Self Studies
Selfstudy
Selfstudy

Biology Test 269

Result Self Studies

Biology Test 269
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    4 / -1

    Stroma in the chloroplasts of higher plant contains

    Solution

    The dark reactions of photosynthesis are purely enzymatic and slower than the primary photochemical reaction. They take place in stroma portion of the chloroplast and are independent of light, i.e., they can occur either in presence or in absence of light provided that assimilatory power is available.

     

  • Question 2
    4 / -1

    The first phase in the breakdown of glucose, in animal cell, is

    Solution

    Glycolysis is the first step of glucose breakdown in both animals and plants. During glycolysis 6-carbon glucose molecule is converted into 2 molecules of 2 carbon pyruvic acid. In this process net gain of 2 ATP and 2 NADH2 occurs. It is a common pathway for both aerobic and anaerobic modes of respiration.

     

  • Question 3
    4 / -1

    End product of glycolysis is

    Solution

    Glycolysis or EMP pathway is the breakdown of glucose to two molecules of pyruvic acid through a series of enzyme mediated reaction releasing energy. Pyruvic acid is a 3-carbon compound. In glycolysis net gain of 2ATP and 2 NADH2 molecules occurs.

    It can be represented in equation form as –

     

  • Question 4
    4 / -1

    Name the plant growth regulator which upon spraying on sugarcane crop, increases the length of stem, thus increasing the yield of sugarcane crop.

    Solution

    Spraying sugarcane crop with gibberellins increases the length of the stem and increasing the yield by as much as 20 tonnes per acre.

     

  • Question 5
    4 / -1

    Which of the following is not an inhibitory substance governing seed dormancy?

    Solution

    Gibberellic acid involved in growth promoting activities, such as cell division, cell enlargement, pattern formation, tropic growth, flowering, fruiting and seed formation and is called plant growth promoters. It breaks seed dormancy which is antagonistic to abscisic acid.

     

  • Question 6
    4 / -1

    A baby boy aged two years is admitted to play school and passes through a dental check-up. The dentist observed that the boy had twenty teeth. Which teeth were absent?

    Solution

    Boy aged two years will have milk teeth. Milk teeth of man include 8 incisors, 4 canines, 8 molars. Premolars are absent.

     

  • Question 7
    4 / -1

    Which cells of ‘Crypts of Lieberkühn’ secrete antibacterial lysozyme?

    Solution

    Paneth cells, present in the bottom of crypts of Lieberkuhn, are rich in zinc and contain acidophilic granules. There is evidence that these cells secrete antibacterial lysozyme. Zymogen cells or peptic cells are present in stomach and secrete pepsinogen. Kupffer cells are present in liver. They are phagocytic in nature and engulf disease causing microorganisms, dead cells, etc. Argentaffin cells, found in crypts of Lieberkuhn, synthesise hormone secretin and 5- hydroxytryptamine.

     

  • Question 8
    4 / -1

    Select the correct events that occur during inspiration.

    (1) Contraction of diaphragm
    (2) Contraction of external inter-costal muscles
    (3) Pulmonary volume decreases
    (4) Intra pulmonary pressure increases

    Solution

    Inspiration is initiated by the contraction of diaphragm that increases the volume of thoracic chamber in the antero-posterior axis. The contraction of external inter-costal muscles lifts up the ribs and the sternum causing an increase in the volume of the thoracic chamber in the dorso-ventral axis. The overall increase in the thoracic volume causes a similar increase in pulmonary volume. An increase in pulmonary volume decreases the intrapulmonary pressure to less than the atmospheric pressure which forces the air from outside to move into the lungs.

     

  • Question 9
    4 / -1

    Tidal volume and expiratory reserve volume of an athlete is 500 mL and 1000 mL respectively. What will be his expiratory capacity if the residual volume is 1200 mL?

    Solution

    Expiratory capacity is the total volume of air a person can expire after normal inspiration. It includes tidal volume (TV) and expiratory reserve volume (ERV).

    EC = TV + ERV = 500 mL + 1000 mL = 1500 mL

     

  • Question 10
    4 / -1

    Name the blood cells, whose reduction in number can cause clotting disorder, leading to excessive loss of blood from the body.

    Solution

    Thrombocytes are called blood platelets. They are minute disc-shaped cell fragments in mammalian blood. They are formed as fragments of larger cells found in red bone marrow; they have no nucleus. They play an important role in blood clotting and release thromboxane A2, serotonin and other chemicals, which cause a chain of events leading to the formation of a plug at the site of the damage, thus preventing further blood loss. A reduction in their number can lead to clotting factors which will lead to excessive loss of blood from the body.

     

Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now