AEROBIC RESPIRATION-GLYCOLYSIS

                                             MECHANISM OF RESPIRATION

Glycolysis or EMP pathway- The process  starts with glucose.In anaerobic respiration initial reactions are common as a result of which pyruvic acid is formed by breakdown of glucose. This process does not require oxygen. 

TCA cycle or Krebs cycle- After glycolysis if oxygen is present there is a complete oxidation of pyruvic acid into water and carbon dye oxide. This process called Krebs cycle.

Anaerobic respiration- If oxygen is absent, pyruvic acid form ethyl alcohol and carbon dye oxide.

Aerobic respiration consists of three steps- Glycolysis
                                                                     Krebs cycle
                                                                     Terminal oxidation
 In this article I am describing the first step- 
                                   Glycolysis.

Glycolysis is given by Embden, Meyerhof and Parnas so it also called EMP Pathway. Glycolysis is a process of breakdown of glucose or similar hexose sugar to molecules of pyruvic acid through a series of enzyme mediated releasing some energy and reducing power.It take place in cytoplasm.


                   

                                               
 Glycolysis held in 10 steps. 
Important points of this pathway is-
1.Function of kinase enzyme is add phosphate molecule.

2.In glycolysis ATP release at 2 places. 
These are-
(a)1,3 Biphosphpglycerte to 3 phosphoglycerate

 (b) phosphoenolpyruvate to pyruvate

3.One NADH release in this step-glycerleldehyde 3 phosphate to    3 phosphoglycerate.

4. Di hydroxy acetone phosphate also changed in glycereldehyde 3 phosphate so 2 molecules of glycereldehyde take part in one glycolysis cycle.
5.Total 4 ATP molecules release in one cycle and 2 molecule of  NADH.

TOTAL ENERGY PRODUCTION IN GLYCOLYSIS-

Direct ATP molecules-2x2= 4 ATP
By NADH molecules- 2x3= 6 ATP 
Total ATP molecules-4+6= 10 ATP
Used ATP in one cycle-  2 ATP
FINALLY GOT ATP MOLECULES- 10 ATP-2 ATP
                                                              = 8 ATP

NOTE-one NADH = 3 ATP.
           According to new rules 1 NADH= 2.5 ATP
           So if you do calculation according to new rules it is also correct.
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 questions about this topic-

1.What is the full form of ATP
Answer- Adinocin tri phosphate

2.What is the structure of pyruvate?
Answer-

3.How many pyruvic acid got in glycolysis?
Answer-2

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EASY NOTES CSIR UGC NET LIFE SCIENCE ABOUT CELL BIOLOGY CAPSULE 4

Today I come with a new capsule the capsule 4.

This capsule has best collection of questions about minor things which are very important for NET LIFE SCIENCE EXAMINATION.
A second thing is that the capsule is very easy to read and mesmerizer.
The capsule has 25 points about cell biology so please read carefully.
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EASY NOTES CSIR NET/JRF LIFE SCIENCE ABOUT CELL BIOLOGY CAPSULE 4
In my next article I  present a question paper of 25 questions you may solve it. The answers will publish after 3 days of question paper.
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IMPORTANT TOPICS FOR NET/JRF LIFESCIENCE EXAMINATION

Hello friends,
you know the NET EXAM date is announced and exam will be held in December month.

In this article I teach you miner skills to attempt the question paper.I you it will be helpful for you. I know some students has low perpetration but don't be worried some miner and quick tricks will be helpful for you.
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The important topics are-
1.Mostly I see that students left there lot of time in question study.I want to tell you that don't get nervous at exam time firstly choose the questions which are belongs to your well prepare topic.Read question 2 time and then select your answer.
2.If you don't complete your foll syllabus please read genetic and cell biology topics properly because these are important topics and many questions are belong to these topics.
3.Solve mind game questions because PART C of NET EXAM has special questions.
4.Solve previews year question papers.this  I suggest you Objective Life Science: MCQs for Life Science Examination (CSIR, ICAR, DBT, ASRB, IARI, NET, SET)2015 

 5.DNA replication and acids effect on cells are very important topics.
6.Applied zoology based questions are important.
7.Ask questions to me and I discuss the answer with description.
8.Special effects of amino acids, proteins are important points.on it helpful for you.
9.chromosome is important topic.
10.structural changes in chromosomes is very important for DEC. NET exam on this topic I will upload a video on this site h
If you have any other problem you can ask I try to give you a satisfied answer.

