Best free csir net life science notes pdf

Dear Students,

Today I share you a best free pdf e book for life science basics, this free e book is made by department of basic education Republic of South Africa.

this free e book is basically make for 12 th grade of south africa but this book not only contains basic of life sciences but it will also give us:-

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3-Give study tips

4-Excellent mind map and mnemonics tips for every life science student.

So this free pdf book is essential and must read for all 12 th to M.sc (life science) students.

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Best free basic life science with memory tips and easy csir net life science notes pdf 

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Free Capsules of csir net life science

CAPSULE 5-ABOUT STRUCTURAL CHANGES IN CHROMOSOMES

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structural changes in chromosome play effective role in genetics.
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CSIR UGC NET/JRF LIFE SCIENCE DECEMBER 2015 EXAM PAPER AND ANSWER KEY

last December CSIR UGC(Human Resource Development Group Council of Scientific & Industrial Research) organized NET/JRF EXAM on 20th December .This exam was held in three shifts A, B and C.

 For all booklets answer keys are different.

Here I am providing all three booklets official exam papers and answer keys. Candidates check there answer key according there booklet number.

CSIR UGC NET/JRF LIFE SCIENCE DEC. 2015 

Booklet A Question paper

Booklet B Question paper

Booklet C Question paper

DOWNLOAD ANSWER KEY FOR A B AND C BOOKLETS HERE-

ANSWER KEY

CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 SOLVED PAPER WITH DESCRIPTION Q.NO. 91 to 95

Click here for-
86 TO 90 QUESTIONS WITH ANSWER OF CSIR UGC NET/JRF EXAMINATION

Q91.For successful fertilization in sea urchin,interaction between the surface of the egg and acrosomal proteins, specifically a 30.5 kDa protein called bindin, is necessary. The following factors could affect this interaction and prevent fertilization-
1.Removal of egg jelly polysaccharides.
2.Removal of bindin receptors on the egg vitelline membrane.
3.Removal of bindin receptors from the egg jelly.
4.Removal of bindin receptors from a single culster on the vitelline membrane.
Which one or the combination above statements is correct?
(A)1 and 4                                      (B)Only 2
(C)1 and 2                                      (D)Only 3
Answer-B
Description-Bindin is a 30.5kDa (Head of spermatozoa acrosomal protein) is a major protein of sea urchin required for fertilization(interaction with the surface of egg via specific receptor on egg surface)
Removal of bindin receptors on the egg vitelline memb. will affect this interaction and prevent fertilization.

Q92.A mutant embryo of Drosophila in which one of the major sex determining gene.sex lethal, can only undergo default splicing. was allowed to develop. The following statements are towards explaining the determination of sex of the embryo-
1.The embryo will develop into a male fly.
2.The embryo will develop into a female fly.
3.sex lethal gene product directly regulates sex specific alternate splicing of double sex RNA.
4. sex lethal  gene product regulates sex specific splicing of transformer RNA which in turn regulates splicing of double sex RNA.
The correct combination of above statements to explain sex determination of the given embryo is-
(A)1 and 3
(B) 1 and 4
(C)2 and 4
(D) 2 and 3
Answer-B
Description-Sxl-gene is sex lethal gene is a master switch gene for somatic cell determination in D. melanogaster.
In XX animals, Sxl become activated and imposes female development,while in XY animals. Sxl remains inactive and male development ensures.
Therefore,default splicing will inactivate Sxl protein product.Hence male develops.
Also sex lethal is an RNA splicing enzyme whose immediate target is transformer mRNA.Since, Sxl are not transcribed in males, its action on Transformer is restricted to females.
sex lethal acts positively in the functional splicing of transformer mRNA. Transformer is another splicing factor. Hence determines female development fate.

Q93.A two celled embryo made of blastomeres A and B. If the two blastomeres are experientially separated, the A blastomere generates all the cells it would normally make. However the B blastomere in isolation makes only a small fraction of cells it would normally make.Based on the above observations only, which one of the following conclusions is correct?
(A) A blastomere is autonomously specified while B blastomere is conditionally specified.
(B) A blastomere is conditionally specified while B blastomere is autonomously specified.
(C) Descendants of A blastomere are autonomously specified.
(D)Descendants of B blastomere can either be autonomously specified or conditionally specified.
ANSWER-A

94.A mutant was experimentally generated which has wings reduced to haltere like structure. The following statements are put forward regarding this phenotype-
1.ultrabithorex gene ectopically expressed in second thoracic segment.
2. antennapedia gene ectopically  expressed in second thoracic segment.
3.A homeotic mutation.
4. A mutation in gap gene.
The following combination of statements will be most appropriate explaining the molecular basis of mutant phenotype-
(A)1 and 2
(B)2 and 3
(C)3 and 4
(D)1 and 3
ANSWER-D
Description- Ultrabithorax (activated in absence of hunchback protein) Ubx gene found in insects.
In D.melanogaster it is expressed in the 3rd thoracic and 1st abdominal segments and repress wing formation. Ubx gene regulates the decisions regarding the no. of wings and legs.
Homeotic mutation causes tissues to alter their normal differentiation pattern producing integrated structures but in unusual locations.

