نبات زراعى 101 E

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    Cairo University Semester :The Second

    Faculty of Agriculture Academic Year:2010/2011

    Agricultural Botany Dept. Allowed Time : 2 hours

    Agricultural Botany/ Physiology 101 ABT English Program

    Your answer must be supported by drawings as you can

    Exam in Two PagesAnswer the following questions.

    Question 1: (15 marks)

    A- Define with drawing Three of the following scientific terms:-(7.5 marks)

    1- Leaf venation:

    It is the distribution of veins (vascular bundles) through the blade

    of leaf

    2- Abscission zone: it is an area located perpendicular to petiole.3- The pit: it is an area of the primary wall does not have secondary wall

    located on.

    4- freely membrane: it is a membrane which allow any substance to pass

    through it

    B- Choose the correct answer between brackets: (7.5 marks)

    1. An organ digests wastes inside the plant cell.

    (pits - Cell vacuole - ribosomes - nuclear pore)

    2. The tissue which consists of vessel, tracheids, parenchyma and fibers.

    (epidermis - cortex - sclerenchyma - xylem) .

    3. The sclerenchyma tissues function is.

    (protein synthesis - starch storage - respiration - supporting)4. Which of these tissues considered as a simple tissue ?

    (xylem - phloem - colenchyma - all of these).

    5- Fibers have bordered pits on the perpendicular wall.

    (sieve plate - tapered end walls - perforation plate - bordered pits)

    Question 2: (15 marks)

    A- Complete the following sentences: (7.5

    marks)

    1- Sclerenchyma tissue consists of two types cells areSclereids and Fibers

    2- Lateral roots are initiated frompericycle

    3- Primary phloem consists ofSieve tube member, sieve cell, parenchyam and fibers

    4- Plastid types are chloroplastid, chromoplastid, andproplastid

    5- Vascular cambium cells arefusiform and ray initials

    B- Complete the following table : (7.5 marks)

    Organ Structure FunctionStomata Stomatal pore, guard cell Gas exchange

    Choloroplastids Double membrane, stroma, grana Photosynthesis

    Mitochonderia Double membrane, matrix Respiration

    Ribosome Spherical body Protein synthesis

    Meristematic cell Cell wall, protoplast & nucleus division

    Question 3: (15 marks)

    A- Give the key word suitable for each of the following statement: (5 marks)

    1

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    1 International Code of Botanical Nomenclature ICBN

    2 The dominant form of Mosses, that contains antherdia and

    archegonia

    gametophyt

    e

    3 Conjunction between male and female gametes each has (N)

    chromosome to form zygote (2N).

    fertilization

    4 The flower that contain androecium and gynoecium and lack ofcalyx and corolla

    Perfect

    5 The type of classification based on morphological similarities of

    vegetative characters of the plant.

    Artificial

    B- Answer the following sentences with (Yes) or (No). (5

    marks)

    1- Double fertilization occurs in gymnosperms. (No)

    2- Ferns are vascular plants. (Yes)

    3- Complete flower lack of androecium and gynoecium. (No)

    4- Each genus belongs to order. (No)

    5- Eukaryotes have true nucleus. (Yes)

    C- Complete the following sentences: (5 marks)

    1- Asexual reproduction types are Fission, vegetative and spores

    2- The order includes group of relatedfamlies.

    3- The common and constant stage of gymnosperms issporophyte

    4- In flowering plants, the ovule consists ofegg cell, synergid cells, pollar

    nuclei and antipodial cells.

    5- Spirogyra belongs to division chlorophyta

    Question 4: (15 marks(

    In glycolysis, the glucose, is broken down into two molecules of a 3-carbon

    molecule called pyruvate. This change is accompanied by a net gain of 2 ATP

    molecules and 2 NADH molecules.

    Second step start at the mitochondria when the pyruvate enter to the matrix and

    called Citric acid or Krebs cycle

    2

    Net Products from ATPs

    4 ATP

    + 10 NADH.H+

    + 2 FADH2

    1 NADH.H+ = 3 ATP

    1 FADH2 = 2 ATP

    4ATP + 30ATP + 4ATP = 38 ATP