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1st semester test in 1999/2000.

Problem–solving tests

Test I.

Translocation of newly synthesized proteins across the membrane of endoplasmicreticulum can be studied in vitro in a translation system containing microsomes. Inthe experiment described in this paper the translocation of a [35S]– methionine–labelled protein synthesized on a purified, specific mRNA was analyzed. Translationwas performed in the presence or absence of microsomes, thereafter both reactionmixtures were divided into 4–4 aliquots, and the samples were treated as follows:

1. No treatment;2. Treatment with a protease;3. Treatment with detergent and protease;4. Disruption of microsomes and treatment with endoglycosidase H (endo H).

(Endo H removes N–linked oligosaccharides attached to the protein in theendoplasmic reticulum.)

Reaction products of the samples were fractionated by SDS–polyacrylamide gelelectrophoresis followed by autoradiography. Figure 1 shows the treatment of thedifferent samples and a schematic drawing of the autoradiogram. Using the data inthis figure, solve the following multiple–choice questions.

+1 2 3 4 5 6 7 8

+++

+––

––

–––– +

++

+––

––

––––

withoutmicrosomes

withmicrosomes

a

bcd

TREATMENTproteasedetergentendo H

Figure 1

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Experiment Analysis

1. ___ The protein encoded by the mRNA is translocated into the lumen of theendoplasmic reticulum.

2. ___ The mature protein contains O–linked oligosaccharides.

3. ___ The mature protein contains N–linked oligosaccharides.

4. ___ The protein is N–glycosylated in the cytosol.

5. ___ The synthesis of the polypeptide chain requires enzymes of the endoplasmicreticulum.

6. ___ The protein is proteolytically processed in the endoplasmic reticulum.

7. ___ In the presence of microsomes, the lifespan of the protein is increased.

8. ___ The protein contains disulfide bonds.

Test II.

The gene coding for the lac repressor had been mutated in an E. coli strain, so thatthe repressor protein is no longer able to bind to DNA at 43°C; DNA–binding isnormal at 30°C (mutant A). In another bacterial strain, temperature–sensitivemutation occured in the structural gene of (3–galactosidase: at 30°C the enzymeworks well, but at 43°C it is inactivated (mutant B). How will these strains behaveunder different culture conditions?

Four–choice Association

A: Mutant A cellsB: Mutant B cellsC: Both of themD: Neither of them

9. ___ At 30°C, in a medium containing glucose, there is hardly any β–galactosidase activity in these cells.

10. __ At 43°C, in a medium containing glucose, there is hardly any β–galactosidase activity in these cells.

11. __ At 30°C, in a medium containing only lactose as carbon–source, the cellsshow high levels of β–galactosidase activity.

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12. __ At 43°C, in a medium containing only lactose as carbon–source, the cellsshow high levels of β–galactosidase activity.

13. __ At 30°C, in a medium lacking both lactose and glucose, these cells havehigh β–galactosidase activity.

14. __ At 43°C, in a medium lacking both lactose and glucose, these cells havehigh β–galactosidase activity.

15. __ These cells are able to efficiently metabolize lactose at 43°C.

Quantitative Comparison

16. __ A: The concentration of cAMP in mutant A cells at 30°C, in the presence ofglucose

B: The concentration of cAMP in mutant B cells at 30°C, in the presence ofglucose

Test III.

Molecular chaperones are protein molecules that play an imporlant role in the foldingand membrane translocation of newly synthesized proteins a well as in the sembly ofpmtein complexes. Hsp70 heat shock proteins are the best characterzed family ofchaperones. The experiments described In this test were performed to study the roleof a cytoplasmic yeast Hsp70 (designated Ssb1) in the iranslation machinery.A crude cytoplasmic extract was prepared from yeast cells and subjected to sucrosedensity gradient centrifugation (Figure 2A). UV absorption at 260 nm was monitoredin the gradient (A260).

80S

80S

80SControl

I II

Ssb1

L16

5 5 510 10 1015 15 15

Sedimentograms

A260

A B C

fractions

+RNase + Na-azid

Western blots

direction of centrifugation

Figure 2

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Five–choice Completion

17. __ Which of the following components contributed most to the UV absorptioncurve of Figure 2A?

A: DNAB: rRNAC: mRNAD: tRNAE: ribosomal proteins

18. __ What is the number of ribosomes in the complexes of peak I?A: 0B: 1C: 2D: 3E: Can not be determined

Aliquots of the gradient fractions were run in an SDS–polyacrylamide gel andWestern blots were performed using antibodies raised againts the Ssb1 protein orthe large ribosomal subunit protein L16. To obtain Figure 2B, the extract wasexposed to mild RNase treatment before centrifugation. Figure 2C shows the resultsof a similar experiment; in this case the extract was prepared from sodium azide–treated cells. (Sodium azide allows translating ribosomes to run off of the mRNA butprevents reinitiation.) What can be the difference between the ribosomes present insamples B and C?

