Category: 1st PUC

You can Download Chapter 1 Duryodhana Vilapa Questions and Answers Pdf, Notes, Summary, 1st PUC Kannada Textbook Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Kannada Textbook Answers Sahitya Sanchalana Chapter 1 Duryodhana Vilapa

Duryodhana Vilapa Questions and Answers, Notes, Summary

You can Download Chapter 1 बड़े घर की बेटी Questions and Answers Pdf, Notes, Summary, 1st PUC Hindi Textbook Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Hindi Textbook Answers Sahitya Vaibhav Chapter 1 बड़े घर की बेटी

बड़े घर की बेटी Questions and Answers, Notes, Summary

I. एक शब्द या वाक्यांश या वाक्य में उत्तर लिखिए:

Bade Ghar Ki Beti Notes KSEEB Solution प्रश्न 1.
ठाकुर साहब के कितने बेटे थे?
उत्तर:
ठाकुर साहब के दो बेटे थे।

Bade Ghar Ki Beti Question Answers KSEEB Solution प्रश्न 2.
बेनीमाधव सिंह अपनी आधी से अधिक संपत्ति किसे भेंट के रूप में दे चुके थे?
उत्तर:
बेनीमाधव सिंह अपनी आधी से अधिक संपत्ति वकीलों को भेंट कर चुके थे।

Bade Ghar Ki Beti Class 11 KSEEB Solution प्रश्न 3.
ठाकुर साहब के बड़े बेटे का नाम क्या था?
उत्तर:
ठाकर साहब के बड़े बेटे का नाम श्रीकंठ सिंह था।

बड़े घर की बेटी प्रश्न उत्तर Pdf KSEEB Solution प्रश्न 4.
श्रीकंठ कब घर आया करते थे?
उत्तर:
श्रीकंठ सिंह शनिवार को घर आया करते थे।

बड़े घर की बेटी Questions And Answers KSEEB Solution प्रश्न 5.
आनंदी के पिता का नाम लिखिए।
उत्तर:
आनंदी के पिता का नाम भूपसिंह था।

Beni Madhav Singh Ke Parivar Ka Parichay Dijiye KSEEB Solution प्रश्न 6.
थाली उठाकर किसने पलट दी?
उत्तर:
लालबिहारी सिंह ने थाली उठाकर पलट दी।

Bade Ghar Ki Beti Question Answer KSEEB Solution प्रश्न 7.
गौरीपुर गाँव के जमीनदार कौन थे?
उत्तर:
बेनीमाधव सिंह गौरीपुर गाँव के जमीनदार थे।

Bade Ghar Ki Beti Class 11 Notes KSEEB Solution प्रश्न 8.
किसकी आँखें लाल हो गयी थी?
उत्तर:
श्रीकंठ की आँखे लाल हो गयी थीं।

बड़े घर की बेटी कहानी Pdf Notes KSEEB Solution प्रश्न 9.
बिगड़ता हुआ काम कौन बना लेती हैं?
उत्तर:
बड़े घर की बेटियाँ बिगड़ता हुआ काम बना लेती हैं।

Bade Ghar Ki Beti Answers KSEEB Solution प्रश्न 10.
‘बड़े घर की बेटी’ कहानी के लेखक कौन हैं?
उत्तर:
‘बड़े घर की बेटी’ कहानी के लेखक उपन्यास सम्राट प्रेमचन्द हैं।

अतिरिक्त प्रश्नः

Bade Ghar Ki Beti Question And Answers KSEEB Solution प्रश्न 11.
बेनीमाधव सिंह किस गाँव के ज़मीनदार थे?
उत्तर:
बेनीमाधव सिंह गौरीपुर गाँव के ज़मीनदार थे।

बड़े घर की बेटी प्रश्न उत्तर KSEEB Solution प्रश्न 12.
ठाकुर साहब के छोटे बेटे का नाम क्या था?
उत्तर:
ठाकुर साहब के छोटे बेटे का नाम लाल बिहारी सिंह था।

Bade Ghar Ki Beti Hindi Notes KSEEB Solution प्रश्न 13.
श्रीकंठ सिंह का किसके साथ ब्याह हो गया?
उत्तर:
श्रीकंठ सिंह का ब्याह आनंदी के साथ हो गया।

प्रश्न 14.
कौन स्वभाव से ही दयावती थी?
उत्तर:
आनंदी स्वभाव से ही दयावती थी।

प्रश्न 15.
बुद्धिमान लोग किनकी बातों पर ध्यान नहीं देते?
उत्तर:
बुद्धिमान लोग मूरों की बातों पर ध्यान नहीं देते।

प्रश्न 16.
मुगदर की जोड़ी किसने बनवा दी थी?
उत्तर:
मुगदर की जोड़ी श्रीकंठ ने बनवा दी थी।

II. निम्नलिखित प्रश्नों के उत्तर लिखिएः

प्रश्न 1.
बेनीमाधव सिंह के परिवार का संक्षिप्त परिचय दीजिए।
उत्तर:
प्रस्तुत कहानी ‘बड़े घर की बेटी’ जिसके लेखक प्रेमचंद हैं।
बेनीमाधव सिंह गौरीपुर गाँव का जमींदार था। उसके पिता किसी समय बड़े धन धान्य सम्पन्न थे। गाँव में भव्य मंदिर एवं पक्का तालाब बनवाया था जिसकी अब मरम्मत भी मुश्किल थी, कहा जाता है कि कभी उनके दरवाजे पर हाथी झूमता था, आज वहाँ बूढ़ी भैंस थी। उनकी वर्तमान आय एक हजार रुपये वार्षिक से अधिक न थी, आधी से अधिक संपत्ति वकीलों को भेंट कर चुके थे। इनके दो पुत्र थे। बड़े बेटे का नाम श्रीकंठ, छोटे बेटे का नाम लालबिहारी सिंह था। श्रीकंठ अध्ययनशील और मेहनती लड़का था। उसने बी.ए. की डिग्री प्राप्त कर, शहर में नौकरी में लग गया था, साथ ही उसे आयुर्वेद औषधियों में अत्यधिक रूचि भी थी। वह भारतीय संस्कृति एवं सभ्यता से प्रेरित था जिसके कारण उसका गाँव में बड़ा सम्मान था। लालबिहारी सिंह दोहरे बदन का नौजवान था जो अपने बड़े भाई की तुलना में अधिक रौबीला एवं स्वस्थ था। वह खूब खाता पीता और मस्ती में व्यस्त रहता था। बड़े बेटे की शादी आनंदी से हुई जो एक सुशील, सम्पन्न परिवार की लड़की थी। छोटे बेटे का विवाह एक साधारण जमींदार की लड़की के साथ हुआ।

प्रश्न 2.
आनंदी ने अपने ससुराल में क्या रंग-ढंग देखा?
उत्तर:
आनंदी एक बड़े घर की बेटी थी। वह अपने घर में सभी सुख-सुविधाओं में पली थी। वे सारी सुख-सुविधाएँ यहाँ ससुराल में नहीं थीं। यहाँ न तो बाग-बगीचे थे, न मकान में खिड़की और जमीन पर फर्श। फिर भी आनंदी ने कुछ ही दिनों में अपने आपको इस घर के अनुकूल बना लिया।

प्रश्न 3.
लालबिहारी आनंदी पर क्यों बिगड़ पड़ा?
उत्तर:
एक दिन लालबिहारी सिंह ने माँस पकाने के लिए कहा। आनंदी ने माँस पकाते समय हाँड़ी में जो घी था, वह सब डाल दिया। जब लालबिहारी ने दाल में घी डालने के लिए कहा, तो आनंदी ने कहा कि माँस पकाने में घी खत्म हो गया। इसी कारण लालबिहारी आनंदी पर बिगड़ गया।

प्रश्न 4.
आनंदी बिगड़ता हुआ काम कैसे बना लेती है?
उत्तर:
श्रीकंठ सिंह के छोटे भाई लालबिहारी सिंह के अभद्र व्यवहार से आनंदी क्रोधित हो जाती है। गुस्से में उसने अपने पति से सारी शिकायतें कीं। झगड़ा इतना बढ़ गया कि घर टूटने तक पहुँच गया। तब बिखरते हुए घर को देखकर आनंदी शांत हो जाती है और लालबिहारी को घर छोड़कर जाने से रोक लेती है। इस प्रकार आनंदी बिगड़ते काम को बना लेती है।

प्रश्न 5.
आनंदी का चरित्र-चित्रण कीजिए।
उत्तर:
आनंदी एक बड़े घर की रूपवान, गुणवान तथा आदर्शवादी महिला है। यद्यपि वह अपने मायके में सुख-सुविधाओं में पली है, फिर भी वह परिस्थितियों के अनुकूल अपने आपको बदल सकती है। वह अपने घर की एकता बनाये रखना चाहती है और बिखरते घर को बचा लेती है। यही उसका बड़प्पन है कि बिगड़ते काम को बना लेती है।

अतिरिक्त प्रश्नः

प्रश्न 6.
श्रीकंठ सिंह के स्वभाव के बारे में लिखिए।
उत्तर:
श्रीकंठ सिंह बेनीमाधव सिंह के बड़े लड़के थे। वैद्यक ग्रंथों पर उनका विशेष प्रेम था। प्राचीन हिंदू सभ्यता का गुणगान उनकी धार्मिकता का प्रधान अंग था। संयुक्त परिवार को वे पसंद करते थे। अंग्रेजी पढ़े होने के बावजूद भी वे अंग्रेजी सामाजिक प्रथाओं के विशेष प्रेमी नहीं थे। दशहरा के उत्सव में वे बड़े उत्साह से भाग लेते थे।

प्रश्न 7.
भूपसिंह का परिचय दीजिए।
उत्तर:
भूपसिंह एक छोटी-सी रियासत के ताल्लुकेदार थे। बड़े उदार-चित्त और प्रतिभाशाली पुरुष थे। उनको एक भी लड़का नहीं था। सात लड़कियां थी। आनंदी उनकी चौथी लड़की थी। उनका विशाल भवन था। एक हाथी, तीन कुत्ते, बाज, बहरी-शिकरे सब थे।

प्रश्न 8.
आनंदी का ब्याह श्रीकंठ सिंह के साथ कैसे हो गया?
उत्तर:
भूपसिंह अपनी चौथी लड़की आनंदी के लिए विवाह देख रहे थे। तभी एक दिन श्रीकंठ सिंह उनके पास नागरी-प्रचार का चंदे का रूपया माँगने आये। भूपसिंह उनके स्वभाव पर रीझ गए और धूमधाम से श्रीकंठ सिंह का आनंदी के साथ ब्याह हो गया।

III. निम्नलिखित वाक्य किसने किससे कहे?

प्रश्न 1.
जल्दी से पका दो, मुझे भूख लगी है।
उत्तर:
यह वाक्य लालबिहारी ने आनंदी से कहा।

प्रश्न 2.
“जिसके गुमान पर भूली हुई हो, उसे भी देखूगा, और तुम्हें भी।”
उत्तर:
यह वाक्य लालबिहारी ने आनंदी से कहा।

प्रश्न 3.
बेटा, बुद्धिमान लोग मूरों की बात पर ध्यान नहीं देते।
उत्तर:
यह वाक्य बेनीमाधव ने श्रीकंठ से कहा।

प्रश्न 4.
लालबिहारी को मैं अब अपना भाई नहीं समझता।
उत्तर:
यह वाक्य श्रीकंठ ने अपने पिता बेनीमाधव सिंह से कहा।

प्रश्न 5.
अब मेरा मुँह नहीं देखना चाहते, इसलिए अब मैं जाता हूँ।
उत्तर:
यह वाक्य लालबिहारी ने आनंदी से कहा।

प्रश्न 6.
भैया, अब कभी मत कहना कि तुम्हारा मुँह न देगा।
उत्तर:
यह वाक्य लालबिहारी ने श्रीकंठ से कहा।

प्रश्न 7.
मुझसे जो कुछ अपराध हुआ क्षमा करना।
उत्तर:
यह वाक्य लालबिहारी ने आनंदी से कहा।

अतिरिक्त प्रश्नः

प्रश्न 8.
“बहू-बेटियों का यह स्वभाव अच्छा नहीं कि मर्दो के मुँह लगे।”
उत्तर:
बेनीमाधव सिंह ने श्रीकंठ से कहा।

प्रश्न 9.
“लाला बाहर खड़े बहुत रो रहे हैं।”
उत्तर:
आनंदी ने अपने पति श्रीकंठ से कहा।

प्रश्न 10.
“इतनी गरम क्यों होती हो, बात तो कहो।”
उत्तर:
श्रीकंठ ने यह वाक्य आनंदी से कहा।

IV. ससंदर्भ स्पष्टीकरण कीजिए:

प्रश्न 1.
अभी परसों घी आया है, इतनी जल्दी उठ गया?
उत्तर:
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ की कहानी ‘बड़े घर की बेटी’ से लिया गया है। इसके लेखक मुंशी प्रेमचन्द हैं।
संदर्भ : इस वाक्य को देवर लाल बिहारी अपनी भाभी आनंदी से कहता है।
स्पष्टीकरण : जब लालबिहारी आनन्दी से माँस पकाने को कहता है तो आनन्दी थोड़ा ही घी होने के कारण सारा घी माँस में डाल देती है। जब लालबिहारी आनन्दी से पूछता है कि दाल में घी क्यों नहीं डाला, तो आनन्दी कहती है कि सारा घी खत्म हो गया। गुस्से में लालबिहारी आकर कहता है कि अभि परसों ही घी खरीदकर लाया था इतनी जल्दी खत्म हो गया?

