Category: Quiz

RNA Polymerase (RNAP)

RNA Polymerase (RNAP)

Q. 1 RNA polymerase recognizes? 
 A

Promoter site

 B Initiation site
 C Regulator site
 D Stop site
Q. 1 RNA polymerase recognizes? 
 A

Promoter site

 B Initiation site
 C Regulator site
 D Stop site
Ans. A

Explanation:

(40)        Promoter site REF: Harper 27`h ed p. 349

RNA polymerase has sigma sub unit which recognises promotor nucleotide sequence. It does not require primer and has no proof reading activity

Types:

  1. r RNA
  2. m RNA, mi RNA, Sn RNA
  3. t- RNA, 5s-rRNA, some SnRNA

Q. 2 To initiate transcription RNA polymerase does not require which one of the following:

 A

Template (ds DNA)

 B

Activated precursors (ATP, GTP, UTP, CTP)

 C

Divalent metal ions (Mn2+, Mg2+)

 D

Primer

Q. 2

To initiate transcription RNA polymerase does not require which one of the following:

 A

Template (ds DNA)

 B

Activated precursors (ATP, GTP, UTP, CTP)

 C

Divalent metal ions (Mn2+, Mg2+)

 D

Primer

Ans. D

Explanation:

DNA polymerase can only elongate existing polynucleotide chains, and thus requires a primer. RNA polymerase can initiate RNA synthesis de novo and hence does not require a primer.

Ref: Molecular Biology of the Gene Watson, 2004, Page 376 ;  Lippincott’s Illustrated Q & A Review of Biochemistry by Michael A. Lieberman, Rick Ricer, 2009, Page 294


Q. 3

The sigma (cr) subunit of prokaryotic RNA polymerase:

 A

Binds the antibiotic rifampicin

 B

Is inhibited by a-amanitin

 C

Specifically recognizes the promoter site

 D

Is part of the core enzyme

Q. 3

The sigma (cr) subunit of prokaryotic RNA polymerase:

 A

Binds the antibiotic rifampicin

 B

Is inhibited by a-amanitin

 C

Specifically recognizes the promoter site

 D

Is part of the core enzyme

Ans. C

Explanation:

C i.e. Specifically recognizes the promoter site

Quiz In Between


Q. 4

RNA polymerase does not require :

 A

Template (ds DNA)

 B

Activated precursors (ATP, GTP, UTP, CTP)

 C

Divalent metal ions (Mn2+, Mg2+)

 D

Primer

Q. 4

RNA polymerase does not require :

 A

Template (ds DNA)

 B

Activated precursors (ATP, GTP, UTP, CTP)

 C

Divalent metal ions (Mn2+, Mg2+)

 D

Primer

Ans. D

Explanation:

D i.e. Primer


Q. 5

The drug inhibiting DNA-dependent RNA polymerase in Mycobacteria is –

 A

INH

 B

Rifampicin

 C

Ciprofloxacin

 D

Ethionamide

Q. 5

The drug inhibiting DNA-dependent RNA polymerase in Mycobacteria is –

 A

INH

 B

Rifampicin

 C

Ciprofloxacin

 D

Ethionamide

Ans. B

Explanation:

Ans. is ‘b’ i.e., Rifampicin

Mechanism of action of important antitubercular drugs

o Rifampicin —> Inhibits DNA dependent RNA synthesis by inhibiting RNA polymerase.

o INH  —> Inhibits synthesis of mycolic acid which is a component of mycobacterial cell wall. o Pyrazinamide —> Inhibits synthesis of mycolic acid.

o Streptomycin —> Inhibits protein synthesis (translation).

o Ethambutol —> Inhibits incorporation ofmycolic acid into bacterial cell wall by inhibiting arbinosyltransferase.


Q. 6

RNA polymerase has which activity

 A

Primase

 B

Helicase

 C

Ligase

 D

Topoisomerase

Q. 6

RNA polymerase has which activity

 A

Primase

 B

Helicase

 C

Ligase

 D

Topoisomerase

Ans. A

Explanation:

 

DNA synthesis cannot commence with deoxyribonucleotides because DNA polymerase cannot add a mononucleotide to another mononucleotide.

  • Thus, DNA polymerase cannot initiate synthesis of complementary DNA synthesis strand of DNA on a totally single stranded template.
  • For this, they require RNA primer, which is a short piece of RNA formed by enzyme primase (RNA polymerase) using DNA as a template.
  • RNA primer is then extended by addition of deoxyribonucleotides.
  • Later on, the ribonucleotides of the primer are replaced by deoxyribonucleotides.
  • Primase is actually a DNA primase which has RNA polymerase activity. This DNA primase is also called DNA polymerase.

Quiz In Between



Post Transcriptional Modification Of Rna

POST TRANSCRIPTIONAL MODIFICATION OF RNA

Q. 1 Which of the following type of RNA has the splicing activity as a function?

 A

mRNA

 B

snRNA

 C

tRNA

 D

rRNA

Q. 1

Which of the following type of RNA has the splicing activity as a function?

 A

mRNA

 B

snRNA

 C

tRNA

 D

rRNA

Ans. B

Explanation:

Small Nuclear RNAs (snRNAs), a subset of the small RNAs, are significantly involved in rRNA and mRNA processing and gene regulation. Of the several snRNAs, U1, U2, U4, U5, and U6 are involved in intron removal and the processing of mRNA precursors into mRNA

 
Ref: Weil P. (2011). Chapter 34. Nucleic Acid Structure & Function. In D.A. Bender, K.M. Botham, P.A. Weil, P.J. Kennelly, R.K. Murray, V.W. Rodwell (Eds), Harper’s Illustrated Biochemistry, 29e.

 


Q. 2

A segment of an eucaryotic gene that is not represented in the mature mRNA, is known as:

 A

Exon

 B

Intron

 C

Plasmid

 D

TATA box

Q. 2

A segment of an eucaryotic gene that is not represented in the mature mRNA, is known as:

 A

Exon

 B

Intron

 C

Plasmid

 D

TATA box

Ans. B

Explanation:

Introns are the RNA sequences in the primary transcript that are not found in the mRNA. Exons RNA sequences in the primary transcript that are found in the mRNA. A TATA box is a DNA sequence that indicates the point at which a genetic sequence can be read and decoded.
 
Transcription: Transcription is the process by which DNA is copied (transcribed) to mRNA, which carries the information needed for protein synthesis. Transcription takes place in two broad steps. First, pre-messenger RNA is formed, with the involvement of RNA polymerase enzymes. Second formation of Messenger RNA.
 
Transcription unit: is a linear sequence of DNA that extends from a transcription start site to a transcription stop site. The transcription unit—that portion of a gene that is copied by RNA polymerase—consists of coding regions of DNA (exons) interrupted by intervening sequences of noncoding DNA (introns). After transcription, during RNA processing, introns are removed and the exons are ligated together to form the mature mRNA that appears in the cytoplasm.
 
1) Transcription Steps: There are three main steps to the process of DNA transcription.
  • Initiation: RNA Polymerase Binds to DNA. DNA is transcribed by an enzyme called RNA polymerase. Specific nucleotide sequences tell RNA polymerase where to begin and where to end. RNA polymerase attaches to the DNA at a specific area called the promoter region.
  • Elongation: Certain proteins called transcription factors unwind the DNA strand and allow RNA polymerase to transcribe only a single strand of DNA into a single stranded RNA polymer called messenger RNA (mRNA). The strand that serves as the template is called the antisense strand. The strand that is not transcribed is called the sense strand.
  • Termination: RNA polymerase moves along the DNA until it reaches a terminator sequence. At that point, RNA polymerase releases the mRNA polymer and detaches from the DNA.
2) Pre mRNA processing:
  • RNA splicing: pre-mRNAs typically include introns.Exons:are the continuous coding regions. Introns: are the  noncoding intervening sequences. They may serve to separate functional domains (exons) of coding information in a form that permits genetic rearrangement by recombination to occur more rapidly than if all coding regions for a given genetic function were contiguous. Introns are removed by RNA processing in which the intron is looped out and cut away from the exons by snRNPs, and the exons are spliced together to produce the translatable mRNA.
Ref: Harper’s illustrated biochemistry, 26th Edition Page  319

 


Q. 3

Poly(A) tail translates into (i.e. on translation give rise to):

 A

Polyproline

 B

Polylysine

 C

Polyalanine

 D

Polyglycine

Q. 3

Poly(A) tail translates into (i.e. on translation give rise to):

