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Sunday, 2 February 2014

Expected Cutoff Marks for UP LEO Exam 2014 DAH पशुधन प्रसार अधिकारी live extension officer

Department of Animal husbandry Uttar Pradesh DAHUP conducted a common written examination form the posts of पशुधन प्रसार अधिकारी / live extension officer or LEO posts. DAHUP LEO / Pashudhan Prasar Adhikari Written Exam was held on 2 February 2014 in various district of Uttar pradesh. After the recruitment process these Live Extension officer will be appointed in the pay scale of rs 5200-20200 and grade pay rs 2800 rs . 90% posts of Pashu dhan Prasar Adhikari will be filled via direct vacancy.

Expected Cutoff marks for Pashudhan Prasar adhikari | LEO posts Cutoff marks |  live extension officer  | DAH merit lists
Our few friends also participated in the written examination of Pashudhan Prasar adhikari posts. They share there exam experience with us. They told that question paper was not so much difficult and easily to attempt question in given time. 
LEO पशुधन प्रसार अधिकारी written exam question paper contains 100 question form various subject. total 1 hour and 30 minutes had given for attempting the question paper. Question was related to Hindi Subject, General study, Zoology, Botany and agriculture subjects. Difficulty level of the question paper was intermediate level. all question was objective type of the nature and those aspirant who cover the intermediate syllabus of Biology (Botany  + zoology ) of Up Board (UP madhymik Shiksha parishad ) they can easily score good marks.
Question of general study GS was related to latest current even and mostly question were related to GS of uttar pradesh.
Question of Hindi subject was objective type. Mostly question was related to Hindi grammar.
On the basis of above analysis and post exam discussion we got a conclusion about expected cutoff marks for पशुधन प्रसार अधिकारी / LEO / live extension officer of DAH. This cutoff marks may be differ to original cutoff so pleas share your own view about what will be expected cutoff marks on the basis of exam experiences
Cutoff marks for General / UR / un reserved category: 70 marks to 80 marks
Cutoff marks for OBC  : 60 to 70 marks
Cutoff marks for SC  : 55 to 65
Cutoff marks for ST : 50 to 60
Maximum marks: 100 , max
What do you think what will be expected cutoff marks LEO posts. ?
As we get Final answer key / Answer sheet and Results news from the official website of DAH /  , Mexam.IN will inform you immediately . so stay in touch.


  1. my view is differ to others i think paper of LEO is too easy therefore cut off marks is too high i think cut of marks for
    General 85-95
    OBC 75-84
    SC 65-74
    ST 55-64

    1. GEN 85 to 76
      OBC 76 to 74
      SC 76 to 72
      ST 76 to 65

    2. tere per agriculture sub hae naa,ishiliye easy lag raha hae

    3. not, yaar i m also bio student, and about 3 que in hindi, 5 que in gk and 15 que in bio are very tipicals, so merit depend on that these 23 que. there are a few lucky students solved them and around 6 to 8 que hardly answered correctly, so 23 - 8 = 15 que not answerd by any student. its my view, final merit may be diffe.r

  2. sahi kah rahe ho yesh bhai but ye to batao result kab tak aayega

  3. • solved questions……… UP LIVESTOCK EXTENTION OFFICER
    1. मनुश्य ईश्वरकी उत्क 2. राम को अनुत्तीर्ण होने की आशंका है 3. बन्दूक एक उपयोगी अस्त्र है 4. रागनी अपने आप चली गई 5. आप अपने घर जाऐं 6. प्रत्युत्पन्नमति 7. अनुग्रह 8. विलाप 9. लक्ष्य 10. आसन 11. बिहार 12.श्री रंगपट्टम 13.फारवर्ड ब्लाक 15.कुम्भ मेला 16.सं रा अमेरिका 16.विषुव 17.राष्टृपती 18.मुम्बई 19.भूकम्प तीव्रता 20.पूणे (महाराष्ट्र) 21.जोरोस्ट्रियन(पारसी )22.टेलीविजन 23.लोकपाल विल 24.CAG 25.national investigation agency 26.सरदार पटेल 27.vitamin D 28.बंधिनी 29.फारुख शेख 30.झारखंड

    31.sporozoite 32.plasmodium-monogenetic 33.euglena 34.paragastric cavity 35.ectodermal cells 36.planula aurelia 37. liver of sheep 38. wuchereria banchrofti-filaria 39.increased absorptin aria of digestive food 40.coelentron 41.ants bees waps 42ommatidia 43.cockroach-excretory organs fly-spongy tipe 45.silver fish 46.mollusca 47.the night 48.tadpole larva of herdmania 49.chordets but not vertibrate 50.amphioxus 52.sexual attraction 53.homoiothermic and amniotic 54.cocoon formation 55.have shelled yolky eggs 56.ratitae flightless birds 57.chitin 58.mitochondria 59.RNA is synthesized on DNA STRAND 60.Kinetochores at centromere = Tt 62.silurian period 63.120mmHg and 80 mmHg 64.blood clotting 65.balancing 66.Holoblastic 67.increase in the incidence of Skin Cancer 68.monosaccharides 69.Diploid 70.parthenocarpy
    71.Nucellus 72.bocal sac and ampelexory pad 73.CO2 74.Nymphyaecae 75.nucleare membrane 76.Mycobacterium tuberculosis-Spirilla 77.WILD FIRE OF TOBACCO 78.viruses grows, multiply and undergo mutation only with in NON LIVING host cells 79.Rocelia tinctoria 80.Nostoc 81.gibberellin causes longation of internodes 82.stroma- dark reaction 83.florigan 84.Cappilary water 85.all of above-intensive propertis 86.increased in both number of ions and ionic mobality of ions 87.velocity of both K+ and NO3- are nearly the same 88.Gracilaria 89.Auxin- responsible for apical dominance 90.TWO O2 required 91.Indole Acetic Acid 92.Mitochondrial matrics- KREB CYCLE
    93.auxin-EPINESTY 94.Gingo biloba 95.ornamental plants 96.Phloem- organic solutes translocation 97.Mesophyll-CAM cycle 98.Transpiration rate 99.Haemoglobin 100. 2 ATP NET Production in glycolysis
    its my according, if any mistake please give sugesions. final answer key may be varry…..