प्रश्न 2.
“स्त्री गालियाँ सह लेती है, मार भी सह लेती है, पर मैके की निंदा उनसे नहीं सही जाती।”
उत्तर:
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ की कहानी ‘बड़े घर की बेटी’ से लिया गया है। इसके लेखक मुंशी प्रेमचन्द हैं।
संदर्भ : इस वाक्य में लेखक प्रेमचंद आनंदी के बारे में टिप्पणी करते हुए पाठकों को उसके चरित्र का परिचय करा रहे हैं।
स्पष्टीकरण : दाल में घी न डालने से तथा घी खत्म हो जाने से लालबिहारी ने गुस्से में भला-बुरा कहा और मायकेवालों के बारे में खरी-खोटी बातें सुनायी। इसी से आनंदी को गुस्सा आ गया। क्योंकि स्त्रियाँ सब कुछ सह सकती है, मार भी खा सकती हैं, पर मायके की निंदा नहीं सह सकती।

प्रश्न 3.
“पर तुमने आजकल घर में यह क्या उपद्रव मचा रखा है।”
उत्तर:
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ की कहानी ‘बड़े घर की बेटी’ से लिया गया है। इसके लेखक मुंशी प्रेमचन्द हैं।
संदर्भ : इस वाक्य को श्रीकंठ अपनी पत्नी आनंदी से कहता है।
स्पष्टीकरण : श्रीकंठ जब शनिवार को घर पहुंचे तो लालबिहारी ने आनंदी के बारे में सबकुछ बता दिया था। आनंदी के बारे में शिकायतें सुनकर जब श्रीकंठ आनंदी के पास जाता है, तो पूछता है – आजकल यह सब क्या हो रहा है? तुमने क्या उपद्रव मचा रखा है?

प्रश्न 4.
“उससे जो कुछ भूल हुई, उसे तुम बड़े होकर क्षमा करो”
उत्तर:
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ की कहानी ‘बड़े घर की बेटी’ से लिया गया है। इसके लेखक मुंशी प्रेमचन्द हैं।
संदर्भ : इस वाक्य को पिता बेनीमाधव सिंह अपने बड़े पुत्र श्रीकंठ से कहते है।
स्पष्टीकरण : लालबिहारी के हाथों अपने स्त्री का अपमान होते देख श्रीकंठ आपे से बाहर हो जाता है। वह अपने पिता बेनीमाधव के सामने अपने क्रोध को प्रकट करता है। बेनीमाधव सिंह अनुभवी आदमी थे। वे इन भावों को ताड़कर अपने बेटे को शांत करने के उद्देश्य से बोलते है कि श्रीकंठ को अपने छोटे भाई को क्षमा करके अपने बड़प्पन का परिचय देना चाहिए।

प्रश्न 5.
“बड़े घर की बेटियाँ ऐसी ही होती हैं। बिगड़ता हुआ काम बना लेती हैं।”
उत्तर:
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ की कहानी ‘बड़े घर की बेटी’ से लिया गया है। इसके लेखक मुंशी प्रेमचन्द हैं। ..
संदर्भ : पिता बेनीमाधव सिंह ने अपने पुत्रों के समक्ष यह वाक्य अपनी बहू की प्रशंसा करते हुए कहते हैं।
स्पष्टीकरण : श्रीकंठ और लालबिहारी के बीच वाद-विवाद होने से तथा उनकी आपसी नाराजगी के कारण भाइयों में प्रेम टूटने तथा घर बिखरने की नौबत आ गई। इससे आनंदी अपने गुस्से को छोड़कर, लालबिहारी को रोक लेती है। सब कुछ ठीक हो जाता है। बड़े घर की बेटियाँ. ऐसे ही बिगड़ा काम बना लेती हैं।

अतिरिक्त प्रश्नः

प्रश्न 6.
“मैके के सामने हम लोगों को कुछ समझती ही नहीं।”
उत्तर:
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ की कहानी ‘बड़े घर की बेटी’ से लिया गया है जिसके कहानीकार मुंशी प्रेमचन्द जी हैं।
संदर्भ : प्रस्तुत वाक्य लालबिहारी सिंह ने श्रीकंठ सिंह से कहा।
स्पष्टीकरण : श्रीकंठ सिंह शनिवार को जब घर आए तो लालबिहारी ने अपनी भाभी की शिकायत करते हुए कहा कि भैया आप भाभी को जरा समझा देना कि मुँह संभाल कर बातचीत किया करें। जब श्रीकंठ सिंह ने पूछा आखिर बात क्या हुई है, तब लालबिहारी कहता है- आप ही आप उलझ पड़ीं। मैके के सामने हम लोगों को कुछ समझती ही नहीं। वह बड़े घर की बेटी है तो हम भी कोई कुर्मी कहार नहीं हैं।

प्रश्न 7.
“तम्हें मेरी सौगंध, अब एक पग भी आगे न बढ़ाना”
उत्तर:
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ की कहानी ‘बड़े घर की बेटी’ से लिया गया है जिसके कहानीकार मुंशी प्रेमचन्द जी हैं।
संदर्भ : प्रस्तुत वाक्य आनंदी लालबिहारी सिंह से कहती है जब लालबिहारी घर छोड़कर जा रहा होता है।
स्पष्टीकरण : लालबिहारी ने देखा कि भैया उस पर क्रोधित हो गए हैं और वे यह सौगन्ध खा चुके हैं कि अब वे लालबिहारी का मुँह तक नहीं देखना चाहते। तब लालबिहारी को भारी ग्लानि हो आती है। वह अपनी भाभी से क्षमा माँगते हुए कहता है- ‘भाभी, भैया से मेरा प्रणाम कह दो। वह मेरा मुँह नहीं देखना चाहते।’ इतना कहकर लौट पड़ा। आनंदी आगे बढ़कर उसका हाथ पकड़ लेती है और कहती है ‘तुम्हें मेरी सौगंध, अब एक पग भी आगे न बढ़ाना।’

प्रश्न 8.
“उन्हें बहुत ग्लानि हो गयी है, ऐसा न हो, कहीं चल दें।”
उत्तर:
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ की कहानी ‘बड़े घर की बेटी’ । से लिया गया है जिसके कहानीकार मुंशी प्रेमचन्द जी हैं।
संदर्भ : आनंदी के अपमान पर श्रीकंठ सिंह का गुस्सा और प्रण देखकर लालबिहारी को अपने किए पर ग्लानि हो आती है।
स्पष्टीकरण : श्रीकंठ सिंह अपने छोटे भाई द्वारा अपनी पत्नी के अपमान की बातें सुनकर गुस्सा हो जाते हैं। वे निर्णय लेते हैं कि इस घर में या तो लालबिहारी रहेगा या वे स्वयं। वे अपने पिता से कहते हैं कि अब वे लालबिहारी का मुँह तक देखना नहीं चाहते।
लालबिहारी अपने भाई के इस प्रण को सुनकर द्रवित हो जाता है और स्वयं घर छोड़ने का फैसला करता है। वह अपनी भाभी से क्षमा माँगता हैं। तब आनंदी दौड़कर अपने पति के पास आती है। वह कहती हैं- ‘उन्हें बहुत ग्लानि हो गयी है, ऐसा न हो, कहीं चल दें।’

V. वाक्य शुद्ध कीजिए।

प्रश्न 1.
उनकी पितामह किसी समय बड़े धन-धान्य सम्पन्न थे।
उत्तर:
उनके पितामह किसी समय बड़े धन-धान्य संपन्न थे।

प्रश्न 2.
स्वयं उनका पत्नी को ही इस विषय में उनसे विरोध थी।
उत्तर:
स्वयं उनकी पत्नी को ही इस विषय में उनसे विरोध था।

प्रश्न 3.
आनंदी अपने नये घर में आया।
उत्तर:
आनंदी अपने नये घर में आयी।

प्रश्न 4.
मुझे जाना दो।
उत्तर:
मुझे जाने दो।

VI. कोष्टक में दिये गये उचित शब्दों से रिक्त स्थान भरिए।

(ऐसी, शनिवार, संतान, गौरीपुर, हाथी)

प्रश्न 1.
इस दरवाजे पर …………. झूमता था।
उत्तर:
हाथी

प्रश्न 2.
………… में रामलीला के वही जन्मदाता थे।
उत्तर:
गौरीपुर

प्रश्न 3.
सुंदर ………. को कदाचित् उसके माता-पिता भी अधिक चाहते थे।
उत्तर:
संतान

प्रश्न 4.
श्रीकंठ सिंह ………… को घर आया करते थे।
उत्तर:
शनिवार

प्रश्न 5.
बड़े घर की बेटियाँ ………… ही होती हैं।
उत्तर:
ऐसी।

VII. अन्य लिंग रूप लिखिए।

प्रश्न 1.
ठाकुर, पति, बेटा, स्त्री, बुद्धिमान, हाथी, भाई।
उत्तरः

• ठाकुर – ठकुराइन
• पति – पत्नी
• बेटा – बेटी
• स्त्री – पुरुष
• बुद्धिमान – बुद्धिमती
• हाथी – हथिनी
• भाई – बहन

VIII. अन्य वचन रूप लिखिए।

प्रश्न 1.
घर, बेटी, भैंस, स्त्री, आँखें, थाली।
उत्तरः

• घर – घर
• बेटी – बेटियाँ
• भैंस – भैंसें
• स्त्री – स्त्रियाँ
• आँखें – आँख
• थाली – थालियाँ

मुहावरें।

• हाथ समेट लेना = खर्च कम करना
• हाथ साफ कर लेना = मार लेना, लूटना, मौके का फायदा उठा लेना
• बाट देखना = इंतजार करना
• मुँह झुलसना = जला लेना
• आड़े हाथ लेना = डाँट-डपट करना।

बड़े घर की बेटी लेखक परिचयः

कहानी सम्राट तथा उपन्यास सम्राट मुंशी प्रेमचन्द का जन्म काशी के निकट लमही नामक गाँव में सन् १८८० (1880) में हुआ था। माता-पिता का नाम था – श्रीमती आनंदीदेवी तथा अजायबराय। प्रेमचन्द का पहला नाम था धनपतराय। प्रेमचन्द ने करीब 13 उपन्यास लिखे हैं और तीन सौ से भी अधिक कहानियाँ लिखी हैं। उपन्यासों में – ‘गोदान’, ‘गबन’, ‘निर्मला’, ‘सेवासदन’, ‘रंगभूमि’, ‘कर्मभूमि’ आदि बहुत प्रसिद्ध हैं। कहानियों में ‘शतरंज के खिलाडी’, ‘नमक का दारोगा’, ‘कफ़न’ आदि प्रसिद्ध हैं। ८ अक्टूबर सन् १९३६ को आपका स्वर्गवास हुआ।

‘बड़े घर की बेटी’ उनकी प्रसिद्ध कहानियों में से एक है। सामाजिक जीवन के सूक्ष्म चित्रकार मुंशी प्रेमचन्द ने इस कहानी में मध्यम वर्ग के गृहस्थ जीवन की एक ऐसी घटना को चित्रित किया है जो हमारे घरों में आये दिन घटती रहती है।

आनंदी बड़े घर की बेटी है। संपन्न घर की सुविधाओं की अभ्यस्थ होने पर भी उसने अपने आपको श्रीकंठ के रुढिग्रस्त तथा अभावों से पूर्ण घर के अनुकूल बना लिया। यही उसका बड़प्पन है। पारिवारिक सदस्यों के आपसी संबंधों के महत्व की कहानी है।

बड़े घर की बेटी Summary in Hindi

उपन्यास सम्राट प्रेमचंद ने बहुत ही अच्छी कहानियाँ लिखी हैं। ‘बड़े घर की बेटी’ उनकी एक श्रेष्ठ सामाजिक कहानी है। आनंदी एक अमीर घराने से मध्यवर्गीय परिवार की बहू बनकर आती है। वहाँ के हालतों के अनुकूल वह अपने को ढाल लेती है। वह अपने देवर से अपमानित होकर भी घर की इज्जत बचा लेती है। आनंदी के उच्च संस्कार उसे सचमुच ‘बड़े घर की बेटी’ बनाते हैं।

आनंदी ताल्लुकेदार भूपसिंह की बेटी थी। वह सुंदर व सुशील थी। वह अपने मायके में बड़े ही लाड़-प्यार से पली थी। आनंदी का विवाह बेनीमाधव सिंह के बड़े बेटे श्रीकंठ से हुआ था। बेनी माधवसिंह एक जमींदार था। उसकी जमीन-जायदाद सब कुछ खत्म हो गई थी। अब वह एक साधारण किसान ही रह गया था।

आनंदी का पति श्रीकंठ पढ़ा-लिखा नौजवान था। शहर में काम करता था। उसका परिवार गाँव में था। उसके घर में सुख-सुविधाएँ बहुत कम थीं। फिर भी आनंदी ने अपने-आप को घर के अनुकूल बनाये रखा। वह खुशी से जीवन-यापन करने लगी। श्रीकंठ पढ़ाई के साथ-साथ, संस्कारित और भारतीय संस्कृति का अनुयायी था। संयुक्त परिवार का हिमायती भी था।

श्रीकंठ के छोटे भाई का नाम लालबिहारी सिंह था। वह अनपढ़ और उजड्ड था। एक दिन लालबिहारी सिंह ने खाना खाते समय घी माँगा। घी खत्म हो चुका था। अतः आनंदी घी नहीं दे पाई। इस कारण लालबिहारी सिंह ने गुस्से में भाभी को तथा उसके मायके को गालियाँ सुना दीं। आनंदी ने भी वापस कुछ खरी-खोटी सुना दी। तब लालबिहारी सिंह ने गुस्से में आकर आनंदी पर खड़ाऊ फेंक कर मार दिया। इससे आनंदी को बड़ा दुःख हुआ।

अब श्रीकंठ शहर से गाँव आया, तो आनंदी ने सारी कहानी उसे सुनाई। श्रीकंठ ने क्रोधित होकर अपने बाप से कहा – इस घर में या तो वह रहेगा, या लालबिहारी। पिता ने पुत्र को समझाया कि घर का बँटवारा न हो। उनका घर गाँव में एक आदर्श था। अगर भाई-भाई के बीच की अनबन की बात गाँव में फैलेगी, तो उनकी बदनामी होगी। लालबिहारी सिंह को अपनी गलती पर पछतावा हुआ। पर उसकी बातें श्रीकंठ को शांत नहीं कर पाईं। वह घर के अलगाव की बात पर अड़े रहे। तब लालबिहारी सिंह ने भाभी के पास जाकर, वह खुद घर से निकल जाने की बात कही। यह सुनकर आनंदी का दिल पिघल गया।

आनंदी बड़े घर की बेटी थी। उसके संस्कार अच्छे थे। उसने अपने पति श्रीकंठ को समझाया कि आगे घर में इस प्रकार के झगड़े नहीं होने देंगे। श्रीकंठ ने अपने भाई को गले लगाया। ठाकुर बेनीमाधव सिंह की आँखें भर आईं। गुणवती आनंदी के संस्कारों के कारण टूटता हुआ घर बच गया। सचमुच बड़े घर की बेटियाँ ऐसी ही होती हैं।

बड़े घर की बेटी Summary in Kannada

बड़े घर की बेटी Summary in English

The descendant of the landed aristocracy, Thakur Beni Madhav Singh was the headman of Gauripur village. During the days of his affluent forefathers, as people say, there stood an elephant in front of their door, adding grandeur to the house. The years rolled by and the elephant was replaced by an old and decrepit buffalo. A substantial chunk of his property was swallowed up by litigations. And his yearly income was reduced to one thousand rupees. Thakur Sahib had two sons. Srikant Singh, the elder one, earned his bachelor’s degree in arts after years of hard labour. He had obtained employment in an office. Lal Bihari Singh, the younger son, was uneducated. He was a strapping young man with a plump face and broad and muscular shoulders. Srikant stood in stark contrast to Lal Bihari.

