{"id":35563,"date":"2022-09-20T10:48:51","date_gmt":"2022-09-20T05:18:51","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35563"},"modified":"2022-09-20T10:48:51","modified_gmt":"2022-09-20T05:18:51","slug":"ap-inter-2nd-year-botany-study-material-chapter-3","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-inter-2nd-year-botany-study-material-chapter-3\/","title":{"rendered":"AP Inter 2nd Year Botany Study Material Chapter 3 Enzymes"},"content":{"rendered":"

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material<\/a> 3rd Lesson Enzymes Textbook Questions and Answers.<\/p>\n

AP Inter 2nd Year Botany Study Material 3rd Lesson Enzymes<\/h2>\n

Very Short Answer Questions<\/span><\/p>\n

Question 1.
\nHow are prosthetic groups different from co-factors?
\nAnswer:
\nProsthetic groups are organic compounds that are tightly bound to the apoenzyme (protein part of the enzyme) whereas cofactors are non-protein parts of the holoenzyme.<\/p>\n

Question 2.
\nWhat is meant by feedback inhibition?
\nAnswer:
\nThe end product of a chain of enzyme-catalyzed reactions inhibits the enzyme of the first reaction as a part of homostatic control of metabolism is called feedback inhibition.<\/p>\n

\"AP<\/p>\n

Question 3.
\nW hy are ‘oxido reductases’ so named?
\nAnswer:
\nEnzyme which catalyse oxidation and reduction of substrates usually involving hydrogen transfer are called oxido reductases.
\n\"AP<\/p>\n

Question 4.
\nDistinguish between apoenzyme and cofactor.
\nAnswer:
\nThe protein part of a holoenzyme is called apoenzyme the non-protein part of a holoenzyme is called cofactor.<\/p>\n

Question 5.
\nWhat are competitive enzyme inhibitors? Mention one example.
\nAnswer:
\nSubstances which are closely resembles the substrate molecules and inhibits the activity of the enzyme are called competitive inhibitors.
\nEg : Inhibition of succinic dehydrogenase by malonate which closely resembles the substrate succinate.<\/p>\n

Question 6.
\nWhat are non-competitive enzyme inhibitors? Mention one example.
\nAnswer:
\nThe inhibitor has no structural similarity with the substrate and bind to an enzyme of locations other than the active sites so that the globular structure of the enzyme is changed are called non-Competitive enzyme inhibitors.
\nEg : Metal ions of Copper, Mercury.<\/p>\n

Question 7.
\nWhat do the four digits of an enzyme code indicate?
\nAnswer:
\nIn enzyme code, the first digit indicates the major class of the enzyme the second digit and the third digit indicate sub class and sub-sub class respectively. The last digit of the code is the serial number of the enzyme.<\/p>\n

Question 8.
\nWho proposed ‘Lock and Key hypotheses’ and ‘Induced fit hypothesis’?
\nAnswer:
\nLock and key hypothesis was proposed by Emil Fisher (1884). Induced fit hypothesis was proposed by Daniel E. Koshland (1973).<\/p>\n

\"AP<\/p>\n

Question 9.
\nDefine Michael’s constant?
\nAnswer:
\nSubstrate cencentration required to cause half the maximal reaction rate is termed as michalis menten constant (km).<\/p>\n

Short Answer Questions<\/span><\/p>\n

Question 1.
\nWrite briefly about enzyme inhibitors.
\nAnswer:
\nThe chemicals that can shut off enzyme activity are called inhibitors. They are of 3 types.
\n1) Competitive inhibitors:
\nSubstances which closely resemble the substrate molecules and inhibits the activity of the enzyme are called competitive inhibitors.
\nEg : Inhibition of succinic dehydrogenase by malonate which closely resembles the substrate succinate.<\/p>\n

2) Non-competitive inhibitors :
\nThe inhibitors which have no structural similarity with the substrate and bind to an enzyme at locations other than the active sites, so that the globular structure of the enzyme is changed are called non competitive inhibitors.
\nEg: Metal ions of Copper, Mercury.<\/p>\n

3) Feedback inhibitors :
\nThe end product of a chain of enzyme catalysed reactions inhibit the enzyme of the first reaction as a part of homoeostatic control of metabolism are called feed back inhibitors.
\nEg : During respiration (Glycolysis) accumalation of Glucose-6 Phosphate occurs, it inhibits the Hexokinase.<\/p>\n