Q95.Following are certain statements regarding the activities of homeotic genes of classes A,B and C involved in floral organ identity-
1.Activity of A alone specifies sepals.
2.Activity of B alone specifies petals.
3.Activity of B and C form stamens.
4.Activity of C alone specifies carpels.
Which one of the following combinations of above statements is correct?
(A)1,2 and 3
(B)1,2 and 4
(C)2,3 and 4
(D)1,3 and 4
ANSWER-D

AEROBIC RESPIRATION-KREBS CYCLE

The molecule of pyruvic acid which is made in glycolysis enters in kerbs cycle(if oxygen is present)
Krebs cycle start with aetyl coA so firstly pyruvic acid changed in acetyl coA. For this TPP co-enzyme and Mg++ are important. In this process oxidation and decarboxilation held together so this process called oxidation decarboxilation. For this process pyruvic dehydrogenase complex is important. This complex is group of three enzymes.These enzymes are-
1.Pyruvate decarboxynilase
2Lipoate reductase trans acetylase
3.De hydrolipoate dehydrogenase

Energy yieldduring oxidative decarboxilation-
2 pyruvic acid+2NAD-2 acetyl coA+ 2NADH2
2NADH2- 6ATP

For first step of aerobic respiration click here-GLYCOLYSIS

KREBS CYCLE/TCA  CYCLE/CITRIC ACID CYCLE
Krebs cycle was discovered by Hans Krebs.It occur inside mitochondria.It is a amphibolic path way because it is a comon path wat for protien,lipids catabolism. Acetyl coA functions as sbstrate entrant for Kreb cycle.This cycle complates in 10 steps.These are-


1.Condensation-Acetyl coA combines with oxalo acetate in the presence of citrate synthetase to form citric acid.
Acetyl coA+OAA→citric acid+CoA-Sh

2.Dehydration-Citrate forming Aconitic acid with releasing of water molecule in the presence of aconitase.
Citrate→cis aconitic acid

3.Hydration-cis aconitate is converted into isocitrate with the addition of water in the presence of aconitase.
Cis aconitate+water molecule→Isocitrate

4.Dehydrogenation-Isocitrate is dehydrogenated to oxalosuccinate in the presence of enzyme isocitrate dehydrogenase and Mn++.NADH2 is produced.
isocitrate+NAD+→oxalosuccinate+NADH2

5.Decarboxilation-Oxalosuccinate is decarboxylated to from α-ketoglutarate through enzyme decarboxylase.
Oxalosuccinate→ α-ketoglutarate

6.Dehydrogenation and Decarboxylation- α-ketoglutarate is both dehydrogenated and decarboxylated by an enzyme complex  α-ketoglutarate dehydrogenase.the enzyme complex contains TPP and lipoic acid.The product combines with coA to from succinyl CoA.
 α-ketoglutarate+CoA+NAD→SuccinylCoA+NADH+H<sup>+

7.Formation of ATP/GTP-Succinyl CoA is acted upon by enzyme succinyl thiokinase to form succinate.The reaction release sufficient energy to form ATP or GTP.</sup>

Succinyl CoA+GDP/ATP→ Succinate+CoA+GTP/ATP

^{8.Dehydrogenation-Succinate undergoes dehydrogenation to form fumarate with the help of a dehydrogenase.FADH2 is produced.}

^{Succinate+FAD→Fumarate+FADH2}

^{9.Hydration-A molecule of water gets added to fumerate to form malate. The enzyme is called fumarase.}

Fumarate+H₂O→Malate

^{10.Dehydrogenation-Malate is Dehydrogenated or oxidised through the energy of malate dehydrogenase to produce oxaloacetate. Hydrogen is accepted by} NADP⁺NAD⁺

Malate=NAD(P)⁺→Oxaloacetate+NAD(P)H⁺

^{Oxaloacetate picks up another molecule of activated acetate to repeat the cycle.}

^{A molecule of glucose yields two molecules of NADH2,2ATP and two pyruvate while undergoing glycolysis. The two molecules of pyruvate are completely degraded in Krebs cycle to form 2 molecules of ATP,8NADH2 and 2FADH2.}

^{Energetics of Krebs cycle reaction                                  No. of ATP}

1.Pyruvic acid to actyle cova(NAD)                                   3ATP
2.Isocitric aid to oxallosuccinic acid(NAD)                        3ATP
3. α-Ketogutric acid to succinyl cova(NAD)                       3ATP
4.Succinic acid to fumaric acid(FAD)                                2ATP
5.Malic acid to OAA(NAD)                                                3ATP
6.Succinyl coA to succinic acid                                         1ATP
                                                                                           =15ATP
Total ATP produced during
 complate oxidation of Pyruvic acid =2x15=30ATP
                       And during glycolysis=8ATP
                                     TOTAL ATP=38ATP
                                   

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

Click here for- CSIR UGC NET/JRF LIFE SCIENCE 2014 SOLVED QUESTION PAPER

LEARN IMPORTANT QUESTIONS IN FORM OF EASY CAPSULES. DOWNLOAD NOW-
EASY LIFE SCIENCE QUESTION CAPSULES

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.
Download CAPSULE 4 here-

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.
click here for-CAPSULE 3

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