Though Srikant had a contemporary English education, he condemned the modern code of ethics and the morals of the west. He was, therefore, held in high esteem by the villagers. He was also an avowed advocate of the extended family system. Anandi, his wife, came from noble stock. Her father, Bhup Singh, was the owner of a small estate, with an enormous mansion. Genial and generous by disposition, he was also bold and brave. He was blessed with seven daughters. Anandi was his fourth daughter. One day Srikant went to Bhup Singh for a donation to some cause. Bhup Singh was highly impressed by his polished manners and in time the marriage of Srikant and Anandi was celebrated with traditional pomp and ceremony.

When Anandi joined Srikant’s family as a daughter-in-law, she was shocked to find her new house was entirely different from her maternal house. There were neither horses or elephants nor a garden. It was a simple house, typical of the countryside. The house had no windows, no carpets and no pictures on the walls. However, Anandi adjusted herself marvellously well with the new life, without even tending to display feelings that she once lived in the lap of luxury

One day Lal Bihari Singh came home with the meat of two wild ducks and asked his bhabhi to cook the meat. When she went to the kitchen to cook the meat, she discovered to her dismay that there was very little ghee. She used all the ghee that was available to cook the meat. When Lal Bihari sat down to take his food, he found that the lentils hadn’t been garnished with ghee. He lost his temper. When he questioned Anandi, she replied curtly that she had used all the ghee that was left in the can. At this Lal, Bihari flew into a rage and taunted her father. Anandi, who could not tolerate the abuse hurled at her parents, said that in her maternal house even barbers and watermen consumed as much ghee every day. Her words stung Bihari. He threw the plate on the floor and snarled at her asking her to hold her tongue. Anandi’s face also turned red with rage and she snapped saying had her husband been there he would have taught him a lesson. Lal Bihari grew reckless; he took one of his wooden sandals and hurled it at Anandi. Anandi parried the blow, safeguarding her head. But her hand was injured. Trembling with anger, she retreated to her room.

Srikant used to visit his village every Saturday. The quarrel had taken place on Thursday. Anandi went without food for two consecutive days and waited impatiently for her husband. As usual, Srikant reached home in the evening and sat in the courtyard, chatting about national and current news, pending court cases and so on until ten in the night. After dinner, he went to his wife’s room where she was sitting tensely in silent rage. She explained what had happened and burst into tears. Srikant was deeply affected by Anandi’s tears. The next day, at the break of, dawn, he went to his father and told him that he and Lal Bihari could no longer pull on together. He then told his father what had happened. Dumbfounded, Beni Madhav fell silent. Beni Madhav tried to calm him down, but he spoke harshly and refused to forgive his brother.

Standing silently near the door, Lal Bihari heard everything that transpired between his father and brother. He respected his brother more than his father. He, therefore, felt dejected and forlorn. He felt penitent and was assailed by a mounting sense of repentance. He went to Anandi and sought her pardon. He stood near the door, his head hung low. Srikant walked past him with his eyes flaming red with anger. At this, Anandi was consumed with guilt and regretted having made the complaint against Lal Bihari. Essentially kind-hearted, she had never imagined that the matter would assume such a serious proportion. As Lal Bihari moved towards the outer door intending to leave the house, Anandi said that she had no grievance against him. She caught hold of him and refused to let him go.

Seeing all this, Srikant’s heart melted. He came out and leapt forward to embrace Lal Bihari. Both of them burst into tears. Beni Madhav Singh emerged from nowhere. The sight of his sons embracing each other gladdened his heart. He remarked, “Such are the daughters of noble families – they guard the families against disintegration.”

कठिन शब्दार्थः

• नंबरदार – गाँव का भूस्वामी;
• मरम्मत – ठीक करना;
• निर्बल – बलहीन, कमज़ोर;
• वैद्यक ग्रंथ – चिकित्सा सम्बन्धी पुस्तक;
• निर्वाह – निभानेवाला;
• बहरी – बाज जैसा एक शिकारी पक्षी;
• शिकरे – बाज से छोटा एक शिकारी पक्षी;
• रीझ – प्रसन्न होना;
• टीमटाम – श्रृंगार, सजावट;
• भावज – भाई की पत्नी, भाभी;
• फिकायत – बचत करना;
• तिनक – चिढ़ना, गुस्सा होना;
• ढिठाई – दुस्साहस;
• उजड्ड – गवार, असभ्य;
• खड़ाऊँ – काठ की बनी खूटीदार पादुका;
• सुधि – ध्यान रहना;
• शऊर – तरीका, ढंग;
• दब्बू – दबकर रहनेवाला;
• हथकंडा – षड्यंत्र;
• घुड़का – डाँटना;
• मुगदर – व्यायाम के लिए लकड़ी की बनी मुंगरी;
• खरल – दवा कूटने के पत्थर की कँडी।

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Karnataka 1st PUC Biology Previous Year Question Paper March 2014 (North)

Time: 3.15 Hours
Max Marks: 70

General Instructions:

1. The question paper consists of four parts A, B, C, and D.
2. All the parts are compulsory.
3. Draw diagrams wherever necessary. Unlabelled diagrams or illustrations do not attract any marks.

Part – A

Answer the following questions in one word / one sentence each: ( 10 × 1 = 10 )

Question 1.
What is a museum?
Answer:
Museums are those places that have collections of preserved animals and plants for taxonomic studies.

Question 2.
Where do you find bulliform cells?
Answer:
The upper epidermis of monocot leaf.

Question 3.
Mention the function of gizzard in cockroaches.
Answer:
Gizzard helps in grinding the food particles and mixing them with digestive enzymes.

Question 4.
Who proposed the Omnis cellula – e cellula?
Answer:
Virchow

Question 5.
Expand PPLO.
Answer:
Pleuropneumonia like organism.

Question 6.
What is Ammonification?
Answer:
The process by which organic nitrogenous compounds are decomposed to produce ammonia is known as ammonification.

Question 7.
What is plasmolysis?
Answer:
Plasmolysis is a phenomenon in which the cell shrinks due to exosmosis when it is surrounded by a hypertonic solution.

Question 8.
Write the dental formula of man.
Answer:
\(\frac{2123}{2123}\) × 2 in adults

Question 9.
What is emphysema?
Answer:
Emphysema is a chronic disorder in which alveolar walls are damaged due to which respiratory surface is decreased it occurs due to cigarette smoking.

Question 10.
State the function of fibrinogen in the blood.
Answer:
Soluble fibrinogen results in the formation of insoluble fibrin which forms a mesh-like network in which the blood components get entangled and results in a blood clot.

Part-B

Answer any FIVE of the following questions in 3-5 sentences each, wherever applicable. (5 × 2 = 10)

Question 11.
Write the important rules of Binomial nomenclature Give one example.
Answer:
(a) Binomial Nomenclature:
This method was introduced by Carolus Linnaeus.
In this method, every organism is given a scientific name, which has two parts, the first is the name of the genus (generic name) and the second is the name of the species (specific epithet).

e.g. Mangifera indica for mangoes and Homo sapiens for human beings. In the above, Mangifera and Homo are generic names, while Indica and sapiens are the names of the species belonging to Mangifera and Homo respectively.

The classification consists of a hierarchy of steps where each step represents a range or category. The various ranges or categories used in classification are called Taxonomic categories.

(b) Guidelines/Principles for nomenclature:

1. Scientific names are, generally, written in Latin or derived from Latin, irrespective of their origin.
2. The scientific names are written in italics or underlined. The first word denotes the name of the genus, and the second word denotes the species.
3. The generic name starts with a capital letter, while the specific name starts with a small letter.
4. The name of the author is written in an abbreviated form after the specific name e.g. Mangifera Indica Linn.-indicates that this species was first described by Linnaeus.
5. The name should be short, precise, and easy to pronounce.

Question 12.
How are viroids different from viruses.
Answer:
Viroids are naked single-stranded infectious RNA molecules causing plant diseases.

Viruses are obligate parasites made up of a protein coat called capsid which encloses genetic material either DNA or RNA. They cause diseases in plants, animals, and bacteria.

Question 13.
Sketch and label cardiac muscle tissue.
Answer:

Question 14.
Write any two differences between mitosis and meiosis.
Answer:

1. Mitosis consists of single nuclear divisions.
   Meiosis consists of two nuclear divisions.
2. In Mitosis, crossing over is absent.
   In Meiosis, crossing over is present.

Question 15.
What is hydroponics? Give one application of this technique.
Answer:
The technique of growing plants in a nutrient solution is known as hydroponics. The application of this technique is that the plants can be grown in a soil-free, defined mineral solution.

Question 16.
Mention the role of hydrochloric acid indigestion.
Answer:
(a) It activates inactive pepsinogen into active pepsin.
(b) It converts inactive prorenin into active renin.
(c) It provides an acidic medium (pH 1.5 – 2.5) for the action of both pepsin, and renin.
(d) HCl kills harmful bacteria and loosens fibrous food.
(e) It inactivates the ptyalin.

Question 17.
Hemoglobin is an oxygen carrier Explain.
Answer:
Hemoglobin is called an oxygen carrier because about 97% of the oxygen is transported by RBCs. RBCs contain hemoglobin. Hemoglobin has an affinity towards oxygen. Hence hemoglobin bonds with oxygen molecules to form oxyhemoglobin.

Oxyhaeinoglobin is a very unstable compound. Hence it dissociates into hemoglobin, and molecular oxygen, when it reaches tissues.

Question 18.
Name anyone gonadotropin hormones, giving a function of each.
Answer:
(a) FSH (Follicle-stimulating hormone).
Function:

1. In females, it promotes the growth and maturation of the Graafian folliçie.
2. In females, it also stimulates the follicle cells of the ovary to secrete female sex hormone or estrogen.

(b) LH (Luteinizing hormone).
Function:

1. In females stimulates final maturation and release of the ovum from the Graafian follicle.
2. It in females helps in the conversion of the Graafian follicle into the corpus luteum. It stimulates the corpus luteum to secrete progesterone.
3. In males, it stimulates the interstitial cells of the testis called Leydig cells to secrete the male sex hormones called testosterone.

PART-C

Answer any FIVE of the following questions in 40-80 words each, wherever applicable: (5 × 3 = 15)

Question 19.
Assign the following to their respective phyla.
(a) Sycon (b) Aurelia (c) Liver fluke
Answer:
(a) Phylum Porifera
(b) Phylum Coelenterata
(c) Phylum Platyhelminthes.

Question 20.
Mention the salient features of the family Solanaceae
Answer:
Vegetative characters: Plants mostly herbs, shrubs, and rarely small trees.
Stem: Herbaceous rarely woody, aerial, erect, cylindrical, branched, solid or hollow, hairy or glabrous, underground stem in potato (Solanum tuberosum).

Solanum nigrum (makoi) plant: (a) Flowering twig (b) Flower (c) L.S. of flower (d) Stamens (e) Carpel (f) Floral diagram

Leaves: alternate, simple, rarely pinnately compound, exstipulate, venation reticulate. Floral Characters:
Inflorescence: Solitary, axillary, or cymose as in Solanum.
Flower: bisexual, actinomorphic.
Calyx: sepals five, united, persistent, valvate aestivation.
Corolla: petals five, united: valvate aestivation.
Androecium: stamens five, epipetalous.
Gynoecium: carpellary, syncarpous, ovary superior, bilocular, placenta swollen with many ovules.
Fruits: berry or capsule.
Seeds: many, endosperms.

Class: Monocotyledonae Order: Liliflorae Family: Liliaceae
Vegetative characters: Perennial herbs with underground bulbs/corms/ rhizomes. Leaves: mostly basal, alternate, linear, exstipulate with parallel venation.
Floral characters:
Inflorescence: solitary / cymose; often umbellate clusters.
Flower: bisexual; actinomorphic.
Perianth: tepal six (3 + 3), often united into a tube; valvate aestivation.
Androecium: stamen six (3 + 3).
Gynoecium: Tricarpellary, syncarpous, ovary superior, trilocular with many ovules; axile placentation.
Fruit: capsule, rarely a berry.
Seed: endosperms

Question 21.
What is a vascular bundle? Explain the types of vascular bundles.
Answer:
The Vascular Tissue System: Vascular bundles carried defined as a compact strand of xylem and phloem with or without cambium. Vascular tissues derived from the procambium, during the primary growth of the plant, are called primary vascular tissues. They include primary xylem and primary phloem. Vascular tissues derived from vascular cambium., during secondary growth of the plants, are called secondary vascular tissues. They include secondary xylem and secondary phloem.

• Protoxylem: It is the first-formed xylem, in which the elements are much elongated and have narrow lumen.
• Metaxylem: It is the later formed xylem, in which the elements are shorter and have wider lumen.
• Exarch xylem: A condition of the xylem in which the protoxylem is facing away from the center is called Exarch xylem.
• Endarch xylem: A condition of the xylem, in which the protoxylem is facing towards the center is called endarch xylem.

Types of Vascular bundles:
1. Radial vascular bundles: The vascular bundles in which xylem and phloem are present in separate bundles and lie on different radii are called radial vascular bundles.

2. Conjoint vascular bundles: The vascular bundles in which xylem and phloem are present in the same bundle on the same radius are called conjoint vascular bundles.