Question 2.
\nExplain different types of co-factors.
\nAnswer:
\nCofactors are two types.
\n1) Metal ion cofactor :
\nMetallic cations get tightly attached to the apoenzyme are called metallo enzymes.
\nEg : Cu+2<\/sup> cytochrome oxidase<\/p>\n

2) Organic cofactors :
\nThey are two types.
\na) Coenzyme :
\nThey are small organic molecules which are loosely associated with the apoenzyme.
\nEg : Thiamine pyrophosphate, Vitamin B.<\/p>\n

b) Prosthetic group :
\nThey are the organic cofactors which are tightly bounded to the apoenzyme.
\nEg : Haeme is the prosthetic group of enzyme peroxidase.<\/p>\n

\"AP<\/p>\n

Question 3.
\nExplain the mechanism of enzyme action.
\nAnswer:
\nThe substrate ‘S’ has to bind the enzyme of its active site with in a given cleft. The substrate has to diffuse towards the active site leads to the formation of ES complex. This is called transition state structure. Very Soon, after the expected bond breaking \/making is completed, the product is released from the active site. In other words, the structure of substrate gets transformed into the structure of products.
\n\"AP<\/p>\n

When we represent this pictorially through a graph by taking potential energy on y-axis and progress of reaction on x-axis, we notice the energy level difference between ‘S’ and ‘P’. If ‘P’ is at lower level than S, it is an exothermic reaction. One need flot supply energy to form the product. However, whether it is an exothermic or an endothermic reaction (Energy requiring reaction) the ‘S’ has to go through a much higher energy state or transition state.<\/p>\n

The difference between average energy content of ‘S’ and that of this transition state is called activation energy. –<\/p>\n

Each enzyme [E] has a substrate [S] binding site in its molecule so that a highly reactive enzyme substrate complex [ES] is produced. This complex is short lived and dissociates into its products [P] and the unchanged enzyme with an intermediate formation of the enzyme product complex [EP].
\nE+S \u2192 ES \u2192 EP \u2192 E+P<\/p>\n

Formation of ES complex has been explained with lock and key hypothesis by Emil Fisher and much later with induced fit hypothesis by Daniel E. Koshland.
\nThe catalytic cycle of an enzyme action \u2018is<\/p>\n

    \n
  1. The substrate binds to the active site of the enzyme.<\/li>\n
  2. The binding of the substrate induces the enzyme to alter its shape, fitting more tightly around the substrate.<\/li>\n
  3. The active site of the enzyme breaks the chemical bonds of the substrate and the new enzyme product complex is formed.<\/li>\n
  4. The enzyme releases the products of the reaction and the free enzyme is ready to bind to another molecule of the substrate.<\/li>\n<\/ol>\n

    Intext Questions<\/span><\/p>\n

    Question 1.
    \nExplain how pH effects enzyme activity with the help of a graphical representation.
    \nAnswer:
    \n\"AP
    \nEnzymes generally function in a narrow range of pH. Each enzyme shows its highest activity at a particular pH called optimum pH.
    \nActivity declines both below and above the optimum level.<\/p>\n

    \"AP<\/p>\n

    Question 2.
    \nExplain the importance of [ES] complex formation.
    \nAnswer:
    \nEach enzyme [E] has a substrate [S] binding site in its molecule so that a highly reactive enzyme substrate complex [ES] is produced. This complex is short lived and dissociates into its products [P] and the unchanged enzyme with an intermediate formation of the enzyme product complex [EP].
    \nE+S \u2192 ES \u2192 EP \u2192 E+P<\/p>\n","protected":false},"excerpt":{"rendered":"

    Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 3rd Lesson Enzymes Textbook Questions and Answers. AP Inter 2nd Year Botany Study Material 3rd Lesson Enzymes Very Short Answer Questions Question 1. How are prosthetic groups different from co-factors? Answer: Prosthetic groups are organic compounds that are tightly bound to the apoenzyme (protein part …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35563"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35563"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35563\/revisions"}],"predecessor-version":[{"id":35564,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35563\/revisions\/35564"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35563"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35563"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35563"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}