 My next capsule about genetic part will publish tomorrow please visit again.

CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 SOLVED PAPER WITH DESCRIPTION Q.NO. 86 to 90

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Q86. The following graph represents the expression of Tryptophan Synthetase (TS) in E.coli cels in absence or presence of Tryptophan in the medium-

If the two trp codons in the leader sequence of trp operon is mutated to ala, which of the following graph will best represent activity of TS in E.coli cells grown in the absence or presence of tryptophan?

(A)

(B)

(C)

(D)

ANSWER-B

Description-42-45 nucleotide coding leader sequence(trpL) contains an attenuator site.

Att-Transcription termination loop.(3:4)

Q87.The lifetime of a peptide bond in protein is very large,which statement below is incorrect with respect to stability of peptide bond?

(A)free energy of hydrolysis is negative

(B)The free energy of hydrolysis is positive and large

(C)The energy barrier to be crossed to go to the hydrolysed state in the large

(D)The peptide bond can be hydrolysed by6N HCl at 100 degree C.

ANSWER-B

Description-Because peptide bond hydrolysis is a spontaneous process. so it have highly negative value.

Q88. Following are certain statements related to euckryotic DNA replication-

1.The genome of multicellular animals contain many potential origins of replication.

2. During early development, when embryos are undergoing rapid cell divisions origin sites are uniformly activated.

3.'Pulse-chase' technique is used to label sites of DNA replication.

4.The rate of elongation of different DNA chains during genome replication varies drastically.

Which one of the following combinations of above statements are correct?

(A)1,2 and 3

(B)1,3 and 4

(C)2,3 and 4

(D)1,2 and 4

ANSWER-A

Description-Because the size of Eukaryotic genome is very large so,it contains multiple ori, uniformly transcribed at each ori, during early development stages.

'Pulse-chase' technique is also used to label the sites of DNA replication.

Method was used to determine the existence and function of Okazaki fragments(DNA replication)

Q89.A researcher wanted to immunize individuals of a particular area with viral infections. The researcher developed two different vaccine types(A and B) with the following properties-

1.When vaccine type A specific for a viral strain is administered to individuals, they develop strong neutralizing antibody response with very poor immunological memory.Hence it has to be administered in repetitive doses.

2.When vaccine type B specific for a viral strain is administered to individuals, they fail to develop circulating antibody response at the time of infection but they develop strong immunological memory.

If two viral strain V1(incubation period-2 days) and V2(incubation period-15days) are likely to infect the area which of the following vaccine combination would provide maximum immunization?

(A)V1 specific type A and V1 specific type B

(B)V1 specific type A and V2 specific type B

(C)V2 specific type A and V1 specific type B

(D)V2 specific type A and V2 specific type B

ANSWER-B

Description-Type A specific type to viral strain, produces strong neutralizing Abs but very poor immunological memory.

Hence repetitive booster doses are required.

But type B vaccine is generating strong memory.

Therefore to control V1 viral strain and V2, a combination of type A and B vaccine should be beneficial.

Q90.The expression of a hypothentical gene was analysed by Northen and Westenbolt hybridizations under control and induced condition.The results are summarized blow-

Expression of gene can be regulated by-

1.control at transcription initiation

2.alternative splicing

3.control the translation initiation

4.protein stability

Which of the above regulatory mechanisms can explain the observations show in the figures?

(A)Only 2

(B)Only 1 and 2

(C)Only 2 and 3

(D)1,2,3 and 4

ANSWER-D

Description-Northern bolt hybridization-RNA

                   Western bolt hybridization-Protiens

Expressions of gene can be regulated at any of following stages-A,B,C and D.

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CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 SOLVED PAPER WITH DESCRIPTION Q.NO. 81 to 85

CSIR UGC NET/JRF ANIMAL PHYSIOLOGY- HAEMOSTASIS

                                                         Haemostasis

When a blood vessel is damaged, loss of blood is stopped and healing occurs in a series of overlapping processes, in which platelets play a vital part, The more badly damaged the vessel wall is, the faster coagulation begins, sometimes as quickly as 15 seconds after injury.

