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Pulse Field Gel Electrophoresis (PFGE) Definition, Procedure and Application

Definition: It is a technique used for the separation of large deoxyribonucleic acid (DNA) molecules  by applying to a gel matrix an electric field that periodically changes direction.
Developed David C. Schwartz and Charles Cantor at Columbia University in 1984.
How is PFGE different from conventional agarose gel electrophoresis?
In conventional gels, the current is applied in a single direction (from top to bottom). But in PFGE, the direction of the current is altered at a regular interval.
Conventional electrophoresis can separate DNA fragments up to 20 kb. DNA molecules larger than 20kb can be separated by PFGE. DNA fragments of up to ~10 Mb can be effectively separated using PFGE.
Image credit 
General procedure is same as conventional agarose gel electrophoresis. The differences are in sample preparation and in direction of current flow.
Sample preparation: High molecular weight DNA t be separated using PFGE is easily cleaved through shearing and has very high solution viscosity. For these reasons, DNA samples for PFGE are generally prepared by embedding in gel medium. Cellular source material is suspended in low gelling agarose and the gelled suspension is poured into molds. All subsequent manipulations (for example, cell lysis, protein removal, and restriction digestion) are performed by diffusing reagents into the resultant gel plugs. The processed gel plugs are then carefully loaded into wells of an agarose gel used for PFGE.*
Direction of current: Instead of constantly running the voltage in one direction as in conventional agarose gel electrophoresis, the voltage is periodically switched among three directions; one that runs through the central axis of the gel and two that run at an angle of 60 degrees either side. 
PFGE Application:
• First used to separate yeast chromosomal DNA. Therefore ideal for separating chromosomal DNAs of different organisms for genome projects
• Pulsed field gel electrophoresis has been used as a means of identifying the genetic defects that cause many hereditary diseases**.
• used for genotyping or genetic fingerprinting.
• It is commonly considered a gold standard in epidemiological studies of pathogenic organisms. PFGE applied as a universal generic method for subtyping of bacteria. Only the choice of the restriction enzyme and conditions for electrophoresis need to be optimized for each species.
Schwartz DC and Cantor CR (1984) Separation of yeast chromosome-sized DNAs by pulsed field gradient gel electrophoresis. Cell 37, 67–75 (original paper on PFGE).
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Chromosome classification based on position of centromere and function

We have discussed a lot about about chromosomes in previous posts. 
Chromosomes based on position of centromere
In this post we are limiting our discussion to classification of chromosomes based on position of centromere and function 
Based on Position of Centromere
1.Metacentric: centromere in the middle forming two equal sized arms
2.Submetacentric:  centromere near the middle forming two unequal arms, one slightly longer than the other
3.Acrocentric:  centromere near the tip forming two unequal arms, one much shorter (p arm) than the other (q arm)
4.Telocentric:  centromere at the tip and a single arm
Based on function
1.Autosomes: contains genes that controls non-sexual or somatic characters

2.Sex chromosomes: contains genes that controls sexual  characters
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CSIR UGC JRF NET Life Sciences Part B Questions and Answers June 2015

11. Which one of the following enzymes is NOT a part of pyruvate dehydrogenase enzyme complex in glycolysis pathway?
a. Pyruvate dehydrogenase
b. Dihydrolipoyl transferase
c. Dihydrolypoyl dehydrogenase
d. Dihydrolypoyl oxidase
Ans: d. Dihydrolypoyl oxidase
Explanation: Pyruvate dehydrogenase enzyme complex (PDC) is a complex of three enzymes that convert pyruvate into acetyl-CoA by a process called pyruvate decarboxylation. The complex consist of Pyruvate dehydrogenase (E1), Dihydrolipoyl transferase (E2)and Dihydrolypoyl dehydrogenase (E3).