Question 22.
Describe the following:
(a) Synapsis (b) bivalent (c) chiasmata
Answer:
(a) The process of the pairing of homologous chromosomes is called Synapsis.
(b) A chromosome containing two chromatids is called a bivalent.
(c) The regions where segments of non-sister chromatids of the bivalents are exchanged are called chiasmata.

Question 23.
How are the minerals absorbed by the plants?
Answer:

1. The process of absorption occurs in two distinct phases.
2. In the first phase, there is a rapid uptake of irons into the ‘free space’ or outer space of the cell called the apoplast. it is a passive process and takes place through non-channels.
3. In the second phase, the ions are taken slowly into the inner space – the symplast of the cells. This is an active process and requires the expenditure of metabolic energy.

Question 24.
Write a note on ultrafiltration.
Answer:
Blood brought by the efferent arteriole is filtered in the glomerular capillaries because of the net effective pressure of 25mm of Hg. Net effective pressure is due to the differences in glomerular hydrostatic pressure, blood colloidal osmotic pressure, and capsular hydrostatic pressure.

This can be represented as,
Ghp – (BCOP + CHP) = NEP
i.e., 75mm Hg – (30mm Hg + 20mm Hg) = 25mm Hg.

During filtration, the blood cells and large organic molecules are retained in the capillaries. The fluid containing dissolved substances is passed into the capsular lumen. This is called glomerular filtrate. It contains a number of useful substances like glucose. amino acids, minerals, etc., as well as waste products.

Question 25.
(a) What is a pacemaker?
Answer:
SAN is also known as the pacemaker of the heart because it initiates the electrical impulses for the heartbeat.

(b) Define a cardiac cycle and the cardiac output.
Answer:
Cardiac cycle (Heart Beat):
The sequential cyclic events of systole (contraction state) and diastole (relaxation state) together are known as one cardiac cycle or heartbeat. In a healthy person, the heart completes 72bcats/min. On average, each complete heartbeat requires 0.8 sec in which systole of ventricles requires 0.3 sec, systole of auricles require 0.1 sec, diastole of the heart requires 0.4 sec.

Cardiac Output:
The amount of blood pumped by the right and left Ventricles per minute is known as cardiac output. It is determined by multiplying the stroke volume with the number of heartbeats/mm.

Cardiac Output = stroke volume × number of heart beats / minute
= 70 ml × 72 beats / min
= 5040 ml of blood.

Question 26.
Describe the different types of movements exhibited by cells of the human body.
Answer:
Types of Movement:
Cells of the human body show three basic types of movement:
1. Amoeboid Movement/Pseudopodial movement:

• Macrophages and leucocytes exhibit amoeboid movement. It is also seen in the amoeba.
• Amoeboid movement is affected by pseudopodia formation by cytoplasmic streaming.
• It also involves cytoskeletal elements like microfilaments.

2. Ciliary Movement:
These movements are caused by fine versatile hair called cilia.

• Ciliary movement occurs in those hollow tubular visceral organs, that is internally lined by ciliated epithelium.
• Ciliary movement in the trachea helps in removing foreign substances or particles.
• Ciliary movement in the fallopian tube causes the ova to move to the uterus.

3. Muscular Movement:
The contractile property of muscles is responsible for muscular movement.

• Movements of jaws, limbs, eyelids, etc. are muscular movements involving striated muscles.
• Movement of food ¡n the alimentary canal and movement of urine in ureters are movements involving smooth muscles.

4. Flagella movement: It is caused by a long whip-like process called flagella. It is seen in spermatozoa, euglena, and sponges.

Part-D (Section – I).

I. Answer any FOUR of the following questions in 200-250 words each. (4 × 5 = 20)

Question 27.
Describe the important characteristics of gymnosperms.
Answer:
Gymnosperms represent primitive phanerogams (flowering plants) or spermatophytes. They are generally regarded as ‘naked seeded plants’ due to the presence of exposed seeds without being protected by the ovary wall as there is no well-defined carpel.

Gymnosperms are distributed throughout the world, represented in tropical and temperate forests, and dominating mountainous ranges. The group comprises nearly seven hundred species. They exhibit diverse habits showing the presence of woody climbers (lianas), shrubs, and trees.

General features:
1. The life cycle has a distinct, dominant, diploid, asexual phase represented by the well-differentiated evergreen woody plant, which is known as the sporophyte.

2. The sporophyte is heterosporous beating microspores and megaspores within microsporangia and megasporangia respectively. These structures occur on leaf-like ¡iicrosporophylls and megasporophylls. These are further organized into fertile structures called strobili or cones.

3. Sporophyte shows the presence of a taproot system which is well developed. The stem possesses branches that are dichotomies. Leaves are well developed and are dimorphic (two types of leaves): viz,

• Green photosynthetic leaves (Foliage).
• Brown-colored scale leaves.

4. Microspore develops into male gametophyte and megaspore produces female gametophyte. These gametophytes represent the haploid phase and are highly inconspicuous in comparison with sporophytic generation.

5. Female gametophyte is enclosed within a megasporangium that in turn is covered by an integument. Such an integument megasporangium possessing the female gametophyte is called the ovule.

Question 28.
Draw a labeled diagram of the Alimentary canal system of cockroaches.
Answer:

Question 29.
Write any five differences between cartilage and bony fishes.
Answer:

1. Chondrichthyes (cartilaginous fishes).
2. Osteichthyes (bony fishes).

Chondrichthyes	Osteichthyes
1. Skeleton is made up of cartilage.	1. Skeleton is made. up of bones.
2. Exclusively marine.	2. Both marine and freshwater forms are seen.
3. They possess placoid scales.	3. Possess cycloid or ctenoid scales.
4. Mouth and nostrils are situated on the ventral side of the body.	4. Mouth and nostril are terminal in their position.
5. Passes 5 pairs of gill slits.	5. Possess 4 pairs of- gill slits which are covered by operculum.
6. Tail is heterocercal.	6. Tail is homocercal.
7. Devoid of air bladders.	7. Possess air bladders.
8. Males usually possess a pair of claspers. eg: Sciiodon (Shark) Narcine (Electric ray) Trygon (Stingray).	8. Claspers are absent. eg: Anguilla (Freshwater Eel) Carias (Catfish) Hippocampus (Sea horse).

Question 30.
A neat labeled diagram explains the structure of mitochondria.
Answer:

Question 31.
Explain the mode of enzyme action.
Answer:
Enzymes are specific in their substrate molecules.
When enzyme (E) and its substrate (S) come together, they make an intermediary complex called the E-S complex, like a key coming with a lock and making a complex. This is the primary step for enzyme action.

So the enzymes are highly specific for the type of reaction they catalyze.
Over a fraction of a time, the enzyme cleaves the substrate into its product.

Since the product is different from the substrate, it cannot remain in contact with the enzyme, the enzyme becomes free.

The free enzyme can take up another substrate molecule.
E + S = ES → E + P,

Question 32.
Describe the transpiration pull model of water transport in plants.
Answer:
It is one of the most successful physical theories of the ascent of sap. Dixon and Jolly proposed
this theory. According to this theory, the ascent of sap is due to three factors namely,
(a) Cohesive force of water
(b) Adhesion of water molecules to the walls of the xylem vessel
(c) Transpiration pull

A strong intermolecular force of attraction exists between water molecules. Thus water molecules are bound to each other forming a continuous column of water.

There is a force of attraction between water molecules on the inner walls of xylem elements.
Therefore water molecules are attached to the wall of the xylem. It is called adhesin.

Due to the adhesive and cohesive properties of water, a continuous column of water is present in the xylem. Transpiration at the leaf surface forces the mesophyll cells of the leaves to draw water from the neighboring xylem elements. This creates a suction force known as Transpiration pull.

This suction force pulls the water column in the xylem upwards. This ascent of sap is due to the combined effect of adhesive, and cohesive properties of water and transpiration pull.

Merits:

1. The ascent of sap is directly proportional to the rate of transpiration.
2. It is a physical process and does not need energy.
3. Strong cohesive, and adhesive forces are sufficient to prevent the rupture of the water columns in xylem vessels.

Demerits:

1. The presence of an air bubble breaks the continuity of the water column.
2. Ascent of sap continues even in the absence of transpiration, as at night times.
3. The strength of the water column is thoughtful against two opposing forces such as gravitation force and transpiration tension.

Section – II

II. Answer any THREE of the following questions in 200-250 words each, wherever applicable. (5 × 3 = 15)

Question 33.
Explain the cyclic photophosphorylation with schematic representation.
Answer:

Cyclic Photophosphorylation :
It is a cyclic path of electrons expelled from chlorophyll through a series of substrates that are arranged in a suitable oxidation-reduction potential. The energy in the electrons is used for the phosphorylation of ADP to ATP.

In PS I the absorbed photons of light excite chlorophylL a 700 to eject energized electrons on makes it positively charged and unstable. Electrons pass through the sequence → FRS → FD → Cyt b6 → Cytf → PC and generate ATP at two places. Finally, an electron from PC returns to chi-a restoring its stability.

Question 34.
Draw a labeled diagram of a multipolar neuron.
Answer:

Question 35.
Give the schematic representation of Kreb’s cycle?
Answer:

Note: To account for two molecules of acetyl CoA produced from ¡ molecule of glucose the entire reaction has to be multiplied by two.

Bloenergetics:

1. Number of ÄTPs produced 2
2. Number of NADH2 produced = 6 (6 × 3 = 18ATPs)
3. Number of FADH2 produced = 2 (2 × 2 = 4 ATPs)
   The total yield of ATPs = 24

Note: Efficiency of Krebs cycle along with its predatory reaction is 30 ATPs.

Question 36.
(a) What is Photoperiodism? What is its significance?
(b) Explain the sigmoid growth curve.
Answer:
Photoperiodism:
Plants, in order to flower, required a particular day length or light period called photoperiod, and the response of the plants to photoperiod ¡n terms of flowering is called Photoperiodism.

Photoperiodism was first studied by W.W. Garner, and HA. Allard.

Based on their photoperiodic responses, plants are classified into the following groups:

1. Long Day Plants: These flower in photoperiod more than critical day length. eg: Wheat, oats, etc.
2. Short Day Plants: These flower ¡n photoperiod less than critical day length. e.g: Tobacco, Chrysanthemums, etc.
3. Day Neutral Plants: These are the plants that are not influenced by the duration of light for their flowering. e.g: Tomato, cucumber, cotton, etc.

Growth curve:
The curve obtained when the growth in size or weight or on entire plant or its individual parts is plotted against time is called growth curve ft is always ‘S’ or sigmoid in shape shows free distinct phases namely.

1. Lag phase.
2. Log phase.
3. Steady phase.

1. Lag phase: This is the initial stage where the growth rate is very slow But it gradually increases with time.

2. lag phase: In this phase, the rate of growth is very rapid. The plant shows a high rate of growth and reaches maximum height. Hence this phase is called the exponential phase or grand of growth.

3. Steady phase or stationary phase: This is the last phase in which plants show no growth. Therefore the curve becomes almost orbital.

Question 37.
Diagrammatically indicates the location of the various endocrine glands in our body.
Answer:

You can Download Chapter 2 Vachanagalu Questions and Answers Pdf, Notes, Summary, 1st PUC Kannada Textbook Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Kannada Textbook Answers Sahitya Sanchalana Chapter 2 Vachanagalu

Vachanagalu Questions and Answers, Notes, Summary

Students can Download 1st PUC Biology Previous Year Question Paper March 2014 (South), Karnataka 1st PUC Biology Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Biology Previous Year Question Paper March 2014 (South)

Time: 3.15 Hours
Max Marks: 70

General Instructions:

1. The question paper consists of four parts A, B, C, and D.
2. All the parts are compulsory.
3. Draw diagrams wherever necessary. Unlabelled diagrams or illustrations do not attract any marks.

Part – A

Answer the following questions in one word / one sentence each: ( 10 × 1 = 10 )

Question 1.
In which group of algae, red pigment r-phycoerythrin is present?
Answer:
Rhodophyta,

Question 2.
Write the biological name of potato.
Answer:
Solanum tuberosum.

Question 3.
Define mesophyll tissue.
Answer:
Photosynthetic tissue present between upper and lower epidermis of leaf is mesophyll.

Question 4.
Name the bond present between monosaccharides in a polysaccharide.
Answer:
Glycosidic bond.

Question 5.
Write the energy currency of the cell.
Answer:
ATP.

Question 6.
Define imbibition.
Answer:
Imbibition is a special kind of diffusion that involves the movement of water molecules along a diffusion gradient from a region of higher concentration onto a suitable solid matrix (adsorption).
or
It is the adsorption of water molecules by hydrophilic colloids.

Question 7.
Define hydroponics.
Answer:
The technique of growing plants in a nutrient solution is known as hydroponics.

Question 8.
Mention the normal blood pressure in humans.
Answer:
120/80 mm of Hg.

Question 9.
Where is the pituitary gland located?
Answer:
Sella tunica (bony caving) is attached to the hypothalamus.

Question 10.
Mention the function of the pericardium.
Answer:
It provides protection to the heart.

PART-B

Answer any FIVE of the following questions in 3-5 sentences each, wherever applicable. (5 × 2 = 10)

Question 11.
Classify the bacteria based on their shape.
Answer:
Bacteria are the smallest and most primitive cellular organisms. They are microscopic and often regarded as germs.

Bacteria are cosmopolitan in distribution and found in all environments i.e, soil, air, and water. They also occur in plants and animals, both in living and dead.

Bacteria are unicellular organisms and they exist in various forms (shapes).

Four major forms of bacteria exist in nature.
1. Coccus form: These are spherical in shape. Cocci forms may occur singly, in pairs, in a clamp or in chains. Accordingly, they are named as:

• Gonococcus: Occur singly (Micrococcus).
• Diplococcus: Occur in pairs (Diplococcus pneumonia).
• Staphylococcus: Occur in groups (Streptococcus pyro genes).

2. Bacillus form: These are rod-shaped. They may occur singly, in pairs, or in chains. i.e. Bacillus, Diplohacilli, Streptobaci III.

3. Spinllum form: These bacteria have spirally coiled bodies. eg: Treponema Rhodo spirillum.

4. Vibrio form: These are comma-shaped. eg: VI brio Cholerae.

Question 12.
Write the excretory organs of
(i) Arthropods
(ii) class Mammalia.
Answer:
(i) Malpighian tubules and green glands
(ii) Metanephric kidneys.

