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CSIR UGC NET JRF Life Sciences Part B (June 2014)

11. During protein synthesis in prokaryotes, the peptidyl transferase activity required for peptide bond formation is due to
a) ribosomal protein
b) aminoacyl tRNA
c) 16S ribosomal RNA
d) 23S ribosomal RNA

12. The distance between bacterial genes as determined from interrupted mating experiments are measured in units of
a)cM
b)bp
c)minutes
d)micrometers

13. Branchiostomaisa
a) deuterostome and schizocoelomate
b) protostome abd schizocoelomate
c) deuterostome and enterocoelomate
d) protostome and anterocoelomate

14. The most important reproductive strategies of big trees in a forest are
a) Earlier age at first reproduction and production of large number of small seeds
b) Earlier age at first reproduction and production of a small number of large seeds
c) Later age at first reproduction and production of a large number of small seeds
d) Later age at first reproduction and production of a small number of large seeds

15. Which one of the following is unfavorable for protein folding?
a)conformational entropy
b) Hydrogen bonding
c) Vander Waals interaction
d)hydrophobic interaction

16. The organs ‘radula’ and ‘citellum’ are found in
a) Mollusca and Annelida, respectively
b) Annelida and Mollusca , respectively
c)Coelenterata and Echinodermata, respectively
d) Echinodermata and Coelenterata, respectively

17.The peptide unit (-CO-NH-) is planar due to
a) Restriction around C-C bond
b) Restriction around C-N bond
c) Restriction around N-C bond
d) Hydrogen bonding between carbonyl oxygen and imino hydrogen of the peptide backbone.

18. Molecules primarily responsible for the formation of lipid raft are
Space-filling models of sphingomyelin (a) and cholesterol (b).

a) sphingolipids and cholesterol
b) phosphatidyl inositol and cholesterol
c) glycosylphosphatidyl inositol and chloesterol
d) phosphatidyl serine and phosphatidyl serine

19. Membrane bound and free ribosomes, are structurally identical, but differ only at a given time in terms of association with
a) Phospholipids
b) nascent proteins
c) glycosylated proteins
d) acetylated proteins

20. A pathogen like mycobacterium, which colonizes inside the cells of the host, is likely to be least affected by which one of the following host immune defence mechanisms?
a) Cytokines
b) cell mediated immune response
c) cd+4 T lymphocytes
d)Humoral immnune response
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Answers
11. d) 23S ribosomal RNA
12. c)minutes
13. c) deuterostome and enterocoelomate
14. d) Later age at first reproduction and production of a small number of large seeds
15. a)conformational entropy
16. a) Mollusca and Annelida, respectively
17. b) Restriction around C-N bond
18. a) sphingolipids and cholesterol
Lipid rafts generally contain 3 to 5-fold the amount of cholesterol found in the surrounding bilayer. Also, lipid rafts are enriched in sphingolipids such as sphingomyelin, which is typically elevated by 50% compared to the plasma membrane.
19. b) nascent proteins
20. d)Humoral immnune response
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Part B CSIR UGC Life Sciences Questions and Answers June 2014

1. The dye used in Gram staining is
a) Giesma
b) Methylene blue
c) Giesma
d) Crystal violet

2. A bioinformatics tool used to find the sequence similarity in the subunit of hemoglobin is
a)FASTA
b)HUMMER
c)BLAST
d)PSI :PLOT

3. Electron microscopes have much higher resolution than the type of light microscope because
a) of their higher magnification
b) of very short wave length of electrons
c) the lenses used are of much higher quality
d) the images are viewed on screen rather than directly using an eye piece or ocular lens

4. Which one of the following is a type of intercellular junction in animal cells?
a) Middle lamella
b) Desmosomes
c) Glycocalyx
d) Plasmodesmata

5. Ethylene signalling pathway is important for a fruit ripening. Which one of the following responses is routinely used to identify ethylene signalling pathway components?
a)Triple response
b) flowering time response
c) lateral root formation response
d)cotyledon expansion response

6. Which one of the following molecular marker types uses combination of both restriction enzyme and PCR techniques?
a)SSR
b)AFLP
c)SNP
d)RAPD

7.During the operation of C2 oxidative photosynthetic cycle, which of the following metabolites is transported from chloroplast to peroxisome?
a)Glycine
b)Glycerate
c)Glycolate
d)Serine

8. The long feather train of a peacock is quoted as an example supporting
The handicap principle
a) Hamilton’s rule
b) Zahavi’s handicap principle
c) The Red Queen hypothesis
d) Haldane’s rule

9. Which of the following is a correct hierarchical sequence for classification of living organisms?
a) Domain – Kingdom-Phylum – Class- Order – Family – Genus - Species
b) Kingdom- Domain – Phylum – Class- Order – Family – Genus - Species
c) Domain – Kingdom-Phylum – Order – Class- Family – Genus - Species
d) Kingdom- Domain – Phylum –Order – Class- Family – Genus – Species