12. What phenotype would you predict for a mutant mouse lacking one of the genes required for site-specific recombination in lymphocytes?
a. Decrease in T cell counts
b. Immunodeficient
c. Increase in T cell count
d. Increase in B cell count
Ans: b. Immunodeficient
Explanation: Site-specific recombination is a type of genetic recombination in which DNA strand exchange takes place between segments possessing only a limited degree of sequence homology. Mutation in genes required for site specific recombination results in reduced or absence of mature T and B cells. Immature cells are still present resulting in immune deficient state.
2015 csir

13. A 1% (w/v) solution of a sugar polymer is digested by an enzyme (20µg, MW=200,000). The rate of monomer sugar (MW=400) liberated was determined to have a maximal initial velocity of 10 mg formed/min. the turnover number (min-1) will be
a. 5 x 104
b. 2.5 x 10-2
c. 4.0 x 10-6
d. 2.5 x 105
Ans: d. 2.5 x 105
Explanation: Kcat is the turn over number  and is calculated as Kcat
Vmax=max velocity; Et= total enzyme concentration.
In the question we should convert the concentration terms to molarity.
Vmax= 10x10-3/400= 0.25x10-4;
Et= 20x10-6/200000=10-10.
Kcat= 0.25x 10-4/10-10=2.5x105/min.

14. Beating of cilia is regulated by
a. actin
b. Myosin
c. cofilin
d. Nexin
Ans: d. Nexin
Explanation: Nexin is highly extensible protein responsible for the beating of cilia. Other options are related to microfilaments. (Refer Cilia vs Flagella)

15. Which one of the following events NEVER activates the G-protein coupled receptor for sequestering Ca2+ release?
a. Interaction of bindin to sperm receptors.
b. Activation of Frizzled by Wnt.
c. Cortical reaction blocking polyspermy
d. DNA synthesis and nuclear envelop breakdown.
Ans: d. DNA synthesis and nuclear envelop breakdown.
Explanation: G-protein coupled receptor ((GPCRs) form a very large group cell surface receptors that are coupled to signal transducting trimeric G proteins. Here, all other options would result in increased concentration of Ca2+ except for DNA synthesis and nuclear envelop breakdown which might occur due to activation of a cAMP dependent kinase.

16. Hydra shows morphallactic regeneration and involves which one of the following signal transduction pathway in its axis formation?
a. Wnt/β-catenin pathway
b. Retinoic acid pathway
c. FGF pathway
d. Delta- Notch pathway
Ans: a. Wnt/β-catenin pathway
Explanation: When a hydra is cut in half, the half containing the head will regenerate a new basal disc, and the half containing the basal disc will regenerate a new head. Moreover, if a hydra is cut into several portions, the middle portions will regenerate both heads and basal discs at their appropriate ends. No cell division is required for this to happen, and the result is a small hydra. This regeneration is morphallactic.
Reference: Developmental Biology by Gilbert S F
Recent detailed analyses of the activation of the Wnt–β-catenin pathway in bisected Hydra shows that the route taken to regenerate a structure as complex as the head varies dramatically according to the level of the amputation.
Galliot, Brigitte, and Simona Chera. "The Hydra model: disclosing an apoptosis-driven generator of Wnt-based regeneration." Trends in cell biology 20.9 (2010): 514-523.

17. The main difference between normal and transformed cells are
a. Immortality and contact inhibition
b. Shorter generation time and cell mobility
c. Apotopsis and tumour suppressor gene hyper function
d. Inactivation of oncogenes and shorter cell cycle duration
Ans: a. Immortality and contact inhibition
Explantion: Apart from check point regulation, one of the crucial mechanism that prevents uncontrolled cell division is contact inhibition, a hall mark character of normal cells. Once transformed, the cells become immortal or cancerous due to the absence of ‘contact inhibition’ mechanism. (Refer: Characteristics of cancer cells)