1. Vasoconstriction-  platelets come into contact with the damaged blood vessel, there surface becomes sticky and they adhere to the damaged wall, then they release serotonin(5-hydroxytrptamine), which constricts the vessel, reducing blood flow through it. other chemicals that cause vasoconstriction, e.g. thromboxanes, are released by the damaged vessel itself.

2. Platelet plug formation- The adherent platelets clump to each other and release, other substances, including adinosine diphosphasate(ADP), which attract more platelets to the site. Passing platelets stick to those already at the damaged vessel and they too release their chemicals. This is a positive feedback system by which many platelets rapidly arrive at the site of vascular damage and quickly form a temporary seal-the platelet plug. Platelet plug formation is usually complete by 6 minutes after injury.

3. Fibrinolysis- After the clot has formed the process of removing it and healing the damaged blood vessel begins.The breakdown of the clot or fibrinolysis, is the first stage. An inactive substance called plasminogen is present in the clot and is converted to the enzyme plasmin by activators released from the damaged endothelial cells.cells. Plasmin initiates the breakdown of fibrin to soluble products that are treated as waste material and removed by phagocytosis. As.As the clot removed the healing process restores the ingratiate blood vessel.

EASY NOTES CSIR UGC NET LIFE SCIENCE ABOUT CELL BIOLOGY CAPSULE 3

Hello friends,

                     I hope you read my last two capsules. I hope you like that capsules please give your reviews and suggestions. I will try to provide you  best quality capsules. please regularly visit the site and study properly.
In my this capsule all questions are related to cell organs. I try to cover all important points about cell organs. May it helpful for you. read the capsule ans mesmerise if you have any question or problem you can ask the question in comment.

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CSIR UGC NET LIFE SCIENCE CELL BIOLOGY CAPSULE 3

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EASY NOTES CSIR UGC NET LIFE SCIENCE ABOUT CELL BIOLOGY CAPSULE 2

 CSIR UGC NET LIFE SCIENCE ABOUT CELL BIOLOGY                                               CAPSULE 2

I hope you read my first capsule about cell biology. 
So now I am come with the second capsule. This capsule also contains 25 questions.
 I want to say you that if you don’t read my capsule 1 so please read this first. It can helpful for your exam preparation.

Apply on-line for CSIR UGC NET EXAM

In my previews capsule I told you about my capsules and I also told that how its work.
 If you are new and you don't know about this you scan visit in this link and also read my first capsule. If you have questions you can ask and I will give answers but please read capsules regularly.

EASY CSIR NET LIFE SCIENCE NOTES CAPSULE 1

So now I starting the capsule second please click on this link to download the capsule 2.
EASY NOTES CSIR UGC NET LIFE SCIENCE ABOUT CELL BIOLOGY CAPSULE 2

The next capsule will be upload on 07.10.2015

EASY NOTES CSIR UGC NET LIFE SCIENCE ABOUT CELL BIOLOGY CAPSULE 1

When you ill you take capsule from doctor to regain your health.

 My unique capsule method for NET preparation is very easy like a doctor treatment. 

I make my notes in Capsule of 25-25 questions. 

How to take my Capsules:- Download Easy notes capsule as pdf file from here and memorize all 25 question of capsule after memorize first capsule download second capsule and memorize them..Continue this method until you memorize all of my 200 capsules which mean you memorize 25x200 = 5000 important points for CSIR UGC NET exam of Life sciences. 
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Now I am starting my capsule first about cell structure in this your will cover all important points about cell.

Download capsule 1 from this link

EASY NOTES CSIR UGC NET LIFE SCIENCE ABOUT CELL BIOLOGY CAPSULE 1

Capsule 2 will be publish on 30.09.2015 So visit again on 30.09.15 

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CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 SOLVED PAPER WITH DESCRIPTION Q.NO. 71 to 80

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CSIR UGC NET LIFE SCIENCE DEC. PAPER OUETION NO, 61 TO 70 WITH ANSWER

Q71. A gene producing red pigment was placed near centromeres  of fission yeast and thus subjected to position effect variegation and produced white colonies. A screen for mutants that increased the red pigment production was undertaken. Which of the following genes, when mutated, is likely to produce this genotype?
(a) Histone deacetylase 
(b) Histone acetylase
(c) RNA polymerase II
(d) TATA binding factor
Answer-A
Description- gene- red pigment
                    gene is placed near centromere: Inactivated due to position effect variegation. moved to centromere. Hence, produces white colonies. Normally deacetylase represses the gene expression but by the increased production of product could occur.

