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Corona virus SARS COV-2 Life Cycle, structure and genome explained

Corona virus is a family of virus that causes diseases in mammals and birds.
According to World Health Organization (WHO), in humans, several coronaviruses are known to cause respiratory infections ranging from the common cold to more severe diseases such as Middle East Respiratory Syndrome (MERS) and Severe Acute Respiratory Syndrome (SARS).
Why the name “corona”?
The most prominent feature of coronaviruses is the club-shape spike projections emanating from the surface of the virion. These spikes are a defining feature of the virion and give them the appearance of a solar corona under two-dimensional transmission electron microscopy (TEM), prompting the name, coronaviruses. (Latin corona meaning “crown or halo”)
This video explains the life cyle of SARS CoV-2 virus that cause COVID-19 disease.

Thank you... Please Stay @ home, stay safe and Save lives.
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Simple Formula to Calculate the number of phosphodiester bond in a DNA molecule with Practice problems


How to calculate the number of phosphodiester bond in a DNA molecule with Practice problems

Simple Formula to Calculate the number of phosphodiester bond in a DNA molecule

1. How many phosphodiester bonds are present in the following DNA molecule?
5'-CTAGAG-3'
3'-GATCTC-5
 a) 6  b) 11  c) 12  d) 10
Answer:
Suppose, no. of base pairs in DNA is ‘n’
You can answer any question like this using the given formula
No. of base pairs =6
No. of phosphodiester bonds in a DNA molecule = 2n-2
Where n=No. of base pairs (bp)
=(2x6)-2=12-2=10
2. How many phosphodiester bonds are present in a DNA with 50 bp?
a) 100  b) 99  c) 200  d) 98
If the no. of base pairs in DNA is ‘n’
No. of base pairs=50
No. of phosphodiester bonds in a DNA molecule = 2n-2
=(2x50)-2=100-2=98

3. In a DNA molecule the no of phosphodiester bonds are 1200. Find the number of base pairs?
a) 600  b) 601  c) 602  d) 1199
If the no. of base pairs in DNA is ‘n’
No. of base pairs=?
No. of phosphodiester bonds in a DNA molecule = 2n-2
       =2n-2=1200
      2n=1200+2=1202
      n=1202/2=601

4. How many phosphodiester bonds are present in a ssDNA with 50 bases?
a) 100 b) 99  c) 200  d) 49
No. of phosphodiester bonds in a single strand of DNA= n-1
Where n is the no. of bases/nucleotides (incase of single stranded DNA or RNA)
ssDNA =n-1 phosphodiester bonds
=50-1=49
Thank you....


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Translation in Prokaryotes and Eukaryotes simple difference between video

Translation is a process in living cells in which the genetic information encoded in messenger RNA(mRNA) called genetic code in the form of a sequence of nucleotide triplets (codons) is translated into a sequence of amino acids in a polypeptide chain during protein synthesis.
Here we are sharing all resources for understanding the concept of translation or protein synthesis

Some useful Resources on translation
Power point presentation


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Chemiosmosis and ATP synthesis in Photosynthesis Simplified steps

How is ATP synthesized in photosynthesis by chemiosmosis?
  • Chemiosmotic Hypothesis by Peter Mitchell 1961
  • Awarded the Nobel Prize for Chemistry in 1978.
  • This process is similar to the ATP synthesis during cellular respiration by Chemiosmosis.

Chemiosmosis and ATP synthesis in Photosynthesis

EXACT SITE OF REACTION
  • Organelle: Chloroplast during light dependent reaction
  • Site of Electron transport chain: Thylakoid membrane of chloroplast
  • Proton (H+) pumped into the thylakoid lumen or thylakoid space
  • ATP synthesis occurs towards the stromal side (see the above figure)
  • ATP produced during light reaction is used to fix carbon dioxide to carbohydrates in Calvin cycle
Chemiosmotic Theory states that Electron transport and ATP synthesis are coupled by a proton gradient across the inner mitochondrial membrane.
Let’s simplify this statement
The 3 major events in chemiosmosis are
1. Electron flow & energy release
2. Proton movement and gradient formation
3. proton motive force (PMF) driven ATP synthesis

Background info: Light absorbing pigments are arranged on the thylakoid membrane as photosystems (PS-I and PSII). Light energy is absorbed by the pigment molecules of the photosystem. The absorbed energy is transferred to nearby pigment molecules finally reaching the reaction centre chlorophyll-a molecule by resonanace transfer (vibratory transfer).