 A

Polyproline

 B

Polylysine

 C

Polyalanine

 D

Polyglycine

Ans. B

Explanation:

B i.e. Polylysine

Codon

AAA

GGG

CCC

TTT/ UUU

Translation product

LysinQ

Glycine

Proline

Phenylalanine

Quiz In Between


Q. 4 Introns are exised by

 A

RNA splicing

 B

RNA editing

 C

Restriction endonuclease

 D

DNAase

Q. 4

Introns are exised by

 A

RNA splicing

 B

RNA editing

 C

Restriction endonuclease

 D

DNAase

Ans. A

Explanation:

A i.e. RNA splicing


Q. 5

A segment of a eukaryotic gene that is not represented in the mature messenger RNA is known as:

 A

Intron

 B

Exon

 C

Plasmid

 D

TATA box

Q. 5

A segment of a eukaryotic gene that is not represented in the mature messenger RNA is known as:

 A

Intron

 B

Exon

 C

Plasmid

 D

TATA box

Ans. A

Explanation:

A i.e. Intron


Q. 6

Splicing Activity is a function of:

 A

mRNA

 B

sn RNA

 C

r RNA

 D

t RNA

Q. 6

Splicing Activity is a function of:

 A

mRNA

 B

sn RNA

 C

r RNA

 D

t RNA

Ans. B

Explanation:

B i.e. snRNA

–   Self splicing introns function as ribozyme and do not require external protein enzymes or high energy cofactors (eg ATP). Intron forms lariat in group II but not in group I self splicing introns.

–  Splicing is process of removing introns (i.e. the segment of gene that is not represented in mature m-RNA) from primary transcript and joining (or ligating together) exons (RNA sequences that code protein)Q. It is mediated by sn RNP or snrups formed from sn RNAQ.

Small nuclear RNAs (sn RNAs) are involved in RNA splicingQ; small nucleolar RNAs (sno RNAs) are involved in r RNA modificationsQ; micro RNAs (mi RNAs) & small temporal RNAs (st RNAs) are involved in inhibition (regulation) of gene expression (gene silencing)Q; small interfering RNAs (si RNAs) are involved in RNA interference (RNAi)Q; and B2 RNA binds to pol II and block transcription of many genes during heat shock

– Mammalian genomes seem to encode more non coding RNAs (nc RNAs) than coding m RNAs. nC RNAs are RNAS that do not encode proteins including r RNAS, t RNAs, mi RNAs, si RNAs, st RNAs and sn RNAs.

Quiz In Between


Q. 7 Defect in Snurps causes ‑

 A

Defect in 5′ – capping

 B

Defect in addition of poly-A tail

 C

Defect in Splicing

 D

Defect in terminal addition of nucleotide

Q. 7

Defect in Snurps causes ‑

 A

Defect in 5′ – capping

 B

Defect in addition of poly-A tail

 C

Defect in Splicing

 D

Defect in terminal addition of nucleotide

Ans. C

Explanation:

 

m-RNA processing

Prokaryotic mRNA is functional immediately upon synthesis, i.e. prokaryotic primary transcript of mRNA is functional. Thus it does not require post-transcriptional modification. In Eukaryotes the primary tran­script of mRNA is the hn RNA (hetrogeneous nuclear RNA). After transcription hnRNA is extensively modi­fied to form functional mRNA. These modifications are as follows.

1) The 5′-capping :- This is the first processing reaction. 51-end of mRNA is capped with 7-methylguansosine. This cap helps in initiation of translation (protein synthesis) and stabilizes the structure of mRNA by protecting from 5′-exonuclease.

2) Addition of poly ‘A’ tail :- As the name suggests, multiple ‘A’ (adenylate) residues are added at 3’end.This poly-A tail is not transcribed from DNA, but rather added after transcription. These tails helps to stabilize the mRNA (by protecting from 3′-exonuclease), facilitate exit from the nucleus, and aid in translation. After mRNA enters the cytosol, the poly-A tail is gradually shortened. Some mRNAs do not have poly-A tail, e.g. mRNAs of histones and some interferons.

3) Removal of introns (splicing) :- Eukaryotic genes contain some coding sequences which code for protein and some intervening non-coding sequences which do not code for protein. The coding sequences are called `exons’ and intervening non-coding sequences are called `introns’. The process by which introns are excised and exons are linked to form functional mRNA is called splicing. Thus mature mRNA does not contain introns.

  • Splicesome Splicesome is an assembly made up of small nuclear RNA (snRNA), some proteins and hnRNA. snRNA combines with proteins to form small nuclear ribnonucleoprotein particles (snRNPs or snurps) that mediate splicing. It is snRNA component of snurps that catalyzes splicing. Snurps are U4, U5 and U6.
  • Only about 1-5% of human DNA has coding sequence (exons). Remaining is non-coding (introns).

4) Alternate splicing :- The hn-RNA molecules from some genes can be spliced in alternative way in different tissues. Thus two or more different mRNA (and therefore 2 or more proteins) can be synthesized from same hnRNA. For example, difference isoforms of tropomyosin in different tissues in due to alternate splicing.


Q. 8 Defect in Snurps causes‑

 A

Sickle cell anemia

 B

Thalassemia

 C

Marfan syndrome

 D

EDS

Q. 8

Defect in Snurps causes‑

 A

Sickle cell anemia

 B

Thalassemia

 C

Marfan syndrome

 D

EDS

Ans. B

Explanation:

 

Defective splicing (defect in snurps) is the most common mutation causing thalassemia.

  • Molecular defect in pathogensis of thalassemia:‑

A) β-Thalassemia

  • Most common type of genetic abnormality in β-thalassemia is point mutation i.e., nonsense.
  • Some may also occur due to deletion or insertion i.e., frame shift mutations.
  • Defect may occur at different steps of β-chain synthesis:

i) Splicing mutations

  • Mutations leading to aberrant splicing are the most common cause of P-thalassemia.

ii) Chain terminator mutations

  • This cause premature termination of mRNA translation.

iii) Promoter region mutations

  • This results in transcription defect.

B) α-Thalassemia

  • The most common cause of reduced a-chain synthesis is deletion of a-globin genes.
  • Rarely nonsense mutation may also cause a-thalassemia.

Q. 9 Splicing is a process of ‑

 A

Activation of protein

 B

Removal of introns

 C

Synthesis of protein

 D

Replication of DNA

Q. 9

Splicing is a process of ‑

 A

Activation of protein

 B

Removal of introns

 C

Synthesis of protein

 D

Replication of DNA

Ans. B

Explanation:

 
In molecular biology and genetics, splicing is a modification of an RNA after transcription, in which introns are removed and exons are joined.

This is needed for the typical eukaryotic messenger RNA before it can be used to produce a correct protein through translation.

For many eukaryotic introns, splicing is done in a series of reactions which are catalyzed by the spliceosome a complex of small nuclear ribonucleoproteins (snRNAs), but there are also self-splicing introns.

Quiz In Between



Disorders of heme biosynthesis

Disorders of heme biosynthesis

Q. 1 A girl on sulphonamides developed abdominal pain and presented to emergency with seizure. What is the probable diagnosis?

 A

Acute intermittent porphyria

 B

Congenital erythropoietic porphyria

 C

Infectious mononucleosis

 D

Kawasaki disease

Q. 1

A girl on sulphonamides developed abdominal pain and presented to emergency with seizure. What is the probable diagnosis?

 A

Acute intermittent porphyria

 B

Congenital erythropoietic porphyria

 C

Infectious mononucleosis

 D

Kawasaki disease

Ans. A

Explanation:

Abdominal pain is the most common symptom in Acute intermittent porphyria and is usually steady and poorly localized but may be cramping. Seizures can be due to neurologic effects or to hyponatremia.
Treatment of seizures is difficult because most anti seizure drugs can exacerbate AIP (clonazepam may be safer than phenytoin or barbiturates).
 
The drugs which precipitate attacks of acute intermittent porphyria are microsomal enzyme inducers.
They are metabolized by p450 enzyme. In response to these drugs the synthesis of cytochrome p450 proteins increases, leading to an enhanced consumption of heme, a component of p450 proteins.
This in turn causes a decrease in the concentration of heme in the liver.
Since the patient is deficient in enzymes synthesizing porphyrins it cannot makeup for the increase in demand of porphyrins.
This leads to increased accumulation of porphyrin precursors precipitating attacks of porphyria.
 
