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ICMR JRF 2016 Notification Apply Online Now



ICMR in collaboration with PGIMER, Chandigarh will hold a National level examination for the award of Junior Research Fellowships (JRF), for Indian national candidates.
Exam DateJuly 17, 2016 (Sunday).
ICMR JRF exam 2016
A total of  120 fellowships awarded in Life Sciences stream (like microbiology, physiology, molecular biology, genetics, human biology, biotechnology, biochemistry, bioinformatics, biophysics, immunology, pharmacology, zoology, environmental sciences, botany, veterinary sciences. 
Another 100 candidates would be selected for consideration for positions of JRF under various research schemes of ICMR (subject to fulfilling the conditions for appointment under the schemes) for the duration of that scheme. These JRFs would also be permitted to complete Ph.D. while working in the scheme, if enrolled.  
The validity of result will be two years for placement in ICMR funded projects.
Important Dates
  • Date of opening of Registration: 09.04.2016
  • Last date for filling of online application: 09.05.2016
  • Late date of fee deposit in Bank: 12.05.2016
  • Date of Entrance Examination (Tentative): 17.07.2016
Duration and Emoluments: The existing value of the fellowship is at present Rs. 25000/- (Rupees Twenty Five  thousand only)  Per  month and an annual contingency grant up to Rs. 20,000/- per annum. The local institution shall review the performance of JRF after two years through an appropriate Review Committee constiuted by the Head of the institution. The fellow may be awarded SRF after successful assessment by the Review Committee.
    Age Limit: The age limit for admission to the eligibility test is 28 years (upper age limit relaxable upto 5 years in case of candidates belonging to SC/ST, physically handicapped (PH) and female candidates, 3 years in the case of OBC category.

    Examination fee  a) SC/ST/PH : Rs. 800/- 
                                    b) All other categories (General & OBC) : Rs. 1,000/-

    Examination Centre :  01) Chandigarh, 02) Chennai, 03) Delhi 04) Kolkata, 05) Mumbai, 06) Hyderabad  07) Guwahati 08) Varanasi 09 Bhopal Code Name of the Centre 10) Bhubaneshwar 11) Sri Nagar 12) Bangaluru

    Syllabus: As prescribed by UGC.

    Subjects covered under Life Sciences include microbiology, physiology, molecular biology, genetics, human biology, biotechnology, biochemistry, bioinformatics, biophysics, immunology, pharmacology, zoology, botany, environmental sciences and veterinary science.


    Free ICMR JRF Preparation Resources
    ICMR-JRF Exam + Question Pattern * ICMR-Model QuestionsSample question paper
    ICMR Previous Questions & Answers ( Set 1Set 2Set 3 /Set 4 / Biochemistry / Biostatistics )
    Scheme of Test
    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.

    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 will be made for each of
    the wrong answer
    . The questions in both the sections would appear in English only.
    The qualifying marks will be 55% obtained in both the sections (A+B or C) for General
    Category and OBC and 50% for SC/ST and physically handicapped.
    First and foremost thing is to begin the preparation now onwards
    “Wishing the very Best"
    Biology Exams 4 U is preparing with U for your exam. Keep Visiting ………
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    10 Quick Facts about Zika Virus

       Fact 1: About the Zika virus
    • Zika fever is caused by the Zika virus (ZIKV), an arthropod-borne virus (arbovirus).
    • The Zika virus is a member of the Flavivirus genus in the family Flaviviridae.



    • RNA virus
    • It is related to dengue, yellow fever, West Nile and Japanese encephalitis, viruses that are also members of the virus family Flaviviridae.


    Fact 2: Why the name ‘Zika’?
    • First identified in Uganda in 1947 in rhesus monkeys.  It is named after the ‘Zika forest’ in Uganda.
    • It was subsequently identified in humans in 1952 in Uganda and the United Republic of Tanzania.


    • The virus is known to circulate in Africa, the Americas, Asia and the Pacific.
        Fact 3: About the vector
    • The vector is Aedes mosquitoes mainly Aedes aegypti 
    Fact 4: Symptoms

    Fact 5: Diagnosis
    Laboratory testing for the presence of Zika virus RNA in the blood or other body fluids, such as urine or saliva.
    Fact 6: Treatment
    • No specific treatment or vaccine currently available. Zika virus disease is usually relatively mild and requires no specific treatment. If symptoms last for a longer period, seek medical attention.
    • The best form of prevention is protection against mosquito bites.
    Fact 7: Why Zika is considered dangerous?
    Potential complications: Two major diseases seem to be associated with Zika disease. They are

    a) Guillain-Barré Syndrome (GBS) is an autoimmune disease in which a person’s own immune system damages their nerve cells, causing muscle weakness and sometimes even paralysis.


    b) Microcephaly is a medical condition in which the brain does not develop properly resulting in a smaller than normal head. Recent report from Brazil suggests increased incidents of babies born with microcephaly from infected mother.
    Fact 8: Mode of Transmission
    • Primarily through the bite of an infected Aedes species mosquito (A. aegypti and A. albopictus)
    • Mother to child: A mother already infected with Zika virus near the time of delivery can pass on the virus to her newborn around the time of birth.
    • Zika virus can be spread by a man to his sex partners.


          The virus can survive in semen longer than in blood.
    Fact 9: “Still largely unknown”
    •  The mechanism of viral infection yet to be discovered.
    Fact 10: Global incidence of Zika?
    •  Many scientists suggest global warming as a reason for transmission of virus to low temperature countries. Rise in temperature favors the survival of mosquitoes, the vector of Zika virus.