Now let as move into the detail
1. Electron flow and Energy release:
High energy electrons are released from the reaction centre chl-a molecule that is transferred through series of electron carriers in cyclic and non cyclic photophosphorylation. During the movement of electrons from high energy level to low energy level, some amount of energy is released

2. Proton movement and gradient formation
The energy released during electron flow is used to pump proton (H+ ions) from stromal side to the thylakoid lumen or thylakoid space of chloroplast (see figure). This creates a proton gradient or (Electrochemical gradient or proton motive force) across the thylakoid membrane (that is higher concentration of H+ ions in the thylakoid space compared to the stroma).

3. Proton motive force (PMF) driven ATP synthesis
The H+ ions should move to stroma to maintain equilibrium (to balance H+ ion concentration). As phopholipid bilayer of thylakoid membrane is impermeable, the only way out is through the protein complex called ATP synthase which spans the thylakoid 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 towards the stromal side which is utilized in Calvin cycle for fixing CO2 to glucose (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 Bio....
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Chemiosmosis and ATP synthesis in Cellular respiration Step by Step Simple Explanation


How is ATP synthesized in Cellular respiration by chemiosmosis?
Chemiosmotic Hypothesis by Peter Mitchell 1961
Awarded the Nobel Prize for Chemistry in 1978.
Chemiosmosis and ATP synthesis in Cellular respiration site
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)

Chemiosmotic Theory states that Electron transport and ATP synthesis are coupled by a proton gradient across the inner mitochondrial membrane.
Lets simplify this statement
The 3 major events in chemiosmosis are
1. Electron flow & energy release
2. Proton movement and gradient formation
3. proton motive force (PMF) driven ATP synthesis

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 Bio....
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What are Carbohydrates? How are they Classified? with simple video

Definition: Carbohydrates are bio molecules which are polyhydroxy aldehydes or ketones made  up of Carbon C, Hydrogen H and Oxygen O in the ratio (1:2:1)
Classification of Carbohydrates
This is a simplified 8 minute video on different classification of carbohydrates
A. Based on No. of Carbohydrate Units
1. Monosaccharides
  • Simplest carbohydrate unit or monomer
  •  that cannot be hydrolysed further
  • General formula (CH2O)n
  • Minimum no of Carbon atom is 3
  • Example: glucose, fructose
2. Disaccharides:
Two monosaccharide units joined by glycosidic bond
Example: Glucose + Glucose=Maltose (Disaccharide)
                   Glucose + fructose= Sucrose
3. Oligosaccharides
Carbohydrates usually made up of 3-10 monosaccharide units joined by glycosydic bond
Eg: Raffinose
4. Polysaccharides
  • Complex Carbohydrates made up of hundreds of monosaccharide unit s joined by glycosydic bond
  • Structural Polysaccharides like Cellulose in plants
  • Storage Polysaccharides like starch in plants and glycogen in animals
B. Based on functional group
  • Aldoses: Carbohydrates with aldehyde(-CHO) functional group Eg: Glucose
  • Ketoses: Carbohydrates with ketone (=C=O) functional group Eg: Fructose
C. Based on no. of carbon atoms
  • Simplest monosaccharide has 3 carbon atom called as Triose. Eg: Glyceraldehyde
  • Tetrose: Monosaccharide with 4 carbon atom Eg: erythrose
  • Pentose: 5 carbon atom Eg: pentose
  • Hexose: 6 carbon atom Eg: Glucose
  • Heptose: 7 carbon atoms Eg: sedoheptulose
D. Based on Rotation (Dextro and Levo)
  • Based on stereochemistry of the highest numbered chiral carbon of the Fischer projection.
  • If the hydroxyl group of the highest numbered chiral carbon is pointing to the right, the sugar is designated as D (Dextro: Latin for on the right side). 
  • If the hydroxyl group is pointing to the left, the sugar is  Levo (Levo: Latin for on the left)
  • Most naturally occurring carbohydrates are of the D-configuration.
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ICMR JRF 2018 Life Science Questions and Answers (Download PDF)

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. 
ICMR JRF 2018 Life Science Questions and Answers (Download PDF)
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 
ICMR JRF 2018 Questions and Answers - Section I- AptitudeICMR JRF 2018 Life Science Questions and Answers

ICMR JRF 2018 Life Science Questions and Answers

ICMR JRF 2018 Life Science Questions and Answers

ICMR JRF 2018 Life Science Questions and Answers

ICMR JRF 2018 Life Science Questions and Answers

ICMR JRF 2018 Life Science Questions and Answers

ICMR JRF 2018 Life Science Questions and Answers
Download pdf
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ICMR JRF 2018 Questions and Answers - Section I- Aptitude

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.
ICMR JRF 2018 Questions and Answers - Section I- Aptitude
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.

Candidates are required to indicate the option for Section B or C in the application form too.
Each question carries one mark. Negative marking @ 0.25 .




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