Question 13.
Write short notes on epithelial tissue.
Answer:
The four main characteristic features are:
(a) The cells are arranged compactly, without interceÍlular spaces.
(b) Epithelial cells always rest on the basement membrane composed of fine fibers and non-living protein-polysaccharide material.
(c) The cells are arranged in the form of a continuous layer or sheet. Such a sheet may cover the external surface of an organ or internal cavity.
(d) The cells may be grouped as tubes or follicles to produce glands.
(e) Epithelial tissues are avascular.

Question 14.
Give any four examples for secondary metabolites.
Answer:
Pigments, Alkaloids, Essential oils, Toxins. …….

Question 15.
Mention any four forms of essential nutrient elements.
Answer:

1. Components of biomolecules energy-related molecules.
2. Components of energy-related molecules.
3. Act as enzyme regulators.
4. Act as osmotic potential regulators in the cytoplasm.

Question 16.
List the different types of valves present in the human heart
Answer:
Semilunar, tricuspid and bicuspid valves.

Question 17.
Write any four characters of skeletal muscles.
Answer:
Skeletal muscle:

1. It occurs in limbs, neck, trunk, etc.
2. Each fiber contains a number of nuclei just below sarcolemma.
3. Muscle fibers are unbranched and run parallel.
4. Cross bands are distinct.
5. It is voluntary in function.
6. They suffer fatigue.

Question 18.
Write short notes : (a) Gout (b) Tetany.
Answer:
Tetany: Rapid spasms (wild contractions) in muscle due to low Ca⁺² in body fluids.
Gout: Inflammation of joints due to accumulation of uric acid crystals.

PART-C

Answer any FIVE of the following questions in 40-80 words each, wherever applicable: (5 × 3 = 15)

Question 19.
Who proposed the five-kingdom classification? Write the names of five kingdoms.
Answer:
R.H. Whittaker.

1. Kingdom monera
2. Kingdom Protista
3. Kingdom Mycota
4. Kingdom Animalia
5. Kingdom Plantae.

Question 20.
Define the following:
(a) Hypogynous flower
(b) Phyllotaxy
(c) Venation
Answer:
Hypogynous flower: A flower possessing a superior ovary.
Phyllotaxy: The mode of arrangement of leaves on the stem axis is known as phyllotaxy.
Venation: The mode of arrangement of veins and veinlets is called venation.

Question 21.
Explain the structure of Stomata.
Answer:

Question 22.
What are chromosomes? Classify the chromosomes based on the position of the centromere.
Answer:
Chromosomes: Chrómosomes are the thread-like colored bodies that are in the intranuclear position which act as the vehicles of heredity and variations.

They are self-replicating and exhibit cyclic changes in size and shape. They are visible during cell division: Waldeyer described them in 1888.

Chromosomes are the self-reproducing components of the nucleus with DNA, which become visible during cell division. They exhibit a cyclic change in size & shape.

1. Chromosom size: The size ‘of the chromosomes measured a the metaphase of mitosis. It is constant for a particular species. Its length varies from 0.2 to 5.2. Diameter 0.2 to 2i. Chromosomes of plants are larger than that of animals. Monocots have very large ones. The more the number of chromosomes per nuçieus, the smaller is the size, the lesser the number bigger in size, the Example for large chromosomes. Oocytes of vertebrates – Lampbrush chromosomes, Insects – Salivary gland or polytene.

2. Chromosome shape: Shape changes with that of the stage of cell division. In interphase, it is chromatin & thin thread-like, but long and slender in prophase.

Chromomeres: The bead-like swelling of prophase chromosomes is called chromomeres. They become thick during metaphase.

Types of Chromosomes:
I. Classification of chromosomes based on the number of centromeres:
(a) Acentric chromosome: It is a chromosome without centromere.
(b) Monocentric chromosome: Here there is one centromere to hold the chromatids together.
(c) Biceiitric chromosome: It is the presence of two centromeres in a chromosome.
(d) Polycentric chromosome: It is more than three centromeres in a chromosome.

II. Classification of chromosomes based on the position of centromere:
(a) Metacentrie type: Here the centromere is exactly at the center of two chromatids. It looks V-shaped during anaphase.
(b) Sub metacentric type: Here the ‘centromere is eccentric iii positions so that one of the chromatids is long and the other is shorter. It looks L – shaped during anaphase.
(c) Acrocentric type: Here the centromere is almost towards one end of the chromatid to form a very long arm and another very short arm. It looks hook-shaped during anaphase.
(d) Telocentric type: Here the centromere is towards one end of the chromatid such that one chromatid is only present. It looks rod-shaped during anaphase.

III. Classification of chromosomes based on their functions:
(a) Autosomes[AA]: They are also called somatic chromosomes that control body characteristics.
(b) Allosomes X or VI: They are also called sex chromosomes that determine the gender of an individual.

Functions of chromosomes:

1. Chromosomes are very important in the higher animals for the phenomenon of sex determination.
2. Chromosomes play an active role in the metabolic process of a cell.
3. They carry the heredity information from parents to offspring ¡n the form of genes.

Question 23.
Write the schematic representation of the nitrogen cycle
Answer:
Nitrogen Cycle:
1. Nitrogen is a limiting nutrient for both natural, and agricultural ecosystems.

2. It exists as two nitrogen atoms held together by strong triple covalent bonds (N = N).

3. The nitrogen cycle involves the following steps:

1. Nitrogen fixation
2. Ammonification
3. Nitrification and
4. Denitrification.

1. Nitrogen Fixation:

• The process of conversion of nitrogen into ammonia and/or other nitrogen compounds is known as nitrogen fixation.
• In nature, lighting and ultraviolet radiation provide energy, to convert nitrogen into nitrogen compounds like N₂O, NO, and NO₂.
• The atmosphere also gets some amount of nitrogen oxides from forest fires, automobile exhausts, industrial combustion, and power generating stations.

2. Ammonification:

• The process by which organic nitrogenous compounds arc decomposed to produce ammonia is known as ammonification.
• Some ammonia volatilizes and re-enters the atmosphere.
• Most of the ammonia is converted into nitrites and nitrates.

3. Nitrification:

• Nitrification is the process of converting ammonia first into nitrite, and then into nitrate.
  2NH + 3O₂ → 2NO₂ + 2H⁺ + 2H₂O
  2NO₂ + O₂ → 2NO₃
• These processes are carried out by soil bacteria that are chemoautotrophs.
• Ammonia is oxidized into nitrite by Nitrosomonas and Nitrococcus.
• Nitrite is oxidized to nitrate by Nitrobacter,
• The nitrates are absorbed by the plants. and reduced to nitrites.
• The nitrites are transported to the leaves and reduced to ammonia, which forms the amino group of amino acids.

4. Denitnfication :

• It is the process of conversion or reduction of nitrates into free nitrogen.
• It is carried out by bacteria like Pseudomonas and Thiobacillus.

Question 24.
Briefly explain any two disorders of the digestive system.
Answer:
Jaundice [Icterus]:
It is the most common digestive disorder. Jaundice relates to the yellow appearance of skin, urine, nails, and cornea of the eye due to an increased Bilirubin pigment (Bile pigment).

Jaundice is of three types:
1. Haemolytic Jaundice or Neonatal Jaundice or Pre hepatic Jaundice: It is commonly found in newly born babies up to one week of age. it is due to excessive hemolysis (Destruction of RBCs).

2. Infective or Hepatocellular Jaundice: It is due to the infection of the hepatitis virus of the liver. So, the liver becomes unable to process Bilirubin. It results in a heavy accumulation of Bilirubin which leads to Hepatocellular Jaundice.

3. Obstructive Jaundice or Extra Hepatic jaundice: It occurs due to the obstruction in the Bile duct. The obstruction of the Bile duct is due to gallstone formation. The bile duct ¡s thus blocked, and bile juice cannot enter into the duodenum.

Causes:

1. Liver damage due to viral infection, and consumption of alcohol.
2. Increased Bilirubin pigment in the body.
3. Excessive hemolysis.
4. Blockage of the bile duct due to gallstone.
5. Presence of tumors in the bile duct.

Effects:

1. Yellow coloration of the skin, eyes, and urine.
2. Fever, loss of appetite, nausea, vomiting, and headache.
3. Infective jaundice leads to bleeding, and shrinkage of the liver [Cirrhosis]
4. Unconsciousness and finally it may lead to death.

Prevention:
It can be prevented by avoiding the consumption of alcohol and also by avoiding the consumption
of contaminated food and water.

Treatment :

1. Vaccines and anti-viral drugs are used against infective jaundice.
2. Obstructive jaundice can be treated by surgery.

Question 25.
Show the diagrammatic representation of the exchange of gases between alveoli and blood.
Answer:

Question 26.
Discuss briefly the functions on PCT, Henle’s loop, and DCT.
Answer:
Proximal convoluted tubule (PCT) :
It is a highly coiled twisted tube. his lined by a brush bordered cuboidal epithelial cells. Selective reabsorption of essential substances takes place in PCT during urine formation.

Distal convoluted tubule (DCT) :
The ascending limb of Henle’s loop is continued with DCT. It is lined by cuboidal epithelium.

It is less coiled and broader than PCE Active secretion of certain ions and water from the blood takes place in this region. Collecting duct: It is a straight tube lined with columnar epithelium.

Several collecting ducts open into the larger tubule which in turn opens into the pelvis of the Kidney. It also reabsorbs water and Na+ ions from the urine.

Part-D (Section – I).

I. Answer any FOUR of the following questions in 200-250 words each. (4 × 5 = 20)

Question 27.
Answer the following.
(a) State the law of limiting factors
Answer:
Blackman’s Law of Limiting Factors:
The law states that “When a process is conditioned to its rapidity by the number of separate factors the role of the process is limited by the pace of the slowest factor”.

(b) Discuss the various factors affecting photosynthesis.
Answer:
Factors affecting photosynthesis:
The following factors influence photosynthesis:
1. Light: Sunlight ¡s used as a primary source of energy. Its three aspects affecting the process are:
(a) Light Intensity: Optimum light intensity for photosynthesis is 2000-2500 foot candles/ Higher light intensity bleaches chlorophyll and is called solarisation.
(b) Light Quality: Photosynthetic rate is maximum in red light, next maximum in blue light, and least in green light.
(c) Duration: Photósynthetic rate is more efficient in intermittent light than in continuous light supply.

2. CO₂: Used as a raw material in photosynthesis. Its concentration in air is 0.03%. Its increase up to 0.5% increases the rate of the process but above 0.5% inhibits photosynthesis.

3.O₂: Liberated as a by-product in photosynthesis. its increase above the normal 21% in air, decreases the rate of the process. It is called the Warburg effect.

4. Temperature: It affects photosynthesis through its influence on the enzyme-controlled dark reaction. The optimum temperature is 18 – 400C for photosynthesis.

5. Water: Used as a raw material in photosynthesis. In the dehydrated state of cells, photosynthesis is inhibited.

Question 28.
Write the schematic representation of glycolysis.
Answer:
It occurs in the cýtoplasm of the cell.
It is an enzymatic reaction, thus temperature sensitive.
It is a common reaction for both aerobic and anaerobic respiration.

Question 29.
Draw a neat labeled diagram of the digestive system of cockroaches.
Answer:

Question 30.
What are auxins? List any four applications of auxins.
Answer:
Auxins:

• Auxins were the first hormones to be discovered in plants.
• F.W. Went confirmed and isolated the auxins from coleoptiles of Avena saliva.
• Chemically it is called indole-3-Acetic Acid and is derived from amino acid tryptophan.
• Auxins can stimulate cell division, cell elongation, and cell maturation.
• It promotes Apical dominance.
• It initiates Root formation.
• Can promote parthenocarpy.
• Prevent premature fall of flowers, buds, and leaves,
• 2, 4D, and 2, 4, 5T are used as selective weedicides.
• Promote phototrophic and hydrotropic movements.
• Promote xylem differentiation.

Question 31.
Explain apoplastic and symplastic pathways of water movement in roots.
Answer:
Plants absorb water by the root hairs in two distinct pathways namely, the apoplast pathway, and symplast pathway.

In the apoplast pathway, the movement of water occurs exclusively through the intercellular spaces and the walls of the cells. This can take place only up to the endodermis because the suberised Casparian strip will block the further movement of water.

In the symplast pathway, the movement of water takes place through the interconnected protoplasts through cytoplasm and plasmodesmata connections. This is the most accepted pathway.

Structure of not hair:
Root hairs are unicellular and arise from some of the epiblema cells. These cells enclose a large vacuole surrounded by thin cytoplasm which extends into root hair. There is a single nucleus in the cytoplasm at the tip of the root hair. The root hair is covered by a thin, and permeable cell wall made up of cellulose and Pectin.

• The total amount of water present in the soil is called hoard.
• The water available to the plants for absorption is called Chard.
• Soil water that can not be absorbed by the plait is Echard.
• The path of absorption of water ¡n roots is as follows:

Root hairs → Epiblema → Cortex → Endodermis (Passage cells) → Pericycle → Xylem

Pathway of water movement in the root

Question 32.
Draw a neat labeled diagram of a neuron.
Answer:

Section – II

II. Answer any THREE of the following questions in 200-250 words each, wherever applicable. (5 × 3 = 15)

Question 33.
Discuss the uses of algae.
Answer:

1. Brown algae like sargassum, laminaria, ficus are good sources of agar. It is used in the preparation of nutrient agar, a microbial growth medium.
2. Algin extracted from brown algae is used in the preparation of Ice creams, jellies, and chocolates.
3. Some algae like Ulva, Dictyota, and sargassum are used as food, as they are rich in mineral ions.
4. Algae are used as fodder.
5. Some algae are used as biofertilizers.