    References:


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    59.75%- The cut off Percentage of CSIR UGC NET JRF Life Sciences December 2015

    To check the result Click this link 1. Result-CSIR UGC NET JRF in Life Sciences December 2015 
    To know your marks Click this link 2. Marks obtained-CSIR UGC NET JRF Life sciences December 2015
    Hearty congrats for the qualifiers..... and congrats for the aspirants as you are so close to your dream... 
    The cutoff percentage for CSIR UGC NET JRF  life sciences for the year 2015 December 
    was 59.75%.
     I just want to stress the point that whatever you learn, learn deep as each question will take you many miles towards your dream of qualifying this exam. If you are expecting a question from geological time scale, I know you need to remember all those eras and organism groups originated during that time period. Still it is a worth as you got an answer right and you are damn sure about it.



    Your strategy should be "Focus on topics where you can expect questions, in depth preparation will definitely give you the result you want. It is not about the quantity or vastness of the topic you have covered, it is all about your in depth understanding in the topics you covered.
    Just sit down, prepare a time table, work out maximum previous question papers and build your confidence.
    Success is the sum of small efforts repeated day in and day out. Spent at least 30 minutes a day for your preparation. Definitely, later on you could sit without any distraction for 1 hour or more. Remember all great journeys begins with a single step.
    You can.............. just go for it. Jump into the depth of the most beautiful and amazing “science of life”…..The following links may help you...
    Happy learning and best wishes for the exam....    From Biologyexams4u team
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    CSIR UGC NET JRF Life Sciences 2016 June Apply Online


    CSIR will hold   the Joint CSIR-UGC Test on 19th June, 2016 for determining the eligibility of the Indian National  candidates for the award of Junior Research Fellowships (JRF) and for determining eligibility for appointment of  Lecturers (NET) in certain subject areas falling under the faculty of Science & Technology. The award of Junior Research  Fellowship (JRF) to the successful eligible candidates will depend on their finding admission/placement in a university/  national laboratory/ institution of higher learning and research, as applicable
    Exam DateCSIR-UGC JRF(Junior Research Fellowship)& NET (Eligibility for Lectureship) :19th June, 2016.
    Important Dates:



    Age Limit & Relaxation:
    • For Junior Research Fellowships  (JRF):Maximum 28 years as on 01-01-2016 (upper age limit may be relaxed up to 5 years in case of candidates belonging to SC/ST/OBC, Physically handicapped/Visually handicapped and female applicants).
    • For LS (NET): No upper age limit.
    Educational Qualification:
    • BS-4 years program/BE/BTech/BPharma/MBBS/Integrated BS-MS/MSc or equivalent degree with at least 55% marks for general and OBC (50% for SC/ST candidates, physically and Visually Handicapped candidates).
    • Candidates enrolled for M.Sc or having completed 10+2+3 years of the above qualifying examination are also eligible to apply in the above subject under the Result Awaited (RA) category on the condition that they complete the qualifying degree with requisite percentage of marks within the validity period of two years to
      avail the fellowship from the effective date of award.
      Such candidates will have to submit the attestation format (Given at the reverse of the application form) duly certified by the Head of the Department/Institute from where the candidate is appearing or has appeared.
    • BSc (Hons) or equivalent degree holders or students enrolled in Integrated MS-PhD program with at least 55% marks for general and OBC candidates; 50% marks for SC/ST candidates, physically and visually handicapped candidates are also eligible to apply.Candidates with bachelor’s degree, whether Science, engineering or any other discipline, will be eligible for fellowship only after getting registered/enrolled for PhD/Integrated PhD program within the validity period of two years.
    • The eligibility for lectureship of NET qualified candidates will be subject to fulfilling the criteria laid down by UGC. PhD degree holders who have passed Master’s degree prior to 19th September 1991, with at least 50% marks are eligible to apply for Lectureship only.
      Syllabus
      The question paper shall be divided into three parts, (A, B & C) as per syllabus & Scheme of Exam.
      Part 'A' shall be common to all subjects including Engineering Sciences. This part shall contain questions pertaining to General Aptitude with emphasis on logical reasoning, graphical analysis, analytical and numerical ability, quantitative comparison, series formation, puzzles etc.
      Part 'B' shall contain subject-related conventional Multiple Choice questions (MCQs), generally covering the topics given in the syllabus.
      Part 'C' shall contain higher value questions that may test the candidate's knowledge of scientific concepts and/or application of the scientific concepts. The questions shall be of analytical nature where a candidate is expected to apply the scientific knowledge to arrive at the solution to the given scientific problem.
      Negative marking for wrong answers, wherever required, shall be applicable as per subject wise scheme of Exam.
      Examination fee:

      Examination Centres: Bangalore, Bhavnagar, Bhopal, Bhubaneshwar, Chandigarh, Chennai, Cochin, Delhi, Guntur, Guwahati, Hyderabad, Imphal, Jammu, Jamshedpur, Karaikudi, Kolkata, Lucknow, Nagpur, Pilani, Pune, Raipur, Roorkee, Srinagar, Thiruvananthapuram, Udaipur and Varanasi.
      Exam Result& Validity period of fellowship:
      The final result of this Single MCQ test may be declared sometime in the month of March/April, 2016 and fellowship to successful candidates will be effective from 1 st January 2017 with the validity period of 2 years for joining the fellowship under CSIR Scheme.
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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.
      PFGE
      Image credit 
      Procedure:
      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.
      References:
      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).
      *http://www.bio-rad.com/en-is/applications-technologies/pulsed-field-gel-electrophoresis
      **http://www.bio.davidson.edu/Courses/Molbio/MolStudents/spring2003/Cobain/method.html
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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
      Explanation:
      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 http://jpeg-optimizer.com, http://www.imageoptimizer.net.
      • 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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