Some important drugs causing acute intermittent porphyria :-
  • Barbiturate
  • Chloramphenicol
  • Sulfonamides
  • Thiopental Na
  • Oral contraceptives
  • Phenylbutazone
  • Rifampicin
 
Ref: Desnick R.J., Balwani M. (2012). Chapter 358. The Porphyrias. In D.L. Longo, A.S. Fauci, D.L. Kasper, S.L. Hauser, J.L. Jameson, J. Loscalzo (Eds), Harrison’s Principles of Internal Medicine, 18e.

Q. 2

A 27-year-old woman goes to an emergency room with severe abdominal pain. She had previously experienced similar episodes of pain that lasted several hours to a few days, but this episode is the most severe. She has also been experiencing nausea, vomiting, and constipation. The physician is left with the impression that she is agitated and somewhat confused, and an accurate history is difficult to elucidate. The patient is sent for emergency laparotomy, but no pathology is noted at surgery. Following the unrevealing surgery, an older surgeon comments that he had once seen a similar case that was actually due to porphyria.The porphyrias are biochemical abnormalities in which of the following pathways?

 A

Glycogen degradation

 B

Heme synthesis

 C

Lipoprotein degradation

 D

Nucleotide degradation

Q. 2

A 27-year-old woman goes to an emergency room with severe abdominal pain. She had previously experienced similar episodes of pain that lasted several hours to a few days, but this episode is the most severe. She has also been experiencing nausea, vomiting, and constipation. The physician is left with the impression that she is agitated and somewhat confused, and an accurate history is difficult to elucidate. The patient is sent for emergency laparotomy, but no pathology is noted at surgery. Following the unrevealing surgery, an older surgeon comments that he had once seen a similar case that was actually due to porphyria.The porphyrias are biochemical abnormalities in which of the following pathways?

 A

Glycogen degradation

 B

Heme synthesis

 C

Lipoprotein degradation

 D

Nucleotide degradation

Ans. B

Explanation:

The porphyrias are a group of rare, related diseases that have in common a block in the heme synthesis pathway. The block is usually partial rather than complete, and thus many of these patients have only intermittent symptoms. Most cases of porphyria present with either a neurovisceral pattern (including both psychiatric symptoms and abdominal pain) or with photosensitive skin lesions. These two patterns are associated with different forms of porphyria.
Must know:

  • Associate abnormalities of glycogen degradation (with the glycogen storage diseases, such as Von Gierke disease, Pompe disease, and Forbes disease.
  • Associate abnormalities of lipoprotein degradation with some forms of hyperlipoproteinemia (notably Type I).
  • Associate abnormalities of nucleotide degradation with Gout and Lesch-Nyhan syndrome.
Ref: Murray R.K. (2011). Chapter 31. Porphyrins & Bile Pigments. In D.A. Bender, K.M. Botham, P.A. Weil, P.J. Kennelly, R.K. Murray, V.W. Rodwell (Eds), Harper’s Illustrated Biochemistry, 29e.

Q. 3 The enzyme deficient in erythropoietic porphyria is:

 A

PBG deaminase

 B

Uroporphyrin II cosynthetase

 C

Coprophyrin

 D

Ferrochelatase

Q. 3

The enzyme deficient in erythropoietic porphyria is:

 A

PBG deaminase

 B

Uroporphyrin II cosynthetase

 C

Coprophyrin

 D

Ferrochelatase

Ans. D

Explanation:

 

–  Erythropoietic protoporphyria is due to deficiency of enzyme ferrochelataseQ.

–  All porphyrias are autosomal dominant (AD) except congenital erythropoietic porphyria (autosomal recessive, AR), ALA dehydratase (porphobilinogen) deficiency (AR) and X-linked protoporphyria (X-linked).

Quiz In Between


Q. 4 Which of the following porphyrias is not inherited as an Autosomal Dominant disorder-

 A

Acute Intermittent Porphyria

 B

Congenital Erythropoietic Porphyria

 C

Porphyria Cutanea Tarda

 D

Hereditary Coproporphyria

Q. 4

Which of the following porphyrias is not inherited as an Autosomal Dominant disorder-

 A

Acute Intermittent Porphyria

 B

Congenital Erythropoietic Porphyria

 C

Porphyria Cutanea Tarda

 D

Hereditary Coproporphyria

Ans. B

Explanation:

Ans. is ‘b’ i.e., Congenital erythropoietic porphyria

Inheritance of Porphyrias

Autosomal dominant                                       Autosomal Recessive                                   X-linked

o Acute intermittent porphyria (AIP)            o ALA dehydratase deficiency                   o X-linked protoporphyria

o Porphyria cutanea Tarda (PCT)                 o Congenital erythropoietic porphyria

o Hereditary coproporphyria (HCP)             o Erythropoietic protoporphyria o Variegate porphyria (VP)


Q. 5 In porphyrias, which of the following enzyme defects does not lead to photosensitivity

 A

Uroporphyrinogen synthase

 B

Uroporphyrinogen decarboxylase

 C

Protoporphyrinogen oxidase

 D

Coproporphyrinogen oxidase

Q. 5

In porphyrias, which of the following enzyme defects does not lead to photosensitivity

 A

Uroporphyrinogen synthase

 B

Uroporphyrinogen decarboxylase

 C

Protoporphyrinogen oxidase

 D

Coproporphyrinogen oxidase

Ans. A

Explanation:

Ans.a. Uroporphyrinogen synthase
(Ref Harper 29/e p3l2-314, 27/e p277)
In porphyrias, uroporphyrinogen synthase enqlme defects lead to Acute Intemittent Porphyria, which shows purely
ne urological manifestation witho ut photosensitivity.


Q. 6

Skin is not involved in ‑

 A

Erythropoietic porphyria

 B

Porphyria cutanea torda

 C

Acute intermittent porphyria

 D

Hereditary coproporphyria

Q. 6

Skin is not involved in ‑

 A

Erythropoietic porphyria

 B

Porphyria cutanea torda

 C

Acute intermittent porphyria

 D

Hereditary coproporphyria

Ans. C

Explanation:

Ans. is ‘c’ i.e., Acute intermittent porphyria

Type of Porphyria  Neuropsychiatric symptoms Skin symptoms/Phototoxicity 

Hepatic porphyrias

Acute intermittent porphyri a

5-ALA dehydratase deficiency

Hereditary coproporphyria

Variegate porphyria

Porphyria cutanea tarda

 
+
+
+
+
_
 
_
 _
+
+
+
 
 Erythropoietic porphyrias

Erythropoietic protoporphyria

Congenital erythropoietic porphyria

X-linked sideroblastic anemia

 
 _
_
_
 
+
+
_

Quiz In Between



Xenobiotics

xenobiotics

Q. 1 In metabolism of xenobiotics all of the following reactions occur in phase one except?

 A

Oxidation

 B

Reduction

 C

Conjugation

 D

Hydrolysis

Q. 1

In metabolism of xenobiotics all of the following reactions occur in phase one except?

 A

Oxidation

 B

Reduction

 C

Conjugation

 D

Hydrolysis

Ans. C

Explanation:

Conjugation[Ref. K.D.T. 7″/e p. 24, 25, 26]

  • Drugs are not intrinsic cellular constituents. They are xenobiotics.
  • They are eliminated from the body mainly through kidney.
  • Drug metabolism means necessary changes in the drug molecule which are essential .for easy excretion from the body.
  • Drug metabolism usually converts lipophilic chemical compounds into more readily excreted polar / water soluble / ionizable compounds.
  • When the drug turns polar it becomes more soluble in the extracellular fluid and thus more filterable in the renal tubules. The drug also becomes highly ionizable so it is less reabsorbale and hence readily excretable from the kidney.
  • The metabolism occcurs more or less in every tissue but mainly in the liver.

Usually the drug metabolism occurs in two phases –

  • Phase I and
  • Phase II
  • Phase I and phase II reactions are biotransformations of chemicals that occur during drug metabolism.

Phase I reactions

  • Phase I reactions usually preceede phase II reactions though not necessarily. During these reactions, polar molecules are either added or unmasked, which results in more polar metabolites of the original chemicals. These metabolites are more water solube than the original compound.
  • Phase I reactions can lead to either activation or inactivation of the drug.

Phase I reactions are usually of the following types :

  • Oxidation
  • Reduction
  • Hydrolysis
  • Cyclization
  • Decyclization
  • If the metabolites of phase I reactions are sufficiently polar they may be readily excreted at this point.
  • However many phase I products are not eliminated rapidly and undergo a subsequent reaction in which an endogenous substrate combines with the newly incorporated functional group to form a highly polar conjugate (Phase H reaction).

