What is Glycosidic Bond in DNA and RNA?

 In DNA, Glycosidic Bond, N-C linkage connects Sugar with Nitrogenous base.

The C1 of deoxyribose forms β-N-Glycosidic bond with nitrogen-9‘ of purine (Adenine and Guanine) bases or with nitrogen-1 of pyrimidine (Thymine and Cytosine or Uracil in RNA) bases

It is called beta as the OH in first C deoxyribose is above the plane.

This 2 minute video will explain How glycosidic bond is formed in DNA?

As water is removed during the reaction, the reaction is called dehydration condensation reaction.

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pUC Vector: Features, Selectable markers, Blue White Screening for Selection of Recombinants using pUC Vector

 What is a pUC vector?

• Vector obtained by modifying pBR 322 vector
• Smaller in size 2686bp with high copy number
• ‘p’=plasmid “UC”: University of California
• Joachim Messing and co-workers (1983)
  
  
  

Regions:

1. Ori: E.coli origin

2 selectable markers: ampicillin (amp ^r)& lacZ’ gene

 This region has genes providing resistance against ampicillin,

LacZ’ codes for β Galactosidase enzyme

3. Restriction sites: lacZ’ gene with multiple cloning sites for 13 Restriction enzymes

Selection of recombinant colonies using Blue White colony screening

After transformation, 3  types of colonies

Grow in a medium containing Agar+Ampicillin + Xgal+IPTG

IPTG is an inducer of Beta galactosidase enzyme. It binds to the repressor activates beta galactosidase transcription

1)    Non-transformed: without vector: Cannot grow in Amp containing medium

     2) Transformed:

a) Transformed with non recombinant: Can grow in Amp medium, Active LacZ gene so can  convert X gal into blue product

    b)Transformed with  recombinant vector : can grow in Amp medium. Lac Z gene is  inactive by insertional inactivation. It is the inactivation of a gene upon insertion of our gene of interest. It cannot convert X gal to Blue product

Recombinant Colonies: White as LacZ gene is inactive by insertional inactivation

Non Recombinant: Blue Colonies LacZ gene is active, the galactosidase formed will convert  Xgal into blue product

This recombinant selection procedure is called Blue White Colony Screening.

Advantages of pUC vector

• High copy number 500-600 copies per cell
• Easy one step selection of recombinant colonies
• Many restriction sites in MCS (Multiple Cloning Sites)

Disadvantage

• Insert size is 15 Kb

Learn More: Insertional-inactivation-in-pBR322
Ideal Characteristics of Gene Cloning Vector

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Reversible Enzyme Inhibition: Competitive, Non Competitive and Uncompetitive Inhibition with examples

Enzymes are Biological catalyst that speeds up the rate of reaction without undergoing any change by itself.

Reversible inactivation: Inhibitor forms an unstable, non-covalently bonded, enzyme inhibitor complex. Lost activity can be regained.

Irreversible inactivation: Inhibitor forms strong covalent bonds so that it cannot be dislodged. Therefore the enzyme activity lost forever.

Three major types of Reversible Inhibition

1. Competitive (Inhibitor binds to Active site)

2. Non Competitive (Inhibitor binds to Allosteric site)

3. Uncompetitive Inhibition (Inhibitor binds to ES complex)

Competitive Inhibition or Isosteric inhibition

• The inhibitor structural homologue competes with the substrate for the same active site of an enzyme.
• Depends on relative concentration of inhibitor and its affinity
• Binds to free enzyme
• Can be reversed by adding more substrate so that inhibitor is out competed.

Lowering Km; decreases affinity (increases Km value) and Vmax remains the same; as Vmax can be achieved by adding more substrate

What is Km of an enzyme?

Michaelis constant (Km): The substrate concentration at which the reaction rate is half of V max

Km value is different for different enzymes

Km refers to the affinity of an enzyme for its substrate

Smaller the value higher is the affinity

Non-Competitive Inhibition

• The inhibitor binds to allosteric site
• It never blocks E-S formation
• This binding causes conformational change preventing the conversion of E-S complex to E-P complex
• It is not fully reversible inhibition some re-activators to  delock non competitive inhibitor
• Km remains  same (no change in affinity) & Vmax decreases as many ES complex will not form the product due to inhibitor binding to the allosteric site
• Eg : Heavy metals like Hg++, Pb++, drugs, pesticides, cyanide on cytochrome oxidase – can be removed by using a chelating agent EDTA

Uncompetitive Inhibition

• The inhibitor binds to E-S complex
• Inhibitor binds to E-S complex and not with free enzyme and facilitates tight E-S binding preventing enzyme action
• Cannot be overcome by increasing substrate concentration
• Decreases Km  value (increases affinity) &Vmax
• Eg: Inhibition of placental alkaline phosphatase by phenylalanine

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ICMR JRF - 2019 - Cell and Molecular Biology Questions

1. Which is the largest human cell?

1) Liver 2) Ovum 3) Spleen 4) Skin

2. Which of the following statement is incorrect regarding DNA methylation?

1) S-Adenosyl Methionine (SAM) is one of the most important methyl donors.