18. Following are three single stranded DNA sequences that form secondary structures.
(a) A T T G A G C G A T C A A T
(b) A T T G A G C G A T A T C A A T
(c) A G GG A G C G A T C CC T
Based on their stability, which one is correct?
a. (a) = (b)=(c)
b. (c)>(a)>(b)
c. (b)>(c)=(a)
d. (b)>(c)>(a)
Ans: b. (c)>(a)>(b)
Explanation: Here, C has the highest GC pairing. So, it shows the highest stability. Three H-bonds are involved in GC pairing. Therefore, more the number of G and C, the more will be the stability.
option (c) has the maximum number of G and C
Option (a) and (b) has same number of G and C, but option (a) is more stable as the total percentage of GC bond is more in option (a) as option (b) sequence length is slightly longer.
For More..Refer  DNA double helix model, Chargaff’s rule, Stability of DNA and RNA
19. The key determinant of the plane of cytokinesis in mammalian cells is the position of
a. Chromosomes
b. Central spindle
c. Centrioles
d. Pre- prophase band
Ans: b. Central spindle
Explanation: Mitotic spindle, more specifically the astral spindle, determines the the site of the cleavage furrow in mammalian cell. (Read more: Cytokinesis in Animal cell vs Cytokinesis in Plant cell)

20. Dark- grown seedlings display ‘triple response’ when exposed to ethylene. Which one of the following is NOT a part of ‘triple response’?
a. Decrease in epicotyl elongation
b. Rapid unfolding and expansion of leaves
c. Thickening of shoot.
d. Horizontal growth of epicotyl.
Ans: b. 2. Rapid unfolding and expansion of leaves
Triple response of seedlings: The hormone (ethylene) prevents the seedling stem and root from elongating. It induces the stem and root to swell radially, thereby increasing in thickness. Together these responses strengthen the seedlings stem and root. The seedling stem bends so that embryonic leaves and the delicate meristem grow horizontally rather than vertically. The bend portion of the stem then pushes through the soil. This bent portions forms as the result of an imbalance of auxin across the stem axis. Ethylene drives this auxin imbalance.
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JEST 2016: Application, Eligibility & Imported Dates

Joint Entrance Screening Test (JEST)
Applicants seeking admission  for a Ph.D / Integrated Ph.D Programme in Physics or Theoretical Computer Science or Neuroscience or Computational Biology in  one  of the Participating Institutes may appear for the Joint Entrance Screening Test (JEST) at one of the Exam Centers.
The Science & Engineering Research Board (SERB) (statutory body established through an Act of Parliament) recognizes JEST as a National Eligibility Test (NET) in their office Memo vide OM No.SB/S9/Z-01/2015.
JEST 2016
Imported dates:
  • Online applications will start from November 5, 2015.
  • Last date for online applications and payments is December 10, 2015.
  • JEST 2016 will be held on Sunday, February 21, 2016.
Participating Institutes have their own eligibility criteria. Applicants who are expected to complete their final examinations by August of each year are also eligible to appear for the JEST exam of that year.
Ph.D. Degree in Neuroscience or Computational Biology - Participating Institutes:
  • IMSc (The Institute of Mathematical Sciences, Chennai):Theoretical Physics, Theoretical Computer Science, and Computational Biology
  • NBRC (National Brain Research Centre, Manesar): Molecular, Computaional and Systems Neuroscience. Sensory & motor systems, learning & memory, language & speech processing, functional neuroimaging: EEG, MEG, fMRI, MRS, stem cells, developmental neurobiology, neurogenetics, neurodegenerative and neurodevelopmental disorders, cancer signaling & glial tumor biology.
 Please note that JEST is meant as a means for students from a Physics background to take up research in Neuroscience or Computational Biology, and interested students have to take the PHYSICS JEST examination. Please visit the NBRC webpage  or the IMSc webpage for more information.
Eligibility Ph.D. Programme -Theoretical Computer Science/Neuroscience/Computational Biology
  • M.Sc./ M.E. / M.Tech. / M.C.A. in Computer Science and related disciplines, and should be interested in the mathematical aspects of computer science. Visit website of IMSc for further details. M.Sc (Physics/ Mathematics), B.E/ B.Tech/ M.C.A in Computer Science for Ph.D. in Neuroscience at NBRC.
  • ​M.Sc./ M.E. / M.Tech. ​/ MCA ​ in any engineering or science discipline, with good mathematical skills and strong interest in biological problems for Ph.D. in Computational Biology at IMSc.
Exam Centers: Ahmedabad, Aligarh, Allahabad, Bangalore, Bardhaman, Bhopal, Bhubaneswar,Chandigarh, Chennai, Delhi, Goa, Guwhati, Hyderabad, Indore, Jaipur, Kanpur, Kharagpur, Kochi, Kolkata, Madurai, Mumbai, Nagpur, Nainital, Patna, Pune, Raipur, Roorkee, Sambalpur, Silchar, Siliguri, Srinagar, Trivandrum, Udaipur, Vishakhapatnam.