Phase II reactions

  • Phase H reactions are usually known as “conjugation reactions”.
  • This involves conjugation with glucuronic acid, sulfonates and glutathione.

The phase II reactions are ?

– Methylation          

– Acetylation

– Glucuronidation

  • The products of conjugation have increased molecular weight and are usually inactive unlike phase I reactions which often produce active metabolites.

Q. 2 All of the following enzymes and their reactions are involved in the metabolism of Xenobiotics, EXCEPT:

 A

Cytochrome oxidase

 B

Cytochrome p 450

 C

Methylation

 D

Hydroxylation

Q. 2

All of the following enzymes and their reactions are involved in the metabolism of Xenobiotics, EXCEPT:

 A

Cytochrome oxidase

 B

Cytochrome p 450

 C

Methylation

 D

Hydroxylation

Ans. A

Explanation:

Cytochrome oxidase is not mentioned as an enzyme in the metabolism of xenobiotics but Cytochrome P450 and Hydroxylation reactions are involved in Phase-1 metabolic reactions and Methylation forms a part of phase II reactions.

 
Ref: Harpers Biochemistry, 27th Edition, Page 633; Lippincott’s Illustrated Reviews: Pharmacology, 4th Edition, Page 14; Goodman and Gilman’s The Pharmacological Basis of Therapeutics, 11th Edition, Page 44

Q. 3 In metabolism of xenobiotics all of the following reactions occur in phase one, EXCEPT:

 A

Oxidation

 B

Reduction

 C

Conjugation

 D

Hydrolysis

Q. 3

In metabolism of xenobiotics all of the following reactions occur in phase one, EXCEPT:

 A

Oxidation

 B

Reduction

 C

Conjugation

 D

Hydrolysis

Ans. C

Explanation:

Xenobiotic is a compound that is foreign to the body.
The principal classes of xenobiotics of medical relevance are drugs, chemical carcinogens, and various compounds that have found their way into our environment by one route or another, such as polychlorinated biphenyls (PCBs) and certain insecticides.

In phase 1, the major reaction involved is hydroxylation, catalyzed mainly by members of a class of enzymes referred to as monooxygenases or cytochrome P450s. Hydroxylation may terminate the action of a drug, though this is not always the case. In addition to hydroxylation, these enzymes catalyze a wide range of reactions, including those involving deamination, dehalogenation, desulfurization, epoxidation, pre oxygenation, and reduction. Reactions involving hydrolysis (eg, catalyzed by esterases) and certain other non-P450-catalyzed reactions also occur in phase 1.
 
In phase 2, the hydroxylated or other compounds produced in phase 1 are converted by specific enzymes to various polar metabolites by conjugation with glucuronic acid, sulfate, acetate, glutathione, or certain amino acids, or by methylation.
 
Conjugation reaction occurs in phase 2 and not in phase 1.
Ref: Murray R.K. (2011). Chapter 53. Metabolism of Xenobiotics. In D.A. Bender, K.M. Botham, P.A. Weil, P.J. Kennelly, R.K. Murray, V.W. Rodwell (Eds), Harper’s Illustrated Biochemistry, 29e

Q. 4 What is metabolised like xenobiotics

 A

Myoglobin

 B

Bilirubin

 C

Biliverdin

 D

Haemoglobin

Q. 4

What is metabolised like xenobiotics

 A

Myoglobin

 B

Bilirubin

 C

Biliverdin

 D

Haemoglobin

Ans. B

Explanation:

B i.e. Bilirubin


Q. 5

Bile salts undergo xenobiotics:

 A

After conjugation with taurine and glycine

 B

After conjugation with lysine

 C

After conjugation with betaglucuronic acid

 D

After conjugation with derived proteins

Q. 5

Bile salts undergo xenobiotics:

 A

After conjugation with taurine and glycine

 B

After conjugation with lysine

 C

After conjugation with betaglucuronic acid

 D

After conjugation with derived proteins

Ans. C

Explanation:

C i.e. Conjugation with betaglucuronic acid

– Phase I biotransformation reactions include Hydrolysis, Hydroxylation, Reduction, Oxidation, Oxidative Deamination, Dealkaylation and Epioxidation. Mn “High RODE”. Whereas phase II reactions include Glucronidation, Sulfation, Conjugation (with glucronic acid, glutathione, sulfate or aminoacids), Acetylation and Methylation. Mn “5G-SCAM”.

Cytochrome p450 monoxygenase is the main enzyme responsible for activation of xenobiotics (phase I hydroxylation reaction)Q.

Bilirubin is metabolized like xenobiotics; it is conjugated with beta glucronic acid in hepatocytes to convert a nonpolar hydrophobic bilirubin to a polar hydrophilic one which is readily excreted in bileQ.


Q. 6

In metabolism of xenobiotics all of the following reactions occur in phase one except?

 A

Oxidation

 B

Reduction

 C

Conjugation

 D

Hydrolysis

Q. 6

In metabolism of xenobiotics all of the following reactions occur in phase one except?

 A

Oxidation

 B

Reduction

 C

Conjugation

 D

Hydrolysis

Ans. C

Explanation:

Ans. is ‘c’ i.e., Conjugation


Q. 7

Xenobiotics are metabolized to ‑

 A

Increase water solubility

 B

Increase lipid solubility

 C

Make them nonpolar

 D

None of the above

Q. 7

Xenobiotics are metabolized to ‑

 A

Increase water solubility

 B

Increase lipid solubility

 C

Make them nonpolar

 D

None of the above

Ans. A

Explanation:

Ans. is ‘a’ i.e., Increase water solubility

BIOTRANSFORMATION (METABOLISM)

  • Most of the drugs are treated by the body as foreign substances (xenobiotics).
  • Like other foreign substances (xenobiotics), body tries to eliminate drugs by various mechanisms for ridding itself of chemical intruders.
  • Biotransformation means chemical alteration of the drug in the body.

Why drug transformation is necessary ?

  • Kidney plays a pivotal role in terminating the activity of drugs.
  • For renal excretion the drug tends to be polar (lipid insoluble/water soluble) so that it can not diffuse back from tubular lumen and can be excreted.
  • But pharmacologically active organic molecules (drugs) tend to be lipophlic (nonpolar) and remains unionized or only partially ionized at physiological pH.
  • Biotransformation is needed to render nonpolar (lipid soluble) compounds polar (water soluble) so that they are not reabsorbed in the renal tubules and are excreted.

Sites and processes of biotransformation

  • Primary site of drug metabolism is liver, others are – kidney, intestine, lung and plasma.

Biotransformation of drugs may lead to :-

LInactivation

  • Most drugs and their active metabolites are rendered inactive.

Active metabolite from an active drug

  • Many drugs are partially converted to one or more active metabolites.
  • The effects observed are the sumtotal of that due to the parent drug and its active metabolite.

Activation of inactive drugs

  • Few drugs are inactive as such and need conversion in the body to one or more active metabolites.
  • Such a drug is called prodrug.


RNA editing

RNA editing

Q. 1 Apoprotein A is found in :

 A

Chylomicrons

 B

VLDL

 C

HDL

 D

LDL

Q. 1

Apoprotein A is found in :

 A

Chylomicrons

 B

VLDL

 C

HDL

 D

LDL

Ans. C

Explanation:

C i.e. HDL


Q. 2

Apo B48 and apo 13,00 are expressed as two different apoproteins because of difference in :

 A

RNA editing

 B

RNA splicing

 C

Chromosomal loci

 D

Apo-B gene

Q. 2

Apo B48 and apo 13,00 are expressed as two different apoproteins because of difference in :

 A

RNA editing

 B

RNA splicing

 C

Chromosomal loci

 D

Apo-B gene

Ans. A

Explanation:

A i.e. RNA editing


Q. 3

APO B48 & APO B100 is synthesized from the same rnRNA; the difference between them is due to:

 A

RNA splicing

 B

Allelic exclusion

 C

 Deamination of cytidine to uridine

 D

Upstream repression

Q. 3

APO B48 & APO B100 is synthesized from the same rnRNA; the difference between them is due to:

 A

RNA splicing

 B

Allelic exclusion

 C

 Deamination of cytidine to uridine

 D

Upstream repression

Ans. C

Explanation:

C i.e. Deamination of cytidine to uridine IRef: Vasudevan

  • Least post translational modification occurs in prokaryotic mRNA, which is generally identical to its primary transcript. Post translational modification of t-RNA includes removal of introns from anticodon loop, trimming of 5′ & 3′ ends, methylation / reduction / deamination / alkylation / rearranging glycolsidic bond to produced modified bases like methylated bases, dihydrouracil (D) & pseudo uracil (W) bases in nucleus, whereas cleavage and attachment of CCA tailing occur in cytoplasmQ.
  • In RNA, gene during processing undergoes nucleoside modifications, nucleoside cleavage and terminal addition but not chemical hydrolysisQ.
  • Post translational modification of mRNA involves 5′ capping , 3′ polyadenylation (addition of poly ‘A’ tail at 3′ end), splicing (removal of non coding intervening or intron sequences and ligation / joining of coding exons) by Sn RNA/ Sn RNPs / Snurps or self splicing d/t ribozyme activity of self splicing introns with formation of lariat intermediates, RNA editing and secondary methylationQ.
  • Apo B-48 and Apo B-100 are synthesized from same Apo B gene and same ApoB- m-RNA. Apo B 100 is a 100 kDa protein synthesized in liver by full length translation of corresponding mRNA of Apo B gene. Apo – B-100 forms part of LDL, IDL and VLDL. Apo B-48 is a 48 KDa protein (48% shorter form of Apo B-100) synthesized in intestine by partial translation of same mRNA of Apo B gene. Apo B 48 forms part of chylomicron & chylomicron remnant. This difference between the sizes of Apo B100 and Apo B48 occurs because post transcriptional processing (editing) of Apo B mRNA , deaminates the cytidine (C) to uracil (U) in intestine at 2153 position. After cyti dine deamination the CAA codon (which codes glutamine in liver) becomes UAA (nonsense or stop codon) in intestine. This results in shorter apo B-48 protein being made in intestine (and incorporated into chylomicron) than is made in the liver full length Apo B-100, incorporated in to VLDL.

Quiz In Between



Transport of Bilirubin

Transport of Bilirubin

Q. 1 Bilirubin is absent in urine because it is :

 A

Distributed in the body fat

 B

Conjugated with glucoronide

 C

Not filterable

 D

Lipophilic.

Q. 1

Bilirubin is absent in urine because it is :

 A

Distributed in the body fat

 B

Conjugated with glucoronide

 C

Not filterable

 D

Lipophilic.

Ans. D

Explanation:

Unconjugated serum bilirubin is always bound to albumin, is not filtered by the kidney (not found in the urine)
In biliary obstruction or hepatocellular diseases, both conjugated and unconjugated bilirubin accumulate in plasma.
In hemolytic jaundice, total plasma bilirubin increases, but the proportion of the unconjugated and conjugated fractions remains unchanged.


Q. 2

Bilirubin is secreted by:

 A

Bile Salts

 B

Bile pigments

 C

Secretin

 D

CCK.

Q. 2

Bilirubin is secreted by:

 A

Bile Salts

 B

Bile pigments

 C

Secretin

 D

CCK.

Ans. A

Explanation:

A i.e. Bile salts

Substances that increase the secretion of bile are called as cholerecticsQ. Bile salts are amongst the most important physiological cholerectionQ


Q. 3 Bilirubin is the degradation product of –

 A

Albumin

 B

Globulin

 C

Heme

 D

Transferrin

Q. 3

Bilirubin is the degradation product of –

 A

Albumin

 B

Globulin

 C

Heme

 D

Transferrin

Ans. C

Explanation:

Ans. is ‘c’ i.e., Heme

Bilirubin metabolism

o Bilirubin is the end product of heme degradation.

o The heme is derived from –

(i)       Senescent erythrocytes by mononuclear phagocytic system in the spleen, liver and bone marrow (major source).

(ii)     Turnover of hemoproteins (e.g. cytochrome p.450).

o Heme is oxidized to biliverdin by heme oxygenase.

o Biliverdin is then reduced to bilirubin by biliverdin reductase.

o Bilirubin is transported to liver in bound form with albumin.

o There is carrier mediated uptake of bilirubin in the liver.

o This bilirubin is conjugated with glucuronic acid by UDP glucuronyl transferase (UGT1A1) to from conjugated bilirubin (bilirubin glucronides).

o Conjugated bilirubin is excreted into bile.

o Most of the conjugated bilirubin is deconjugated and degraded to urobilinogen.

o The most of the urobilinogen is excreted in the feces.

o Approximately 20% of the urobilinogen is reabsorbed in the ileum and colon and is returned to the liver, and promptly rexcreted into bile —> Enterohepatic circulation.

o The small amount that escapes this enterohepatic circulation is excreted in urine.

Quiz In Between


Q. 4 Bilirubin is absent in urine because it is‑

 A

Distributed in the body fat

 B

Conjugated with glucoron

 C

Not filtered

 D

None

Q. 4

Bilirubin is absent in urine because it is‑

 A

Distributed in the body fat

 B

Conjugated with glucoron

 C

Not filtered

 D

None

Ans. C

Explanation:

Answer is C (Not filtered)

Normal urine does not contain bilirubin because normal blood contains bilirubin in the uncongugated form. Unconjugated bilirubin is lipid soluble or lipophilic (water insoluble) because it is transported in the blood as a complex with albumin (albumin-bilirubin complex) which is not allowed to filter through the glomerulus.

Although conjugated bilirubin is water soluble and filterable at the glomurulus, conjugated bilirubin is not present in the blood normally and hence does not filter to appear in the urine.

Unconjugated bilirubin present in blood is complexed with albumin to make it soluble in blood and transport it to the liver. However, the glomerulus does not allow the albumin bilirubin complex to filter and hence bilirubin does not appear in urine.

Conjugated bilirubin is formed in the liver and directly excreted into the GIT through bile where it is reduced to urobilinogen and stercobilinogen. Conjugated bilirubin does not normally circulate in the blood at all and hence despite being filterable and water soluble it does not appear in urine.


Q. 5

Urobilinogen is formed in the:   
September 2006

 A

Liver

 B

Kidney

 C

Intestine

 D

Spleen

Q. 5

Urobilinogen is formed in the:   
September 2006

 A

Liver

 B

Kidney

 C

Intestine

 D

Spleen

Ans. C

Explanation:

Ans. C: Intestine

Urobilinogen is a colourless product of bilirubin reduction. It is formed in the intestines by bacterial action. Some urobilinogen is reabsorbed, taken up into the circulation and excreted by the kidney. This constitutes the normal “enterohepatic urobilinogen cycle”.

Urobilinogen content is determined by a reaction with Ehrlich’s reagent, which contains para-Dimethyl amino benzaldehyde and may be measured in Ehrlich units

Quiz In Between


Q. 6 Bilirubin bound inside hepatocyte to ‑

 A

Albumin

 B

Ubiquinone

 C

Ligandin

 D

Globulin

Q. 6

Bilirubin bound inside hepatocyte to ‑

 A

Albumin

 B

Ubiquinone

 C

Ligandin

 D

Globulin

Ans. C

Explanation:

Ans. is ‘c’ i.e., Ligandin

Bilirubin metabolism

Bilirubin is the end product of heme degradation.

The heme is derived from –

i) Senescent erythrocytes by mononuclear phagocytic system in the spleen, liver and bone marrow (major source).

ii) Turnover of hemoproteins (e.g. cytochrome p.450).

Heme is oxidized to biliverdin by heme oxygenase.

Biliverdin is then reduced to bilirubin by biliverdin reductase.

Bilirubin is transported to liver in bound form with albumin.

Bilirubin is transferred to hepatocytes where it is bound to ligandin.

There is carrier mediated uptake of bilirubin in the liver.

This bilirubin is conjugated with glucuronic acid by UDP glucuronyl transferase (UGT1A1) to from conjugated bilirubin (bilirubin glucronides).

Conjugated bilirubin is excreted into bile.

Most of the conjugated bilirubin is deconjugated and degraded to urobilinogen.

The most of the urobilinogen is excreted in the feces.