Q72. In order to prove that liposome can serve as a model membrane (mimicking cellular plasma membrane) and can be used as a target for complement- mediated immunolysis, an experiment as below is designed. To initiate such experiment, hepten- conjugated liposomes are made and loaded with umbelliferyl phosphate(UMP) ; hydrolysed product of UMP is umbelliferone and is fluorescent). Such loaded hapten conjugated liposomes in 10 mM Tris buffered saline,pH 7.4 were mixed with anti hapten antibodies and fresh guinea pigserum (as a source of complement) to induce immunolysis of lipsomal membrane. To quanyify only the membrane lysis component which of the assay sequences below is most appropriate?
(a) Mixture is ultra centrifuged and the supernatant reacted with alkaline phosphatase and fluorescence measured
(b) Mixture is sequentially reacted with phospholipase and alkaline phosphatase followed by fluorescence measurements
(c)Mixture is directly subjected to fluorescence measurement
(d) Mixture is treated with Triton X-100 and fluorescence measured
Answer-A
Description- If  the liposomal membrane is lysed, UMP fluorescence will be measured within the supernatant reacted with Alkaline phosphatase.

Q73.A null  mutation is created in a gene which is responsible for specific phosphorylation at 6th carbon position of mannose on acid hydrolases occurring in cis-Golgi. The following statements are given towards explaining the effect of this mutation-
(1) The lysosome will be devoid of lysosomal enzymes.
(2) lysosomal enzyme will be secreted out.
(3) Lysosomal enzymes will be get localized in cytoplasm.
which statement or combination of statements will explain the effect of mutation if the acid hydrolases in the mutant do not get degraded?
(a)1 and 3
(b) 2 and 3
(c)n 3 only
(d) 1 and 2
Answer-D
Description-Within cis-golgi an enzyme Glc-N-Ac phosphotransferase is responsible for adding a Glc-N-Ac-1-phosphate residue on the Mannose sugar through the formation of a phosphodiestet bond: Man-phosphate-Glc-N-Ac. Once formed the lysosomal enzymes are translocated to the trans golgi.

Q74.A newly identified sequence was experrimentally tested by in vitro transport assay using a radiolabelled protein containing the sequence to test import into mitochondria. Transport assay was done for a short time with or without membrane potential and after the assay, the mitochondria were either treated on not treated with proteinase K. At the end of the assay the mitochondria were pelleted and total protein of the pellet was isolated and separated on SDS-PAGE and autoradiographed. A representative auto-radiogram is shown below. Based on this experimental data, which of the following statements is not correct?
(A) The protiens goes into the matrix
(b)Not all the added protein was imported
(c) The protein requires membrane potential for import
(d) The protein is associated with the outer mitochondrial membrane
Answer-D
Description- The proteins inserted into mitochondria requires the membrane potential. The presence of proteinase K will be able to degrade the protein outside the mitochondria.

Q75. Acetyl-(Ala)18-CONH2 exists in alpha-helical conformation in solution. Most of the backbone dihedral angles(ᵠ and ᵩ) will be-
(A) -60 degree and -30 degree
(b) 60 AND 3. Degree
(c)-60 and -30 degree(50%) and 60 and 3. degree(50%)
(D)-80 and -120 degree
ANSWER-A


Q76. Enzyme parameters of four isoenzymes is given below-
          Isozyme                   Km micromolar                              Vmax
              A                                0.1                                               15
              B                                1.5                                               45
              C                                4.0                                               100
              D                                0.01                                             10
         These isoenzymes are localized in different tissues. In liver the substrate concentration is 0.2 micromolar. The liver isozyme is likely is to be-
(a)A                                        (b) B
(c) C                                       (d) D
Answer-A
Description- Km of an enzyme is the substrate concentration at which the reaction reaches half of its maximum.
At               [S] = Km;  V=Vmax/2
The Km of an enzyme tends to be similar to the maximum conc. of its substrate so the liver isozyme is likely to be A.