  4. vaishe mere 78-79 theek hain,per mujhe agriculture walo se problem hae,govt ko agriculture aur bio walo ki alag alag merit banani chahiye

  5. yaar AG walo ko bhi 80 percent solve karna muskil hoga, result kabtak ayega

  6. hello mr. Bdn
    i am pleasure to see your answer sheet but i think some answer in your sheet is wrong. i think the
    question no 100
    write answer is 8 ATP because in glycolysis total 8 ATP is formed. while the end product of glycolysis is 2ATP and 2 pyruvic acid.
    Question no 90
    write answer is 1 molecule because for conversion of pyruvic acid in to acetyal co-A 1 molecule of oxygen is required (refrence to sharma and sharma UP board book page no. 625) lakin mae god say pray kerta hu ke aap ka answer he sahe ho kyu ke by mistake mae 2 molecule answer ker k aya hu.
    Question no 69
    write answer is (n) heploid. because in alternation of generation of bryophytes the main plant body is (n ) gametophyte which are develops spore mother cell there fore answer is (n) rest of plant main plant body is 2n
    Question no 62
    Write answer is Cambrian period. because silurian period is the ages of fish.
    Question no 35
    Answer is interstitial cell. because it have ability of totipotancy. they can repair and regenrate all type of cell which are essential for colentrates
    i am also bio student belong to OBC . u know mujko 5 question k answer nahe pata chal pa rahae thay jo aap nae bata diya jismae sae 2 sahe ho gayae lakin 2 question aisae bhe thay jinko mae sahe saamaj raha tha magar aap k answer sae mae satisfied honae k baad mae samajh gaya ke wo galat ho gaya.
    abhe to chunao achar sanhits lag hae to result to ab june - july k baad he aa payega tab tak animal husbandary mae sab jugar chal raha hoga . sab ko jugar ka badia time mil gaya hae.

    1. thanks dear for ur suggestions,
      ur answers for 62. cambrian period may be right.
      but dear 69.spore mother cells in bryo are diploid, after meiosis every spore mother cell formed 4 haploid spores, these spores are show first stage of gametophyte. in 100. read care fully question is NET production of ATP in glycolysis, so net (end product) production of ATPs are 2. in que 35. nematocyst always developed from ectoderm in coelentrates

    2. if any confusion, u may contect to me 7499478130

  7. question no 100
    what is the net production of ATP during glycolytic conversion of ONE MOLECULE OF GLUCOSE TO TWO MOLECULE OF PYRUVIC ACID
    so the write answer is 8ATP
    Question is asking net (total) no's of ATP during 1 glucose to 2 pyruvic acid it not asking about end product of glycolysis
    Question no 35
    read the Dr. Ramesh gupta book old version before 2012 write answer is interstitial cell. otherwise tell me what is the function if interstitial cell
    Question no 69
    your answer is may be write because spore mother cell is produce by moss capsule and it is 2n structure

    1. que.35 nematocyst developed from interstitial cells is right, nimatocyst found in ectoderm but developed from interstitial cells of ectoderm.
      93. epinasty caused by both ethylene and auxin so answer may be ....
      100. net production ATP = total produced ATP - consumed ATP so answer is 2ATP

  8. I have no confusion that's why i am not calling u. sorry but your Question no 53 is also wrong the write answer is pokliothermic.and amniotic your question no 93 write answer is ethyline not the auxine hormones.

  9. Question no 100. The total no of ATP during glycolysis is 10 ATP while 2ATP is consumed in this process their fore
    Net production of ATP= 10 ATP-2ATP
    = 8ATP

  10. Question no 100
    Total no of ATP produced by glycolysis is 10 ATP, while 2ATP is consumed in this process therefore net gain of total ATP during this process is 8ATP so the write answer is 8ATP
    Net ATP= Total produced ATP-Total consumed ATP
    = 10 ATP-2ATP
    = 8ATP

  11. Result aa gaya hai..mera written me selection ho gaya hai dosto..mere 73 ques correct the.

    1. Congrates dear, i have also selected for interveiw

    2. plz tell me syllabus nd book which is helpfull in this exm

    3. hi; plz tell me syllabus nd book which is helpfull in this exm