Q. 7 All are involved in bilirubin metabolism except‑

 A

ALA synthase

 B

Heme oxygenase

 C

Biliverdine reductase

 D

Glucuronyl transferase

Q. 7

All are involved in bilirubin metabolism except‑

 A

ALA synthase

 B

Heme oxygenase

 C

Biliverdine reductase

 D

Glucuronyl transferase

Ans. A

Explanation:

Ans. is ‘a’ i.e., ALA synthase 

Quiz In Between



Hyperbilirubinemias

hyperbilirubinemia

Q. 1

Following are causes of unconjugated hyperbilirubinemia, except:
 A Hemolytic anemia

 B

Large hematoma

 C

Rotor syndrome

 D

Megaloblastic anemia

Q. 1

Following are causes of unconjugated hyperbilirubinemia, except:

 A Hemolytic anemia

 B

Large hematoma

 C

Rotor syndrome

 D

Megaloblastic anemia

Ans. C

Explanation:

Answer is C (Rotor syndrome)
Rotor’s syndrome is an Autosomal recessive inherited disorder characterized by a deject in biliary excretion leading to conjugated hyperbilirubinemia:

Indirect hyperbilirubinemia                                                              

Direct hyperbilirubinemia

A.   Hemolytic disorders

A.   Inherited conditions

1.    Inherited

1.     Dubin-Johnson syndrome

a.   Sperocyteosis, elliptocytosis

2.     Rotor’s syndrome

Glucose-6-phosphate dehydrogenase and pyruvate kinase deficiencies

b. Sickle cell anemia

 

2.    Acquired                         _

a. Microangiopathic hemolytic anemias

b. Paraoxysmal nocturnal hemoglobinuria

c. Immune hemolysis

 

B.    Ineffective erythropoesis

 

1.    Cobalamin, folate, thalassemia, and severe iron deficiencies

 

C. Drugs

 

1.    Rifampicin, probenbecid, ribavirin

 

D.   Inherited conditions

 

1.    Crigler-Najjar types I and II

 

2.    Glibert’s syndrome

 


Q. 2 Conjugated hyperbilirubinemia is seen in all EX­CEPT:
March 2013

 A

Dubin Johnson syndrome

 B

Rotor syndrome

 C

Gilbert syndrome

 D

None of the above

Q. 2

Conjugated hyperbilirubinemia is seen in all EX­CEPT:
March 2013

 A

Dubin Johnson syndrome

 B

Rotor syndrome

 C

Gilbert syndrome

 D

None of the above

Ans. C

Explanation:

Ans. C i.e. Gilbert syndrome
Gilbert syndrome presents with unconjugated hyperbilirubinemia


Q. 3

Unconjugated hyperbilirubinemia is seen in all of the following except:   
March 2010

 A

Crigler Najjar Syndrome

 B

Physiological jaundice

 C

Dubin-Johnson syndrome

 D

Gilbert syndrome

Q. 3

Unconjugated hyperbilirubinemia is seen in all of the following except:   
March 2010

 A

Crigler Najjar Syndrome

 B

Physiological jaundice

 C

Dubin-Johnson syndrome

 D

Gilbert syndrome

Ans. C

Explanation:

Ans. C: Dubin-Johnson Syndrome

Dubin-Johnson syndrome is an autosomal recessive disorder that causes an increase of conjugated bilirubin without elevation of liver enzymes (ALT, AST).

This condition is associated with a defect in the ability of hepatocytes to secrete conjugated bilirubin into the bile.

The conjugated hyperbilirubinemia is a result of defective endogenous and exogenous transfer of anionic conjugates from hepatocytes into the bile.

Pigment deposition in lysosomes causes the liver to turn black.

Other causes of conjugated/direct hyperbilirubinemia:

  • Hepatocellular diseases:

– Hepatitis:

  • Neonatal idiopathic hepatitis
  • Viral (Hepatitis B, C, TORCH infections)
  • Bacterial (E. colt, urinary tract infections)

–        Total parenteral nutrition

–        Hepatic ischemia (post-ischemic damage)

–        Erythroblastosis fetalis (late, “Inspissated Bile Syndrome”)

Metabolic disorders:

  • Alpha-1 antitrypsin deficiency
  • Galactosemia, tyrosinemia, fructosemia
  • Glycogen storage disorders
  • Cystic fibrosis

Biliary tree abnormalities:

–         Extrahepatic biliary atresia: In first 2 weeks, unconjugated bilirubin predominates; elevated conjugated bilirubin is late.

–        Paucity of bile ducts

–        Choledochal cyst

–        Bile plug syndrome

Causes of unconjugated/indirect hyperbilirubinemia:

  • Increased lysis of RBCs (i.e., increased hemoglobin release)

–        Isoimmunization (blood group incompatibility: Rh, ABO and minor blood groups)

–        RBC enzyme defects (e.g., G6PD deficiency, pyruvate kinase deficiency)

–        RBC structural abnormalities (hereditary spherocytosis, elliptocytosis)

–        Infection (sepsis, urinary tract infections)

–        Sequestered blood (e.g., cephalohematoma, bruising, intracranial hemorrhage)

–        Neonatal Jaundice

–        Polycythemia

–        Shortened life span of fetal RBCs

Decreased hepatic uptake and conjugation of bilirubin

–        Immature glucuronyl transferase activity in all newborns: term infants have 1% of adult activity, preterm infants have 0.1%.

–        Gilbert Syndrome

–        Crigler Najjar Syndrome (Non-hemolytic Unconjugated Hyperbilirubinemia): inherited conjugation defect (very rare)

–        Breastmilk Jaundice (pregnanediol inhibits glucuronyl transferase activity)

Increased enterohepatic reabsorption

–        Breastfeeding jaundice (due to dehydration from inadequate milk supply)

Quiz In Between


Q. 4 Conjugated hyperbilirubinemia

 A

Dubin johnson syndrome

 B

Criggler naj jar syndrome

 C

Breast milk jandice

 D

Gilbert syndrome

Q. 4

Conjugated hyperbilirubinemia

 A

Dubin johnson syndrome

 B

Criggler naj jar syndrome

 C

Breast milk jandice

 D

Gilbert syndrome

Ans. A

Explanation:

Ans. is ‘a’ i.e., Dubin johnson syndrome
Breast milk jaundice –

  • Decrease bilirubin uptake across hepathocyte membrane.
  • Entero-hepatic recirculation.
  • Leads to indirect hyperbilirubinemia.

Crigler naj jar & Gilbert syndrome (deficiency of glucuronyl transferase)

  • Decrease conjugation leads to Indirect hyperbilirubinemia.
  • Defect in hepatocyte secretion of conjugated bilirubin.
  • Leads to direct hyperbilirubinemia

Q. 5 Causes of unconjugated hyperbilirubinemia include?

 A

Sepsis

 B

Criggler-Najar syndrome

 C

Rotor syndrome

 D

Gilbert syndrome

Q. 5

Causes of unconjugated hyperbilirubinemia include?

 A

Sepsis

 B

Criggler-Najar syndrome

 C

Rotor syndrome

 D

Gilbert syndrome

Ans. A:B:D:E

Explanation:

Answer- A, B, D, E, Sepsis, Criggler-Najar syndrome, Gilbert syndrome, Intravascular hemolysis
Unconjugated hyperbilirubinemia:-

  • Increased production of bilirubin from hemoglobin, So that the capacity of liver to conjugate bilirubin is overwhelmed by increased production, e.g.
  1. Hemolytic anemia (both intravascular and extamascular)s Hereditary sphnocytosis, G6PD defciency.
  2. Inefrective erythropoiesis- Thalassemia, Pernicious anemia.
  3. Reduced hepatic uptake of bilirubin from bilirubin – albumin complex > Drugs,
  4. Infections:- Sepsis, UTI
  5. Impaired hepatic conjugation.