Q77. DNA is not hydrolysed by alkali whereas RNA is readily hydrolysed. This is due to-
(A) The double helix structure of DNA
(B) The presence of uridine in RNA
(C) Due to features observed in RNA such as stem-loop structures
(D) The presence of 2'-OH group in RNA
Answer-D


Q78. Two homologous proteins were isolate from a psychrophile(p) and a thermophile (T). The purified proteins were subjected to denaturation, protease digestion and circular dichroism (CD). Following observations were made-
1. The CD spectra of P and T proteins are identical
2.Their amino acid composition is 95% identical
3. T and P are equally susceptible to proteolysis in the presence or absence of reducing agent
4. T has higher midpoint of thermal denatu ration than P
The reason for enhanced stability in T is due to-
(A) Altered secondary structure
(B) Increased number of disulfieds in T
(C) Increase in water of hydration
(D) Increase in number of salt bridges
Answer-D
Description- The thermophile protein (T) is more stable as compared to psychrophile (P) as it possess large no. of salt bridges.

Q79. Binding of two ligands to their binding proteins were investigated. Following binding isotherms were obtained-
which of the following statement is correct?
(A) A is a obtained with an oligomeric protein and B is obtained with a monomeric protein
(B) B is obtained with protein with positive cooperativity
(C) A and B were obtained by the same protein at two different temperatures
(D) The profile B is not possible
ANSWER-B
Description- According to scatchard Plot, ligand binds specifically and non covalently with the receptor. A receptor  may have one or more binding site over itself.

Q80. Which of the following statement best described archaebacteria?
(A) Mostly autotrophic, cell wall contains peptidoglycan, 60S ribosomes, live in extreme environment
(B) Divide by fission not susceptible to lysozyme, live in extreme environments, mostly autotrophic
(C) Not susceptible to lysozyme, contain Golgi and linear chromosomes
(D) Chitinous cell wall, obligate aerobic, circular chromosomes
ANSWER-B
Description- The archaebacteria are the organisms that divide by fission, are not susceptible to lysozyme, can survive at extreme environmental conditioning and mostly these are autotrophic in nature.
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CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 SOLVED PAPER WITH DESCRIPTION Q.NO. 81 to 85

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CSIR UGC/ JRF NET life science December exam paper Q.No. 71-80

Q81.Puromycin is an antibiotic used to inhibit protein  synthesis. Given below are few statements about the antibiotic.

1.It enters the E-site of the ribosome where it prevents the release of deacetylated tRNA after the action of peptidyl transferase.

2.It blocks the translocation process by binding to the translocation factor EF-G

3. Puromycin resembles the initiatior tRNA, tRNA_i^f-met 

^{4.It resembles the aminoacyl tRNA and binds to the A-site of the ribosome}

^{5.Puromycin inhibits only prokaryotic protein synthesis.}

^{6.Puromycin inhibits both prokaryotic and eukaryotic protein synthesis.}

^{Which of the above statements are true?}

^{(A)1 and 5}

^{(B) 2 only}

^{(C) 4 and 6}

^{(D) 3 and 5}

ANSWER-C

Description- Puromycin is a secondary metabolite of Streptomyces alboniger which is a structural analog of aminoacyl t-RNA and binds to A-site on the ribosome . Peptidyl transferase enzyme catalyse the covalent linkage of puromycin to growing peptide chain covalently ultimately inhibiting protein biosynthesis in P.K and E.K

Q82.Total RNA was isolated separately from cytosol and nuclei of human cells growing in a cell culture. Each sample was mixed with a purified denatured fragment of a DNA corresponding to a large intron of a house keeping gene incubated under renaturating condition. given below are the statements amde about the outcome of the experiment.

1.RNA isolated from nuclei will from RNA-DNA duplexes because of the presence of introns in the primary RNA.

2.Cytosolic RNAs usually will not from RNA-DNA duplexes.

3. Both cytosolic and nuclear RNA will not from RNA-DNA duplexes as transcription and splicing occur simultaneously.

4.cytosolic RNA will from RNA-DNA duplexes because unspliced cytosolic RNAs are exceptionally stable.

Which of the above statements are most likely to be true?

(A) only 3

(B) 1 and 4

(C) 1 and 2

(D) only 4

ANSWER-C

Description-hnRNA in nucleus do not undergo ant splicing.Hence contains intronic sequences along with exonic part. After splicing that takes places in Nucleus introns are removed and exons are ligated to from m RNA molecule, ultimately m RNA is transported into cytosol i.e., out of the nucleus.