Four–choice Association

A: Ribosomes in the RNase–treated extractB: Ribosomes from sodium azide–treated cellsC: Both of themD: Neither of them

19. __ Consist of a small and a large subunit.

20. __ Are mostly present in a polysomal form.

21. __ Contain the L16 protein.

22. __ tRNA molecules are attached to them.

23. __ May carry nascent polypeptide chains.

24. __ Ssb1 protein is bound to them.

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In the second part of the experiment the effect of puromycin treatment was studiedon the binding of Ssb1 protein to the polysomes. (Puromycin is an aminoacyl–tRNAanalog that is incorporated into the C–terminus of the growing polypeptide chain,stops its synthesis and causes the release of the nascent chain from the ribosome.)Figure 3 shows the results of an experiment in which yeast extracts were either leftuntreated (Figure 3A) or treated with puromycin (Figure 3B) and then analyzed asdescribed for Figure 2. Compare the two charts and solve the following multiple–choice questions.

80S 80S

2 24 46 6

A

A260

–puromycin +puromycin

Ssb1Western blots

Sedimentograms

B

fractions

direction of centrifugation

Figure 3

Four–choice Association

A: Untreated extractB: Puromycin–treated extractC: Both of themD: Neither of them

25. __ Contains mRNA.

26. __ Contains polyribosomes.

27. __ Ssb1 protein binds to its ribosomes.

Five–choice Completion

28. __ Based on the evidence presented in this paper, determine the most likelybinding site of Ssb1 during protein synthesis:

A: The peptidyl transferase enzymeB: The small ribosomal subunitC: The nascent polypeptideD: mRNA sequences bound to the ribosomesE: mRNA sequences between ribosomes

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Test IV.

Study the following picture and answer the questions.

Five–choice Completion

29. __ Which of the following items contain site of RNA polymerase III function?A: Item 21B: Item 9C: Item 19D: Item 23E: Item 14

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30. __ Which of the following items contain importin?A:Item 9B:Item 6C: Item 16D:Item 8E: Neither of them

Four–choice Association

A: Item 18B: Item 20C: Both of themD: Neither of them

31. __ It contains 0–linked glycosylated protein.

32. __ Its function can be studied by autoradiography after [3H]–uridineincorporation.

33. __ It can be found in the microsomal fraction after cell fractionation.

34. __ It has the same function as item 21.

35. __ It contains RNA.

36. __ It contains proteins.

Multiple Completion

37. __ Which of the following micrographs were photographed by transmissionelectron microscopy?

1. Micrograph containing item 1–22. Micrograph containing item 243. Micrograph containing item 154. Micrograph containing item 3–4

38. __ Which of the following items were visualized by heavy metal shadowing?1. Item 12. Item 63. Item 244. Item 15

39. __ Which of the following items contain proteins synthesized by freepolysomes?

1. Item 32. Item 123. Item 64. Item 22

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40. __ Which of the following items contain chromatosomes?1. Item 82. Item 223. Item 54. Item 2

Traditional tests

Five–choice Completion

41. ___Using anion–exchange chromatography compound "A" is eluted first, followedby the elution of compound "B". Which one of the following statements is truefor sure?

A: "A" is positively charged, "B" is negatively chargedB: "A" is larger than "B"C: in the case of using a cation–exchanger column – at the same pH

– "B" would be eluted firstD: "A" and "B" can be separated from each other by gel filtrationE: none of the above statements is necessarily true

42. ___Which one of the following steps of translation requires GTP as the energysource?

A: translocation of the ribosomeB: the peptydil–transferase reactionC: association of the mRNA with the small ribosomal subunitD: all of themE: none of them

43. ___A cell culture is treated with colchicine, another one is subjected to serumstarvation. After mixing the cells their DNA is stained, thereafter flow cytometryis performed. Which one is NOT true of the following statements?

A: some cells of the sample are diploidB: some cells contain two–chromatid chromosomesC: [3H]–thymidine labeling would lead to the appearance of grains

above the nuclei of some cellsD: in some cells active MPF can be detectedE: cells can be separated into two groups based on the difference of

their DNA amount

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Multiple Completion

44. ___It is true for the inner mitochondrial membrane:

1. protons are pumped through it to the intermembranal space2. ions cannot be passively transferred through it3. it contains F0/F1 complex4. it contains cardiolipin

45. ___It is synthesized on free ribosomes:

1. steroid receptor2. histone3. porin4. lysosomal enzymes

46. ___Bacterial cells are kept in a solution containing both glucose and lactose.What can be the consequence of this?