2) It is catalysed by enzymes

3) Occurs at CpG islands

4) Mainly G is methylated

3. Alu elements in human genome represent:

1) Exons

2) Introns

3) Nucleotide repeats

4) Transposable elements

4. Which of the following structures is known to maintain the shape of a cell?

1) Ribosomes

2) Microtubules

3) Nucleus

4) Mitochondria

 5.The specific DNA sequences to which the transcription factors bind are referred to as

1) Replication elements

2) Blocking factors

3) Transcription factors

4) Regulatory elements

6. Which of the following is not an example of post translational modification?

1) Addition of prosthetic groups

2) Proteolytic Processing

3) mRNA splicing

4) Loss of signal sequences

Practice Makes Perfect ----MCQ Corner

7 . Calmodulin contributes to signal transduction by binding to

1) CAMP

 2) Calcium

3) Magnesium

4)Sodium

8. Crossing over occurs in which phase

1) Prophase I

2) Telophase I

3) Anaphasel

4) Metaphase I

9. Which of the following growth media would you expect to result in synthesis of high levels of mRNA for the enzymes of the E. coli lac operon?

1) High glucose, high lactose

2) Low glucose, low lactose

3) High glucose, low lactose

4) No glucose, high lactose

10. Which of the following statement is incorrect about the genetic code is?

1) A codon is a triplet of nucleotides that codes for a specific amino acid

2) A specific first codon in the sequence establishes the reading frame

3) A codon specifies more than one amino acid

4) Nucleotide triplets are read in a successive, non-overlapping fashion 

11. The original codon changes to stop codon in which type of mutation

 1) Sense mutation

2) Mis-sense mutation

3) Non-sense mutation

4) Reverse mutation

12. What is the mode of action of exonuclease III?

1) Exonuclease III acts on single stranded DNA in 3'-5'direction

2) Exonuclease III acts on double stranded DNA in 5'-3'direction

3) Exonuclease III acts on single stranded DNA in 5'-3'direction

4) Exonuclease III acts on double stranded DNA in 3'-5'direction 

13. Human telomeres consist of Tandem repeats of sequence

1) (TTAGGG)n

2) (TTAAGGG) n

3) (MAAGG) n

4) (1TAAAGG)n

ICMR JRF QUESTIONS: Biochemistry Biostatistics 

Answers: 

1. 2) Ovum

2. 4) Mainly G is methylated

3. 4) Transposable elements

4. 2) Microtubules

5. 4) Regulatory elements

6. 3) mRNA splicing

7.  2) Calcium

8. 1) Prophase I

9. 4) No glucose, high lactose

10. 3) A codon specifies more than one amino acid

11. 3) Non-sense mutation

12. 4) Exonuclease III acts on double stranded DNA in 3'-5'direction

13. 1) (TTAGGG)n

Notes Cell Biology Cell Cycle , Mitosis, Meiosis Cell Biology MCQ

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2019 ICMR JRF Questions and Answers -Biochemistry

1.  Which is true about enzymes?

1) All enzymes are proteins

2) All enzymes are vitamins

3) All enzymes are not proteins

4) All proteins are enzymes

Ans: 3) All enzymes are not proteins

2. Which of the following type of enzyme inhibition is also called as end-product inhibition?

1) Substrate regulation

2) Feedback inhibition

3) Competitive inhibition

4) Non-competitive inhibition

Ans: 2) Feedback inhibition

3. Which vitamin is only found in animal products?

1) Vitamin A

2) Vitamin B3

3) Vitamin B12

4) Vitamin C

Ans: 3) Vitamin B12

4. Cholesterol does not act as the precursor for

1) Cardiolipin

2) Progesterone

3) Cortisol

4) Estradiol

Ans: 1) Cardiolipin

5. Prussic acid is another name of

1) Sulphuric acid

 2) Nitric oxide

3) Oxalic acid

 4) Hydrogen cyanide

Ans:  4) Hydrogen cyanide

 6. Wavelength range of absorption peptide bond is

1) 190-230 nm

2) 240-270 nm

3) 160-180 nm

 4) 250-280 nm

Ans: 1) 190-230 nm

7. Pick up the amino acid, which is present in the body but not found in proteins

1) Arginine

2) 4-Hydroxyprotine

3) Ornithine

4) Selenocysteine

Ans: 3) Ornithine

8. Allopurinol is used for the treatment of gout. It is an inhibitor of

1) Thymidylate synthase

2) Xanthine oxidase

3) Hypoxanthine-guanine phosphoribosyl transferase

4) Adenosine phosphoribosyl transferase

Ans: 2) Xanthine oxidase

9. The highly repetitive DNA in the eukaryotes occupies the WHICH fraction of the Cot Curve

1) Slow

 2) Intermediate

3) Fast

4) All of the above

Ans: 3) Fast

10. α-oxidation of fatty acids takes place in

1) Endoplasmic reticulum

2) Cytosol

3) Mitochondria

4) Peroxisomes

Ans: 4) Peroxisomes

11. Which of the following enzyme participates in both the citric acid cycle and the electron

transport chain?