Q83. If a proteasome inhibitor is added to synchronously cycling human cells in G2 phase which one of the following events is likely to happen?

(A) Induce re-replication of DNA

(B) Arrest cells in G2 phase

(C) Arrest cells in Anaphase

(D) Block chromatin condensation

ANSWER-C

Description-proteasome- multi subunit enzyme complex that play a central role in the regulation of proteins that control cell cycle progression and apoptosis, an imp. target of cancer therapy.in G2 phase cell growth continues and proteins are synthesized in preparation for mitosis.

the spindle- Assembly checkpoint monitors the alignment of chromosomes the mataphase spindle.

Q84. A promoter deletion study was done in order to determine the binding sites for a transcription factor on the promoter, which is activated on treatment with the drug 'X'. The following constructs were made-

Luciferase assay revealed the following results-

The following statements can be made-

1. Region between -1800 ans -1210 contains a binding site for the activator

2.Region between -868 and -1210 contains a binding site for a repressor

3.Region between -868 and -432 contains a binding site for a repressor

4. Region between -1210 ans -868 contains a binding site for the activator

Which of above are true?

(A) 1 and 3

(B) 2 and 3

(C)1 and 4

(D) 2 only

ANSWER-C

Description-Activator protein binding region lies upstream to the transcription start site of sense strand.Promoter region is defined as DNA sequence, where transcription of a gene by RNA pol. begins. Both RNA pol. & necessary transcription factors(T.F) binds to the promoter sequences.

Q85. A pharmacy student designed a drug to specifically target the receptors for retinoic acid in order to prevent stem cell differentiation. After in vitro trial, the investigator found that the drug seemed to be underwent differentiation and the drug seemed to be ineffective. The following reasons were given by the student-

1.The size of drug exceeded the size of molecules that could cross the membrane.

2.The drug was small in size but hydrophobic in nature.

3.the drug did not bind to its receptors.

Which one of the above cold be the probable reason for drug ineffectiveness?

(A) Only 3

(B)1 and 3

(C) 1,2 and 3

(D)Only 2

Answer-B

Description-Retinoic acid receptor is a type of nuclear receptor which can also act as a T.F. that is activated by both all trans retinoic acid. so if the drug small+ hydrophobic.

Therefore easily cross PM and nuclear membrane.

CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 SOLVED PAPER WITH DESCRIPTION Q.NO. 61 to 70

CSIR UGC NET LIFE SCIENCE EXAM. DEC.  QUESTION NO.51 TO 60 WITH SOLUTION

Q61. The degree of genetic relatedness between the offspring and their parents-
(a) higher than that between sister and brother
(b) lower than that between sister and brother
(c) the same as that between sister and brother
(d) dependent on the number of siblings
Answer-C

Q62. You want to purify a recombinant protein of your interest. You can use affinity chormatograpty of purity as you have nickel columns available in the laboratory. With what molecule will you tag the protein to purity using those columns?
(a) GST
(b) Histidine
(c) Histamine
(d) proline
Answer-B

Q63. During which geological period was there an explosive increase in the number of many marine invertebrate phyla?
(a) Ordovician
(b) Devonian
(c) Permian
(d) Cambrian
Answer-D
Q64. An example of the species interaction called commensalism is-
(a) nitrogen- fixing bacteria in association with legume plants roots
(b) microbes in living human gut
(c) female mosquito deriving nourishment from human blood
(d) orchid plant growing on the trunk of a mango tree
Answer-D

Q65. In which ecosystem is the detrital pathway of energy flow most important?
(a) Lakes
(b) Grasslands
(c) Tropical rain forests
(d) Oceans
Answer-C

Q66. What parameter, plotted on Y- axis against generation time, would yield the curve slown in the figure?
(a) Survivorship
(b) Body size
(c) Lifespan
(d) Intrinsic rate of increase
Answer-D

Q67. In an experiment of detect a new protein in fixed cells, no secondary antibody tagged with fluorescence dye is available. what should be the best choice out of the following to detect the proteins?
(a) Protein A-FITC
(b) Protien A- Sepharose
(c) Biotin FITC
(d) Avidin-FITC
Answer-A