Application Procedure (Before filling the ON-LINE application form, candidate is requested to read the following instructions carefully)
ON-LINE Application
To apply online, applicants must create an account using a valid email-id. This account will remain valid from the time of account creation till one month after the date of declaration of results, for that year. Email-ids are linked to the created account as well as  to the application, and will be used for all communications with the applicants. Therefore, the email-id should be active and must not be changed during this period. To submit application online,  applicants must login to the created account and complete the application form. After submission, an unique submission number will be sent to your email-id. Please save all the emails sent to you and the unique submission number must be quoted by the applicant for any communication with the JEST authority.
The steps for submission of ON-LINE application form are
  1. Register yourself before you login. It is assumed that you already have an email-id. Registrations will start from November 5, 2015.
  2. To register, the candidate has to click on the "register to appear for the exam" link in the Homepage. You will receive an email with  your account information after registering with a valid email-id. Follow the instructions given in the email for resetting your password and further login.
  3. Login to the JEST homepage with your username (your email-id) and password. Fill the form and save it, or, modify an already saved form.  Press the "Submit Application" button ONLY when you have inserted and checked all the mandatory entries.
  4. Upload jpg files of your Passport size photo and scanned signature.
  5. Pay the application fee online using the State Bank of India i-collect facility. Please check the FAQ for details of the payment process.
  6. After ON-LINE submission of the application form and successful e-payment of the application fee, your submission procedure will be considered complete. NO modification will be allowed after this.
E-Photo and E-Signature
Photo - Signature
  • While filling up the form, you will be asked to upload jpeg files of your passport size photograph and your signature. The size of the individual files should not exceed 50 Kilo bytes. Crop out the extra white portion from the background surrounding the image to retain clarity.
  • Before uploading your photo and signature files, make sure you have given it a good name.  File names should use  alphabets, underscores, and numbers only.  Special characters such as spaces,  (, #, &,  punctuation, etc., should not be used in filenames.  If you use a logical naming system, like loginname_date_photo.jpg, loginname_date_sig.jpg then it will be printed without any errors.
  • In case the jpeg file of your photo which has been taken with camera exceeds the limit, you should compress the file size using compression softwares available in the websites, such as,
  • Your signature can be scanned with a digital scanner, and saved as jpeg file for uploading. In case this file size produced by scanner also exceeds 50 kB limit, you can compress it using the same tools as explained above.
  • Please make sure the image type is jpeg. For checking, right click and go to the properties of the file being uploaded.
  • Dimension of the image files should be the following
    • Minimum of 100 X 125 pixels or maximum of 240 x 300 pixels (width x height) for photo and 150 x 30 pixels (width x height) for signature
  • Please check ALL the properties of these image files before uploading.
Application Fee
An Application Fee of Rs. 300 (Rs 150 for SC/ST applicants ONLY) is required to be paid along with the application which can be paid online using Debit/Credit cards and Net-banking. Payment by any other mode will NOT be accepted. Click Here for Available Payment Options.
Post-Submission of Application form
  • On successful submission of ON-LINE application form you will receive an E-mail with following information from JEST Website.
• Application data after final submission containing  Unique Submission Number and filled in data.
  • Please save this e-mail for further correspondence with the JEST authority.
  • If you do not receive the above-mentioned e-mail after submission of application form and the successful e-payment of the application fee, contact JEST authority.
  • Admit card will be available in the JEST webpage after 15th of January, 2016. Bring a printout of the downloaded admit card and one of your photo-ids for entry into the examination centre. Admit cards will NOT be sent by post/email.
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CSIR UGC JRF NET Life Sciences JUNE 2015 Solved Questions and Answers