Question 34.
Enumerate salient characters of class Mammalia.
Answer:
Class Mammalia:

1. Class Mammalia includes warm-blooded and highly organized animals possessing mammary (milk) glands for suckling their young ones. They live almost in all the environments of the world and undergo secondary adaptation to suit their habitats. They are also known as hair quadrupeds.
2. Mammals are all homoeothermic (warm-blooded) animals.
3. The body shape varies, and it is covered by hairs.
4. The body is divisible into the head, neck, trunk, and tail.
5. Skin is glandular and includes sebaceous, and sweat glands in both the sexes, mammary glands in the females.
6. The Head region possesses a pair of eyes protected by eyelids and eyelashes, a pair of external ears or pinnae, and a pair of external nares.
7. They possess two pairs of pentadactyl legs, which are variously modified to perform different functions [like walking, running, climbing, swimming, burrowing, and flying]. Digits bear horny claws, nails, or hoofs.
8. Teeth are placed in sockets [Thecodont]. Teeth are heterodont [different types of teeth i.e, all are not of the same morphology], and diphyodont [i.e, 2 sets of teeth milk teeth, and permanent teeth].
9. There is well-developed voice apparatus, the larynx.
10. There is a muscular diaphragm separating the thorax and abdomen.
11. Respiration is by lungs.

Question 35.
Explain the structure of mitochondria with a neat labeled diagram.
Answer:

Section through mitochondrion (diagrammatic)
The elementary particles have been shown only on one crista.

It is present in eukaryotes except for mammalian RBC, and absent in prokaryotes. Its shape is oval, or sausage. Its number per cell depends on the metabolic state of the cell.

Structure: The mitochondria are bounded by two lipoprotein unit membranes namely the outer membrane and inner membrane. In between, they lie per mitochondrial matrix containing water, minerals, and enzymes.

The outer membrane is smooth and unfolded while the inner membrane is folded and it produces an inward finger-like process called Castle. Along the inner surface of the inner membrane, there are numerous tiny tadpole-like structures called elementary particles or F particles, Racker’s particles, or exosomes. F1 particle contains basal piece. stalk and head. Sites of ATPase are between adjacent elementary particles.

The inner membrane contains an electron transport system. it is made up of a chain of co-enzymes in the order of NAD, FAD, cytochrome B, cyt C, cyt A, cyt A3. The inner space of mitochondria is filled with a dense fluid called mitochondria matrix containing water, proteins, lipids, all enzymes of Kreb’s cycle, circular DNA, and 80s ribosomes.

Functions:

1. Mitochondria are the centers of aerobic respiration.
2. They are the sites of synthesis and storage of energy as ATP. Hence called the powerhouse of the cell.
3. As they have circular DNA & ribosomes, they synthesize a few proteins for their own requirement. Hence they are called ‘Semi-autonomous cell organelles’.

Question 36.
Describe the various stages of the I prophase of I meiosis.
Answer:
Prophase I: It lasts for quite a long period and is studied under five substages.

1. Leptotene
2. Zygotene
3. Pachytene
4. Diplotene
5. Diakinesis.

1. Leptotene: (Gr leptons slender; one – thread)
The chromosomes are thin, long, and uncoiled. Each is a double chromosome consisting of two chromatids. These chromatids are held firmly.
Each chromosome appears as a string of beads, the beads are the chromomeres.

2. Zygotene: (Gr: Zygon-inale; tene- thread)
During this stage, the pairing of homologous chromosomes (half of the maternal and half of them paternal). This pairing is called synapsis. The pairs so formed are called bivalents.

Each bivalent consists of four chromatids and is therefore called a tetrad. The two chromatids of the same chromosomes are called sister chromatids and the belonging to two different chromosomes of a homologous pair are termed non-sister chromatids.

3. Pachytene: (Gr: Pachus – thick tene – thread) .
Crossing over takes place by breakage and reunion of chromatid segments. After crossing over, the two chromatids of a chromosome become dissimilar.

The points of interchange are X-shaped and are called chiasmata. (sing chiasma)

4. Diplotene: (Gr: Diplo – double; tene – thread)

• Repulsion between homologous chromosomes begins.
• Each tetrad now appears in different shapes i.e, X-shaped, ‘8’ shaped, or ‘O’ shaped.
• Repulsion results in the criminalization of chiasmata (sliding of chiasmata towards the ends of chromosomes.

5. Diakinesis: (Gr: dia-cross; kinesis- movement).

• The nucleolus breaks down and disappears.
• Spindle fibers appear.
• The nuclear membrane breaks down.
• Chromosomes are released into the cytoplasm.

Question 37.
Discuss the role of pancreatic hormones in the regulation of blood glucose levels.
Answer:
Pancreas:
The pancreas is both an exocrine and an endocrine gland. For its endocrine function, it has islets of Langerhans. These islets have β (beta) or B cells, α (alpha) or A cells, and (delta) or D cells. The B cells secrete insulin and A cells secrete glucagon. Both are proteinaceous hormones regulating blood glucose levels.

The D cells secrete a hormone called somatostatin. It inhibits the digestion and absorption of nutrients. It also inhibits the secretion of insulin and glucagon.

1. Glucagon:

• It is a peptide hormone.
• It is a hyperglycemic hormone, i.e., it increases the level of glucose.
• It acts on the liver cells and hepatocytes in the blood.
• It stimulates glycogenolysis and increases the level of glucose in the blood.
• It stimulates gluconeogenesis and the synthesis of glucose from amino acids.

2. Insulin:

• It is a peptide hormone.
• It is a hypoglycemic hormone, i.e., it reduces the level of glucose in the blood by stimulating the uptake and utilisation of glucose by tissue cells.
• It acts on the hepatocytes and stimulates the conversion of glucose into fats.
• It also acts on adipocytes and stimulates the conversion of glucose into fats.
• Deficiency of insulin causes hyperglycemia and prolonged hyperglycemia results in diabetes mellitus

Students can Download 1st PUC Biology Previous Year Question Paper March 2015 (North), Karnataka 1st PUC Biology Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Biology Previous Year Question Paper March 2015 (North)

Time: 3.15 Hours
Max Marks: 70

General Instructions:

1. The question paper consists of four parts A, B, C, and D.
2. All the parts are compulsory.
3. Draw diagrams wherever necessary. Unlabelled diagrams or illustrations do not attract any marks.

Part – A

Answer the following questions in one word / one sentence each: ( 10 × 1 = 10 )

Question 1.
Name the basic unit of classification.
Answer:
Species are the basic unit of classification.

Question 2.
Define Phyllotaxy.
Answer:
The mode of arrangement of leaves on the stem axis is known as phyllotaxy.

Question 3.
Mention the function of Nephridia.
Answer:
Excretion.

Question 4.
Name the bacterial Ribosome.
Answer:
70s Ribosome

Question 5.
Define Osmosis.
Answer:
It is a special type of diffusion in which only the solvent or water molecules move from a region of their higher concentration to a region of their lower concentration through a semi-permeable membrane.

Question 6.
Which is the non – digesting enzyme of Intestinal Juice?
Answer:
Enter Kinase.

Question 7.
Name the structural and functional unit of the Kidney.
Answer:
Nephron.

Question 8.
Expand ECG
Answer:
Electrocardiogram.

Question 9.
What is Transpiration?
Answer:
It is the phenomenon by which excess water is eliminated, in the form of vapors through the aerial parts of the plant body.

Question 10.
Mention any Inclusion bodies.
Answer:
Reserve food material in prokaryotic cells is stored in the cytoplasm in the form of inclusion bodies. These are not bound by any membranes.

Part-B

Answer any FIVE of the following questions in 3-5 sentences each, wherever applicable. (5 × 2 = 10)

Question 11.
Write any four characteristics of Fungi.
Answer:
Kingdom Fungi is divided into five classes, based on the following characters:
(a) Morphology of mycelium
(b) Mode of spore formation.
(c) Mode of sexual reproduction and
(d) Nature of fruiting bodies.

Question 12.
Mention the uses of Diatomaceous Earth.
Answer:
Diatomaceous Earth is used:

• as an absorbent for liquid nitroglycerin to make explosives.
• for filtering the liquids in sugar factories.
• as an inert extender in paints, as an insulator in boilers and blast furnaces.
• in the powdered form as abrasive in silver polish and in the manufacture of toothpaste.

Question 13.
Differentiate Bilateral and Radial Symmetry.
Answer:
Germ layers or germinal layers are primary layers that differentiate in the embryo at the time of gastrulation. Animals have a minimum of two germinal layers and a maximum of three germinal layers. All the tissues and organs of the body develop from germinal layers.

Types:
(a) Diploblastic Animals: Animals having two germinal layers viz, outer ectoderm and inner endoderm are called diploblastic animals.
(b) Triploblastic Animals: Animals having three germinal layers viz, outer ectoderm, middle mesoderm, and inner endoderm are called Triploblastic animals.
eg: Platyhelminthes, Aschelminthes, Annelida, Arthropoda, Mollusca, Echinodermata, and Chordata.

Question 14.
Write any four Biological significance of Carbohydrates.
Answer:
(a) Carbohydrates particularly cellulose, chitin, and pectin provide structural support and protection to plants and fungi.
(b) The major food storage molecules are starch in plants and glycogen in animals.
(c) Pentoses form an integral part of nucleic acids and thus help in heredity.
(d) Chitin is an important exoskeleton in arthropods.

Question 15.
Draw a labeled diagram of Mitochondria.
Answer:

Question 16.
Mention the role of HCl in the Stomach.
Answer:
(a) It activates inactive pepsinogen into active pepsin.
(b) It converts inactive prorenin into active renin.
(c) It provides an acidic medium (pH 1.5 – 2.5) for the action of both pepsin, and renin.
(d) HCl kills harmful bacteria and loosens fibrous food.
(e) It inactivates the ptyalin.

Question 17.
Define Inspiration and Expiration.
Answer:

Inspiration	Expiration
1. It is a process of drawing air oxygen into the lungs from outside.	1. It is a process of leaving out carbon dioxide from the lungs to the outside.
2. Diaphragm and intercostal muscles contract.	2. Diaphragm and intercostal muscles relax.
3. Due to this the volume of the thorax, increases, and the lungs expand.	3. Due to this the volume of the thorax, decreases, and bings get compressed.
4. Pulmonary pressure falls.	4. Pulmonary pressure increases.

Question 18.
Briefly explain the transport of CO₂ by blood.
Answer:
As carbon dioxide from the tissues enters into the blood, a small quantity of about 45% enters into RBCs. In the RBCs, carbon dioxide combines with hemoglobin and forms carbamino hemoglobin. The carbamino-hemoglobin on reaching the lungs dissociates into hemoglobin, and carbon dioxide.
CO₂ + HbNH₂ → HbNHCOOH.

PART-C

Answer any FIVE of the following questions in 40-80 words each, wherever applicable: (5 × 3 = 15)

Question 19.
Write any three economic importance of Cyanobacteria.
Answer:
Economic Importance:
Many of them are used as bio-fertilizers because. of their nitrogen-fixing capacity. e.g. Nostoc, Aulosira, Anabaena.

Spirulina is unicellular and spiral-like; ¡t is cultured and used as cattle-feed and human food, for it is rich in proteins. it is called a single called protein.

Question 20.
Mention briefly respiration in Frog.
Answer:
Respiratory System: Frogs respire on Land and in the water by two different methods. In water, their skin acts as the aquatic respiratory organ, and it is exchanged through the skin by diffusion.

Question 21.
List any three functions of Parenchyma.
Answer:
Parenchyma is defined as a living, simple, permanent, and storage tissue. It is the basic and packing tissue that occurs in all parts of the plant body such as cortex, pith, pericycle, etc.

Characters:

1. Cells appear usually oval or spherical.
2. Cells are compactly arranged without intercellular spaces except for the lacunar type.
3. The cell wall is made up of cellulose and pectin.
4. The cell walls show uneven thickening, especially at corners.
5. Cells have cytoplasm and nucleus and hence are called living cells.
6. Cells have large and prominent vacuoles.

Types:
Based on the pectin deposition, collenchyma is of the following three types.
(a) Angular collenchyma – Here pectin is deposited in the angles or intercellular spaces of cells, eg: Datura, Ficus, etc.
(b) Lamellar collenchyma – Here pectin is deposited in between the cellular layers in the form of sheets. eg: Radish, Sunflower, etc.
(c) Lacunar collenchyma – Here pectin is deposited along the sidewalls around the intercellular space leaving a small space in the center. eg: Malva, Althea, etc.

Functions:

1. Collenchyma gives mechanical support.
2. Cells possess chloroplasts in young stems and help in photosynthesis.
3. Cells become meristematic and help in secondary growth.
4. In young parts of dicot plants, it gives resistance against the pulling and bending actions of wind.

Question 22.
What is the Cell Cycle? Mention the phases of the cell cycle.
Answer:
The sequence of events that occur between the formation of a cell and its division into daughter cells is called a cell cycle.

It consists of 2 phases:
(a) Interphase
(b) Mitotic phase

Question 23.
Explain briefly Mass flow Hypothesis.
Answer:
The accepted mechanism used for the translocation of sugars from source to sink is called the pressure-flow hypothesis. The glucose prepared by photosynthesis is converted to sucrose and is then moved in the form of sucrose into companion cells and into sieve tube cells by active transport.

This loading creates a hypertonic condition in the phloem & water in the adjacent xylem moves into the phloem by osmosis. As osmotic pressure builds up the phloem sap will move to areas of lower pressure. Again by active transport sucrose is transported out of phloem sap into the cells where it is used to convert into energy, starch, or cellulose.

Question 24.
Mention the formed elements of Blood.
Answer:
(a) Erythrocytes
(b) Leucocytes
(c) Platelets

Question 25.
Explain the role of Saliva.
Answer:
It is secreted in an inactive form called prorenin. it is activated into rennin by HCl.

Rennin acts on the milk protein casein, and converts it into Para casein.

Para casein combines with calcium ions to form solid casein or curd.
Para casein + Ca²⁺ → Calcium paracaseinate (Solid Casein).
Calcium Para casemate is acted upon by pepsin and converted into protests, peptones, and polypeptides

Question 26.
What is Double Circulation? Mention the stages in double circulation.
Answer:
The human heart acts as a double pump.

The right side of the heart receives, and pumps deoxygenated blood and the left side of the heart receives and pumps oxygenated blood and hence both the routes are kept completely separate and there is no mixing of oxygenated and deoxygenated blood. This type of circulation is said to be complete double circulation.

Double circulation involves shorter pulmonary circulation and longer systematic circulation.

Part-D (Section – I).