10. A black Labrador homozygous for the dominant alleles (BBEE) is crossed with a yellow Labrador homozygous for the recessive alleles (bbee). On intercrossing the F1, the F2 progeny was obtained in the following ratio : 9 black : 3 brown: 4 yellow. This is an example of
a) recessive epistasis where allele e is epiststic to B and b
b) Recessive epistasis where allele e is epiststic to e and E
c) Dominant epistasis where allele E is epistatsic to B and b
d)complementary epistasis where allele b is epistastic .
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Answers
1. d) Crystal violet Refer: Gram staining Principle
2. c)BLAST Refer: Bioinformatics MCQ Quiz
3. b) of very short wave length of electrons Refer: Electron microscope vs Light Microscope
4. b)Desmosomes Refer: An overview of Intercellular junction
5. a)Triple response Refer: Major functions of Ethylene
6. b)AFLP
7. c)Glycolate Refer : C2 cycle pathway
8. b) Zahavi’s handicap principle
  • Zahavi's handicap principle is an argument for sexual selection in which the costliness of the male character, such as a peacock's tail for example, is positively useful to the female. The argument, originally suggested by the biologist Amotz Zahavi.
9. a) Domain – Kingdom-Phylum – Class- Order – Family – Genus - Species
10. a) recessive epistasis where allele e is epiststic to B and b
Refer: Example of Epistasis : Dominanat and Receessive Epistasis
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Simple Protocol on DNA Extraction from Onion Cells

An easy protocol for demonstration of DNA to students. Not even a centrifuge is required.
Aim: To isolate DNA from onion
Materials Required:
  • Onion bulbs
  • Extraction solvents: 50ml liquid detergent (labolene), 11.25g NaCl in 1000ml distilled water
  • Isopropyl alcohol or 95% ethanol kept on ice
  • Cheese cloth (4 thicknesses)
  • Water bath
  • Ice bucket
  • Beaker, glass rod, test tubes etc
dna exaction from onion
Procedure:
1. 10g of fresh peeled onion is weighed out and ground with mortar and pestle to breakdown the cell wall.
2. Add 20 ml extraction and grind well to release cell contents.
3. Transfer to a test tube and Incubate in a water bath at 50-600C for 15 minutes
4. Immediately incubate in ice water for 5 minutes
5. Filter through 4 layers of cheese cloth
6. To 3 ml filtrate, add 5 ml ice cold iso-propanol along the sides of the test tube without shaking
Keep this test tube undisturbed for 3-5 minutes in ice
White strands of DNA migrate from the filtrate to the alcohol layer
6. Using a glass rod spool the DNA out
7. Transfer the DNA to micro centrifuge tube and stored in TE buffer
Observation: Cloudy white strands of DNA appear in alcohol layer
Note: This protocol is not suitable for getting pure DNA for molecular biology experiments
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Function of SDS in DNA extraction and in SDS PAGE

Sodium lauryl sulfate or Sodium dodecyl sulphate (SDS). SDS is a strong anionic detergent used to denature proteins.
Applications:
  • Used in SDS-PAGE and  in DNA extraction procedure.
  • SDS-PAGE (Sodium dodecyl sulphate –polyacrylamide gel electrophoresis) is a technique for separating proteins based on size.
Mechanism of action:
function of SDS
  • In DNA extraction procedure, SDS is used for cell lysis and release of cell contents
  • In SDS PAGE, SDS has 2 function
         1. It denatures the protein
         2. It provides an overall negative change to the protein so that separation is based on size of the protein in SDS-PAGE.
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CSIR UGC NET JRF Life Science June 2014 Answer Key

This is the answer key for CSIR UGC NET JRF Life science June 2014 released by CSIR.

Please check your booklet series (A or B or C), then go through the key.

If you are new to the question paper work it out and go through the key..

Related:
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CSIR UGC NET JRF Life Sciences Exam Notification 2014 December

CSIR will hold   the Joint CSIR-UGC Test on 21st December, 2014 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.
CSIR UGC NET exam
Exam DateCSIR-UGC JRF( Junior Research Fellowship )And NET (Eligibility for Lectureship) : 21st December, 2014.
Important Dates:

                                     Exam Opportunities after Post Graduation in Life Sciences
Age Limit & Relaxation:
  • For Junior Research Fellowships  (JRF):Maximum 28 years as on 01-01-2014 (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.
                                                         Tips for CSIR UGC NET JRF exam
    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.
    csir fee details
    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 Cut-off 2013
    csir Exam Cut-off

    Exam Result& Validity period of fellowship:
    The final result of this Single MCQ test may be declared sometime in the month of March/April, 2015 and fellowship to successful candidates will be effective from 1 st April, 2015 with the validity period of 2 years for joining the fellowship under CSIR Scheme. 
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    Cut off marks for CSIR UGC NET june 2013 Life science

    The cutoff percentage for CSIR UGC NET JRF  life sciences for the year 2013 was 51%. 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. You can.............. just go for it
    Happy learning and best wishes for the exam
    From BE4U 
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    Uses of PET (Position Emission Tomography)

    Positron Emission Tomography (PET) scanning is a type of radionuclide scanning in which a radioactive substance is introduced into the body to assess structure and functions of tissues. It was developed in 1970. Unlike other scanning technique like CT scanning, PET does not produce good structural images; instead it gives information about the chemical activity of tissues or organs. It may also be used to assess blood flow.
    PET SCAN
    In this technique, the molecules that are to be taken up by the tissue or organ are labeled with a radioactive substance (radionuclide) before they are introduced into the body.  PET uses radionuclides that emit particles called positrons. The radiation produced by the particles is detected by PET scanner. The radiation produced by the particles is detected by PET scanner. The number of positrons emitted by an area of tissue or organ indicates how much radionuclei it has taken up.
    Uses
    · PET is used to study the heart and brain.
    · It is used to locate the origin of epileptic activity in the brain.
    · It is also used to detect tumours.
    · It is also used to study the mental health problems such as depression.
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