1. activity of adenylate cyclase is increased in the cells2. CAP binds the the promoter region3. transcription of the structural genes is increased4. repressor protein does not bind to the operator region

47. ___Which of the following components are involved in the nuclear transport?

1. homeodomain proteins2. importin3. perichromatin granules4. nuclear localisation signal

48. ___Phoshorylation may lead to the change of:

1. the isoelectric point of the target protein2. the tertiary structure of the target protein3. activity of the target protein4. the mobility of the target protein during SDS polyacrylamide

gelelectrophoresis

49. ___Which of the following methods can be used for the synchronization of a cellculture?

1. mitotic shake off2. inhibition of the S phase3. inhibition of the M phase4. [3H]–thymidine labeling

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50. Which of the followings is true for the expression vector?

1. its product can be detected by Western blotting2. its product can be detected by Northern blotting3. its product can be detected by immunocytochemistry4. its presence can be detected by Southern blotting

Four–choice Association

A: esther bondB: peptide bondC: both of themD: neither of them

51. ___It is composed of four different atoms.

52. ___It is formed between the α and β phosphate groups of nucleotides

53. ___It is formed by the release of a water molecule.

54. ___Protein kinases catalyze the formation of this kind of bond.

A: primaseB: DNA polimerase I.C: both of themD: neither of them

55. ___It is involved in the synthesis of Okazaki fragments.

56. ___It can be inhibited by streptomycin.

57. ___It possesses a 5' → 3' exonuclease activity.

58. ___It is involved in the degradation of the primer.

59. ___It can be inhibited by α–amanitine.

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Five–choice Association

A: formaldehyde–agarose gelelectrophoresisB: SDS polyacrylamide gelelectrophoresisC: isoelectric focusingD: all three of themE: none of them

60. ___The separation is independent of the molecular weight of the molecules to beseparated.

61. ___Separated molecules can be visualized by ethidium–bromide.

62. ___The migration of separated molecules stops reaching a certain pH value of thegel.

63. ___This is a step of Northern blotting.

64. ___This method is used in DNA sequencing.

A: Xeroderma pigmentosumB: Familial hypercholesterolemiaC: Pituitary dwarfismD: Systemic lupus erythematosus (SLE)E: none of the above

65. ___The failure of excision repair is the reason of the disease.

66. ___Mutation of the Pit–1 transcription factor gene may lead to the disorder.

67. ___Abnormal mRNA splicing is involved in the pathomechanism of the disease.

68. ___Lack of lysosomal enzymes is characteristic of the disease.

69. ___The disorder may lead to heart attack even in the early childhood.

70. ___Skin cancer frequency is very high in the disorder.

Quantitative Comparison

71. ___A: Amount of nucleic acid in the cDNA library of a cellB: Amount of nucleic acid in the genomic library of the same kind of a cell

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72. ___A: LDL concentration of the blood in the heterozygous form of familialhypercholesterolemia

B: LDL concentration of the blood in the homozygous form of familialhypercholesterolemia

Variation Relationship

73. ___A: The length of primers used in PCRB: The speed of amplification

74. ___A: Phosphorylation of the H1 histoneB: Condensation of chromatin

75. ___A: The length of introns inside a geneB: The length of the polypeptide chain encoded by the same gene

Relationship Analysis

76. ___The two chains of mitochondrial DNA can be separated from each other byrate–zonal centrifugation, BECAUSE the base constitution of the chainsshows a large difference.

77. ___Mutations of genes of cytochrome P450 enzymes might be lethal for the cell,BECAUSE in the absence of normal cytochrom P450 enzymes the formationof carcinogenic benzpyrene compound is increased.

78. ___The result of the Kornberg experiment would be the same with using either (γ–32P)CTP or (α–32P)CTP, BECAUSE the DNA polymerase is able toincorporate both kinds of nucleotides into the synthesized chain.

79. ___One of the symptoms of testicular feminization is the lack of menstruation,BECAUSE in the absence of testosteron the testosteron receptor cannottranslocate to the nucleus.

80. ___The Trp operon is active in the presence of Trp, BECAUSE in the absence ofTrp the repressor protein binds to the operator region.

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Correct answers for seminar test in1999/2000. year

0 1 2 3 4 5 6 7 8 9

0 - A C A B B A C C C

1 C C A D A A C B D C

2 D C A A A C C A C D

3 E D C D D C C A B E

4 A E A C E A D C E A

5 E B D C A C D B B D

6 C D C A B A C D E B

7 A B B C A C D C D B

8 E