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

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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.

FOR PREVIEWS 18-85 QUESTION ANSWERS CLICK HERE-

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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