1) NADH dehydrogenase

2) Malate dehydrogenase

3) Succinate dehydrogenase

4) Isocitrate dehydrogenase

 Ans: 3) Succinate dehydrogenase

12. Which of the following molecule yields maximum number of ATPs upon oxidation?

1) Glutamate

 2) Pyruvate

3) Palmitate

 4) Glucose

Ans: 3) Palmitate

Practice Makes Perfect ----MCQ Corner

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ICMR JRF Life Science 2019 Question Paper

 The test will consist of one paper of 2 hours duration. The paper will consist of 2 Sections.

The Aptitude Section (Section A) will have 50 questions on

(i) scientific phenomenon in everyday life;

(ii) general knowledge in sciences; and

(iii) common statistics.

All these questions would be compulsory with each question carrying 1 mark. The subject Specific Section (Section B & C) would pertain to (B) Life Sciences and (C) Social Science. 

The candidate may attempt questions in either of the two areas. Each area of section B & C would have 100 questions and the candidate may attempt any 75 questions in the predesigned area of Section B or C. Each question carries one mark. Negative marking @ 0.25
 

Practice Makes Perfect ----MCQ Corner  Notes Videos

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Electron Transport Chain (ETC) in Cellular Respiration: Definition, Location and Steps Simplified

Cellular respiration is a catabolic process which involves the intracellular oxidation of glucose or organic molecules  through series of enzymatic reaction producing energy in the form of ATP with the release of CO₂ and H₂O as byproducts.

This is the summarized video on Electron transport chain

3 stages of cellular respiration:

1. Glycolysis (Glyco=Glucose; lysis= splitting) is the oxidation of glucose (C 6) to 2 pyruvate (3 C) with the formation of ATP and NADH.

2.The Krebs cycle, Citric acid cycle or TCA cycle is an eight step cyclic reactions in which acetyl CoA is oxidized producing CO2, reduced coenzymes (NADH + H+ and FADH2), and ATP.

3. Electron Transport Chain: 

ETC is the step by step transfer of high energy electrons through a series of electron carriers located in multienzyme complexes, finally reducing molecular O₂ to form water with the formation of ATP by chemiosmosis.

EXACT SITE OF REACTION

• Organelle: Mitochondrion
• Site of Electron transport chain: Mitochondrial Inner membrane
• Proton (H+) pumped into the intermembrane space creating proton gradient
• ATP synthesis occurs towards the matrix region (see the above figure)

Background info: At the beginning of electron transport chain we have NADH and FADH2 synthesized during Kerb’s cycle and glycolysis. Approximately only 4 ATP are synthesized directly (2 from glycolysis and 2 from Krebs cycle) from a glucose molecule. The rest ~32-34 ATP are synthesized during Electron transport chain (ETC) by chemiosmosis.

Now let as  move into the detail

1. Electron flow and Energy release:

NADH and FADH2 donates high energy electrons that pass through different protein complexes and electron carriers in the ETC.

As the electrons moves from high energy to low energy level, some amount of energy is released. The final electron acceptor is O₂ which splits and takes up H+ to form water (H₂O)

 

2. Proton movement and gradient formation

The energy released during electron flow is used to pump proton (H+ ions) from matrix side to the intermembrane space of mitochondrion. (see figure). This creates a proton gradient or (Electrochemical gradient or proton motive force) across the inner mitochondrial membrane (that is higher concentration of H+ ions in the intermembrane space compared to the matrix).

3. Proton motive force (PMF) driven ATP synthesis

The H+ ions should move to matrix to maintain equilibrium (to balance H+ ion concentration). As phopholipid bilayer of inner mitochondrial membrane is impermeable, the only way out is through the protein complex called ATP synthase which spans the inner mitochondrial membrane and has a proton channel.

The flow of H+ ions through ATP synthase provides energy for the addition of phosphate to ADP thus forming ATP. (just like turbine in hydroelectric power plant where water forces turbine movement, here flow of H+ ions drives ATP synthesis)

The proton gradient (Proton motive force) driven ATP synthesis is called Chemiosmosis.

Hope things are clear. Watch the video for better understanding. Thank you and enjoy learning Biology

10 steps of Glycolysis Video: https://youtu.be/XcdL9o3yidU

8 Steps of Krebs cycle:  https://youtu.be/W05eIbXeiMA

How Hans Krebs Discovered Krebs cycle: https://youtu.be/sLu7oGGy2cw

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How is Ester bond formed in Fats?

Fats are macronutrients with highest energy storing potential. They are insoluble in water.

Ester bond formation in fats?

Watch this simple 2 minute video

Fats are made up of two types of smaller molecules: glycerol and fatty acids

Glycerol is a 3-C alcohol with a hydroxyl group attached to each carbon

A fatty acid consists of a carboxyl group attached to a long hydrocarbon skeleton

In a fat, 3 fatty acids are joined to 1 glycerol by ester linkage, forming triacylglycerol, or triglyceride

Here the OH of the carboxyl group (-COOH) group of each fatty acid bonds with 1st, 2nd and 3rd carbon OH of the glycerol by ester linkage relaesing 3 molecules of water (H2O)

This is how ester bond is formed in fats.

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