.
c. The level of Erythropoeitin (EPO) is likely to be elevated in patients suffering from polycythemia vera but not in
individual with polycythemia caused by an increased oxygen affinity of their hemoglobin.
d. Methemoglobinopathy that is caused by homozygous mutation in both beta globin alleles, but not in any of the alpha
globin alleles is lethal.
e. It is expected that a patient with permanent and full kidney loss of function will develop erythrocytic anemia.
36. Accumulation in erythrocytes of the following adnormal hemoglobin(s) characterizes the specified matched
thalassemia(s):
a. HbH but not Hb Bart is accumulated in erythrocytes in β° thalassemia.
b. Both HbH and Hb Bart are accumulated in erythrocyctes in β° thalassemia.
c. Free chains of α globin are accumulated in erythrocytes in α° but not α α thalassemia.
d. alpha1 abnormal hemoglobin is accumulated in erythrocytes in β° thalassemia.
e. gamma1 abnormal hemoglobin is accumulated in erythrocytes in β° thalassemic adults.
f. Free chains of alpha globin are accumulated in erythrocytes of both β° and β thalassemia.
37. The following (theoretical) treatment could potentially alleviate clinical manifestations of sickle cell anemia in an HbS homozygote:
a. Elevation of the concentration in erythrocyte of 2,3 biphosphoglycerate (2,3 BPG).
b. Administration of a drug that increases the activity of the enzyme carbonic anhydrase.
c. Administration of a drug that selectively increases the volume of the erythrocytes.
d. Administration of a drug that selectively kills F-cells.
38. Mutation in the β globin gene caused cyanosis. The following hemoglobin is likely to be present in the erythrocytes of the affected individual.
a. Hb Bart
b. Unstable hemoglobin that aggregates into Heinz bodies.
c. Methemoglobin
d. HbS
e. Hemoglobin having a decreased oxygen affinity
f. Hemoglonin having an increased oxygen affinity
39. A new drug was discovered by a bright young TeAM student engaged in tropical diseases research project which was
funded by the Bill and Melinda Gates Foundation. The drug prevented leakage of potassium (K+) ions from erythrocytes
and thus an unchanged normal intracellular concentration of this ion was maintained. Administration of the drug is
expected to result in:
a. Increased survival in the cells of Plasmodium falciparum which entails higher Malaria morbidity and mortality.
b. Diminished survival in the cells of Plasmodium falciparum which results in increased protection against Malaria.
c. Erythrocytes of sickle cell anemia homozygotes will resist sickling even under low oxygen pressure becuase potassium
leakage is the mandatory cause of cell sickling.
d. The erythrocytes will become anhydrated due to increased loss of water to the outside and thus HbS polymerization will
be facilitated.
40. A battery of newly minted drugs was considered for the treatment/therapy of β thalassemia. Bearing in mind their respective
mechanisms of action, an expert hematologist concluded that only one of the offered drugs might improve the clinical state of β thalassemic individuals.
Which Drug is it?
a. A drug that stimulates the activity of cytosine methyl transferase (an enzyme that methylates cytosine residues in CpG pairs in DNA)
b. A drug that inhibits the activity of histone deacetylase (an enzyme that removes acetyl groups from acetylated histones).
c. A drug that decreases the nuclease hypersensitivity of the Locus Control Region (LCR) of the β globin gene cluster.
d. A drug that specifically blocks transcription of the ε globin and γ globin.
e. A drug that specifically blocks transcription of the ε globin and δ globin.
41. Which of the following mutations is likely to result in β° thalassemia?
a. AAUAAA to AACAAA muation in the 3' untranslated region (3' UTR) of the β globin gene.
b. A nonsense (termination) mutation in the codon for amino acid 144 of the β globin chain
c. A C -> T mutation in the 5° nucleotide downstream from the 5° junction between the first exon and the first intron of the β globin gene.
d. A mutation that renders inactive the Locus Control Region (LCR) of the β globin gene cluster.
e. A (G)U -> (G)A mutation in the 2nd nucleotide downstream from the 5° junction between the first exon and the first intron of the β globin gene [the G in
the GU dinucleotide is the 1st intronic nucleotide downstream from the junction].
1
C
2
A
3
E
4
D
5
D
6
A
7
C
8
E
9
D
10 D
11 D
12 C
13 D
14 E
15 B
16 D
17 E
18 C
19 E
20 E
21 C
22 C
23 A
24 E
25 F
26 C
27 C
28 D
29 E
30 D
31 C
32 A
33 E
34 D
35 E
36 F
37 C
38 C
39 A
40 B
41 E