I. Answer any FOUR of the following questions in 200-250 words each. (4 × 5 = 20)

Question 27.
Write the general characters of the Phylum Annelida.
Answer:
General characters:

1. Annelids are mostly aquatic-either marine or freshwater forms. Some are terrestrial burrowing, free-living, or sedentary forms. A few are ectoparasites.
2. Annelids are bilaterally symmetrical, triploblastic, coelomate, and segmented animals.
3. The body is soft, cylindrical, or dorsoventrally flattened, uniform with cephalization at the anterior end.
4. The body is protected by a thin cuticle which is secreted by the underlying epidermal cells.
5. The body wall is dermo muscular (dermis with muscle cells or Libres), therefore can twist and turn or be highly contractile in nature.
6. The body cavity is a true coelom fined by coelomic epithelium, the coelom is formed by splitting of embryonic mesoderm, and hence it is described as ‘schizócoetous’ or ‘Schizocoelic’ made of formation and the coelom as ‘Schizocoel’.
7. The digestive system is a straight tubular structure and is complete with a mouth and anus. it also shows regional differentiation into parts like the buccal cavity, pharynx, esophagus, gizzard, intestine, and rectum.
   The digestive system shows segmental specialization.
8. Annelids are the first group to have a blood vascular system. A definite circulatory system or blood vascular system is present which consists of blood and blood vessels. Blood flows only indefinite blood vessels, hence the circulatory system is described as the ‘closed type’ of the circulatory system. Blood is red in color due to the presence of hemoglobin in the plasma.
9. Respiration occurs by simple diffusion through the body surfaõe or through respiratory structures called glib.
10. The excretory system consists of coiled ectodermal tubes called nephridia, through which excretion takes place.
11. The nervous system is well developed consisting of a nerve ring and a double ventral nerve cord.
12. Characteristic locomotory structures are usually present, either in the form of chitinous setae or fleshly lateral extensions called parapodia.
13. Reproduction is usually done in sexual mode. Members may be either bisexual or unisexual.
14. Development may be direct or indirect, where the larval stage present is represented by a trochophore larva.

Question 28.
Enumerate any five characteristic features of Gymnosperms.
Answer:
1. The life cycle has a distinct, dominant, diploid, asexual phase represented by the well-differentiated evergreen woody plant, which is known as the sporophyte.

2. The sporophyte is heterosporous bearing microspores and megaspores within microsporangia and megasporangia respectively. These structures occur on leaf-like microsporophylls and megasporophylls. These are further organized into fertile structures called stroll or cones.

3. Sporophyte shows the presence of a taproot system which is well developed. The stein possesses branches that are dichotomies. Leaves are well developed and are dimorphic (two types of leaves): viz,

• Green photosynthetic leaves (Foliage).
• Brown-colored scale leaves.

4. Microspore develops into male gametophyte and megaspore produces female gametophyte.
These gametophytes represent the haploid phase and are highly inconspicuous in comparison with sporophytic generation.

5. Female gametophyte is enclosed within a megasporangium that in turn is covered by an integument. Such an integument megasporangium possessing the female gametophyte is called the ovule.

Question 29.
Draw a neat labeled diagram of the Digestive system of the Cockroach.
Answer:

Question 30.
Explain Fluid Mosaic Model? Mention the functions of the Plasma membrane.
Answer:
Fluid mosaic Model By Singer & Nicholson:
(a) According to this model, the plasma membrane consists of a double layer of lipid molecules and globular protein molecules and sterols distributed at random. In comparative terms, it can be said that the plasma membrane is formed of ‘protein icebergs in the sea of lipids’.

(b) The protein molecules are globular proteins and these proteins penetrate or lie at the periphery to form a mosaic pattern.

(c) Heads of the phospholipid molecuLes of the two layers are directed in the opposite directions while tails of the two layers face each other.

(d) In animal cells glycolipids and cholesterol are also present along with proteins.

Modifications of the plasma membrane.
To perform specific functions and exhibit flexibility, PM undertakes the following modifications:

1. Microvilli: These are minute folding of the plasma membrane to increase surface area for absorption.
2. Desmosomes: the inner surface of adjacent plasma membranes has thickened areas called desmosomes which help in cell adhesion.
3. Pinocytic vesicles: Invaginations of the plasma membrane into cytoplasm and formation of fields that contain fluids.
4. Mesosomes: In prokaryotes, invaginations of the plasma membrane are associated with the respiratory system enzymes and help in respiration and are called mesosomes.

The functions of the plasma membrane are:

1. It maintains the size and shape of the cell.
2. Osmosis: Osmosis is the process by which water molecules pass through a semi-permeable membrane (here it is the plasma membrane) from the region of its higher concentration- to one of lower concentration.
3. Active transport: It is energy (from ATP) dependent transport of molecules or ions across a semi-permeable membrane against the concentration (electrochemical gradient). it can also be called “metabolically linked transport”.
   e.g: Sodium – Potassium Pump: (Revolving door model):

The moving machinery of Na & K through active transport is called ‘Sodium – Potassium pump’.

Question 31.
Write the differences between Mitosis and Meiosis.
Answer:

Mitosis	Meiosis
(a) occurs in all somatic cells and germ cells.	(a) occurs in germ cells of organisms that reproduce sexually.
(b) It is equational cell division.	(b) It is reductional cell division.
(c) Two daughter cells are produced.	(c) Four daughter cells are produced.
(d) Daughter cells have the same genetic constitution as the parent cells.	(d) Genetic constitution of daughter cells is different than parental cell due to crossing over
(e) Both the homologous chromosomes are present in daughter cells.	(e) The daughter cells contain one chromosome of each homologous pair

Question 32.
Give schematic representation of Kreb’s Cycle.
Answer:

Note: To account for two molecules of acetyl CoA produced from 1 molecule of glucose the entire reaction has to be multiplied by two.

Bloenergetics:
1. Number of ArPs produced = 2 .
2. Number of NADH₂ produced = 6 (6 × 3 = 18ATPs)
3. Number of FADH₂ produced = 2 (2 × 2 = 4 ATPs)
The total yield of ATPs = 24.
Note: Efficiency of Krebs cycle along with its predatory reaction is 30 ATPs.

Section – II

II. Answer any THREE of the following questions in 200-250 words each, wherever applicable. (5 × 3 = 15)

Question 33.
Explain the physiological roles of Cytokinins.
Answer:

• Discovered by Skoog and Miller
• Chemically these are 6-fur furyl amino purine.
• Synthesized in young fruits, leaves, buds, rost tips and translocated through the xylem.
• Cytokinins stimulate cell division.
• Cytokinins prevent early aging or senescence by stabilizing proteins and chlorophyll. This
  the phenomenon is called Richmond – Lang effect.
• Cytokinins break seed dormancy and induce germination.

Question 34.
Write the Z scheme of Light reaction.
Answer:
Non-cyclic photophosphorylation is also called ‘Z’ scheme electron transport.
In this cycle, the electrons lost by chlorophyll pigment do not return to the same molecule.
Both PSI and PS II are utilized.
One molecule of ATP is formed.
Two molecules of NADPH₂ are formed.
One molecule of O₂ is formed.

Question 35.
Explain the mechanics of Ascent of Sap.
Answer:
It is one of the most successful physical theories of the ascent of sap. Dixon and Jolly proposed this theory. According to this theory, the ascent of sap is due to three factors namely,
(a) Cohesive force of water.
(b) Adhesion of water molecules to the walls of the xylem vessel
(c) Transpiration pull

A strong intermolecular force of attraction exists between water molecules. Thus water molecules are bound to each other forming a continuous column of water.

There is a force of attraction between water molecules on the inner walls of xylem elements. Therefore water molecules are attached to the wall of the xylem. It ¡s called adhesion.

Due to the adhesive and cohesive properties of water, a continuous column of water is present in the xylem.

Transpiration at the leaf surface forces the mesophyll cells of the leaves to draw water from the neighboring xylem elements. This creates a suction force known as Transpiration pull.

This suction force pulls the water column ¡n the xylem upwards. This ascent of sap is due to the combined effect of adhesive, and cohesive properties of water and transpiration pull.

Merits:

1. The ascent of sap is directly proportional to the rate of transpiration.
2. It is a physical process and does not need energy.
3. Strong cohesive, and adhesive forces are sufficient to prevent the rupture of the water columns in xylem vessels.

Demerits:

1. The presence of an air bubble breaks the continuity of the water column.
2. Ascent of sap continues even in the absence of transpiration, as at night times.
3. The strength of the water column is thoughtful against two opposing forces such as gravitation force and transpiration tension.

Question 36.
Draw a neat labeled diagram of V.S. of Human Heart.
Answer:

Question 37.
List any Five hormones of the Pituitary gland and write one function of each.
Answer:
The pituitary gland weighs about 0.5 gm and is a pea-sized endocrine gland that lies on the ventral surface of the brain attached to the hypothalamus by a nervous stalk called the infundibulum.

It is called the master gland or conductor of the endocrine orchestra as several of its hormones control other endocrine glands directly. The hormones of the pituitary that influence other endocrine glands are called tropins or trophic hormones. However, the pituitary gland itself works under the influence of the hypothalamus through releasing factors. The pituitary gland is called the master gland because it controls the functions of other endocrine glands by secreting hormones.

Based on origin, it is divided into two parts namely Adenohypophysis and Neurohypophysis.

I. Adenohypophysis (Anterior Pituitary): It accounts for nearly 75% of the total weight of the gland. It is derived from the buccal cavity in the form of a projection called Rathke’s pouch.

Adenohypophysis is further divided into three regions: Pars distalis, Pars intermedia, and Pars tubular. Pars intermedia degenerates .during fetal development and occurs only as a small strip in adults. Pars tubular has no functional significance. Adenohypophysis cells secrete seven major hormones.

1. Somatotrophic hormone or Human growth hormone (STH or HGH): Promotes the growth of muscles. It stimulates the uptake of amino acids by tissues and their synthesis into proteins.

• Hypo secretion in childhood causes Pituitary dwarfism. Such an individual will be abnormally dwarf and is called a midget.
• Hypersecretion in childhood causes Pituitary gigantism. Such an individual will be abnormally tall.
• Hypersecretion in adolescence causes acromegaly which is characterized by the formation of disproportionately large hands, feet, cheekbones, jaws, etc.

2. Thyroid-stimulating Hormone (TSH): Controls secretion of thyroid hormones by the thyroid gland and also regulates iodine intake by the thyroid gland.

• Hypo secretion leads to goiter, cretinism, and myxoedema.
• Hypo secretion leads to hyperthyroidism.

3. Adrenocorticotrophic hormone (ACTH): It controls the( secretion of hormones (Cortisone) by the adrenal cortex.

• Hypersecretion of ACTH causes Cushing’s syndrome.
• Hypo secretion causes Addison’s disease.

4. Follicle Stimulating Hormone (FSH): In females the induces growth and maturation of Graffian follicle and stimulates follicular secretion of estrogen. In males, it stimulates the testis to produce sperm.

5. Luteinizing Hormone or Interstitial Cell Stimulating Hormone (LH or ICSH): In females, it stimulates ovulation and the formation of the corpus luteum. In males, it stimulates interstitial cells in the testis to secrete testosterone.

Students can Download 1st PUC Biology Previous Year Question Paper March 2015 (South), Karnataka 1st PUC Biology Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Biology Previous Year Question Paper March 2015 (South)

Time: 3.15 Hours
Max Marks: 70

General Instructions:

1. The question paper consists of four parts A, B, C, and D.
2. All the parts are compulsory.
3. Draw diagrams wherever necessary. Unlabelled diagrams or illustrations do not attract any marks.

Part – A

Answer the following questions in one word / one sentence each: ( 10 × 1 = 10 )

Question 1.
What is a taxon?
Answer:
Classification unit in Linnae& hierarchy ¡s called taxon.

Question 2.
Write the binomial name of the plant mango.
Answer:
Mangifera indica.

Question 3.
What is an actinomorphic flower?
Answer:
A flower can be divided into two equal radial halves in any radial plane passing through the center.

Question 4.
Define a cell.
Answer:
The cell is the fundamental, structural and functional unit of all living organisms.

Question 5.
Name the proteins that are involved in the transport of molecules across the membrane.
Answer:
Porins.

Question 6.
Define solute potential.
Answer:
Decrease in water potential due to dissolution of solute into the solvent.

Question 7.
Which is the most abundant plant pigment present in the world?
Answer:
Chlorophyll-a.

Question 8.
What is chlorosis?
Answer:
Yellowing of leaf due to lack of chlorophyll pigment.

Question 9.
Name the cells present in the mucosal epithelium of the small intestine that secretes mucus.
Answer:
Goblet Cells.

Question 10.
Name the hormone that is secreted by the pineal gland.
Answer:
Melatonin.

Part-B

Answer any FIVE of the following questions in 3-5 sentences each, wherever applicable. (5 × 2 = 10)

Question 11.
Write the four universal rules of binomial nomenclature.
Answer:

1. Scientific names are, generally, written in Latin or derived from Latin, irrespective of their origin.
2. The scientific names are written in italics or underlined. The first word denotes the name of the genus, and the second word denotes the species.
3. The generic name starts with a capital letter, while the specific name starts with a small letter.
4. The name of the author is written in an abbreviated form after the specific name e.g. Mangifera Indica Linn.-indicates that this species was first described by Linnaeus.
5. The name should be short, precise, and easy to pronounce.

Question 12.
Name two symbiotic associations of Lichens.
Answer:
Algae and Fungi.

Question 13.
What are diploblastic animals? Name the phylum that exhibits diploblastic condition.
Answer:
Animals whose body wall is made up of two germ layers are called diploblastic animals. eg: Hydra.

Question 14.
Differentiate between the open and closed types of vascular bundles.
Answer:
Open type of vascular bundles:
If the cambium is present in between the xylem and phloem, then the vascular bundle is said to be open type. it is closed for secondary growth.

Closed type of vascular bundles:
If the cambium is absent in between the xylem and phloem, then the vascular bundle is said to be closed type.

Question 15.
Name four main types of tissues present in animals.
Answer:

1. Epithelial tissue
2. Connective tissue
3. Muscular tissue
4. Neutral tissue

Question 16.
Mention four types of chromosomes based on the position of the centromere.
Answer:
I. Classification of chromosomes based on the number of centromeres:
(a) Acentric chromosome: It is a chromosome without centromere.
(b) Monocentric chromosome: Here there is one centromere to hold the chromatids together.
(c) Iiiceiitric chromosome: It is the presence of two centromeres in a chromosome.
(d) Polycentric chromosome: It is more than three centromeres in a chromosome.