Q. 6 Congenital hyperbilirubinemia is/are seen in: 

 A

Prematurity

 B

Hypoalbuminaemic state

 C

Hepatitis

 D

Sepsis

Q. 6

Congenital hyperbilirubinemia is/are seen in: 

 A

Prematurity

 B

Hypoalbuminaemic state

 C

Hepatitis

 D

Sepsis

Ans. A:C:D:E

Explanation:

Ans. a. Prematurity; c. Hepatitis; d. Sepsis; e. Polycythemia
Albumin less than 3.0 mg/dl is risk for hyperbilirubinemia neurotoxicity
Two other groups of disorders are associated with hyperbilirubinemia:
(1) Unconjugated hyperbilirubinemia seen in,

  • Breast milk jaundice
  • Blood group incompatibility
  • Lucey-Driscoll syndrome
  • Congenital hypothyroidism
  • Upper intestinal obstruction
  • Gilbert disease
  • Crigler-Najjar syndrome
  • Hereditary spherocytosis
  • Non-spherocytic hemolytic anemia
  • Drug-induced hyperbilirubinemia

(2) Conjugated hyperbilirubinemia present in,

  • Dubin-Johnson syndrome
  • Rotor syndrome
  • Biliary atresia
  • Neonatal hepatitis

Quiz In Between



Polymerase Chain Reaction (Pcr)

PCR

Q. 1 PCR is done for:       
UP 11; UPSC 13; AIIMS 13

 A

Cloning of DNA in cells

 B

Replication of DNA in vitro

 C

Sequencing of DNA

 D

Both A and B

Q. 1

PCR is done for:       
UP 11; UPSC 13; AIIMS 13

 A

Cloning of DNA in cells

 B

Replication of DNA in vitro

 C

Sequencing of DNA

 D

Both A and B

Ans. B

Explanation:

Ans. Replication of DNA in vitro


Q. 2

PCR does not require:
AIIMS 07

 A

Primer

 B

DNA-fragments

 C

DNA polymerase

 D

Radio-labeled DNA probe

Q. 2

PCR does not require:
AIIMS 07

 A

Primer

 B

DNA-fragments

 C

DNA polymerase

 D

Radio-labeled DNA probe

Ans. D

Explanation:

Ans. Radio-labeled DNA probe


Q. 3

Not a component of PCR ‑

 A

Primer

 B

Taq polymerase

 C

DNA Polymerase

 D

Restriction enzyme

Q. 3

Not a component of PCR ‑

 A

Primer

 B

Taq polymerase

 C

DNA Polymerase

 D

Restriction enzyme

Ans. D

Explanation:

 

Steps in PCR

PCR uses DNA polymerase to repetitively amlify targeted portion of DNA. Each cycle doubles the amout of DNA in the sample, leading to exponential increase with repeated cycles of amplification. Thus amplification after ‘n’ number of cycle in (2)”. Twenty cycles provide an amplification of 106 (million) and 30 cycles of 109 (billion).

PCR occurs in following steps –

i)      Isolation of target DNA sequence :- 

ii)     Primers construction:- 

iii)    Denaturation of DNA :- 

iv)    Annealing of primers to single stranded DNA :- 

v)     Chain extension:- 

Thus following are required in PCR :- Target double stranded DNA, two specific primers, a thermostable DNA polymerase (Taq polymerase), deoxyribonucleotides (dNTP).

Quiz In Between


Q. 4 In PCR, DNA polymerase is derived from‑

 A

Experimental E coli

 B

Thermus aquaticus

 C

Retroviruses

 D

Bacteriophages

Q. 4

In PCR, DNA polymerase is derived from‑

 A

Experimental E coli

 B

Thermus aquaticus

 C

Retroviruses

 D

Bacteriophages

Ans. B

Explanation:

Ans. is ‘b’ i.e., Thermus acquaticus

PCR is a method of enzymatic amplification of a target sequence of DNA.

It is sensitive, selective (specific) and extremely rapid means of amplifying any desired sequence of double stranded DNA, which can be as short as 50-100 base pairs (bp) and as long as 10 kbp.

In PCR, the DNA to be amplified is replicated by DNA polymerase of Thermus aquaticus (Taq).

Taq polymerase is used because it is thermostable, not denatured at a temperature upto 95°C (in PCR DNA is to be heated to 94°-95° C for separation of strands).


Q. 5

Enzyme used in PCR is ‑

 A Reverse transcriptase 

 B

Tag polymerase

 C

RNA polymerase

 D

None

Q. 5

Enzyme used in PCR is ‑

 A

Reverse transcriptase 

 B

Tag polymerase

 C

RNA polymerase

 D

None

Ans. B

Explanation:

Ans. is ‘b i.e., Taq polymerase 

PCR is a method of enzymatic amplification of a target sequence of DNAe.

  • It is sensitive, selective (specific) and extremely rapid means of amplifying any desired sequence of double stranded DNAe, which can be as short as 50-100 base pairs (bp) and as long as 10 kbp.
  • In PCR, the DNA to be amplified is replicated by DNA polymerase of Thermus aquaticus (Taq). Taq polymerase is use because it is thermostable0, not denatured at a temperature upto 95°C (in PCR DNA is to be heated to 94°-95° C for separation of strands).
  • For amplifying a desired DNA sequence in DNA, we have to know short flanking sequences on either side of the target segue nce so that complementary primers can be prepared.
  • Primers0 are the synthetic oligonucleotides of 20-35 sequence, which have sequence complementary to flanking sequence, i.e. sequence of flanking region of target DNA sequence.
  • Primers are amplified to produce desired sequence of DNA.

Q. 6 Real Time PCR is used for:

 A

Multiplication of RNA

 B

Multiplication of specific segments of DNA

 C

Multiplication of Proteins

 D

To know how much amplification of DNA has occurred

Q. 6

Real Time PCR is used for:

 A

Multiplication of RNA

 B

Multiplication of specific segments of DNA

 C

Multiplication of Proteins

 D

To know how much amplification of DNA has occurred

Ans. D

Explanation:

Ans. (d) To know how much amplification of DNA has occured PF 1,m,etz. 25/e 714, 27/e, p 124; Greenwood 18/e 79

Real Time. PCR

  • It is the molecular detection technique that discriminates real time amplification from conventional PCR assays.
    • The real-time polymerase chain reaction (PCR) uses fluorescent reporter molecules to monitor the production of amplification products during each cycle of the PCR reaction.
    • This combines the DNA amplification and detection steps into one homogeneous assay and obviates the requirement for gel electrophoresis to detect amplification products.
    • Its simplicity, specificity, and sensitivity, together with its, more reliable instrumentation, and improved protocols, has made realtime                     oenLhmark technology for lite ocLuLtion of DNA.
    • Real time PCR is extremely useful in medical microbiology, with greatest impact on virology.

Quiz In Between


Q. 7 Which is not a step of PCR ‑

 A

Annealing

 B

Extension

 C

Transformation

 D

Denaturation

Q. 7

Which is not a step of PCR ‑

 A

Annealing

 B

Extension

 C

Transformation

 D

Denaturation

Ans. C

Explanation:

Ans. is ‘c’ i.e., Transformation [Ref Lippincott’ s 5thle p. 479-83; Harper 28th/e p. 395] Steps in PCR

  • Isolation of target DNA sequence→  Primer construction → Denaturation of DNA→ Annealing of primers to single stranded DNA→ Chain extension.

Q. 8 Which of the following enzymes have proof reading function in PCR [Polymerase Chain Reaction] 

 A

Taq polymerase

 B

PFU Polymerase

 C

Thermos thermophilus

 D

Thermal flavus (Replinase)

Q. 8

Which of the following enzymes have proof reading function in PCR [Polymerase Chain Reaction] 

 A

Taq polymerase

 B

PFU Polymerase

 C

Thermos thermophilus

 D

Thermal flavus (Replinase)

Ans. B:E

Explanation:

Ans. is ‘b’ i.e., PFU Polymerase; & ‘e’ i.e., T-7 polymerase [Ref Textbook of PCR by Mike McPherson]

  • The use of high fidelity DNA polymerases in PCR is essential for reducing the introduction of amplification errors in PCR products.
  • Several thermostable DNA polymerases with 3′ → 5′ exonuclease – dependent proofreading activity have been introduced for high.
  • Pfu DNA polymerase → Derived from Pyrococcus fusarious.
  • Pwo DNA polymerase → Isolated from Pyrococcus woesei.
  • KOD HiFi DNA polymerase → Isolated from Thermococcus Kodakaraensis.
  • T7 DNA polymerase.