Q68. Lower limits of detection by sensord is important. Which method of detection is more sensitive than glass electrode used for pH measurement?
(a) Absorption spectroscopy
(b) Refractive index
(c) Circular dichroism
(d)Fluorescence spectroscopy
Answer-D

Q69. If  a researcher intends to identify a specific brain area activity linked to a cognitive function in human subjects, which one of the following techniques should be used?
 (a) CAT
(b) MRI
(c) fMRI
(d) Patch clamp
Answer-C

Q70. Which of the following statement is incorrect for fluorescence in situ hybridization (FISH) technique?
(a) A fluorescence or confocal microscope is used for detection of signal
(b) A labelled sequence of nucleotides are used
(c) Specific fluorescence tagged antibodies are used
(d) A stringent washing step is essenyial to remove appearance of non specific signal
Answer-C

CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 SOLVED PAPER WITH DESCRIPTION Q.NO. 51 to 60

CSIR UGC NET LIFE SCIENCE EXAM. DEC.  QUESTION NO.41 TO 50 WITH SOLUTION

 Q51. An interrupted mating experiment was performed between Hfr Str^s a⁺  b⁺ c^{+  and} f^- Str^r a^- b^- c^{- s}    strains. The genotype of majority of streptomycin resistant exconjugant after 10,20 and 30 minutes of interrupted mating is given below-
The most probable gene order would be-
(a) a b c                      (b) c a b
(c) b a c                      (d) a c b
Answer-D

Q52.Which one of the following functions is not served by the plasma proteins?
(a) Blood clotting
(b) O2 transport
(c) Hormone binding and transport
(d) Buffering capacity of blood
Answer-B

Q53. Two plants with white flower are crossed. White flower arise due top recessive mutation all F1 progeny have red flowers. When the F1 plants are selfed, both red and white flowered progeny are observed. In what ratio will red- flowered plants and white flowered plants occur?
(a)1:1           (b) 3:1
(c)9:7           (d) 15:1
Answer-C

Q54. The population density of an insect species increases from 40 to 46 in one month. If the birth rate during that period is 0.4 what is the death rate?
(a)0.25           (b)0.15
(c)0.87           (d)0.40
Answer-A

Q55. Two 18-residue helical peptides A and B are enantiomers. They can be distinguished by-
(a) recording there MALDI mass spectrum
(b) hydrolysis followed by amino acid analysis
(c) sequencing by Edman's method
(d) examining their circular dichroism spectra
Answer-D

Q56. Schizocoelous coelom formation, mouth formation from embryonic blastopore, spiral and determinate cleavage are characteristics of-
(a) deuterostomes
(b) pseudocoelomates
(c) protists
(d) protostomes
Answer-D

Q57.Which species concept utilizes morphological and molecular characters to distinguish between species?
(a) Evolutionary
(b) Ecological
(c) Biological
(d) Phylogenetic
Answer-D

Q58. Worker bees, instead of themselves reproducing, help the queen reproduce. This behaviour is explained as an example of-
(a) kin selection
(b) group selection
(c) sexual selection
(d) natural selection
Answer-A

Q59.Which of the following is a correct match of the animal with its taxonomic group?
(a)Hirudinea-leech, Chelicerata-Horse shoe creb, Cestoda-Tapeworm Echinoidea-sea urchins, Earthworm Oligochaeta-Earthworm
(b)Hirudinea-Earthworm,Chelicerata-Horse shoe creb,Cestoda-Octopus, Echinoidea-Tapeworm,Cephalopoda-Earthworm,Oligochaeta-leech,
(c)Hirudinea-Tapeworm,Chelicerata-leech, Cestoda-Tapeworm,Echinoidea-Horse shoe creb,  Cephalopoda-Earthworm,Oligochaeta- Octopus,
(d)Hirudinea-leech, Chelicerata-Tapeworm,Cestoda-Earthworm,Echinoidea-sea urchins, Cephalopoda- Octopus,Oligochaeta-Horse shoe creb, 
Answer-A

Q60.The wings of insect and the wings of bats represent a case of-
(a) divergent evolution
(b) convergent evolution
(c) parallel evolution
(d) neutral evolution
Answer-B

CSIR UGC/JRF NET LIFE SCIENCE DEC. EXAM PAPER QUESTION NO. 61 TO 70 WITH SOLUTION