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Q81.Puromycin is an antibiotic used to inhibit protein synthesis. Given below are few statements about the antibiotic.
1.It enters the E-site of the ribosome where it prevents the release of deacetylated tRNA after the action of peptidyl transferase.
2.It blocks the translocation process by binding to the translocation factor EF-G
3. Puromycin resembles the initiatior tRNA, tRNAif-met
4.It resembles the aminoacyl tRNA and binds to the A-site of the ribosome
5.Puromycin inhibits only prokaryotic protein synthesis.
6.Puromycin inhibits both prokaryotic and eukaryotic protein synthesis.
Which of the above statements are true?
(A)1 and 5
(B) 2 only
(C) 4 and 6
(D) 3 and 5
ANSWER-C
Description- Puromycin is a secondary metabolite of Streptomyces alboniger which is a structural analog of aminoacyl t-RNA and binds to A-site on the ribosome . Peptidyl transferase enzyme catalyse the covalent linkage of puromycin to growing peptide chain covalently ultimately inhibiting protein biosynthesis in P.K and E.K
Q82.Total RNA was isolated separately from cytosol and nuclei of human cells growing in a cell culture. Each sample was mixed with a purified denatured fragment of a DNA corresponding to a large intron of a house keeping gene incubated under renaturating condition. given below are the statements amde about the outcome of the experiment.
1.RNA isolated from nuclei will from RNA-DNA duplexes because of the presence of introns in the primary RNA.
2.Cytosolic RNAs usually will not from RNA-DNA duplexes.
3. Both cytosolic and nuclear RNA will not from RNA-DNA duplexes as transcription and splicing occur simultaneously.
4.cytosolic RNA will from RNA-DNA duplexes because unspliced cytosolic RNAs are exceptionally stable.
Which of the above statements are most likely to be true?
(A) only 3
(B) 1 and 4
(C) 1 and 2
(D) only 4
ANSWER-C
Description-hnRNA in nucleus do not undergo ant splicing.Hence contains intronic sequences along with exonic part. After splicing that takes places in Nucleus introns are removed and exons are ligated to from m RNA molecule, ultimately m RNA is transported into cytosol i.e., out of the nucleus.
Q83. If a proteasome inhibitor is added to synchronously cycling human cells in G2 phase which one of the following events is likely to happen?
(A) Induce re-replication of DNA
(B) Arrest cells in G2 phase
(C) Arrest cells in Anaphase
(D) Block chromatin condensation
ANSWER-C
Description-proteasome- multi subunit enzyme complex that play a central role in the regulation of proteins that control cell cycle progression and apoptosis, an imp. target of cancer therapy.in G2 phase cell growth continues and proteins are synthesized in preparation for mitosis.
the spindle- Assembly checkpoint monitors the alignment of chromosomes the mataphase spindle.
Q84. A promoter deletion study was done in order to determine the binding sites for a transcription factor on the promoter, which is activated on treatment with the drug 'X'. The following constructs were made-
Luciferase assay revealed the following results-
The following statements can be made-
1. Region between -1800 ans -1210 contains a binding site for the activator
2.Region between -868 and -1210 contains a binding site for a repressor
3.Region between -868 and -432 contains a binding site for a repressor
4. Region between -1210 ans -868 contains a binding site for the activator
Which of above are true?
(A) 1 and 3
(B) 2 and 3
(C)1 and 4
(D) 2 only
ANSWER-C
Description-Activator protein binding region lies upstream to the transcription start site of sense strand.Promoter region is defined as DNA sequence, where transcription of a gene by RNA pol. begins. Both RNA pol. & necessary transcription factors(T.F) binds to the promoter sequences.
Q85. A pharmacy student designed a drug to specifically target the receptors for retinoic acid in order to prevent stem cell differentiation. After in vitro trial, the investigator found that the drug seemed to be underwent differentiation and the drug seemed to be ineffective. The following reasons were given by the student-
1.The size of drug exceeded the size of molecules that could cross the membrane.
2.The drug was small in size but hydrophobic in nature.
3.the drug did not bind to its receptors.
Which one of the above cold be the probable reason for drug ineffectiveness?
(A) Only 3
(B)1 and 3
(C) 1,2 and 3
(D)Only 2
Answer-B
Description-Retinoic acid receptor is a type of nuclear receptor which can also act as a T.F. that is activated by both all trans retinoic acid. so if the drug small+ hydrophobic.
Therefore easily cross PM and nuclear membrane.
I hope you read my first capsule about cell biology. 