Q. 9

Which of the following is/are true about PCR except:

 A

Uses heat labile DNA polymerase

 B Uses heat stable DNA polymerase

 C

Is technique for DNA amplification

 D

Used to yield multiple copies of DNA

Q. 9

Which of the following is/are true about PCR except:

 A

Uses heat labile DNA polymerase

 B

Uses heat stable DNA polymerase

 C

Is technique for DNA amplification

 D

Used to yield multiple copies of DNA

Ans. A

Explanation:

Ans: a. Uses heat labile DNA polymerase,[Ref Harper 30th/458-59; Lippincott 6th/479-83, 5th/497-83; Chatterjea er Shinde7th/267-272]

  • SpecificityQ is based on the use of two oligonucleotide primers that hybridize to complementary sequence on opposite strands of DNA & flank the target sequence Double stranded DNA can be disrupted by heat or high pH, giving rise to single stranded DNA. The single stranded DNA serves as a template for synthesis of a complementary strand by replicating enzymes, DNA polymerase.
  • Early PCR reaction used an E. coli DNA polymerase that was destroyed by each heat denaturation cycle. Substitution of a heat-stable DNA polymerase (Taq polymeraseY from Thermus aquaticus, obviates this problem & has made possible automation of the reaction, since the polymerase reactions can be run at 70°C

Quiz In Between



Hepatorenal Syndrome

Hepatorenal syndrome

Q. 1 Albumin treatment along with antibiotic in the setting of SBP(spontaneous bacterial peritonitisis indicated to prevent the development of hepatorenal syndrome is in all , EXCEPT:

 A

Serum creatine is > 1 mg/dl

 B

BUN > 30mg/dl

 C

Total bilirubin is > 4 mg/dl

 D

INR > 2

Q. 1

Albumin treatment along with antibiotic in the setting of SBP(spontaneous bacterial peritonitisis indicated to prevent the development of hepatorenal syndrome is in all , EXCEPT:

 A

Serum creatine is > 1 mg/dl

 B

BUN > 30mg/dl

 C

Total bilirubin is > 4 mg/dl

 D

INR > 2

Ans. D

Explanation:

In patients with SBP along  with cefotaxime albumin infusion is indicated in the setting , when

1.Serum creatine is  > 1 mg/dl
2. BUN >   30 mg/dl
3. Total bilirubin is > 4 mg/dl
 
Dose o f albumin: 1. g/Kg within 6 hours of antibiotic treatment and 1 g/kg  on day 3.
A decrease in mortality from 30%to 10 % is noted.
Ref: AASLD practice  guidelines:Hepatology, Vol.49 ,No.6 ,2009.

Q. 2 Features of Hepatorenal syndrome are

 A

Urine sodium < 10 meq/1

 B

Normal renal histology

 C

Renal function abnormal even after liver become normal

 D

a and b

Q. 2

Features of Hepatorenal syndrome are

 A

Urine sodium < 10 meq/1

 B

Normal renal histology

 C

Renal function abnormal even after liver become normal

 D

a and b

Ans. D

Explanation:

Answer is A & B (urine Na < 10 meq/l and Normal Renal Histology)

Hepatorenal syndrome is associated with normal renal histology and supported by a urine sodium excretion l0meq/L

Hepatorenal syndrome

  • Hepatorenal syndrome is defined as a state of functional renal failure (Reduced GFR) in patients with severe liver disease
  • Structurally /Histologically the kidneys are normal and recover function after successful liver transplantation.
  • The pathogenetic hallmark of hepatorenal syndrome is intense renal vasoconstriction with coexistent systemic vasodilatation
  • The diagnosis of hepatorenal syndrome is considered in accordance with the following diagnostic criteria.

Diagnostic of Hepatorenal Syndrome

Major criteria

  • Low glomerular filtration rate. as indicated by serum creatinine > 1.5 mg/dL or 24-hr creatinine clearance < 40 mL/min
  • Absence of shock, ongoing bacterial infection, fluid losses, and current treatment with nephrotoxic drugs
  • No sustained improvement in renal function (decrease in serum creatinine to 1.5 nig/dL or increase in creatinine clearance to 40 mL/min) after diuretic withdrawal and expansion of plasma volume with 1.5L of a plasma expander
  • Proteinuria mg/d1, and no uhrasonographic evidence of obstructive uropathy or parenchymal renal disease Additional criteria
  • Urine volume < 500 mL/d
  • Urine sodium < 10 meq/L
  • Urine osmolality greater than plasma osmolality
  • Urine red blood cells <50/high- power. field
  • Serum sodium concentration < 130 niEqL

Note: All major criteria must be present for the diagnosis of hepatorenal syndrome.

Additional criteria are not necessary for the diagnosis but provide supportive evidence.


Q. 3 Which of the following statements is incorrect with regard to Hepatorenal syndrome in a patient with cirrhosis

 A

Createnine clearance < 40 ml/min

 B

Urinary sodium < 10mq/L

 C

Urine osmolality lower than plasma osmolality

 D

No sustained improvement in renal function after volume expansion.

Q. 3

Which of the following statements is incorrect with regard to Hepatorenal syndrome in a patient with cirrhosis

 A

Createnine clearance < 40 ml/min

 B

Urinary sodium < 10mq/L

 C

Urine osmolality lower than plasma osmolality

 D

No sustained improvement in renal function after volume expansion.

Ans. C

Explanation:

Answer is C (Urine osmolality is lower than plasma osmolality):

Hepatorenal syndrome is associated with urine osmolality greater than plasma osmolality (and not lower than plasma osmolality).

creatinine clearance < 40 ml/minute and poor response to volume expansion are major diagnostic features of hepatorenal syndrome while urinary sodium of less than 10 mmol/L is an additional criteria that provides supportive evidence.

Quiz In Between


Q. 4

Hepatorenal syndrome is characterized by all of the following except:     
March 2005

 A

Reduction in creatinine clearance

 B

Managed effectively by renal vasodilating agents.

 C

Proteinuria less than 500 mg/ d

 D

Normal intrinsic kidney

Q. 4

Hepatorenal syndrome is characterized by all of the following except:     
March 2005

 A

Reduction in creatinine clearance

 B

Managed effectively by renal vasodilating agents.

 C

Proteinuria less than 500 mg/ d

 D

Normal intrinsic kidney

Ans. B

Explanation:

Ans. B: Managed effectively by renal vasodilating agents.

HRS is defined as worsening azotemia with avid sodium retention and oliguria in the absence of identifiable specific cause of renal dysfucntion in setting of acute or advanced chronic liver disease

No specific tests establish the diagnosis of HRS.

Diagnosis of HRS is based on the presence of a reduced GFR in the absence of other causes of renal failure in patients with chronic liver disease. The following criteria help to diagnose HRS:

Major criteria: All major criteria are required to diagnose HRS.

  • Low GFR, indicated by a serum creatinine level higher than 1.5 mg/ dL or 24-hour creatinine clearance lower than 40 mL/ min
  • Absence of shock, ongoing bacterial infection and fluid losses, and current treatment with nephrotoxic medications
  • No sustained improvement in renal function (decrease in serum creatinine to40 mL/ min) after diuretic withdrawal and expansion of plasma volume with 1.5 L of plasma expander
  • Proteinuria less than 500 mg/ d and no ultrasonographic evidence of obstructive uropathy or intrinsic parenchymal disease

Additional criteria: Additional criteria are not necessary for the diagnosis but provide supportive evidence.

  • Urine volume less than 500 mL/d
  • Urine sodium level less than 10 mEq/ L
  • Urine osmolality greater than plasma osmolality
  • Urine red blood cell count of less than 50 per high-power field
  • Serum sodium concentration less than 130 mEq/ L

The best therapy for HRS is liver transplantation.


Q. 5 All are true about hepatorenal syndrome except:

 A

Creatinine level raised

 B

Albumin infusion given

 C

Liver transplantation improves renal functions

 D

May occur in cirrhosis

Q. 5

All are true about hepatorenal syndrome except:

 A

Creatinine level raised

 B

Albumin infusion given

 C

Liver transplantation improves renal functions

 D

May occur in cirrhosis

Ans. E

Explanation:

Answer- E. Low dose dopamine infusion is very effective

  • The hepatorenal syndrome (HRS) is a form of functional renal failure without renal pathology that occurs in about 10% of patients with advanced cirrhosis or acute liver failure.
  • There are marked disturbances in thc arterial renal circulation in Patients with HRS.
  • TyPe I HRS- a significant reduction in creatinine clearance within 1-2 weeks of presentation.
  • Type 2 HRS- an elevation of serum creatinine level.
  • HRS is often seen in patients with refractory ascites.

Treatment-

  • dopamine or prostaglandin analogues were used as renal vasodilating medications.
  • Patients are treated with midodrine, an alpha-agonist, along with octreotide and intravenous albumin.
  • The best theragy for HRS is liver transplantation.

Quiz In Between



Interstitial Cystitis

Interstitial Cystitis

Q. 1 Cells involved in interstitial cystitis?

 A

Lymphocytes

 B

Neutrophils

 C

Macrophages

 D

Mast cells

Q. 1

Cells involved in interstitial cystitis?

 A

Lymphocytes

 B

Neutrophils

 C

Macrophages

 D

Mast cells

Ans. D

Explanation:

Ans. is ‘d’ i.e., Mast cells

  • Mast cells are often seen in the mucosa, lamina propria, and muscularis propria in interstitial cystitis.

Quiz In Between



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