1. When bacteria growing at 20o C are warmed at 37o C, they are most likely to synthesize membrane lipids with more
a. Short chain saturated fatty acids
b. Short chain unsaturated fatty acids
c. Long chain saturated fatty acids
d. Long chain unsaturated fatty acids

2. Which one of the following compounds is generally translocated in the phloem?
a. Sucrose
b. D-Glucose
c. D-Mannose
d. D-Fructose

3. Given below are some statements about prokaryotic and eukaryotic mobile genetic elements or transposons.
A. Most mobile genetic elements in bacteria transpose via an RNA intermediate.
B. Most mobile genetic elements in bacteria are DNA
C. Mobile genetic elements in eukaryotes are only retrotransposons.
D. Both, RNA and DNA transposons, are found in eukaryotes.
Choose correct combination.
a. A and C
b. B and C
c. A and D
d. B and D

CSIR UGC JRF NET Life Sciences JUNE 2015 Solved Questions and Answers
4. Which one of the following statements about LEAFY (LFY) a regulatory gene in Arabidopsis thaliana, is correct?
a. LEAFY (LFY) is involved in floral meristem identity.
b. LEAFY (LFY) is involved in leaf expansion.
c. LEAFY (LFY) is involved in root meristem identity.
d. LEAFY (LFY) is responsible for far-red light mediated seedling growth.

5. Nitrogen gas is reduced to ammonia by nitrogen fixation method. In order to execute the process, which one of the following compounds is usually required?
a. ATP
b. GTP
c. UDP
d. ADP

6. The quantum yield of oxygen evolution during photosynthesis drastically drops in far-red light. This effect is known as:
a. Far red drop.
b. Red drop.
c. Blue drop.
d. Visible spectrum drop.

7. In an alpha helical polypeptide, the backbone hydrogen bonds are between
a. NH of n and CO of n+4 amino acids.
b. CO of n and NH of n+3 amino acids.
c. CO of n and NH of n+4 amino acids.
d. NH of n and CO of n+3 amino acids.

8. The S wave of normal human ECG originates due to
ECG normal sinus rhythm
a. Septal and left ventricular depolarization.
b. Late depolarization of the ventricular walls moving back toward the AV junction.
c. Left to right septal depolarization.
d. Repolarization of atrium.

9. Cystic fibrosis transmembrane conductance regulator (CFTR) is known to control the transport of which ion?
a. Ca+2
b. Mg+2
c. HCO3-
d. Cl-

10. In type II splicing
a. a ‘G-OH’ from outside makes a nucleophilic attack on 5’P of first base of intron
b. a free 2’O of an internal adenosine makes a nucleophilic attack on 5’P of first phase of intron
c. a 3’O of an internal adenosine makes a nucleophilic attack on 5’P of first base of intron
d. the hydrolysis of last base of exon is carried out by U2/U4/U6

Answers with Explanation
1. c. Long chain saturated fatty acids
Explanation: Long chain saturated fatty acids will decrease the fluidity of the plasma membrane. Bacteria synthesise such fatty acid to survive under high temperature  (Refer: Plasma membrane and membrane fluidity)

2. a. Sucrose
Explanation: Sucrose is translocated through phloem in plants.