II. Classification of chromosomes based on the position of centromere:
(a) Metacentric type: Here the centromere is exactly at the center of two chromatids. it looks V-shaped during anaphase.
(b) Sub metacentric type: Here the entrance is eccentric in position so that one of the chromatids is long and the other is shorter. It looks L – shaped during anaphase.
(c) Acrocentric type: Here the centromere is almost towards one end of the chromatid to form a very long arm and another very short arm. It looks hook-shaped during anaphase.
(d) Telocentric type: Here the centromere is towards one end of the chromatid such that one chromatid is only present. It looks rod-shaped during anaphase.

III. Classification of chromosomes based on their functions:
(a) Autosomes[AA]: They are also called somatic chromosomes that control body characteristics.
(b) Allosomes IX or VI: They are also called sex chromosomes that determine the gender of an individual.

Functions of chromosomes:

1. Chromosomes are very important in the higher animals for the phenomenon of sex determination.
2. Chromosomes play an active role in the metabolic process of a cell.
3. They carry the heredity information from parents to offspring ¡n the form of genes.

Question 17.
Give the meaning of the terms (a) Photoperiodism (b) Vernalisation
Answer:
Photoperiodism:
Plants, in order to flower, require a particular day length or light period called photoperiod, and the response of the plants to photoperiod in terms of flowering is called photoperiodism.

Photoperiodism was first studied by W.W. Garner, and HA. Allard.
Based on their photoperiodic responses, plants are classified into the following groups:

1. Long Day Plants: These flower in photoperiod more than critical day length. eg: Wheat, oats, etc.
2. Short Day Plants: These flower in photoperiod less than critical day length. e.g: Tobacco, Chrysanthemums, etc.
3. Day Neutral Plants: These are the plants that are not influenced by the dûration of light for their flowering. e.g: Tomato, cucumbër, cotton, etc.

Vernalisation:
Besides correct photoperiod, somç plants require low-temperature treatment for flowering. This treatment is known as vernalization.

Vernalization prevents precocious reproductive development late in the growing season and enables the plant to have sufficient time to reach maturity.

Certain food plants like wheat, barley, and rye have two varieties called,

1. Spring variety, and
2. Winter variety.

The spring variety is planted in spring, and it completes the lifecycle before the growing season.
The winter variety ¡s normally planted in autumn or spring and is harvested by mid-summer.

Question 18.
Name two kinds of contractile proteins present in myofibrils.
Answer:
(i) Actin
(ii) Myosin

PART-C

Answer any FIVE of the following questions in 40-80 words each, wherever applicable: (5 × 3 = 15)

Question 19.
Sketch and label the structure of the chloroplast.
Answer:

Electron microscopic structure of the chloroplast

Question 20.
How would you identify the following stages of meiosis-I?
(a) Leptotene (b) Zygotene (c) Pachytene
Answer:
(a) Leptotene: Chromosomes are clearly visible as they undergo condensation.
(b) Zygotene: Two identical chromosomes are pairing called synapsis.
(c) Pachytene: Crossing over occurs between non-sister chromatids of homologous chromosomes.

Question 21.
What are proteins? Name the most abundant protein present in plants and animals.
Answer:
Proteins are polypeptides made up of a chain of amino acids.
Abundant proteins found in
Plants – RUBISCO / RUBP Carboxylase
Animals – Collagen

Question 22.
Explain three phases of the sigmoid curve.
Answer:
Growth Curve:
The curve obtained when the growth in size or weight or on entire pLant or its individual parts is plotted against time is called the growth curve. It is always ‘S’ or sigmoid in shape. it shows three distinct phases namely.

1. Lag phase.
2. Log phase.
3. Steady phase.

1. Lag phase: This is the initial stage where the growth rate is very slow. But it gradually increases with time.
2. Log phase: In this phase, the rate of growth is very rapid. The plant shows a high rate of growth and reaches maximum height. Hence this phase is called the exponential phase or grand period of growth.
3. Steady phase or stationary phase: This is the last phase in which plants show no growth. Therefore the growth curve becomes almost horizontal.

Question 23.
Give the meaning of the terms (a) Antiport (b) Osmosis (c) Guttation
Answer:
(a) Antiport: Movement of molecules across the membrane in both directions.
(b) Osmosis: It is a special type of diffusion in which only the solvent or water molecules move from a region of their higher concentration to a region of their lower concentration through a semi-permeable membrane.
(c) Guttation: The loss of water in liquid form through specialized structures called water stomata or hydathodes is called guttation.

Question 24.
Describe briefly the digestion of three types of food by the enzymes of the pancreas.
Answer:

Question 25.
Name the three layers of cranial meninges present between skull and brain.
Answer:
(a) Outer Dura mater.
(b) lyiiddle Arachnoidmater
(c) Inner Pia mater

Question 26.
(a) Mention the types of antigen and antibody present in different blood groups of man.
Answer:

(b) What is uremia?
Answer:
Uraemia is a condition, where urea gets accumulated in the blood due to the failure of nephrons in the kidney.

Part-D (Section – I).

I. Answer any FOUR of the following questions in 200-250 words each. (4 × 5 = 20)

Question 27.
Write five general characteristic features of bryophytes.
Answer:

1. They are non-vascular plants, exhibiting amphibious habitats.
2. The gametophytic phase of the life cycle is representeð by multicellular, haploid thallus, which is called gametophyte.
3. Gametophytic plant body bearing male and female sex organs, represents the dominant phase of life cycle unlike in higher plants.
4. The gametophyte is autotrophic and independent in nature.
5. The gametophyte produces sex organs, the antheridia and the archegonia which are multicellular.
   Antheridia may be embedded within the thallus or may be specifically located at the tips of the gametophyte. It produces biciliate curved antherozoids.

Archegonia is a flask-like structure, it bears a basal bulbous vendor and a narrow, hallow structure at it is summit called the neck. Within the venter, the female gamete or egg is situated along with a venter canal cell. The neck shows neck cells, neck canal cells, and cover cells.

Question 28.
Sketch and label the alimentary canal of the cockroach.
Answer:

Question 29.
Classify enzymes into five types based on the type of reactions they catalyze.
Answer:
Classification of enzymes:

• Oxidoreductases: Act on many chemical groupings to add or remove hydrogen atoms.
• Transferases: Transfer functional groups between donor and acceptor molecules. Kinases are specialized transferases that regulate metabolism by transferring phosphate from ATP to other molecules. ,
• Hydrolases: Add water across a bond, hydrolyzing it.
• Lyases: Add water, ammonia, or carbon dioxide across double bonds, or remove these elements to produce double bonds.
• Isomerases: Carry out many kinds of isomerization: L to D isomerizations. Mutase reactions (shifts of chemical groups) and others.
• Ligases: Catalyze reactions in which t chemical groups are joined (or ligated) with the use of energy from ATP.

Question 30.
Explain the major events of the photochemical phase of photosynthesis.
Answer:
It is a totally light-dependent process.
It includes:
(a) Photoexcitation of chlorophyll.
(b) Photo phosphorylation – cyclic and noncyclic.
(c) Photolysis of water (photoionization of water).

(a) Photoexcitation of chlorophyll: Chlorophyll absorbs light energy as photons. Photoexcited chlorophyll ejects energized electrons, and becomes positively charged, and remains unstable. It regains stability only when its lost electrons are replaced.

(b) Photo phosphorylation:
The extra energy of ejected electrons during photoexcitation is used for phosphorylation of ADP into ATP1-Ìence it is called photophosphorylation. It is of two types namely,
(a) Cyclic Photophosphorylation
(b) Non-cyclic Photophosphorylation.

(a) Cyclic Photophosphorylation: It is a cyclic path of electrons expelled from chlorophyll through a series of substrates that are arranged in a suitable oxidation-reduction potential. The energy in the electrons is used for the phosphorylation of ADP to ATP.

In PS I the absorbed photons of light excite chlorophyll-a 700 to eject energized electrons on makes it positively charged and unstable. Electrons pass through the sequence → FRS → FD → Cyt b6 → Cytf → PC and generate ATP at two places. Finally, an electron from PC returns to chi-a restoring its stability.

(b) Non-cyclic Photo phosphorylation:
1. It is the noncyclic path of electrons from PS Il, and PS Ito release ATP and NADPH2, with the help of protons and electrons from photolysis of water

2. In PS I the absorbed photons of light excite chlorophyll-a 100 to eject energized electrons which pass through FRS, FD, and finally get locked up with NADP which becomes negatively charged, and partially reduced, while PSI becomes positively charged and unstable.

3. At the same time in PS II, the absorbed photons of light excite chlorophyll-a 680 to eject energized electrons which pass through in sequence, pheophytin, PQ, to b6, cytf, plastocyanin, and generate ATP. The electrons finally join P.S Ito replace its lost electrons and restore its stability. Now P.S Il becomes positively charged, and unstable. Meanwhile, electrons of photolyzed water join with PS Jito to replace its lost electrons and restore its stability. Protons join with the already ionized NADP and complete its reduction to NADPH₂.

Question 31.
Write the schematic representation of glycolysis.
Answer:
It occurs ¡n the cytoplasm of the cell.
It is an enzymatic reaction, thus temperature sensitive.
It ¡s a common reaction for both aerobic and anaerobic respiration.

Question 32.
Describe the mechanism of breathing.
Answer:
It is a process of intake of oxygen and leaving out carbon dioxide from the lungs. The mechanism of breathing involves.
1. Inspiration (Inhalation): It is the process of drawing air into the lungs from the outside atmosphere (intake of air). During this process, muscles of the diaphragm contract, which increases the length of the thorax. In meanwhile intercostal muscles also contract, pulling the ribs outward. This phenomenon increases the width of the thorax, resulting in the expansion of the lungs. As a consequence pulmonary pressure falls. So oxygen-rich air rushes into the lungs and fills alveoli.

2. Exchange of gases: The alveoli are closely surrounded by a thin wallet! epithelium, having a capillary network. The oxygen is under high pressure in alveoli, because of its higher concentration. Similarly, carbon dioxide concentration and pressure will be more in the capillaries containing impure blood.

So, the exchange occurs by diffusion through the capillary walls. The oxygen is drawn into the blood, and carbon dioxide is pushed into the alveoli.

3. Expiration (exhalation): lt is the process of throwing out carbon dioxide from the lungs to the outside. During this process, muscles of the diaphragm and intercostal muscles relax causing the collapse of the rib cage. This decreases the volume of the thorax and lungs.

As a consequence pulmonary pressure increases. So, the carbon dioxide-rich air of the alveoli is drawn out through the respiratory passage.

Section – II

II. Answer any THREE of the following questions in 200-250 words each, wherever applicable. (5 × 3 = 15)

Question 33.
(a) Assign the following animals to their respective phylum. (3M)
(i) Comb jellies
(ii) Pila
(iii) Planaria
(iv) Ascaris
(v) Starfish
(vi) Sycon.
Answer:
(i) Ctenophora
(ii) Mollusca
(iii) Platybelminthes
(iv) Aschelminthes
(v) Echinodermata
(vi) Porifera

(b) Describe four important features of Aves.
Answer:

1. Presence of feathers.
2. Presence of pneumatic bones: Pneumatic bones are long hollow bones with air cavities.
3. Forelimbs are modified into wings to help in flight.
4. Hind limbs generally have scales helping them to clasp the prey in-flight or a tree branch.
5. Excretion of uric acid and feces is through a single opening, and excretion uses a very little amount of water, to reduce body weight.
6. An aerodynamically built body helps in flying (by reducing friction).

Question 34.
What is aestivation? Mention any four types with one example for each.
Answer:
The mode of arrangement of sepals, petals, or even tepals in a flower bud is called aestivation. The different kinds of aestivation are as follows:
1. Valvate aestivation: When sepals, petals, or tepals are not overlapping.

2. Imbricate aestivation: When out of the total number of sepals. petals or tepals, one is completely out, one is completely in and the rest are in and out.

3. Descending imbricate aestivation: When the standard petals are large and overlap the two wing petals, which in turn overlap the keel petals. It is technically known as vexillary aestivation.

Note: It is characteristic of the members of the subfamily Papilionoideae (Papilionaceae). e.g: Pea, Bean, Indigofera, Tephrosia, etc.

4. Ascending imbricate aestivation: When the small standard petal is completely ¡n and is overlapped by the lateral wing petals which in turn are overlapped by the keel petals.

Note: It is characteristic of sub-family Caesalpinioideae (Caesalpinae).
eg: Caesalpinia puicherhima, Delonix regia, etc.

Question 35.
With the schematic representation explain Calvin Cycle.
Answer:
It occurs in the grana of the chloroplast.
It utilizes the assimilatory powers i.e., ATP and NADPH₂ produced in light reaction.

Steps:

1. Phosphorylation
2. Carbon dioxide fixation
3. Phosphorylation and reduction.
4. Regeneration and RUMP
5. Glucose formation.

Note: To complete a Calvin cycle, 18 ATPs and 12 NADPH₂ molecules are required.

Question 36.
Sketch and label the structure of the nephron.
Answer:

Question 37.
What is ECG? Explain it with the graphical presentation of a standard ECG
Answer:
ECG has 5 waves or deflections which are conventionally designated as PQRST. The five waves have a horizontal part in the beginning as well as inside. It is called isopotential or baseline. PR and T waves lie above the horizontal line. They are called positive waves. Q and S waves lie below the horizontal line and are therefore called negative waves. P represents the development of action potential at the sinoatrial node and the spread of impulse throughout the atria.

The atrial muscles become depolarized. After a fraction of a second, the next complex of QRS begins with a small downward or negative reflection of Q. QRS represents depolarization of atrial muscles and depolarization of ventricular muscles. Its different components are PQ, QR, and RS. PQ indicates contraction of atria.

QR provides information about the spread of cardiac impulses from the SA node to the AV node. RS gives information about the spread of impulse from the AV node to Purkinje fibers and ventricular depolarization. It initiates the contraction of the ventricles. The ventricular contraction continues during the ST part. It is followed by the relaxation of ventricles and the development of the repolarisation wave.