3. d. B and D
Explanation: Most transposons in bacteria are DNA and eukaryotes have both DNA and RNA

4. a  LEAFY (LFY) is involved in floral meristem identity.
Explanation: LEAFY is a floral meristem identity gene. (Refer: ABC model of Flower Development)

5. a. ATP
Explanation: The process is coupled to the hydrolysis of 16 equivalents of ATP and is accompanied by the co-formation of one molecule of H2.

6. b. Red drop.
Explanation: The sudden fall in the photosynthetic yield in far-red region of the solar radiation spectrum is called Emerson’s “red drop”.

7. c. CO of n and NH of n+4 amino acids.

Explanation: The alpha helix is a common secondary structure of proteins. To fotm the helix, the bond is between C=O group of the first amino acid with NH group of the fifth amino acid ( n+4 hydrogen bonding). (Refer: Bonds in Protein structure)

8 b. Late depolarization of the ventricular walls moving back toward the AV junction.
Explanation: The S wave is the first negative deflection after the R wave. It represents the late ventricular depolarisation.

9. d. Cl-
Explanation: Cystic fibrosis transmembrane conductance regulator (CFTR) is a membrane protein and chloride channel in vertebrates that is encoded by CFTR gene. (Refer: Autosomal Recessive  Disorders - Cystic fibrosis)
10. b. a free 2’O of an internal adenosine makes a nucleophilic attack on 5’P of first phase of intron
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Viruses: Definition and Classification

Viruses are infectious intracellular obligate parasites or infectious nucleoproteins.
It consists of nucleic acid (RNA/DNA) enclosed in a protein coat called capsid.
Nucleic acid along with protein coat is called as nucleocapsid.
In some cases apart from capsid, a membranous envelope may be present. Such viruses are called as enveloped viruses.

Virion: Intact infectious viral particle with nucleocapsid.

  • D. Ivanowski (1892) discovered Tobacco mosaic virus or TMV the causative agent of tobacco mosaic disease.
  • Later Stanley (1935) crystallized TMV and these crystals retain its infectivity even after keeping for long time.
  • F.W.Twort (1915) and F.H.Herelle (1917) discovered bacteriophages or viruses that infects bacteria (bacteria eaters)


  • Smallest viruses ~0.02 μm (20 nm) like Polio virus
  • large viruses ~0.3 μm (300 nm) like Small pox virus

Classification of viruses according to International committee of Viral Nomenclature
Classification of virus
Viral Genome
It may be linear or circular, double- or single-stranded DNA, Double- or single-stranded RNA
Depending on its type of nucleic acid, a virus is called as DNA virus or an RNA virus
Viral genome encodes genes for capsid synthesis and synthesis of essential enzymes for successful multiplication inside the host
Genome size 2 Kbp (Circoviruses) to 1.2 Mbp (Mimiviruses)
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Viral Symmetry or Capsid Symmetry

Classification of virus based on viral symmetry
Viruses are infectious intracellular obligate parasites or infectious nucleoproteins.
It consists of nucleic acid (RNA/DNA) enclosed in a protein coat called capsid.
Viral Symmetry The viral symmetry is determined by the orientation of capsid protein coat that protects the viral genome.
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Virus Capsid and Envelope: Structure and Function

Capsid is the protein coat around the viral genome. It differs in size, shape and symmetry. See classification based on viral symmetry.
Individual protein units are called capsomeres
Capsid proteins are encoded by the viral genome
Adeno virus
1. Protects nucleic acid from host nuclease degradation
2. Helps in introduction of viral genome to the host cell
3. Determines the antigenic specificity of virus
Capsid and Envelope
Viral envelope
In some cases apart from capsid, a membranous envelope may be present. Such viruses are called as enveloped viruses.
It is made of lipid and proteins rarely glycoproteins
Sometimes it may be a modified host plasma membrane or internal membranes
Eg: Herpes simplex, chickenpox virus, Influenza virus etc
Projections from the envelope are known as spikes or peplomers.
Function: attachment of the virus to the host cell.
HIV virus uses its spikes for this purpose.
See the difference between enveloped and non-enveloped viruses
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